The function $u(x,y)$ satisfies the partial differential equation

$$\nabla^{2}u=\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0\text{ in }0<y<a, -\infty<x<\infty$$

and the boundary conditions $u \to 0$ as $x \to \pm\infty$, $\frac{\partial u}{\partial y}=0$ on $y=0$, $u=e^{-|x|}$ on $y=a$ where a is strictly positive constant.

The questions are :

a) Use Fourier transforms to show that $$\frac{\partial^{2}\tilde{u}}{\partial y^{2}}-k^{2}\tilde{u}=0,$$ where $\tilde{u}(k,y)$ is the Fourier transform of $u$ with respect to $x$.

b) Find the boundary conditions satisfied by $\tilde{u}(k,y)$ on $y=0$ and $y=a$

c) Hence show that $$u(x,y)=\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\cosh(ky)}{\cosh(ka)} \frac{e^{ikx}}{1+k^{2}}dk$$.

For part a) I got $\tilde{u}(k,y) = \int_{-\infty}^{\infty}u(x,y)e^{-ikx}dx$ and by transforming the equation I got $\frac{\partial^{2}\tilde{u}}{\partial y^{2}} = -(ik)^{2}\tilde{u}$ which simplifies to $\frac{\partial^{2}\tilde{u}}{\partial y^{2}} -k^{2}\tilde{u}=0$.

However I get stuck when it comes to the boundary conditions bit.


Your boundary conditions are

$$\frac{\partial \hat{u}}{\partial y}(k,0) = 0$$ $$\hat{u}(k,a) = \frac{2}{1+k^2}$$

The latter results from taking the FT of $e^{-|x|}$.

The general solution to the FT'ed equation is

$$\hat{u}(k,y) = A e^{k y} + B e^{-k y}$$

The first BC implies that $A=B$. The general solution is then

$$\hat{u}(k,y) = 2 A \cosh{k y}$$

The 2nd BC implies that

$$2 A = \frac{2\text{sech}{(k a)}}{1+k^2}$$


$$\hat{u}(k,y) = \frac{\cosh{k y}}{\cosh{k a}} \frac{2}{1+k^2}$$

The result follows.


To compute the FT of $u(x,a)=e^{-|x|}$:

$$\begin{align}\hat{u}(k,a) &= \int_{-\infty}^{\infty} dx \: e^{-|x|} e^{- i k x}\\ &= \int_{-\infty}^{0} dx \: e^{(1-i k) x} + \int_{0}^{\infty} dx \: e^{-(1+i k) x}\\ &= \int_{0}^{\infty} dx \: \left ( e^{-(1-i k) x} + e^{-(1+i k) x} \right ) \\ &= \frac{1}{1-i k} + \frac{1}{1+i k} \\ &= \frac{2}{1+k^2}\end{align}$$

| cite | improve this answer | |
  • $\begingroup$ I don't get how to compute the integral of $\tilde{u}(k,a)$ I know you have to split it into two seperate integral avioding having 0 in the middle $\endgroup$ – Adam Mar 19 '13 at 9:57
  • $\begingroup$ That's right, although I would not say that $0$ is avoided. I will illustrate. $\endgroup$ – Ron Gordon Mar 19 '13 at 9:59

For the first part you probably just need to transform the boundary conditions. For $y = a$:

$$ \bar{u}(k) = \int e^{-|x|}e^{-ikx}dx $$

which you can easily evaluate by splitting the range of integration in half; i.e., $(-\infty,+\infty) = (-\infty,0] \cup [0,+\infty)$.

For $y=0$:

$$ \overline{\frac{\partial u}{\partial y}}(k) = 0 = \frac{\partial\bar{u}}{\partial y} $$

Then your answer to part a gives you the general form for $\bar{u}$:

$$ \bar{u}(k,y) = f(k)e^{ky} + g(k)e^{-ky} $$

The boundary conditions from part b, then, give you the form of $f$ and $g$, which allows you the conclusion of part c.

| cite | improve this answer | |
  • $\begingroup$ I'm stuck when it comes to evaluating the first integral but I understand where you got it from $\endgroup$ – Adam Mar 19 '13 at 9:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.