Causality and the inhomogeneous wave equation with a moving source

nickthequick
Messages
39
Reaction score
0
Hi,

I am confused about my solutions to the following governing equation:

u_{tt}-c^2u_{xx}=F_{xx}

For F=A(x)sech^2\left(\frac{x-c_gt}{B}\right)


Where c,c_g,B \in \mathbb{R} and A(x) is a linear function. Also, we have c_g<c. Substituting physical values for the parameters, I can solve the equation using a 2-d integrator to implement the method of Duhamel. My solutions are unclear to me though.

I expect to see the intensity of the response increase with position, as it does along the ray traveling at c_g. I do not, however, see the same response out of the ray traveling at c. I also do not see a ray traveling at -c, as I naively would expect. In the plot attached, the ray traveling at speed c is the lower one, since this is an x-t diagram.

I also would think that the ray traveling at speed c would constantly be generated at the location of the forcing, otherwise it is unclear to me how the information from the forcing would get to the faster wave traveling at speed c.

I think my numerical implementation is fine. Perhaps the fact that I'm working on the half domain, x \ge 0 is messing things up?

Any help is appreciated,

Nick
 

Attachments

  • testplot.jpg
    testplot.jpg
    17 KB · Views: 525
Physics news on Phys.org
Unfortunately, I can not grasp the physical sense of your problem out of hand from your picture, but it seems to me that you may analyse the analytical solution of your problem.

For linear A(x)=ax+b the general solution to your PDE is relatively easy:

u(t,x) = C_1(x+ct)+C_2(x-ct)+4\{[-a(c^2-c_g^2)x-bc^2+c_g^2(aB+b)]\exp[-2(x-c_gt)/B]+aBc_g^2\}\{exp[-2(x-c_gt)/B]+1\}^{-2}(c^2-c_g^2)^{-2},

where C_1,C_2 are arbitrary functions.

I hope that the like solution can be found for more complicated A(x).

Then you have to impose the initial conditions of your physical problem to obtain particular solution which then can be analysed under your goals.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top