POTW Neumann Boundary Value Problem in a Half Plane

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Find all bounded solutions to the PDE ##\Delta u(x,y) = 0## for ##x\in \mathbb{R}## and ##y > 0## with Neumann boundary condition ##u_y(x,0) = g(x)##.
 
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Let ##\hat{u}(k,y) = \int_{-\infty}^\infty u(x,y)e^{ikx}\, dx##, the Fourier transform of ##u## with respect to the first variable. Applying this Fourier transform to the PDE with respect to the first variable yields ##\hat{u}_{yy} - k^2 \hat{u} = 0## with ##\hat{u}_y(k,0) = \hat{g}(0)##. We have general solution ##\hat{u}(k,y) = A(k)e^{ky} + B(k) e^{-ky}##. Assuming boundedness of ##u##, ##A(k) \equiv 0## if ##k > 0## and ##B(k) \equiv 0## if ##k < 0##. So we express ##\hat{u}(k,y) = C(k)e^{-|k|y}##. The condition ##\hat{u}_y(k,0) = \hat{g}(k)## forces ##-|k| C(k) = \hat{g}(k)##. Therefore ##\hat{u}_y(k,y) = -|k| C(k) e^{-|k|y} = \hat{g}(k) e^{-|k|y}##. By the convolution theorem we obtain $$u_y(x,y) = \frac{y}{\pi} \int_{-\infty}^\infty \frac{g(x-t)}{t^2 + y^2}\, dt$$ Integrating with respect to ##y## produces solution ##u(x,y) = \frac{1}{2\pi} \int_{-\infty}^\infty g(x-t) \log(t^2 + y^2)\, dt + c##.
 
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