Solve the PDE by Fourier Transforms

[wxstall]

Homework Statement

Solve:

∂u/∂t = k ∂²u/∂x² - ζu

with the initial condition

u(x,0) = f(x)

where k and ζ are constants.

x is on an infinite domain.

Homework Equations

Define Fourier transforms:

f(x) = ∫[-∞,∞]F(w)e^-iwxdw

F(w) = 1/2∏ ∫[-∞,∞]f(x)e^iwxdxFrom tables of Fourier Transforms:

∂²f/∂x² = (-iw)²F(w)

The Attempt at a Solution

I have little experience with transforms so please don't berate me if this is completely wrong.

Began by taking transform of entire eqn:

F(∂u/∂t) = k F(∂²u/∂x²) - ζF(u)

I will call F(u) = U*

∂U*/∂t = w²k U* - ζU*

∂U*/∂t = (w²k - ζ) U*

So the general solution is:

U* = C(w) e^(w2k-ζ)t

Where C(w) = 1/2∏ ∫[-∞,∞]f(x)e^iwxdx

Is this even remotely correct? It seems too easy, but then again, I suppose that is the point of transforms, to put something into a simpler form.

 

[mighty2000]

Your approach is definitely on the right track! Here are a few things to consider:

1. When taking the Fourier transform, you need to specify the variable of integration. In this case, it should be t, since we are taking the transform with respect to t.

2. Remember that the Fourier transform of a derivative is given by multiplying by iw, not just w. So your equation should be ∂U*/∂t = -w^2kU* - ζU*.

3. Your general solution is almost correct, but you should include the initial condition in your expression for U*. This would give you U* = C(w)e^(-(w^2k+ζ)t) + F(w).

4. Finally, don't forget to take the inverse Fourier transform to get the solution in terms of u(x,t) instead of U*(w). This would give you u(x,t) = 1/2∏ ∫[-∞,∞]C(w)e^(-(w^2k+ζ)t)e^(iwx)dw + f(x).

Overall, you are definitely on the right track and have a good understanding of using Fourier transforms to solve differential equations. Keep up the good work!