solve laplace equation by fourier tranform

Question

If $$ ∇^2 u=0$$ ,for $$ x≥0$$ and if $$u=f(y)$$on $$x=0$$ show that $$u(x,y)=x/π ∫_-∞^∞]〖f(ξ)/(x^2+(y-ξ)^2 ) dξ〗$$ solve by fourier tranform

Answer 1

Based on your proposed solution, you must be working in the plane, even though you did not state as much.

Solutions are not unique unless you impose some sort of boundedness condition. This is because the function $v(x,y)=x$ is 0 on $x = 0$ and is harmonic. For the problem at hand, it is enough to assume that there is a constant $M$ such that $$ \int_{-\infty}^{\infty}|u(x,y)|^{2}\,dy \le M,\;\;\; x \ge 0. $$ This is also a good fit with the Fourier transform. But it does require that your boundary data $f(y)=u(0,y)$ be square integrable on the real line. For convenience, assume that all of the first and second order derivatives of $u$ are similarly bounded, at least for half spaces $x \ge \delta > 0$. This allows you to transform $u$ in $y$. $$ \hat{u}(x,s)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}u(x,y)e^{-isy}dy. $$ The equation $u_{xx}+y_{yy}$ becomes $$ \frac{\partial^{2}}{\partial x^{2}}\hat{u}= \hat{u_{xx}}(x,s)=-\hat{u_{yy}}(x,s)=s^{2}\hat{u}. $$ The only reasonable solution is $C(s)e^{-|s|x}$ because of the boundedness for $x > 0$. But you know that $\hat{u}(0,s)=\hat{f}(s)$. So $C(s)=\hat{f}(s)$. The final solution is the convolution of $f$ with the inverse Fourier transform of $e^{-x|s|}$. For $x > 0$, the inverse Fourier transform of $e^{-x|s|}$ is $$ \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{isy}e^{-xs}ds-\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-isy}e^{-xs}ds = \frac{1}{\sqrt{2\pi}}\left[\frac{1}{x-iy}+\frac{1}{x+iy}\right] = \sqrt{\frac{2}{\pi}}\frac{x}{x^{2}+y^{2}}. $$ The final solution is a convolution integral in the $y$ variable: $$ u(x,y)=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f(y')x}{x^{2}+(y-y')^{2}}dy'. $$