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I need to solve \begin{eqnarray} u_{xx} + u_{yy} = 0 \quad \quad y>0 \quad -\infty < x< \infty \end{eqnarray} With boundary condition \begin{eqnarray} \frac{\partial u(x,0)}{\partial y} = h(x) \mbox{ and } |u(x,y)|<\infty \end{eqnarray} I call $U(w,y)$ the Fourier transorm of $u(x,t)$. I call $H(w)$ the Fourier transform of $h(x)$. Substituting $U(w,y)$ in the partial differential equation I get: \begin{eqnarray} \frac{\partial^2 U}{\partial y^2} = w^2U \end{eqnarray} This differential equation has solution \begin{eqnarray} U(w,y) = C(w)e^{-|w|\cdot y} \end{eqnarray} Differentiating gives \begin{eqnarray} U_y = -|w| C(w) e^{-|w|\cdot y} \end{eqnarray} Plugging in the boundary condition $U_y(w,0) = H(w)$ gives \begin{eqnarray} U(w,y) = -\frac{H(w)}{|w|}e^{-|w|y} \end{eqnarray} Is this step correct? And how do I need to continue if it is? I need to transform back, such that \begin{eqnarray} u(x,y) = \frac{1}{2\pi} \int_{-\infty}^{\infty} -\frac{H(w)}{|w|}e^{-|w|y} e^{iwx}dw \end{eqnarray} How can I solve this integral? Does convolution apply?

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consider the Fourier transform of $u$ $u_{xx}$ ,$u_{yy}$ and $h(x)$ $$u(x,y)=\sum_{n=1}^{\infty}u_n(y)\sin(\frac{n\pi x}{l})$$$$u_{xx}=\sum_{n=1}^{\infty}w_n(y)\sin(\frac{n\pi x}{l})$$$$u_{yy}=\sum_{n=1}^{\infty}v_n(y)\sin(\frac{n\pi x}{l})$$ $$h(x)=\sum_{n=1}^{\infty}h_n(y)\sin(\frac{n\pi x}{l})$$such that $$u_n(y)=\frac{2}{l}\int_{0}^{l}u(x,y)\sin(\frac{n\pi x}{l})$$$$v_n(y)=\frac{2}{l}\int_{0}^{l}u_{xx}\sin(\frac{n\pi x}{l})$$$$w_n(y)=\frac{2}{l}\int_{0}^{l}u_{yy}\sin(\frac{n\pi x}{l})$$$$h_n(y)=\frac{2}{l}\int_{0}^{l}h(x)\sin(\frac{n\pi x}{l})$$ and we conclude following statments $$1.\begin{eqnarray} \frac{\partial^2 U_n(y)}{\partial y^2} = v_n(y) \end{eqnarray}$$$$2.v_n(y)+w_n(y)=0$$$3.$ and integrate $w_n(y)=\frac{2}{l}\int_{0}^{l}u_{yy}\sin(\frac{n\pi x}{l})$ by part and you take $w_n(y)$respect to $ u_n(y)$ therefore by use of $1,2,3$ you can easily find $ u_n(y)$ then you can find$ u(x,y ) $

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    $\begingroup$ best answer$\Large\color{green}{✔}^\color{red}{✿}$ @Maisam Hedyelloo $\endgroup$ – Software Jun 3 '13 at 15:01
  • $\begingroup$ █▓▒░$\Large\color{red}{ℕѺ}$░▒▓█ , I haven't.@Meisam Hedyelloo $\endgroup$ – Software Jun 4 '13 at 0:24
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You can do that last integral if you replace $H(w)$ with its definition as the Fourier transform of $H(x)$. You should then be able to rearrange the (double) integral into the form $$ \int dx'\, h(x') G(x,y,x') $$ where $G(x,y,x')$ is some function that should be able to find in closed form (it is also the solution to your problem in the case $h(x)=\delta(x - x')$).

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