T-II-1: Difference between revisions

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<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> <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
# Find the solution of the latter equation and compute the velocity of the wave
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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>:
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>
<math>
v(u) \simeq c_1 u +v_0   
v(u) \simeq c_1 u +v_0   
</math>
</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:
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>
<math>
\partial_t u(x,t) =  u(x,t) \partial_x u(x,t).
\partial_t u(x,t) =  u(x,t) \partial_x u(x,t). \quad \quad\quad (1)
</math>
</math>
</center>


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


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# 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  <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:  
# Show that  two other solutions exist, namely:  
<math>
<center><math>
u_{\pm}(x,t)= \frac{1}{4 t^2} \left( 1 \pm \sqrt{\Delta(x,t)}\right)^2-1
u_{\pm}(x,t)= \frac{1}{4 t^2} \left( 1 \pm \sqrt{\Delta(x,t)}\right)^2-1
</math>
</math></center>
 
where  <math> \Delta(x,t)= 4 t^2 -4 x t +1  </math>.
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 physically acceptable.  
# Show that for <math> t<0.5  </math> only <math> u_0(x,t)</math> and <math> u_-(x,t)</math> are physically acceptable.  
# Draw the full solution for  <math> t=0.5  </math> and compute its derivative at  <math> x=1  </math>
# 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>
# Show that for <math> t>0.5  </math> the solution is a multivalued  function <math> u_{\text{mult}}(x,t)</math>
 
<center>
<math>
<math>
u_{\text{mult}}(x,t)= \begin{cases}  u_0(x,t)  \quad \text{ for} \;  | x | > 1 \\  
u_{\text{mult}}(x,t)= \begin{cases}  u_0(x,t)  \quad \text{ for} \;  | x | > 1 \\  
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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}
  \end{cases}
</math>
</math></center>
 
and draw some examples of solutions for various values of <math> t</math>.
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
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>
<math>
u_{\text{single}}(x,t)= \begin{cases}  u_0(x,t)  \quad \text{ for} \;    x  <-1  \\  
u_{\text{single}}(x,t)= \begin{cases}  u_0(x,t)  \quad \text{ for} \;    x  <-1  \\  
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  \end{cases}
  \end{cases}
</math>
</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.  
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>.
# Find the conservation law for the wave and determine <math> x_{\text{shock}}(t) </math>.

Revision as of 13:58, 20 September 2011


EDP: 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:

where is the height of the wave with respect to the unperturbed sea level, and c is constant. We will use the following initial condition


  1. Find the solution of the latter equation and compute the velocity of the wave
  2. 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 . On the contrary, when approaching the sea shore this is not anymore true and the velocity becomes proportional to :

where is a constant and is the velocity of the bottom of the wave. Let us take and write the equation for the wave motion in the frame of the bottom of the wave, which yields:

This is the non linear equation originally introduced by Burgers. In Burgers' derivation a viscous term was also present

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 .

The Method of characteristics

  1. Write the solution of Eq.1 with the initial condition of Eq.0.
  2. Show that is alway a solution for . Denote by this solution.
  3. Show that two other solutions exist, namely:

where .

  1. Show that for only and are physically acceptable.
  2. Draw the full solution for and compute its derivative at
  3. Show that for the solution is a multivalued function

and draw some examples of solutions for various values of .

We look now for a solution which is a single-valued function. For this function should display a discontinuity called shock. We will show in the second part of this exercise that the solution is always unstable. The single valued solution for takes the form

where indicates the shock location. The position is imposed by a conservation law that should be valid at all times.

  1. Find the conservation law for the wave and determine .