- #1
- 1,796
- 33
Suppose I have the PDE:
[tex]
\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}=\frac{1}{c^{2}} \frac{\partial^{2}u}{\partial t^{2}}
[/tex]
with
[tex]
u(0,x,y)=\partial_{t}u(0,x,y)=0
[/tex]
along with [itex]u(t,0,y)=f(y)[/itex] With [itex]x\geqslant 0[/itex]. My initial thoughts were to take the Laplace transform in t and the Fourier transform in y to get:
[tex]
\frac{\partial^{2}\hat{u}_{L}}{\partial x^{2}}-\left( k^{2}+\frac{s^{2}}{c^{2}}\right) \hat{u}_{L}=0
[/tex]
The hat donates the Fourier transform and the subscript L denotes the Laplace transform. Treating this as a standard ODE to obtain:
[tex]
\hat{u}_{L}=Ae^{x\sqrt{k^{2}+s^{2}c^{-2}}}+Be^{-x\sqrt{k^{2}+s^{2}c^{-2}}}
[/tex]
Setting x=0 will show us that:
[tex]
A+B=\hat{f}_{L}
[/tex]
I have no other way of determining the other constant (this comes from a question in a book) and I have no idea how to get another boundary condition. I am also concerned about how to do my contour integral when I compute my inverse Laplace transform.
Any suggestions.
[tex]
\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}=\frac{1}{c^{2}} \frac{\partial^{2}u}{\partial t^{2}}
[/tex]
with
[tex]
u(0,x,y)=\partial_{t}u(0,x,y)=0
[/tex]
along with [itex]u(t,0,y)=f(y)[/itex] With [itex]x\geqslant 0[/itex]. My initial thoughts were to take the Laplace transform in t and the Fourier transform in y to get:
[tex]
\frac{\partial^{2}\hat{u}_{L}}{\partial x^{2}}-\left( k^{2}+\frac{s^{2}}{c^{2}}\right) \hat{u}_{L}=0
[/tex]
The hat donates the Fourier transform and the subscript L denotes the Laplace transform. Treating this as a standard ODE to obtain:
[tex]
\hat{u}_{L}=Ae^{x\sqrt{k^{2}+s^{2}c^{-2}}}+Be^{-x\sqrt{k^{2}+s^{2}c^{-2}}}
[/tex]
Setting x=0 will show us that:
[tex]
A+B=\hat{f}_{L}
[/tex]
I have no other way of determining the other constant (this comes from a question in a book) and I have no idea how to get another boundary condition. I am also concerned about how to do my contour integral when I compute my inverse Laplace transform.
Any suggestions.