How to Solve Laplace-Transformed ODE on Infinite Domain?

[timman_24]

Homework Statement

I'm working with a heat equation that requires a Laplace transform. I performed the transform and ended up with a basic ODE with a particular solution. I solved for the particular solution and then realized I was working on an infinite domain in my spatial dimension. Maybe I missed this part in class, but how do I go about solving this problem? Let me show you what I have:

ODE:
\frac{\partial^2 T (x,s)}{\partial^2 x }-\frac{s}{\alpha}T(x,s)=-sin(x)

Solution:
U(x,s)=c_1Exp[\sqrt{\frac{s}{\alpha}}x]+c_2Exp[-\sqrt{\frac{s}{\alpha}}x]+\frac{\alpha}{\alpha+s}sin(x)

The problem does not give me any boundary conditions other than what the range is which is:

-\infty<x<\infty

How do I go about solving the coefficients in this problem or solving in general? The only other thing I was given was the IC, but it was absorbed in the Laplace transform. The equation blows up at both limits, so I feel something is wrong.

 

[Coto]

The general way of working with an infinite spatial domain after doing a Laplace Transform is to reason that for the solution to remain physical, it must remain bounded as x goes to infinity (either positive or negative in your case). The only way this is possible is if both c_1,c_2 are 0. That is how the boundary conditions have presented themselves.

To see if this works out appropriately, do an inverse LT after setting the constants to 0, and plug in this solution to the original PDE to see if it satisfies everything.

 

[timman_24]

Okay, so the general solution will vanish leaving the particular, then transform that back using an inverse transform. I'll give it a go and see if it makes sense physically.

Thanks for the advice!

 

[timman_24]

Thanks Coto, it worked great. I took the inverse laplace transform and got this answer:

\alpha exp(-\alpha t)sin(x)

This answers makes sense physically.

 

[Coto]

Of course :). Glad it worked.