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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?

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  • $\begingroup$ This is just a straight forward application of the Fourier transform $$\hat{u}(w,y) = \frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} u(x,y)e^{-iwx} dx$$ Substitute $u_{xx}$ for $u$ in the integral, integrate by parts in $x$ and then use that $u$ and $u_{x}$ must be integrable. You will then get that \begin{align} \frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} u_{xx}(x,y)e^{-iwx} dx &= -w^{2} \cdot \frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} u(x,y)e^{-iwx} dx \\ &= -w^{2} \hat{u}(w,y) \end{align} $\endgroup$ – mattos Sep 27 '17 at 21:14
  • $\begingroup$ @Mattos I understood that, but what I'm really confused by is what to do with the double partial derivative of y, because I believe we can't do the same thing for $u_{yy}$. $\endgroup$ – Geir Sep 27 '17 at 21:21
  • $\begingroup$ \begin{align} \frac{ 1 }{ \sqrt{2 \pi} } \int_{ \mathbb{R} } \frac{ \partial^{2} u(x,y)}{ \partial y^{2} } e^{-i w x} dx &= \frac{ \partial^{2} }{ \partial y^{2} } \left[\frac{ 1 }{ \sqrt{2 \pi} } \int_{\mathbb{R}} u(x,y) e^{-i w x} dx \right] \\ &= \frac{ \partial^{2} }{ \partial y^{2} } \hat{u}(w,y) \end{align} $\endgroup$ – mattos Sep 27 '17 at 21:25
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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 $$ \frac{\partial^2}{\partial y^2}\hat u(\omega,y)-\omega^2\hat u(\omega,y)=0 $$

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  • $\begingroup$ So by transforming $u_{xx} + u_{yy} = 0$ to $\frac{\partial^2 \hat{u}}{\partial y^2} - w^2\hat{u} = 0$, and by using $\hat{u}$ to to that. Why can we say that we have proved that $\hat{u}$ is a solution that satisfies the PDE? Is it because we have assumed that $u_{xx} + u_{yy} = 0$ is true? $\endgroup$ – Geir Sep 27 '17 at 21:33
  • $\begingroup$ I've added the starting point $\endgroup$ – alexjo Sep 27 '17 at 21:35
  • $\begingroup$ What I am wondering is why we can now, after what we have done, say that we have proved that $\hat{u}$ satisfies the PDE. Could you please explain that? Maybe It is because I don't understand what it means to "satisfy" a PDE. $\endgroup$ – Geir Sep 27 '17 at 21:42
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    $\begingroup$ Actually we've transformed $u_{xx} + u_{yy} = 0$ to an equivalent but easier equation $\hat u_{yy}- \omega^2 \hat u = 0$ and if $\hat u$ is solution of the transformed equation, then $u=\mathcal{F}^{-1}\{\hat u\}$ will be a solution of the original equation. Now you have to find $\hat u$ and then $u$. $\endgroup$ – alexjo Sep 27 '17 at 21:50

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