For the 1D wave equation, it is possible to greatly simplify the governing PDE
$$u_{tt}(x,t) = c^2 u_{xx}(x,t)$$
By applying a laplace transform to each term in $t$ and expressing in Laplace space:
$$s^2U(s) = s u(x,0)-u_t(x,0) = c^2U_{xx}(x,s)$$
Provided the initial conditions are equal to $0$ this will become:
$$U(x,s) = A(s)\exp(sx/c)+B(s)\exp(-sx/c)$$
An ODE easily solvable for $U(x,s)$ using Laplace transform of the boundary conditions. The original solution can be retrieved by applying the inverse Laplace transform to the resulting expression for $U(x,s)$
Now I would like to extend this to the 2D wave equation. The governing equation is:
$$u_{tt} = c^2(u_{xx}+u_{yy})$$
Applying the same idea (I think) one would end up with:
$$U_{xx}(x,y,s)+U_{yy}(x,y,s)=\frac{1}{c^2}(s^2 U(x,y,s)-su(x,y,0)-u_t(x,y,0))$$
This becomes another PDE but this time only in variables $x$ and $y$ as follows:
$$U_{xx}+U_{yy}-\frac {s^2} {c^2}U(x,y,s)+\frac s {c^2} u(x,y,0)+\frac 1 {c^2} u_t(x,y,0) = 0$$
Assuming the initial condition is a surface function of $x$ and $y$ one would obtain:
$$U_{xx}(x,y,s)+U_{yy}(x,y,s)-\frac {s^2} {c^2}U(x,y,s)+\frac s {c^2} f(x,y) +\frac 1 {c^2} g(x,y) = 0$$
To me this looks very much like the form
$$Au_{xx}+2Bu_{xy}+Cu_{yy}+Du_x+Eu_y+F = 0$$ with $A=1, B=0, C=1, D = 0, E = 0$ and $F = -\frac {s^2} {c^2}U(x,y,s)+\frac s {c^2} f(x,y) +\frac 1 {c^2} g(x,y)$
Although I am not really sure about it. I have a few questions:
- Have I developed the Laplace transform of the 2D Wave equation correctly. IE is the resulting PDE correct?
- What method can I use to solve the resulting PDE after laplace transformation? It seems that the equation is elliptic with $B^2-4AC <0$ but I am not sure how to solve it....