Laplace's equation via Fourier transformation

Question

I have a Laplace equation with some data along the $y$-axis: $$ \begin{cases} u_{xx} + u_{yy} &= 0 \\ u(0,y) &= f(y) \\ u_x(0,y) &= g(y). \end{cases} $$ There is no information of any region, properties of the solution or properties of $f$ and $g$ other than that we may assume $f$ and $g$ to have well-defined Fourier transforms.

So, Fourier transforming everything with respect to the $y$-variable we obtain: $$ \begin{cases} \hat{u}_{xx} -\xi ^2 \hat{u} &= 0 \\ \hat{u}(0,\xi ) &= \hat{f}(\xi ) \\ \hat{u}_x(0,\xi ) &= \hat{g}(\xi ). \end{cases} $$ Thus $\hat{u}(x,\xi ) = A(\xi )e^{-\xi x} + B(\xi ) e^{\xi x}$.

I suppose this is where I become puzzled. Clearly $\hat{u}$ behaves very badly as $\xi \to \pm \infty $, depending on the sign of $x$. To fix this I first restricted to the case $x>0$ and simply assumed that $A(\xi ) = 0$ for $\xi <0$ and $B(\xi ) = 0$ for $\xi > 0$ so that $$ \hat{u}(x,\xi ) = C(\xi )e^{-|\xi | x} = \hat{f}(\xi )e^{-|\xi | x}. $$ This is easy to invert since $\mathcal{F}^{-1}[e^{-|\xi |x}](x,y)$ is well-known. But now it seems there is no room for $g$ (or $\hat{g}$) to fit in anymore.

I suppose I must have gone wrong when I restricted to $x>0$ and singled out certain solutions, but how can I avoid this?

Answer 1

You cannot put $A(\xi)=0$ given that boundary conditions. So, $$ \hat u(0,\xi)=A(\xi)+B(\xi)=\hat f(\xi) $$ and $$ \hat u_x(0,\xi)=-\xi A(\xi)+\xi B(\xi)=\hat g(\xi). $$ This is a simple system you have to solve to get $A$ and $B$.

Answer 2

In fact you don't need to solve this problem by using Fourier transform, since you can just seckilling this problem by using D’Alembert’s formula:

$u(x,y)=\dfrac{f(y+ix)+f(y-ix)}{2}-\dfrac{i}{2}\int_{y-ix}^{y+ix}g(t)~dt$

Even you have to solve this problem by using Fourier transform:

$\mathcal{F}_{y\to\xi}\{u_{xx}(x,y)\}+\mathcal{F}_{y\to\xi}\{u_{yy}(x,y)\}=0$

$\hat{u}_{xx}(x,\xi)-\xi^2\hat{u}(x,\xi)=0$

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

$u(x,y)=\mathcal{F}^{-1}_{\xi\to y}\{A(\xi)e^{-x\xi}\}+\mathcal{F}^{-1}_{\xi\to y}\{B(\xi)e^{x\xi}\}$

$u(x,y)=\dfrac{1}{2\pi}\int_{-\infty}^\infty A(\xi)e^{-x\xi}e^{iy\xi}~d\xi+\dfrac{1}{2\pi}\int_{-\infty}^\infty B(\xi)e^{x\xi}e^{iy\xi}~d\xi$

$u(x,y)=\dfrac{1}{2\pi}\int_{-\infty}^\infty A(\xi)e^{i(y+ix)\xi}~d\xi+\dfrac{1}{2\pi}\int_{-\infty}^\infty B(\xi)e^{i(y-ix)\xi}~d\xi$

$u(x,y)=C_1(y+ix)+C_2(y-ix)$

Substitute $u(0,y)=f(y)$ and $u_x(0,y)=g(y)$ to the above equation and follow the procedure that similar in http://en.wikipedia.org/wiki/D%27Alembert%27s_formula, you will get $u(x,y)=\dfrac{f(y+ix)+f(y-ix)}{2}-\dfrac{i}{2}\int_{y-ix}^{y+ix}g(t)~dt$ .

Answer 3

What might get you out of trouble here is that existence of the Fourier transforms of $f,g$ will imply decay conditions at infinity ($|\xi|\to\infty$) on those Fourier transforms $\hat{f}, \hat{g}$. This is the essential content of the Riemann-Lebesgue lemma: see here and page 308 here (from Applied Analysis by Hunter and Nachtergaele). You probably need to say more about the functions $f,g$ in order to check that (i) the lemma applies and (ii) the decay is sufficiently rapid that you can invert $\hat{u}=\hat{f}\cosh(\xi x) +\xi^{-1}\hat{g}\sinh(\xi x)$.

Just to note: your problem is a Cauchy initial value problem for the Laplace equation. Such problems are not generally well-posed. See Hadamard's example e.g. here.