- #1

joshmccraney

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## Homework Statement

Find the solution ##u(x, t)## to the semi-infinite interval problem

$$ u_t = u_{xx} - 4u, \hspace{2 mm} 0 < x < \infty, \hspace{2 mm} t>0\\

u_x(0,t) = -1, \hspace{2 mm} t>0\\

\lim_{x \to \infty}u(x,t) = 0, \hspace{2 mm} t>0\\

u(x,0) = e^{-x}, \hspace{2 mm} 0 \leq x < \infty.$$

## Homework Equations

fourier cosine series, which I shall denote ##C(u)## when in variable ##x## operating over some function ##u##.

## The Attempt at a Solution

after taking Fourier cosine series of the governing PDE i arrive at $$

\frac{\partial}{\partial t}C(u) = \frac{2}{\pi} - \omega^2 C(u) -4C(u) \implies\\

C(u) = \frac{2}{\pi(\omega^2 + 4)} + C_1e^{-(\omega^2 +4)t}$$

Notice ##C(u(x,0)) = C(e^{-x}) = \frac{2}{\pi(1+\omega^2)}##. This implies ##C_1 = \frac{2}{\pi(1+\omega^2)} - \frac{2}{\pi(\omega^2 + 4)} ##. Thus we arrive at $$

C(u) = \frac{2}{\pi(\omega^2 + 4)} + \frac{2}{\pi}\left[ \frac{1}{(1+\omega^2)} - \frac{1}{(\omega^2 + 4)} \right] e^{-(\omega^2 +4)t}$$

I have skipped some in-between steps, but I think the idea is here. My question is, how do I undo this Fourier cosine transform, if I have arrived to a correct solution?

Thanks so much!

Josh