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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.

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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

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

Then

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

The result follows.

ADDENDUM

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

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

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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