Variable coefficient Wave Equation

In summary, the problem is that a is not constant, which complicates the boundary conditions. JC suggests using a numerical solution, but warns that the data may not be smooth enough.
  • #1
vargasjc
28
0
Hello exalted ones. I am working on a set of differential equations for my research and there is one that is becoming mortal.

I am solving a mechanical system whose behavior eq. is that of a one dimensional wave PDE. Namely:

[tex]u_{tt}=a^{2}u_{xx}[/tex]

For which I would derive two parametrized equations in terms of eigenvalues defined by my boundary conditions. Now my problem is that "a" is not constant, but actually a function of both time and space. Plainly:

[tex]a(x,t)=\frac{(E_{m} x+E_{0}) e^{\frac{C_{0} x^{2}}{T(t)}}}{C_{1}}[/tex]

So I have an e^ in function of both variables. I've almost given up trying to look for a closed-form solution.

As boundary conditions go (let's call them North, South, East, West), North is variable but known (input), South is always zero (fixed end). The first derivative of North in terms of x is also zero. Time increases from West to East and the displacement from South to North.

Would you advise me to pursue a numerical solution? What would be your advice on the matter?

JC
 
Physics news on Phys.org
  • #2
Just to clarify, the wave equation becomes:

[tex]
\frac{\partial^{2} u}{\partial t^{2}}=\frac{\partial}{\partial x} (a^{2} \frac{\partial u}{\partial x})
[/tex]

When a is not constant. Any thoughts?
 
  • #3
That looks like two different equations to me. How about first just the first one:

[tex]u_{tt}=a(t,x)^2 u_{xx}[/tex]

with:

[tex]a(t,x)=-\frac{E_m x+E_0}{c_1} e^{c_0 x^2/T(t)}[/tex]

I don't understand your boundary conditions but just for fun, I solved numerically, the IBVP:

[tex]u_{tt}=a(t,x)^2 u_{xx},\quad 0\leq x\leq 1,\quad 0\leq t\leq 1[/tex]

[tex]u(t,0)=0,\quad u(t,1)=0[/tex]

[tex]u(0,x)=\sin(\pi x),\quad u_t(0,x)=0[/tex]

[tex]T(t)=t+2[/tex]

[tex]E_m=1,\quad E_0=1,\quad c_1=1,\quad c_0=1[/tex]

using the Mathematica code below. Perhaps you can adapt it to your particular problem.

Code:
em = 1; 
c1 = 1; 
c0 = 1; 
e0 = 1; 
g[t_, x_] := (((-em)*x + e0)/c1)*
   Exp[c0*(x^2/(t + 2))]
mysol = NDSolve[{D[u[t, x], t, t] == 
     g[t, x]^2*D[u[t, x], x, x], 
    u[t, 0] == 0, u[t, 1] == 0, 
    u[0, x] == Sin[Pi*x], 
    (D[u[t, x], t] /. t -> 0) == 0}, 
   u[t, x], {t, 0, 1}, {x, 0, 1}]

Plot3D[u[t, x] /. mysol, {t, 0, 1}, 
  {x, 0, 1}]
 
Last edited:
  • #4
I don't use Mathematica, but I do have access to Matlab. I don't know there capabilities/limitations though.

The following equation would only be valid for a^2 constant:

[tex]
u_{tt}=a^{2}u_{xx}
[/tex]

Otherwise I'm stuck with this one:

[tex]
\frac{\partial^{2} u(x,t)}{\partial t^{2}}=\frac{\partial}{\partial x} (a(x,t)^{2} \frac{\partial u(x,t)}{\partial x})
[/tex]

Which increases the problem's difficulty. The boundaries are harder too, let me try to jot them down properly:

[tex]
u(0,t)=0,\quad u(L,t)=S, \quad u_{x}(L,t)=0
[/tex]

[tex]
u(x,0)\neq0
[/tex]

[tex]
0\leq x\leq L,\quad T_{0}\leq t\leq T_{1}
[/tex]

S is known and variable, it represents the data read from an accelerometer (choppy data). T0 is known (let's say zero), T1 is known (let's say one). There is a funny procedure to obtain L for this material, but let's just go ahead and say that it is known (let's say L=1 to follow your lead).

Now, I like the idea of a numerical solution because I'm short on time and I could later keep researching for a closed form. I've used finite differences before but I think in this case I need much more raw power.

JC
 
  • #5
Alright, three questions:

1. Can you smooth out S? How about a Fourier Transform or just a least-square fit for now. If you post the raw data, I could perhaps reduce it down to a smooth function.

2. you say u(x,0) is not zero. What is it precisely then? More choppy data? If so, how about smoothing it out too.

3. What exactly is T(t)?
 
  • #6
1. It can be either. No troubles there. Sometimes I use Fourier, sometimes a constrained poly regression.
2. I don't have the resolution to determine what u(x,0) is. It is impossible to measure. The only thing I can say for sure is that u(L,0)=S(0) and that u(0,0)=0. So u(x,0) is some function that takes me from 0 to S(0). S(0) is never zero. In fact, u(x,t) for x > 0 is never zero.
3. Right now T(t) is a temperature variation in time.
 
  • #7
Any thoughts?
 

Related to Variable coefficient Wave Equation

1. What is the variable coefficient wave equation?

The variable coefficient wave equation is a partial differential equation that describes the propagation of waves in a medium where the wave speed or other properties vary with position or time. It is written in the form of u_tt = (a(x,t)u_x)_x + b(x,t)u_t + c(x,t)u, where u represents the wave function, a(x,t) is the coefficient of the second derivative, b(x,t) is the coefficient of the first derivative, and c(x,t) is the coefficient of the wave function itself.

2. What are some real-life applications of the variable coefficient wave equation?

The variable coefficient wave equation has many applications in physics and engineering, including in the study of electromagnetics, acoustics, and fluid dynamics. It is also used in fields such as seismology, where the wave speed varies with depth in the Earth's crust, and in optics, where the refractive index of a medium can vary with position.

3. How is the variable coefficient wave equation solved?

The variable coefficient wave equation can be solved using various techniques, such as separation of variables, the method of characteristics, or numerical methods. The specific approach used depends on the specific form of the equation and the boundary conditions of the problem.

4. What is the relationship between the variable coefficient wave equation and the standard wave equation?

The variable coefficient wave equation is a generalization of the standard wave equation, which assumes a constant wave speed and no other variations in the medium. The standard wave equation can be derived from the variable coefficient wave equation by setting all coefficients to be constant.

5. How does the variation of coefficients in the wave equation affect the behavior of waves?

The variation of coefficients in the wave equation can have a significant impact on the behavior of waves. For example, changes in the coefficients can lead to variations in the speed, amplitude, and shape of the wave. This can result in phenomena such as wave dispersion, where different parts of a wave travel at different speeds, and wave attenuation, where the amplitude of the wave decreases over time.

Similar threads

Replies
4
Views
1K
  • Differential Equations
Replies
2
Views
2K
  • Differential Equations
Replies
4
Views
2K
  • Differential Equations
Replies
7
Views
409
  • Differential Equations
Replies
3
Views
2K
  • Differential Equations
Replies
19
Views
3K
  • Differential Equations
Replies
2
Views
1K
  • Differential Equations
Replies
11
Views
1K
  • Differential Equations
Replies
7
Views
2K
  • Differential Equations
Replies
6
Views
2K
Back
Top