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So the question is:

$$u_{xx} + u_{yy} = 0$$

$$-\infty < x < \infty$$

$y>0$ , $u(x,0)=f(x)$

As $y \to \infty$ $u \to 0$

Express the solution as a convolution.

What i've done:

$$\hat{u}(k,y) = \int_{-\infty}^{\infty} dx \, u(x,y) e^{i k x}$$

so $$-k^2 \hat{u} + \frac{\partial^2 \hat{u}}{\partial y^2} = 0$$

which means that

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

subject to

$$\hat{u}(k,0) = \hat{f}(k)$$

This leads to

A + B = $\hat{f}(k)$

But for U to go to $0$ as y goes to infinity A must equal $0$. Hence,

$$\hat{u}(k,y) = \hat{f}(k)e^{-k y}$$

Taking the inverse fourier transform of this i then get

$$u(x,y) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \hat{f}(k)e^{k(ix- y)} $$

= $f(ix-y)$

So yeah im not sure whether i've done something wrong or if i need to do something to display this as a convolution. Any help would be appreciated!

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    in the last line, when you apply the inverse fourier transform, should it be $\int_{-\infty}^{\infty}dk\hat{f}(k)e^{-ky}e^{-ikx}=\int_{-\infty}^{\infty}dk\hat{f}(k)e^{-ik(x-iy)}=f(x-iy)$? – 2017-02-24
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    Ah yeah you're right, although it should be ik(iy+x) right – 2017-02-24
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    But i don't really know how to display it as a convolution – 2017-02-24
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    What about $f(x) *i\delta(x-iy)$ – 2017-02-24
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    hmm that's probably right but we haven't covered the dirac delta function much so i doubt that's what he's looking for – 2017-02-24

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ A 'particular solution' $\ds{\exp\pars{\ic k_{x}}\exp\pars{-\verts{q}y}}$ $\ds{\pars{~\mbox{with}\ k_{x}, q \in \mathbb{R}~}}$ satisfies $$ \pars{-k_{x}^{2} + q^{2}}\exp\pars{\ic k_{x}}\exp\pars{-\verts{q}y} = 0 \quad\implies\quad \bbx{\ds{k_{x} = \pm q}} $$ So, the solution can be written as \begin{align} \mrm{u}\pars{x,y} & = \int_{-\infty}^{\infty}\hat{\mrm{u}}\pars{q} \expo{\ic qx - \verts{q}y}\,\dd q \\[5mm] \mrm{f}\pars{x} = \mrm{u}\pars{x,0} & = \int_{-\infty}^{\infty}\hat{\mrm{u}}\pars{q} \expo{\ic qx}\,\dd q \implies \hat{\mrm{u}}\pars{q} = \int_{-\infty}^{\infty}\mrm{f}\pars{x}\expo{-\ic qx}\,{\dd x \over 2\pi} \end{align}

Then, \begin{align} \mrm{u}\pars{x,y} & = \int_{-\infty}^{\infty}\bracks{% \int_{-\infty}^{\infty}\mrm{f}\pars{x'}\expo{-\ic qx'}\,{\dd x' \over 2\pi}} \expo{\ic qx - \verts{q}y}\,\dd q \\[5mm] &= {1 \over 2\pi}\int_{-\infty}^{\infty}\mrm{f}\pars{x'} \int_{-\infty}^{\infty}\expo{-\ic q\pars{x - x'} - \verts{q}y}\,\dd q\,\dd x' \\[5mm] & = {1 \over \pi}\,\Re\int_{-\infty}^{\infty}\mrm{f}\pars{x'} \int_{0}^{\infty}\expo{-\ic q\pars{x' - x} - qy}\,\dd q\,\dd x' = {1 \over \pi}\,\Re\int_{-\infty}^{\infty}\mrm{f}\pars{x'} {1 \over y + \pars{x' - x}\ic}\,\dd x' \\[5mm] & = \bbx{\ds{\int_{-\infty}^{\infty}\mrm{f}\pars{x'} {y/\pi \over \pars{x - x'}^{2} + y^{2}}\,\dd x'}} \end{align}