ESPCI Wiki - User contributions [en]

T-II-3

2011-11-29T11:15:46Z

Groupe 01 13: /* Preliminaries: the central limit */

__FORCETOC__

= Extreme Statistics =

Generically, finding the distribution of the maximum of a set of random variables is a non-trivial problem, which appears in many contexts ranging from the maximal height of water in a river to fluctuations in stock markets
We consider ''N'' independent random variables <math>(x_1,...,x_N)</math> drawn from the same distribution <math>p(x)</math>.
We denote
<center><math>y_N=\max(x_1,...,x_N)</math></center>

It is useful to use the following notations for the cumulative distributions
<center><math>P^<(x)=\int_{-\infty}^x dx' p(x')\qquad\qquad P^>(x)=\int_x^{+\infty} dx' p(x') </math></center>

Let us denote by <math>q_N(y)</math> the distribution of <math>y_N</math> and by <math>Q_N(y)=\text{Prob}(y_N<y)</math> its cumulative distribution.

* Write <math>Q_N(y)</math> in terms of <math>P^<(y) </math>. (Help: Start to write this relation for <math>N=2,3,...</math>).
This is the fundamental relation of Extreme statistics and we analyze its consequences in the large ''N'' limit where, analogously to the central limit theorem, extremes statistics display universal features.
* In particular shows that in the large ''N'' limit we can write
<center><math>Q_N(y) \sim \exp\left(-N P^>(y)\right) </math></center>

In the present exercise, we first study the case of the exponential distribution. In a second step we generalize our results to a larger class of distributions.

==Exponential distribution==

The exponential distribution is one of the fundamental continuous distributions, and already for this reason worthy of study. Among many other places, it appears in the Poisson process. The distribution writes:

<center><math>p(x) = \lambda \exp(-\lambda x)</math></center>
where both <math> \lambda</math> and <math>x</math> are positive numbers.
===Preliminaries: the central limit===

* compute the mean value and the variance of this distribution
* consider <math>X_N</math>, the sum of ''N'' independent, exponentially distributed, random variables. How <math>X_N</math> is distributed?

We write <math>X_N</math> in a more convenient way
<center><math>X_N = a_N + b_N z </math></center>
where <math>a_N </math> the location of the distribution and <math>b_N </math> is the width of the distribution of <math>X_N </math>. Both numbers depend on <math>N </math>. Finally, <math>z </math> is a random number and its distribution, <math>\pi(z)</math> becomes independent of <math>N </math> in the large "N" limit. In other words this means that the distribution of <math>X_N</math> is significantly different from zero when the value of <math>X_N </math> is around <math>a_N </math>, in a region of size <math>b_N </math>.

* From the central limit theorem which is the natural choice for <math>a_N </math> and <math>b_N </math>? Write the distribution <math>\pi(z)</math>

===The Maxima===
Consider now the case <math>\lambda=1</math>
* Write <math>P^>(x)</math> and <math>P^<(x)</math>. (Remember that <math>x</math> is a positive number.)
* Write <math>Q_N(y)</math> and <math>q_N(y)</math>.
* Plot <math>q_N(y)</math> for different values of ''N''.

We want now to give a natural definition for the number <math>a_N</math> and <math>b_N</math>.

Consider <math>P^>(\tilde y)=\frac 12</math>. If you draw N independent exponential variables, how many variables (in average) will be greater than <math>\tilde y</math>? Repeat the same exercise with <math>\tilde \tilde y</math> such that
<math>P^>( \tilde \tilde y)=\frac 23</math>

* Justify that <math>a_N</math> can be estimated from
<center><math>P^>(a_N)=\frac 1N</math></center>
* Compute <math>a_N</math> for the exponential distribution and justify that
<center><math>Q_N\left(y=a_N+z\right)</math></center>

In the large ''N'' limit, the distribution <math>\pi(z)</math> becomes <math>N</math> independent.

* Show that in this limit its cumulative takes the from
<center><math>\Pi(z)= e^{-e^{-z}}</math></center>
This is the cumulative distribution of the famous Gumbel distribution.

Let us remark that the precise definition of <math>a_N</math> and <math>b_N</math> fix the mean and the variance of the rescaled distribution <math>\pi(z)</math>
At variance with the central limit case the mean will be different from zero and the variance different from one.
* Compute the mean, the variance and the asymptotic behavior of the Gumbel distribution. Draw the distribution. Explain why <math>z=0</math> is a special point

== Generic case: Universality of the Gumbel distribution ==

The Gumbel distribution is the limit distribution of the maxima of a large class of function. We can say that the Gumbel distribution plays, for extreme statistics, the same role of the Gaussian distribution for the central limit theorem.

By contrast the behavior of <math>a_N</math> and <math>b_N</math> as a function of <math>N</math> strongly depend on the particular distributions <math>p(x)</math>. We discuss here a family of distribution characterized by a fast decay for large <math>x</math>
<center><math>p(x) \sim c e^{- x^\alpha}</math></center>
where <math>\alpha>0</math>
The key point is to be able to determine <math>A(x)</math> such that
<center><math>P^>(x)=\exp(-A(x))</math></center>
* For <math>p(x) = e^{- x}</math> shows <math>A(x)=x</math>
Otherwise <math>A(x)</math> should be determined asymptotically for large <math>x</math>
* Show that <math>A(x)=x^\alpha +(\alpha-1) \log x+...</math>
* Show that in general <math>A(a_N)= \log N+...</math> and compute <math>a_N</math> as a function of <math> \alpha </math> for large <math> N </math>.
* Show that the maximum distribution take the form
<center><math> \lim _{N\to \infty } Q_N(y)=\left( y= a_N+ \frac{z}{A'(a_N)} \right)</math></center>
with <math> z </math> Gumbel distributed
* Identify <math>b_N</math> and discuss its behavior as a function of <math> \alpha </math>

If the distribution <math>p(x)</math> is defined on the entire real axis and is characterized by the same fast decay, it is easy to generalize this result also for the distribution of the minima.

*Write the Gumbel distribution for the minima

==Minimum of exponential random numbers==
The Gumbel distribution is not the only distribution for the extremes. Consider the simple case of the minima of the exponential distribution
* Show analytically that the distribution function for the minimum of <math>N</math> exponential random numbers <math>x = \min(x_1, \dots, x_N) </math> with parameters <math>\lambda_1, \dots \lambda_N</math> is again an exponential random number with parameter <math>\lambda_1 + \cdots + \lambda_N</math>:<br>
<math> \pi(x)= (\lambda_1 + \cdots + \lambda_N) \exp(-(\lambda_1 + \cdots + \lambda_N)x)</math><br> Program this in Python, produce a histogram and compare with the exact result.
* Look on the web which are the possible extreme distributions for independent and identically distributed variable

T-II-2

2011-11-15T14:49:52Z

Groupe 01 13: /* Potential Energy */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===

[[File:image009.jpg]] [[File:image008.jpg]]

We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs as a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> .
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math>V_0\, {\mathcal{V}(\vec r,t)} - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>
where <math>\mathcal V(\vec r,t)</math>is the density of potential energy.

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy, <math>W=\int \, d \vec{r} \, \mathcal V(\vec r,t)</math> writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas, <math>\partial_t \rho=-\rho_0 \vec{\nabla} \cdot \partial_t \vec{u}(\vec{r},t)</math>, show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar field behaves exaclty like an elastic scalar field.
* Determine the sound velocity.

T-II-2

2011-11-15T14:48:20Z

Groupe 01 13: /* Potential Energy */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===

[[File:image009.jpg]] [[File:image008.jpg]]

We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs as a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> .
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math>V_0\, {\mathcal{V}(r,t)} - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>
where <math>\mathcal V(r,t)</math>is the density of potential energy.

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy, <math>W=\int \, d \vec{r} \, \mathcal V(r,t)</math> writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas, <math>\partial_t \rho=-\rho_0 \vec{\nabla} \cdot \partial_t \vec{u}(\vec{r},t)</math>, show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar field behaves exaclty like an elastic scalar field.
* Determine the sound velocity.

T-II-2

2011-11-15T14:47:57Z

Groupe 01 13: /* Potential Energy */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===

[[File:image009.jpg]] [[File:image008.jpg]]

We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs as a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> .
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math>V_0\, {\mathcal{V}(r,t)} - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>
where <math>\mathcal V(r,t)</math>is the density of potential energy.

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy, <math>W=\int \, d \vec{r} \, \mathcal V(r,t)</math> writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas, <math>\partial \rho=-\rho_0 \vec{\nabla} \cdot \partial_t \vec{u}(\vec{r},t)</math>, show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar field behaves exaclty like an elastic scalar field.
* Determine the sound velocity.

T-II-2

2011-11-15T14:45:54Z

Groupe 01 13: /* Elasticity */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===

[[File:image009.jpg]] [[File:image008.jpg]]

We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs as a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> .
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math>V_0\, {\mathcal{V}(r,t)} - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>
where <math>\mathcal V(r,t)</math>is the density of potential energy.

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy, <math>W=\int \, d \vec{r} \, \mathcal V(r,t)</math> writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar field behaves exaclty like an elastic scalar field.
* Determine the sound velocity.

T-II-2

2011-11-15T14:44:50Z

Groupe 01 13: /* Potential Energy */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===

[[File:image009.jpg]] [[File:image008.jpg]]

We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs a a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> .
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math>V_0\, {\mathcal{V}(r,t)} - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>
where <math>\mathcal V(r,t)</math>is the density of potential energy.

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy, <math>W=\int \, d \vec{r} \, \mathcal V(r,t)</math> writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar field behaves exaclty like an elastic scalar field.
* Determine the sound velocity.

T-II-2

2011-11-15T14:43:48Z

Groupe 01 13: /* Potential Energy */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===

[[File:image009.jpg]] [[File:image008.jpg]]

We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs a a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> .
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math>V_0 {\mathcal{V}} - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>
where <math>\mathcal V</math>is the density of potential energy.

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy, <math>W=\int \, d \vec{r} \, \mathcal V</math> writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar field behaves exaclty like an elastic scalar field.
* Determine the sound velocity.

T-II-2

2011-11-15T14:42:27Z

Groupe 01 13: /* Potential Energy */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===

[[File:image009.jpg]] [[File:image008.jpg]]

We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs a a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> .
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math>V_0 {\cal{V}} - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar field behaves exaclty like an elastic scalar field.
* Determine the sound velocity.

T-II-2

2011-11-15T14:42:14Z

Groupe 01 13: /* Potential Energy */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===

[[File:image009.jpg]] [[File:image008.jpg]]

We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs a a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> .
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math>V_0 {\cal V} - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar field behaves exaclty like an elastic scalar field.
* Determine the sound velocity.

T-II-2

2011-11-15T14:40:43Z

Groupe 01 13: /* Dynamics */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===

[[File:image009.jpg]] [[File:image008.jpg]]

We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs a a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> .
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math> - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar field behaves exaclty like an elastic scalar field.
* Determine the sound velocity.

T-II-2

2011-11-15T14:40:13Z

Groupe 01 13: /* Gravity */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===

[[File:image009.jpg]] [[File:image008.jpg]]

We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs a a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> in the small oscillation approximation
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math> - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar field behaves exaclty like an elastic scalar field.
* Determine the sound velocity.

File:Image009.jpg

2011-11-15T10:46:05Z

Groupe 01 13:

T-II-2

2011-11-15T10:45:49Z

Groupe 01 13: /* Elasticity */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===

[[File:image009.jpg]] [[File:image008.jpg]]

We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs a a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

Method 1:

In general the gravitational potential can be written as
<center> <math> E_{g} = g \int_0^1 dx\, \sqrt{1+ (\partial_x y)^2}\, \rho(x)\, y(x).</math></center>
For an elastic string the mass density <math>\rho(x)</math>is not uniform in <math>x</math>. Suppose that each spring has a mass $M/N$ and no mass is accumulated between springs.
* Compute <math>\rho_i</math> for each spring and take the continuum limit.

Method 2:

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> in the small oscillation approximation
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math> - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar field behaves exaclty like an elastic scalar field.
* Determine the sound velocity.

File:Image008.jpg

2011-11-15T10:45:01Z

Groupe 01 13:

T-II-2

2011-11-15T10:44:37Z

Groupe 01 13: /* Elasticity */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===

[[File:image008.jpg]]

We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs a a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

Method 1:

In general the gravitational potential can be written as
<center> <math> E_{g} = g \int_0^1 dx\, \sqrt{1+ (\partial_x y)^2}\, \rho(x)\, y(x).</math></center>
For an elastic string the mass density <math>\rho(x)</math>is not uniform in <math>x</math>. Suppose that each spring has a mass $M/N$ and no mass is accumulated between springs.
* Compute <math>\rho_i</math> for each spring and take the continuum limit.

Method 2:

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> in the small oscillation approximation
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math> - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar field behaves exaclty like an elastic scalar field.
* Determine the sound velocity.

File:Image008.pdf

2011-11-15T10:38:01Z

Groupe 01 13: uploaded a new version of "File:Image008.pdf"

File:Image008.pdf

2011-11-15T10:37:23Z

Groupe 01 13:

T-II-2

2011-11-15T10:36:16Z

Groupe 01 13: /* Elasticity */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===

[[File:image008.pdf]]

We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs a a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

Method 1:

In general the gravitational potential can be written as
<center> <math> E_{g} = g \int_0^1 dx\, \sqrt{1+ (\partial_x y)^2}\, \rho(x)\, y(x).</math></center>
For an elastic string the mass density <math>\rho(x)</math>is not uniform in <math>x</math>. Suppose that each spring has a mass $M/N$ and no mass is accumulated between springs.
* Compute <math>\rho_i</math> for each spring and take the continuum limit.

Method 2:

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> in the small oscillation approximation
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math> - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar field behaves exaclty like an elastic scalar field.
* Determine the sound velocity.

T-II-2

2011-11-15T10:35:27Z

Groupe 01 13: /* Lagrangian description */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===
We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs a a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

Method 1:

In general the gravitational potential can be written as
<center> <math> E_{g} = g \int_0^1 dx\, \sqrt{1+ (\partial_x y)^2}\, \rho(x)\, y(x).</math></center>
For an elastic string the mass density <math>\rho(x)</math>is not uniform in <math>x</math>. Suppose that each spring has a mass $M/N$ and no mass is accumulated between springs.
* Compute <math>\rho_i</math> for each spring and take the continuum limit.

Method 2:

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> in the small oscillation approximation
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math> - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar field behaves exaclty like an elastic scalar field.
* Determine the sound velocity.

T-II-2

2011-11-15T10:34:42Z

Groupe 01 13: /* Lagrangian description */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===
We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs a a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

Method 1:

In general the gravitational potential can be written as
<center> <math> E_{g} = g \int_0^1 dx\, \sqrt{1+ (\partial_x y)^2}\, \rho(x)\, y(x).</math></center>
For an elastic string the mass density <math>\rho(x)</math>is not uniform in <math>x</math>. Suppose that each spring has a mass $M/N$ and no mass is accumulated between springs.
* Compute <math>\rho_i</math> for each spring and take the continuum limit.

Method 2:

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> in the small oscillation approximation
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math> - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + P_0 \vec{\nabla} \cdot \vec{u} -\frac{\gamma}{2} P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar fiel behaves exaclty like an elastic field.
* Determine the sound velocity.

T-II-2

2011-11-15T10:34:10Z

Groupe 01 13: /* Dynamics */

__FORCETOC__

=Variational methods=

==Elastic line==

During the class you determined the shape of an non-extensible string in a gravitational potential.
Here, we study a different situation: the case of an elastic line. Hooke's law states that Elastic energy is proportional to the square of the global elongation.

===Elasticity===
We write in two Step the elastic energy of a string with an elastic constant <math> k </math>. The shape of the string is defined by the function <math> y(x) </math>, where <math> 0<x <1</math> and
<math> y(0)=y(1)=0</math>.

Step one: the discrete model
* Write the elastic energy of a chain of N identical springs a a function of their position <math> y_1, y_2,\ldots,y_N </math> (See figure).
* Determine the elastic constant of each spring in order to have a a string with an elastic constant <math> k </math>

Step two: The continuum limit
* Take the limit <math> N \to \infty</math> and express the elastic energy as a functional of <math> y(x) </math>. (Input: use <math> 1/N \to \text{d} x</math> and <math>\frac{y_{i+1}-y_i}{1/N} \to \frac{\text{d} y(x)}{\text{d} x }</math>)

===Gravity===
We write now the gravitational potential as a functional of <math>y(x)</math>. We will use two different methods. We suppose that the total mass of the line in <math>M</math>.

Method 1:

In general the gravitational potential can be written as
<center> <math> E_{g} = g \int_0^1 dx\, \sqrt{1+ (\partial_x y)^2}\, \rho(x)\, y(x).</math></center>
For an elastic string the mass density <math>\rho(x)</math>is not uniform in <math>x</math>. Suppose that each spring has a mass $M/N$ and no mass is accumulated between springs.
* Compute <math>\rho_i</math> for each spring and take the continuum limit.

Method 2:

* Suppose that springs are massless and <math>N</math> identical masses are located at positions <math>y_1,y_2,\ldots</math>

* Combine your finding for elasticity and gravity and show that the shape of the elastic line is parabolic

=== Dynamics ===

We want to study the motion of the elastic line in absence of gravity and in presence of small oscillations. The shape of the line is now time dependent <math> y(x,t)</math>.

*Write the kinetic energy associated to the <math> N </math> masses located at <math>y_1(t),y_2(t),\ldots, y_N(t) </math>. Take the continuum limit.

The action associated to the elastic chain is
<center><math> \int_{t_1}^{t_2} d t \int_0^1 d x \, \mathcal{L} (x,t) </math></center>
where <math> {\mathcal L}(x,t) </math> is the Lagrangian denisty
* Write explicitly <math> {\mathcal L}(x,t) </math> in the small oscillation approximation
* Write the equation of motion of the chain using the minimum action principle. This equation is the D'Almbert equation.

==Sound waves in ideal gas==

Consider a uniform ideal gas of equilibrium mass density <math>\rho_0 </math> and equilibrium pressure <math>P_0</math>. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves.

===Kinetic Energy===

In a first time we can start from the discrete description of a system of N atoms at positions <math> {\vec u}_1(t), {\vec u}_2(t) ,\ldots , {\vec u}_N(t) </math>.
* Write the kinetic energy, <math> T </math>, for this system as a function of mass of the atom <math>m </math>.
* Consider the limit of small oscillations. Take the continuum limit: <math> {\vec u}_i(t) \to \vec u(\vec r,t) </math> and show that the kinetic energy writes
<center><math> T= \frac{\rho_0}{2} \int \, d \vec{r} \, \dot{\vec{u}}^2 (\vec r,t) </math> </center>
===Potential Energy===
In order to write the full Lagrangian we need to find the potential energy associated to the oscillations. During the oscillations the mass density can have small fluctuations
<center><math> \rho(\vec r,t)=\rho_0+\delta \rho(\vec r,t)=\rho_0\left( 1+\sigma(\vec r,t)\right)</math> </center>
The scalar field <math> \sigma(\vec r,t)</math> represents the relative fluctuation of the density.

As a consequence of these fluctuations, the volume which contains a given mass <math>M </math> expands from the initial value <math>V_0 </math> to a new value <math>V_0 +\delta V</math>. The pressure inside the domain also fluctuates <math>P=P_0+\ldots </math>. The variation of potential energy is given by the work associated to the volume expansion:
<center><math> - \int_{V_0}^{V_0+\delta V} \, P\, d \vec{r}=-P_0 \delta V- \frac{1}{2} \left( \frac{\partial P }{\partial V}\right)_0 \delta V^2+\ldots </math> </center>

Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,
<center><math>P V^\gamma= </math> constant</center>
where <math>\gamma</math> the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately <math>1.4 </math> for ordinary air.
* Show that the total potential energy writes
<center><math> W= P_0 \int \, d \vec{r} \, \left( \sigma + \frac{\gamma}{2} \sigma^2\right) </math> </center>
* Using the continuity equation of the gas show that <math> \sigma(\vec{r},t)= - \vec{\nabla} \cdot \vec{u}(\vec{r},t) </math> and write the potential energy as a function of <math> \vec{u}(\vec{r},t) </math>

===Lagrangian description===
* Show that the Lagrangian density writes
<center><math> \mathcal{L} (\vec{r},t) = \frac{\rho_0}{2} \dot{\vec{u}}^2 + 2 P_0 \vec{\nabla} \cdot \vec{u} -\gamma P_0 (\vec{\nabla} \cdot \vec{u})^2 </math></center>
* Show that the associated equation of motion
<center><math> \rho_0 \frac{\partial^2 \vec{u}}{\partial t^2} -\gamma P_0 \vec{\nabla} \vec{\nabla} \cdot \vec{u}=0 </math></center>
* Using the continuity relation <math> \sigma= - \vec{\nabla} \cdot \vec{u} </math> show that the latter equation corresponds to the D'Alembert Equation for the scalar field <math> \sigma </math>
This scalar fiel behaves exaclty like an elastic field.
* Determine the sound velocity.

Cameroun

2011-11-03T17:14:58Z

Groupe 01 13: /* The Cole Hopf Transformation */

__FORCETOC__

= Waves breaking, Burgers and Shocks=

Far from the beach , i.e., when the sea is deep, water waves travel on the surface of the sea and their shape is not deformed too much. When the waves approach the beach their shape begins to change and eventually breaks upon touching the seashore.

Here we are interested in studying a simple model that captures the essential features of this physical system. More precisely we will examine the Burgers equation, a non-linear equation introduced as a toy model for turbulence. This equation can be explicitly integrated and, in the limit of zero viscosity, its solution generally displays discontinuities called shocks.

=Model 0 : Deep sea=
Consider the following equation for the shape of the wave:

<center><math>\partial_t u(x,t) + v \, \partial_x u(x,t) = 0 </math></center>

where <math>u(x,t) </math> is the height of the wave with respect to the unperturbed sea level, and <math> v </math> is constant. We will use the following initial condition <math>u(x,t=0)=u_{\text{init}}(x) </math>

<center> <math>u_{\text{init}}(x) \begin{cases} 0 \quad \text{for} \; |x|>1 \\ 1-x^2 \quad \text{for} \; |x|<1 \end{cases} \quad \quad \quad (0) </math></center>

* Find the solution of the latter equation and compute the velocity of the wave
* Discuss why this model may mimic wave motion in deep sea.

=Model 1 : Approaching the sea shore=

It is possible to show that, when the height of the wave is negligible as compared to the sea depth, the velocity of the wave does not depend on <math>u</math>. On the contrary, when approaching the sea shore this is not anymore true and the velocity becomes proportional to <math>u</math>:
<center>
<math>
v(u) \simeq c_1 u +v_0
</math>
</center>
where <math>c_1</math> is a constant and <math>v_0</math> is the velocity of the bottom of the wave. Let us take <math>c_1=1</math> and write the equation for the wave motion in the frame of the bottom of the wave, which yields:

<center>
<math>
\partial_t u(x,t) + u(x,t) \partial_x u(x,t) = 0 \quad. \quad \quad\quad (1)
</math>
</center>

This is the non linear equation originally introduced by Burgers. In Burgers' derivation a viscous term was also present
<center>
<math>
\partial_t u(x,t) + u(x,t) \partial_x u(x,t) = \nu \partial_x^2 u(x,t), \quad \quad\quad (2)
</math>
</center>
<math>\nu</math> being the viscosity, and Eq. (1) is called the inviscid limit of Eq. 2.

Our strategy will be to investigate the properties of Eq1 by resorting to the method of characteristics. Moreover, we will integrate Eq 2 by using the Cole Hopf transformation, and then recover Eq.1 by taking the limit <math>\nu \to 0</math>.

== The Method of characteristics==

* Using the method of characteristics show that the solution of Eq.1 writes:
<center> <math>\begin{cases} u (x,t) =u_{\text{init}}(x_0,0)\\ x(t)=x_0+ u_{\text{init}}(x_0,0)\, t \end{cases} </math></center>
* Give an interpretation for <math> x_0</math>
* Show that <math> u(x,t)=0</math> is alway a solution for <math> | x | > 1 </math>. Denote by <math> u_0(x,t)</math> this solution.
* Show that two other solutions exist, namely:
<center><math>
u_{\pm}(x,t)= 1-\frac{1}{4 t^2} \left( 1 \pm \sqrt{\Delta(x,t)}\right)^2
</math></center>
where <math> \Delta(x,t)= 4 t^2 -4 x t +1 </math>.
* Show that for <math> t<0.5 </math> only <math> u_0(x,t)</math> and <math> u_-(x,t)</math> are acceptable.
* Draw the full solution for <math> t=0.5 </math> and compute its derivative at <math> x=1 </math>
* Show that for <math> t>0.5 </math> the solution is a multivalued function <math> u_{\text{mult}}(x,t)</math>
<center>
<math>
u_{\text{mult}}(x,t)= \begin{cases} u_0(x,t) \quad \text{ for} \; | x | > 1 \\
u_-(x,t) \quad \text{ for} \; -1< x < \frac{1}{4 t}+t \\
u_+(x,t) \quad \text{ for} \; 1< x < \frac{1}{4 t}+t
\end{cases}
</math></center>
and draw some examples of solutions for various values of <math> t</math>.

We look now for a solution which is a single-valued function. For <math> t>0.5 </math> this function should display a discontinuity called shock. We will show in the second part of this exercise that the solution <math> u_+(x,t)</math> is always unstable. The single valued solution for <math> t>0.5 </math> takes the form
<center>
<math>
u_{\text{single}}(x,t)= \begin{cases} u_0(x,t) \quad \text{ for} \; x <-1 \\
u_-(x,t) \quad \text{ for} \; -1< x < x_{\text{shock}}(t) \\
u_0(x,t) \quad \text{ for} \; x > x_{\text{shock}}(t)
\end{cases}
</math>
</center>
where <math> x_{\text{shock}}(t) </math> indicates the shock location. The position <math> x_{\text{shock}}(t) </math> is imposed by a conservation law that should be valid at all times.

* Find the conservation law for the wave and determine <math> x_{\text{shock}}(t) </math>.

== The Cole Hopf Transformation==

Consider now Eq. (2) with <math> \nu>0 </math>. This equation can be integrated via the Cole Hopf transformation which amounts to perform the following
change of variable:

<center>
<math>
u = -2\nu \frac{\partial}{\partial x} \ln W
</math>
</center>
This can be done in three steps :
(i) <math> u(x,t)=-2\nu \partial_x h(x,t) </math>,
(ii) integrate the equation that you have obtained,
(iii) substitute back <math> h(x,t) = \ln W(x,t) </math>.

The diffusion equation satisfied by <math> W(x,t) </math> can be solved:

<math>W(x,t)=\int d y e^{-\frac{(x-y)^2}{2 \nu t}} W(y,0) </math>

* Obtain an explicit expression for <math> u(x,t) </math>, solution of Eq. (2) with initial condition in Eq. (0).

* Show that, in the limit <math> \nu \to 0 </math>, the profile <math> u(x,t) </math> can be written as
<center>
<math>
u(x,t)= \frac{x-y_{\text{min}}(x)}{t}
</math>
</center>
where <math>y_{\text{min}}(x) </math> is the location of the global minimum of the function
<center>
<math>
E_{x,t}(y)= \frac{1}{2 t} \,(x-y)^2+ V(y)
</math>
</center>
with <math> \partial_y V(y) = u_{\text{init}}(y) </math>. The initial condition <math> u_{\text{init}}(y) </math> is given in Eq.0.
* Show that the solutions <math> u_0 </math>, <math> u_+ </math> and <math> u_- </math> can be interpreted as the points where <math>
E_{x,t}(y)
</math> is stationary, i.e.
<center>
<math>
\partial_y E_{x,t}(y)= 0
</math>
</center>
*Draw <math>
E_{x,t}(y)
</math> at <math>t=1</math> for different value of <math>x</math>:
<math>x_1< -1, \quad -1< x_2<1, \quad 1<x_3<x_s(1), \quad x_4=x_s(1), \quad x_s(1)< x_5<5/4, \quad x_6>5/4 </math>

*Comment on the obtained results.

Cameroun

2011-11-03T14:03:33Z

Groupe 01 13: /* The Cole Hopf Transformation */

__FORCETOC__

= Waves breaking, Burgers and Shocks=

Far from the beach , i.e., when the sea is deep, water waves travel on the surface of the sea and their shape is not deformed too much. When the waves approach the beach their shape begins to change and eventually breaks upon touching the seashore.

Here we are interested in studying a simple model that captures the essential features of this physical system. More precisely we will examine the Burgers equation, a non-linear equation introduced as a toy model for turbulence. This equation can be explicitly integrated and, in the limit of zero viscosity, its solution generally displays discontinuities called shocks.

=Model 0 : Deep sea=
Consider the following equation for the shape of the wave:

<center><math>\partial_t u(x,t) + v \, \partial_x u(x,t) = 0 </math></center>

where <math>u(x,t) </math> is the height of the wave with respect to the unperturbed sea level, and <math> v </math> is constant. We will use the following initial condition <math>u(x,t=0)=u_{\text{init}}(x) </math>

<center> <math>u_{\text{init}}(x) \begin{cases} 0 \quad \text{for} \; |x|>1 \\ 1-x^2 \quad \text{for} \; |x|<1 \end{cases} \quad \quad \quad (0) </math></center>

* Find the solution of the latter equation and compute the velocity of the wave
* Discuss why this model may mimic wave motion in deep sea.

=Model 1 : Approaching the sea shore=

It is possible to show that, when the height of the wave is negligible as compared to the sea depth, the velocity of the wave does not depend on <math>u</math>. On the contrary, when approaching the sea shore this is not anymore true and the velocity becomes proportional to <math>u</math>:
<center>
<math>
v(u) \simeq c_1 u +v_0
</math>
</center>
where <math>c_1</math> is a constant and <math>v_0</math> is the velocity of the bottom of the wave. Let us take <math>c_1=1</math> and write the equation for the wave motion in the frame of the bottom of the wave, which yields:

<center>
<math>
\partial_t u(x,t) + u(x,t) \partial_x u(x,t) = 0 \quad. \quad \quad\quad (1)
</math>
</center>

This is the non linear equation originally introduced by Burgers. In Burgers' derivation a viscous term was also present
<center>
<math>
\partial_t u(x,t) + u(x,t) \partial_x u(x,t) = \nu \partial_x^2 u(x,t), \quad \quad\quad (2)
</math>
</center>
<math>\nu</math> being the viscosity, and Eq. (1) is called the inviscid limit of Eq. 2.

Our strategy will be to investigate the properties of Eq1 by resorting to the method of characteristics. Moreover, we will integrate Eq 2 by using the Cole Hopf transformation, and then recover Eq.1 by taking the limit <math>\nu \to 0</math>.

== The Method of characteristics==

* Using the method of characteristics show that the solution of Eq.1 writes:
<center> <math>\begin{cases} u (x,t) =u_{\text{init}}(x_0,0)\\ x(t)=x_0+ u_{\text{init}}(x_0,0)\, t \end{cases} </math></center>
* Give an interpretation for <math> x_0</math>
* Show that <math> u(x,t)=0</math> is alway a solution for <math> | x | > 1 </math>. Denote by <math> u_0(x,t)</math> this solution.
* Show that two other solutions exist, namely:
<center><math>
u_{\pm}(x,t)= 1-\frac{1}{4 t^2} \left( 1 \pm \sqrt{\Delta(x,t)}\right)^2
</math></center>
where <math> \Delta(x,t)= 4 t^2 -4 x t +1 </math>.
* Show that for <math> t<0.5 </math> only <math> u_0(x,t)</math> and <math> u_-(x,t)</math> are acceptable.
* Draw the full solution for <math> t=0.5 </math> and compute its derivative at <math> x=1 </math>
* Show that for <math> t>0.5 </math> the solution is a multivalued function <math> u_{\text{mult}}(x,t)</math>
<center>
<math>
u_{\text{mult}}(x,t)= \begin{cases} u_0(x,t) \quad \text{ for} \; | x | > 1 \\
u_-(x,t) \quad \text{ for} \; -1< x < \frac{1}{4 t}+t \\
u_+(x,t) \quad \text{ for} \; 1< x < \frac{1}{4 t}+t
\end{cases}
</math></center>
and draw some examples of solutions for various values of <math> t</math>.

We look now for a solution which is a single-valued function. For <math> t>0.5 </math> this function should display a discontinuity called shock. We will show in the second part of this exercise that the solution <math> u_+(x,t)</math> is always unstable. The single valued solution for <math> t>0.5 </math> takes the form
<center>
<math>
u_{\text{single}}(x,t)= \begin{cases} u_0(x,t) \quad \text{ for} \; x <-1 \\
u_-(x,t) \quad \text{ for} \; -1< x < x_{\text{shock}}(t) \\
u_0(x,t) \quad \text{ for} \; x > x_{\text{shock}}(t)
\end{cases}
</math>
</center>
where <math> x_{\text{shock}}(t) </math> indicates the shock location. The position <math> x_{\text{shock}}(t) </math> is imposed by a conservation law that should be valid at all times.

* Find the conservation law for the wave and determine <math> x_{\text{shock}}(t) </math>.

== The Cole Hopf Transformation==

Consider now Eq. (2) with <math> \nu>0 </math>. This equation can be integrated via the Cole Hopf transformation which amounts to perform the following
change of variable:

<center>
<math>
u = -2\nu \frac{\partial}{\partial x} \ln W
</math>
</center>
This can be done in three steps :
(i) <math> u(x,t)=-2\nu \partial_x h(x,t) </math>,
(ii) integrate the equation that you have obtained,
(iii) substitute back <math> \h(x,t) = \ln W(x,t) </math>.

The diffusion equation satisfied by <math> W(x,t) </math> can be solved:

<math>W(x,t)=\int d y e^{-\frac{(x-y)^2}{2 \nu t}} W(y,0) </math>

* Obtain an explicit expression for <math> u(x,t) </math>, solution of Eq. (2) with initial condition in Eq. (0).

* Show that, in the limit <math> \nu \to 0 </math>, the profile <math> u(x,t) </math> can be written as
<center>
<math>
u(x,t)= \frac{x-y_{\text{min}}(x)}{t}
</math>
</center>
where <math>y_{\text{min}}(x) </math> is the location of the global minimum of the function
<center>
<math>
E_{x,t}(y)= \frac{1}{2 t} \,(x-y)^2+ V(y)
</math>
</center>
with <math> \partial_y V(y) = u_{\text{init}}(y) </math>. The initial condition <math> u_{\text{init}}(y) </math> is given in Eq.0.
* Show that the solutions <math> u_0 </math>, <math> u_+ </math> and <math> u_- </math> can be interpreted as the points where <math>
E_{x,t}(y)
</math> is stationary, i.e.
<center>
<math>
\partial_y E_{x,t}(y)= 0
</math>
</center>
*Draw <math>
E_{x,t}(y)
</math> at <math>t=1</math> for different value of <math>x</math>:
<math>x_1< -1, \quad -1< x_2<1, \quad 1<x_3<x_s(1), \quad x_4=x_s(1), \quad x_s(1)< x_5<5/4, \quad x_6>5/4 </math>

*Comment on the obtained results.

Cameroun

2011-11-03T12:42:04Z

Groupe 01 13: /* The Cole Hopf Transformation */

__FORCETOC__

= Waves breaking, Burgers and Shocks=

Far from the beach , i.e., when the sea is deep, water waves travel on the surface of the sea and their shape is not deformed too much. When the waves approach the beach their shape begins to change and eventually breaks upon touching the seashore.

Here we are interested in studying a simple model that captures the essential features of this physical system. More precisely we will examine the Burgers equation, a non-linear equation introduced as a toy model for turbulence. This equation can be explicitly integrated and, in the limit of zero viscosity, its solution generally displays discontinuities called shocks.

=Model 0 : Deep sea=
Consider the following equation for the shape of the wave:

<center><math>\partial_t u(x,t) + v \, \partial_x u(x,t) = 0 </math></center>

where <math>u(x,t) </math> is the height of the wave with respect to the unperturbed sea level, and <math> v </math> is constant. We will use the following initial condition <math>u(x,t=0)=u_{\text{init}}(x) </math>

<center> <math>u_{\text{init}}(x) \begin{cases} 0 \quad \text{for} \; |x|>1 \\ 1-x^2 \quad \text{for} \; |x|<1 \end{cases} \quad \quad \quad (0) </math></center>

* Find the solution of the latter equation and compute the velocity of the wave
* Discuss why this model may mimic wave motion in deep sea.

=Model 1 : Approaching the sea shore=

It is possible to show that, when the height of the wave is negligible as compared to the sea depth, the velocity of the wave does not depend on <math>u</math>. On the contrary, when approaching the sea shore this is not anymore true and the velocity becomes proportional to <math>u</math>:
<center>
<math>
v(u) \simeq c_1 u +v_0
</math>
</center>
where <math>c_1</math> is a constant and <math>v_0</math> is the velocity of the bottom of the wave. Let us take <math>c_1=1</math> and write the equation for the wave motion in the frame of the bottom of the wave, which yields:

<center>
<math>
\partial_t u(x,t) + u(x,t) \partial_x u(x,t) = 0 \quad. \quad \quad\quad (1)
</math>
</center>

This is the non linear equation originally introduced by Burgers. In Burgers' derivation a viscous term was also present
<center>
<math>
\partial_t u(x,t) + u(x,t) \partial_x u(x,t) = \nu \partial_x^2 u(x,t), \quad \quad\quad (2)
</math>
</center>
<math>\nu</math> being the viscosity, and Eq. (1) is called the inviscid limit of Eq. 2.

Our strategy will be to investigate the properties of Eq1 by resorting to the method of characteristics. Moreover, we will integrate Eq 2 by using the Cole Hopf transformation, and then recover Eq.1 by taking the limit <math>\nu \to 0</math>.

== The Method of characteristics==

* Using the method of characteristics show that the solution of Eq.1 writes:
<center> <math>\begin{cases} u (x,t) =u_{\text{init}}(x_0,0)\\ x(t)=x_0+ u_{\text{init}}(x_0,0)\, t \end{cases} </math></center>
* Give an interpretation for <math> x_0</math>
* Show that <math> u(x,t)=0</math> is alway a solution for <math> | x | > 1 </math>. Denote by <math> u_0(x,t)</math> this solution.
* Show that two other solutions exist, namely:
<center><math>
u_{\pm}(x,t)= 1-\frac{1}{4 t^2} \left( 1 \pm \sqrt{\Delta(x,t)}\right)^2
</math></center>
where <math> \Delta(x,t)= 4 t^2 -4 x t +1 </math>.
* Show that for <math> t<0.5 </math> only <math> u_0(x,t)</math> and <math> u_-(x,t)</math> are acceptable.
* Draw the full solution for <math> t=0.5 </math> and compute its derivative at <math> x=1 </math>
* Show that for <math> t>0.5 </math> the solution is a multivalued function <math> u_{\text{mult}}(x,t)</math>
<center>
<math>
u_{\text{mult}}(x,t)= \begin{cases} u_0(x,t) \quad \text{ for} \; | x | > 1 \\
u_-(x,t) \quad \text{ for} \; -1< x < \frac{1}{4 t}+t \\
u_+(x,t) \quad \text{ for} \; 1< x < \frac{1}{4 t}+t
\end{cases}
</math></center>
and draw some examples of solutions for various values of <math> t</math>.

We look now for a solution which is a single-valued function. For <math> t>0.5 </math> this function should display a discontinuity called shock. We will show in the second part of this exercise that the solution <math> u_+(x,t)</math> is always unstable. The single valued solution for <math> t>0.5 </math> takes the form
<center>
<math>
u_{\text{single}}(x,t)= \begin{cases} u_0(x,t) \quad \text{ for} \; x <-1 \\
u_-(x,t) \quad \text{ for} \; -1< x < x_{\text{shock}}(t) \\
u_0(x,t) \quad \text{ for} \; x > x_{\text{shock}}(t)
\end{cases}
</math>
</center>
where <math> x_{\text{shock}}(t) </math> indicates the shock location. The position <math> x_{\text{shock}}(t) </math> is imposed by a conservation law that should be valid at all times.

* Find the conservation law for the wave and determine <math> x_{\text{shock}}(t) </math>.

== The Cole Hopf Transformation==

Consider now Eq. (2) with <math> \nu>0 </math>. This equation can be integrated via the Cole Hopf transformation which amounts to perform the following
change of variable:

<center>
<math>
u = -2\nu \frac{\partial}{\partial x} \ln W
</math>
</center>
This can be done in three steps :
(i) <math> u(x,t)=-2\nu \partial_x h(x,t) </math>,
(ii) integrate the equation that you have obtained,
(iii) substitute back <math> \psi(x,t) = \ln W(x,t) </math>.

The diffusion equation satisfied by <math> W(x,t) </math> can be solved:

<math>W(x,t)=\int d y e^{-\frac{(x-y)^2}{2 \nu t}} W(y,0) </math>

* Obtain an explicit expression for <math> u(x,t) </math>, solution of Eq. (2) with initial condition in Eq. (0).

* Show that, in the limit <math> \nu \to 0 </math>, the profile <math> u(x,t) </math> can be written as
<center>
<math>
u(x,t)= \frac{x-y_{\text{min}}(x)}{t}
</math>
</center>
where <math>y_{\text{min}}(x) </math> is the location of the global minimum of the function
<center>
<math>
E_{x,t}(y)= \frac{1}{2 t} \,(x-y)^2+ V(y)
</math>
</center>
with <math> \partial_y V(y) = u_{\text{init}}(y) </math>. The initial condition <math> u_{\text{init}}(y) </math> is given in Eq.0.
* Show that the solutions <math> u_0 </math>, <math> u_+ </math> and <math> u_- </math> can be interpreted as the points where <math>
E_{x,t}(y)
</math> is stationary, i.e.
<center>
<math>
\partial_y E_{x,t}(y)= 0
</math>
</center>
*Draw <math>
E_{x,t}(y)
</math> at <math>t=1</math> for different value of <math>x</math>:
<math>x_1< -1, \quad -1< x_2<1, \quad 1<x_3<x_s(1), \quad x_4=x_s(1), \quad x_s(1)< x_5<5/4, \quad x_6>5/4 </math>

*Comment on the obtained results.

Cameroun

2011-11-03T12:41:22Z

Groupe 01 13: Created page with "__FORCETOC__ = Waves breaking, Burgers and Shocks= Far from the beach , i.e., when the sea is deep, water waves travel on the surface of the sea and their shape is not deform..."

__FORCETOC__

= Waves breaking, Burgers and Shocks=

Far from the beach , i.e., when the sea is deep, water waves travel on the surface of the sea and their shape is not deformed too much. When the waves approach the beach their shape begins to change and eventually breaks upon touching the seashore.

Here we are interested in studying a simple model that captures the essential features of this physical system. More precisely we will examine the Burgers equation, a non-linear equation introduced as a toy model for turbulence. This equation can be explicitly integrated and, in the limit of zero viscosity, its solution generally displays discontinuities called shocks.

=Model 0 : Deep sea=
Consider the following equation for the shape of the wave:

<center><math>\partial_t u(x,t) + v \, \partial_x u(x,t) = 0 </math></center>

where <math>u(x,t) </math> is the height of the wave with respect to the unperturbed sea level, and <math> v </math> is constant. We will use the following initial condition <math>u(x,t=0)=u_{\text{init}}(x) </math>

<center> <math>u_{\text{init}}(x) \begin{cases} 0 \quad \text{for} \; |x|>1 \\ 1-x^2 \quad \text{for} \; |x|<1 \end{cases} \quad \quad \quad (0) </math></center>

* Find the solution of the latter equation and compute the velocity of the wave
* Discuss why this model may mimic wave motion in deep sea.

=Model 1 : Approaching the sea shore=

It is possible to show that, when the height of the wave is negligible as compared to the sea depth, the velocity of the wave does not depend on <math>u</math>. On the contrary, when approaching the sea shore this is not anymore true and the velocity becomes proportional to <math>u</math>:
<center>
<math>
v(u) \simeq c_1 u +v_0
</math>
</center>
where <math>c_1</math> is a constant and <math>v_0</math> is the velocity of the bottom of the wave. Let us take <math>c_1=1</math> and write the equation for the wave motion in the frame of the bottom of the wave, which yields:

<center>
<math>
\partial_t u(x,t) + u(x,t) \partial_x u(x,t) = 0 \quad. \quad \quad\quad (1)
</math>
</center>

This is the non linear equation originally introduced by Burgers. In Burgers' derivation a viscous term was also present
<center>
<math>
\partial_t u(x,t) + u(x,t) \partial_x u(x,t) = \nu \partial_x^2 u(x,t), \quad \quad\quad (2)
</math>
</center>
<math>\nu</math> being the viscosity, and Eq. (1) is called the inviscid limit of Eq. 2.

Our strategy will be to investigate the properties of Eq1 by resorting to the method of characteristics. Moreover, we will integrate Eq 2 by using the Cole Hopf transformation, and then recover Eq.1 by taking the limit <math>\nu \to 0</math>.

== The Method of characteristics==

* Using the method of characteristics show that the solution of Eq.1 writes:
<center> <math>\begin{cases} u (x,t) =u_{\text{init}}(x_0,0)\\ x(t)=x_0+ u_{\text{init}}(x_0,0)\, t \end{cases} </math></center>
* Give an interpretation for <math> x_0</math>
* Show that <math> u(x,t)=0</math> is alway a solution for <math> | x | > 1 </math>. Denote by <math> u_0(x,t)</math> this solution.
* Show that two other solutions exist, namely:
<center><math>
u_{\pm}(x,t)= 1-\frac{1}{4 t^2} \left( 1 \pm \sqrt{\Delta(x,t)}\right)^2
</math></center>
where <math> \Delta(x,t)= 4 t^2 -4 x t +1 </math>.
* Show that for <math> t<0.5 </math> only <math> u_0(x,t)</math> and <math> u_-(x,t)</math> are acceptable.
* Draw the full solution for <math> t=0.5 </math> and compute its derivative at <math> x=1 </math>
* Show that for <math> t>0.5 </math> the solution is a multivalued function <math> u_{\text{mult}}(x,t)</math>
<center>
<math>
u_{\text{mult}}(x,t)= \begin{cases} u_0(x,t) \quad \text{ for} \; | x | > 1 \\
u_-(x,t) \quad \text{ for} \; -1< x < \frac{1}{4 t}+t \\
u_+(x,t) \quad \text{ for} \; 1< x < \frac{1}{4 t}+t
\end{cases}
</math></center>
and draw some examples of solutions for various values of <math> t</math>.

We look now for a solution which is a single-valued function. For <math> t>0.5 </math> this function should display a discontinuity called shock. We will show in the second part of this exercise that the solution <math> u_+(x,t)</math> is always unstable. The single valued solution for <math> t>0.5 </math> takes the form
<center>
<math>
u_{\text{single}}(x,t)= \begin{cases} u_0(x,t) \quad \text{ for} \; x <-1 \\
u_-(x,t) \quad \text{ for} \; -1< x < x_{\text{shock}}(t) \\
u_0(x,t) \quad \text{ for} \; x > x_{\text{shock}}(t)
\end{cases}
</math>
</center>
where <math> x_{\text{shock}}(t) </math> indicates the shock location. The position <math> x_{\text{shock}}(t) </math> is imposed by a conservation law that should be valid at all times.

* Find the conservation law for the wave and determine <math> x_{\text{shock}}(t) </math>.

== The Cole Hopf Transformation==

Consider now Eq. (2) with <math> \nu>0 </math>. This equation can be integrated via the Cole Hopf transformation which amounts to perform the following
change of variable:

<center>
<math>
u = -2\nu \frac{\partial}{\partial x} \ln W
</math>
</center>
This can be done in three steps :
(i) <math> u(x,t)=-2\nu \partial_x h(x,t) </math>,
(ii) integrate the equation that you have obtained,
(iii) substitute back <math> \psi(x,t) = \ln W(x,t) </math>.

The diffusion equation satisfied by <math> W(x,t) </math> can be solved:
<math>W(x,t)=\int d y e^{-\frac{(x-y)^2}{2 \nu t}} W(y,0) </math>

* Obtain an explicit expression for <math> u(x,t) </math>, solution of Eq. (2) with initial condition in Eq. (0).

* Show that, in the limit <math> \nu \to 0 </math>, the profile <math> u(x,t) </math> can be written as
<center>
<math>
u(x,t)= \frac{x-y_{\text{min}}(x)}{t}
</math>
</center>
where <math>y_{\text{min}}(x) </math> is the location of the global minimum of the function
<center>
<math>
E_{x,t}(y)= \frac{1}{2 t} \,(x-y)^2+ V(y)
</math>
</center>
with <math> \partial_y V(y) = u_{\text{init}}(y) </math>. The initial condition <math> u_{\text{init}}(y) </math> is given in Eq.0.
* Show that the solutions <math> u_0 </math>, <math> u_+ </math> and <math> u_- </math> can be interpreted as the points where <math>
E_{x,t}(y)
</math> is stationary, i.e.
<center>
<math>
\partial_y E_{x,t}(y)= 0
</math>
</center>
*Draw <math>
E_{x,t}(y)
</math> at <math>t=1</math> for different value of <math>x</math>:
<math>x_1< -1, \quad -1< x_2<1, \quad 1<x_3<x_s(1), \quad x_4=x_s(1), \quad x_s(1)< x_5<5/4, \quad x_6>5/4 </math>

*Comment on the obtained results.