Show that the fourier transform of $u$ satisfies the PDE

Question

We are supposed to solve the Laplace equation in the upper half-plane. $$u_{xx} + u_{yy} = 0$$ With $y>0$, $u(x,0) = f(x)$, etc. But first I need to show that the Fourier transform of $u$, $\hat{u}(w,y) = \mathcal{F}(u)(w,y)$, with respect to $x$ satisfies the PDE $$\frac{\partial^2 \hat{u}}{\partial y^2} - w^2\hat{u} = 0$$ But how do I prove that $\hat{u}$ does indeed satisfy the PDE? I have tried to insert $\hat{u}$ in the equation above but can't really figure out how to make the left side equal to zero. I am also able to transform the Laplace equation to the PDE above by taking the Fourier transform of it, but I believe that won't prove that $\hat{u}$ satisfies the PDE... or does it?

Answer 1

HINT $$u_{xx}+u_{yy}=0$$ and taking the Fourier transform we have $$\mathcal{F}\{u_{xx}\}+\mathcal{F}\{u_{yy}\}=0$$ Using the inverse Fourier transform we can write $$ u(x,y)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat u(\omega,y)\mathrm e^{i \omega x}\mathrm d \omega $$ So we have $$ u_x(x,y)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(i\omega)\,\hat u(\omega,y)\mathrm e^{i\omega x}\mathrm d \omega $$ that is $\mathcal{F}\{u_x\}=i\omega\mathcal{F}\{u\}$ and $$ u_{xx}(x,y)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(i\omega)^2\,\hat u(\omega,y)\mathrm e^{i\omega x}\mathrm d \omega $$ that is $\mathcal{F}\{u_{xx}\}=-\omega^2\mathcal{F}\{u\}$.

For $\mathcal{F}\{u_{yy}\}$ we have $$ u_{yy}(x,y)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{\partial^2}{\partial y^2}\hat u(\omega,y)\mathrm e^{i\omega x}\mathrm d\omega $$ that is $\mathcal{F}\{u_{yy}\}=\frac{\partial^2}{\partial y^2}\mathcal{F}\{u\}$ and finally you'll find $$ \frac{\partial^2}{\partial y^2}\hat u(\omega,y)-\omega^2\hat u(\omega,y)=0 $$