The following are the general steps used in all the problems below :
Green function on infinite domain, which is the solution to \[ \nabla ^{2}G\left ( \vec{x},\vec{x}_{0}\right ) =\delta \left ( \vec{x}-\vec{x}_{0}\right ) \] Is given by\begin{align*} G_{\infty }\left ( \vec{x},\vec{x}_{0}\right ) & =\frac{1}{2\pi }\ln \left ( r\right ) \\ & =\frac{1}{2\pi }\ln \left ( \left \vert \vec{x}-\vec{x}_{0}\right \vert \right ) \\ & =\frac{1}{2\pi }\ln \left ( \sqrt{\left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}}\right ) \\ & =\frac{1}{4\pi }\ln \left ( \left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}\right ) \end{align*}
By placing a negative impulse at location \(\vec{x}_{0}^{\ast }=\left ( x_{0},-y_{0}\right ) \), the Green function for semi-infinite domain is obtained
\begin{align} G\left ( \vec{x},\vec{x}_{0}\right ) & =\frac{1}{2\pi }\ln \left ( r_{1}\right ) -\frac{1}{2\pi }\ln \left ( r_{2}\right ) \nonumber \\ & =\frac{1}{2\pi }\ln \left ( \left \vert \vec{x}-\vec{x}_{0}\right \vert \right ) -\frac{1}{2\pi }\ln \left ( \left \vert \vec{x}-\vec{x}_{0}^{\ast }\right \vert \right ) \nonumber \\ & =\frac{1}{4\pi }\left ( \ln \left ( \left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}\right ) -\ln \left ( \left ( x-x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}\right ) \right ) \nonumber \\ & =\frac{1}{4\pi }\ln \frac{\left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}}{\left ( x-x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}} \tag{1} \end{align}
The following is 3D plot of the above Green function, showing the image impulse and showing that \(G=0\) at the line \(y=0\) (marked as red)
The Green function in (1) is now used to solve \(\nabla ^{2}u\left ( \vec{x}\right ) =f\left ( \vec{x}\right ) \), with \(u\left ( x,0\right ) =h\left ( x\right ) \). Starting with Green formula for 2D\begin{align*}{\displaystyle \iint } u\left ( \vec{x}\right ) \nabla ^{2}G\left ( \vec{x},\vec{x}_{0}\right ) -G\left ( \vec{x},\vec{x}_{0}\right ) \nabla ^{2}u\left ( \vec{x}\right ) dA & ={\displaystyle \oint } \left ( u\left ( \vec{x}\right ) \nabla G\left ( \vec{x},\vec{x}_{0}\right ) -G\left ( \vec{x},\vec{x}_{0}\right ) \nabla u\left ( \vec{x}\right ) \right ) \cdot \hat{n}\ ds\\ & ={\displaystyle \oint } \left ( u\left ( \vec{x}\right ) \nabla G\left ( \vec{x},\vec{x}_{0}\right ) -G\left ( \vec{x},\vec{x}_{0}\right ) \nabla u\left ( \vec{x}\right ) \right ) \cdot \left ( -\hat{\jmath }\right ) \ ds\\ & ={\displaystyle \oint } \left ( G\left ( \vec{x},\vec{x}_{0}\right ) \nabla u\left ( \vec{x}\right ) -u\left ( \vec{x}\right ) \nabla G\left ( \vec{x},\vec{x}_{0}\right ) \right ) \cdot \hat{\jmath }\ ds\\ & ={\displaystyle \int \limits _{-\infty }^{\infty }} \left ( G\left ( \vec{x},\vec{x}_{0}\right ) \frac{\partial u\left ( \vec{x}\right ) }{\partial y}-u\left ( \vec{x}\right ) \frac{\partial G\left ( \vec{x},\vec{x}_{0}\right ) }{\partial y}\right ) _{y=0}\ dx \end{align*}
But \(\nabla ^{2}G\left ( \vec{x},\vec{x}_{0}\right ) =\delta \left ( \vec{x},\vec{x}_{0}\right ) \) and \(\nabla ^{2}u\left ( \vec{x}\right ) =f\left ( \vec{x}\right ) \), therefore the above becomes\[{\displaystyle \iint } u\left ( \vec{x}\right ) \delta \left ( \vec{x},\vec{x}_{0}\right ) dA-{\displaystyle \iint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) dA={\displaystyle \int \limits _{-\infty }^{\infty }} \left ( G\left ( \vec{x},\vec{x}_{0}\right ) \frac{\partial u\left ( \vec{x}\right ) }{\partial y}-u\left ( \vec{x}\right ) \frac{\partial G\left ( \vec{x},\vec{x}_{0}\right ) }{\partial y}\right ) _{y=0}\ dx \] Since \(\int \int u\left ( \vec{x}\right ) \delta \left ( \vec{x},\vec{x}_{0}\right ) dA=u\left ( \vec{x}_{0}\right ) \), the above reduces to\begin{align} u\left ( \vec{x}_{0}\right ) -{\displaystyle \iint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) dA & ={\displaystyle \int \limits _{-\infty }^{\infty }} \left ( G\left ( \vec{x},\vec{x}_{0}\right ) \frac{\partial u\left ( \vec{x}\right ) }{\partial y}-u\left ( \vec{x}\right ) \frac{G\left ( \vec{x},\vec{x}_{0}\right ) }{dy}\right ) _{y=0}\ dx\nonumber \\ u\left ( \vec{x}_{0}\right ) & ={\displaystyle \iint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) \ dA+{\displaystyle \int \limits _{-\infty }^{\infty }} \left ( G\left ( \vec{x},\vec{x}_{0}\right ) \frac{\partial u\left ( \vec{x}\right ) }{\partial y}-u\left ( \vec{x}\right ) \frac{\partial G\left ( \vec{x},\vec{x}_{0}\right ) }{\partial y}\right ) _{y=0}\ dx\tag{2} \end{align}
And since \(G\left ( \vec{x},\vec{x}_{0}\right ) =0\) at \(y=0\), therefore\[ u\left ( \vec{x}_{0}\right ) ={\displaystyle \iint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) \ dA+{\displaystyle \int \limits _{-\infty }^{\infty }} \left ( -u\left ( \vec{x}\right ) \frac{\partial G\left ( \vec{x},\vec{x}_{0}\right ) }{\partial y}\right ) _{y=0}\ dx \] And since \(u\left ( \vec{x}\right ) =h\left ( x\right ) \) at \(y=0\), then\begin{equation} u\left ( \vec{x}_{0}\right ) ={\displaystyle \iint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) \ dA-{\displaystyle \int \limits _{-\infty }^{\infty }} h\left ( x\right ) \left ( \frac{\partial G\left ( \vec{x},\vec{x}_{0}\right ) }{\partial y}\right ) _{y=0}\ dx\tag{3} \end{equation} \(\left ( \frac{dG\left ( \vec{x},\vec{x}_{0}\right ) }{dy}\right ) _{y=0}\) is now evaluated to complete the solution. Using \(G\left ( \vec{x},\vec{x}_{0}\right ) \) in equation (1), therefore\begin{align*} \frac{dG\left ( \vec{x},\vec{x}_{0}\right ) }{dy} & =\frac{1}{4\pi }\frac{d}{dy}\left ( \ln \left ( \left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}\right ) -\ln \left ( \left ( x-x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}\right ) \right ) \\ & =\frac{1}{4\pi }\left ( \frac{2\left ( y-y_{0}\right ) }{\left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}}-\frac{2\left ( y+y_{0}\right ) }{\left ( x-x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}}\right ) \end{align*}
Evaluating the above at \(y=0\) gives\begin{align*} \left ( \frac{dG\left ( \vec{x},\vec{x}_{0}\right ) }{dy}\right ) _{y=0} & =\frac{1}{4\pi }\left ( \frac{-2y_{0}}{\left ( x-x_{0}\right ) ^{2}+y_{0}^{2}}-\frac{2y_{0}}{\left ( x-x_{0}\right ) ^{2}+y_{0}^{2}}\right ) \\ & =\frac{-1}{\pi }\left ( \frac{y_{0}}{\left ( x-x_{0}\right ) ^{2}+y_{0}^{2}}\right ) \end{align*}
Replacing the above into (3) gives \[ u\left ( \vec{x}_{0}\right ) ={\displaystyle \iint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) dA+\frac{y_{0}}{\pi }\int _{-\infty }^{\infty }\frac{h\left ( x\right ) }{\left ( x-x_{0}\right ) ^{2}+y_{0}^{2}}\ dx \] Using the expression for \(G\left ( \vec{x},\vec{x}_{0}\right ) \) from (1), the above result becomes\[ u\left ( x_{0},y_{0}\right ) =\frac{1}{4\pi }{\displaystyle \int \limits _{x=-\infty }^{\infty }}{\displaystyle \int \limits _{y=0}^{\infty }} \ln \frac{\left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}}{\left ( x-x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}}f\left ( x,y\right ) dydx+\frac{y_{0}}{\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} \frac{h\left ( x\right ) }{\left ( x-x_{0}\right ) ^{2}+y_{0}^{2}}\ dx \] And finally, order of \(\vec{x},\vec{x}_{0}\) is reversed giving\[ u\left ( x,y\right ) =\frac{1}{4\pi }{\displaystyle \int \limits _{x_{0}=-\infty }^{\infty }}{\displaystyle \int \limits _{y_{0}=0}^{\infty }} \ln \frac{\left ( x_{0}-x\right ) ^{2}+\left ( y_{0}-y\right ) ^{2}}{\left ( x_{0}-x\right ) ^{2}+\left ( y_{0}+y\right ) ^{2}}f\left ( x_{0},y_{0}\right ) dy_{0}dx_{0}+\frac{y}{\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} \frac{h\left ( x_{0}\right ) }{\left ( x_{0}-x\right ) ^{2}+y^{2}}\ dx_{0}\]
This is similar to part (a), and the image is placed on the same location as shown above, but now the boundary conditions are different. Starting from equation (2) in part (a)
\begin{equation} u\left ( \vec{x}_{0}\right ) ={\displaystyle \iint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) dA+{\displaystyle \int \limits _{-\infty }^{\infty }} \left ( G\left ( \vec{x},\vec{x}_{0}\right ) \frac{\partial u\left ( \vec{x}\right ) }{\partial y}-u\left ( \vec{x}\right ) \frac{\partial G\left ( \vec{x},\vec{x}_{0}\right ) }{\partial y}\right ) _{y=0}\ dx\tag{1} \end{equation} But now \(\frac{\partial G\left ( \vec{x},\vec{x}_{0}\right ) }{\partial y}=0\) at \(y=0\) and not \(G\left ( \vec{x},\vec{x}_{0}\right ) =0\) as in part (a). This means the image is a positive impulse and not negative as in part (a). Therefore \(G\left ( \vec{x},\vec{x}_{0}\right ) \) becomes the following\begin{align} G\left ( \vec{x},\vec{x}_{0}\right ) & =\frac{1}{2\pi }\ln \left ( r_{1}\right ) +\frac{1}{2\pi }\ln \left ( r_{2}\right ) \nonumber \\ & =\frac{1}{2\pi }\ln \left ( \left \vert \vec{x}-\vec{x}_{0}\right \vert \right ) +\frac{1}{2\pi }\ln \left ( \left \vert \vec{x}-\vec{x}_{0}^{\ast }\right \vert \right ) \nonumber \\ & =\frac{1}{4\pi }\left ( \ln \left ( \left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}\right ) +\ln \left ( \left ( x-x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}\right ) \right ) \tag{2} \end{align}
The following is 3D plot of the above Green function, showing that showing that \(\frac{\partial G\left ( \vec{x},\vec{x}_{0}\right ) }{\partial y}=0\) at \(y=0\) (marked as red)
Since\(\ \frac{\partial G\left ( \vec{x},\vec{x}_{0}\right ) }{\partial y}=0\) at \(y=0\) then (1) becomes \[ u\left ( \vec{x}_{0}\right ) ={\displaystyle \iint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) dA+{\displaystyle \int \limits _{-\infty }^{\infty }} \left ( G\left ( \vec{x},\vec{x}_{0}\right ) \frac{\partial u\left ( \vec{x}\right ) }{\partial y}\right ) _{y=0}\ dx \] But \(\frac{\partial u\left ( \vec{x}\right ) }{\partial y}=h\left ( x\right ) \) at \(y=0\), hence the above reduces to\begin{equation} u\left ( \vec{x}_{0}\right ) ={\displaystyle \iint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) dA+{\displaystyle \int \limits _{-\infty }^{\infty }} G\left ( \vec{x},\vec{x}_{0}\right ) _{y=0}h\left ( x\right ) \ dx\tag{3} \end{equation} Evaluating \(G\left ( \vec{x},\vec{x}_{0}\right ) \) at \(y=0\) gives\begin{align*} G\left ( \vec{x},\vec{x}_{0}\right ) _{y_{0}=0} & =\frac{1}{4\pi }\left ( \ln \left ( \left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}\right ) +\ln \left ( \left ( x-x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}\right ) \right ) _{y=0}\\ & =\frac{1}{4\pi }\left ( \ln \left ( \left ( x-x_{0}\right ) ^{2}+y_{0}^{2}\right ) +\ln \left ( \left ( x-x_{0}\right ) ^{2}+y_{0}^{2}\right ) \right ) \\ & =\frac{1}{4\pi }\ln \left ( \left ( \left ( x-x_{0}\right ) ^{2}+y_{0}^{2}\right ) ^{2}\right ) \\ & =\frac{1}{2\pi }\ln \left ( \left ( x-x_{0}\right ) ^{2}+y_{0}^{2}\right ) \end{align*}
Substituting the above in RHS of (3) gives\[ u\left ( \vec{x}_{0}\right ) ={\displaystyle \int \limits _{x=-\infty }^{\infty }}{\displaystyle \int \limits _{y=0}^{\infty }} G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( x,y\right ) dydx+\frac{1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} \ln \left ( \left ( x-x_{0}\right ) ^{2}+y_{0}^{2}\right ) h\left ( x\right ) \ dx \] Reversing the role of \(\vec{x}_{0},\vec{x}\) gives\[ u\left ( \vec{x}\right ) ={\displaystyle \int \limits _{x_{0}=-\infty }^{\infty }}{\displaystyle \int \limits _{y_{0}=0}^{\infty }} G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( x_{0},y_{0}\right ) dy_{0}dx_{0}+\frac{1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} \ln \left ( \left ( x_{0}-x\right ) ^{2}+y^{2}\right ) h\left ( x_{0}\right ) \ dx_{0}\]
In infinite 3D domain, the Green function for Poisson PDE is given by\[ G\left ( \vec{x},\vec{x}_{0}\right ) =\frac{-1}{4\pi r}\] Where \(r\) is given by\[ r=\sqrt{\left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}+\left ( z-z_{0}\right ) ^{2}}\] \(\vec{x}_{0}=\left ( x_{0},y_{0},z_{0}\right ) \) is the location of the impulse. Since \(\frac{\partial G}{\partial y}=0\), then the same sign impulse is located at \(x_{0}^{\ast }=\left ( x_{0},-y_{0},z_{0}\right ) \), and the Green function becomes\begin{align} G\left ( \vec{x},\vec{x}_{0}\right ) & =\frac{-1}{4\pi r_{0}}-\frac{1}{4\pi r_{0}^{\ast }}\nonumber \\ & =\frac{1}{4\pi }\left ( -\frac{1}{\sqrt{\left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}+\left ( z-z_{0}\right ) ^{2}}}-\frac{1}{\sqrt{\left ( x-x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}+\left ( z-z_{0}\right ) ^{2}}}\right ) \tag{1} \end{align}
Using Green formula in 3D gives\begin{align*}{\displaystyle \iiint } u\left ( \vec{x}\right ) \nabla ^{2}G\left ( \vec{x},\vec{x}_{0}\right ) -G\left ( \vec{x},\vec{x}_{0}\right ) \nabla ^{2}u\left ( \vec{x}\right ) dV & ={\displaystyle \iint } \left ( u\left ( \vec{x}\right ) \nabla G\left ( \vec{x},\vec{x}_{0}\right ) -G\left ( \vec{x},\vec{x}_{0}\right ) \nabla u\left ( \vec{x}\right ) \right ) \cdot \hat{n}\ dxdz\\ & ={\displaystyle \iint } \left ( u\left ( \vec{x}\right ) \nabla G\left ( \vec{x},\vec{x}_{0}\right ) -G\left ( \vec{x},\vec{x}_{0}\right ) \nabla u\left ( \vec{x}\right ) \right ) \cdot \left ( -\hat{\jmath }\right ) \ dxdz\\ & ={\displaystyle \iint } \left ( G\left ( \vec{x},\vec{x}_{0}\right ) \nabla u\left ( \vec{x}\right ) -u\left ( \vec{x}\right ) \nabla G\left ( \vec{x},\vec{x}_{0}\right ) \right ) \cdot \hat{\jmath }\ dxdz\\ & =\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\left ( G\left ( \vec{x},\vec{x}_{0}\right ) \frac{\partial u\left ( x,y,z\right ) }{\partial y}-u\left ( \vec{x}\right ) \frac{\partial G\left ( x,y,z,\vec{x}_{0}\right ) }{\partial y}\right ) _{y=0}\ dxdz \end{align*}
But \(\nabla ^{2}G\left ( \vec{x},\vec{x}_{0}\right ) =\delta \left ( \vec{x},\vec{x}_{0}\right ) \) and \(\nabla ^{2}u\left ( \vec{x}\right ) =f\left ( \vec{x}\right ) \), and the above becomes\[{\displaystyle \iiint } u\left ( \vec{x}\right ) \delta \left ( \vec{x},\vec{x}_{0}\right ) dV-{\displaystyle \iiint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) dv={\displaystyle \int \limits _{-\infty }^{\infty }}{\displaystyle \int \limits _{-\infty }^{\infty }} \left ( G\left ( \vec{x},\vec{x}_{0}\right ) \frac{\partial u\left ( \vec{x}\right ) }{\partial y}-u\left ( \vec{x}\right ) \frac{\partial G\left ( \vec{x},\vec{x}_{0}\right ) }{\partial y}\right ) _{y=0}\ dxdz \] But \({\displaystyle \iiint } u\left ( \vec{x}\right ) \delta \left ( \vec{x},\vec{x}_{0}\right ) dV=u\left ( \vec{x}_{0}\right ) \), hence\begin{equation} u\left ( \vec{x}_{0}\right ) -{\displaystyle \iiint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) dV={\displaystyle \int \limits _{-\infty }^{\infty }}{\displaystyle \int \limits _{-\infty }^{\infty }} \left ( G\left ( \vec{x},\vec{x}_{0}\right ) \frac{\partial u\left ( \vec{x}\right ) }{\partial y}-u\left ( \vec{x}\right ) \frac{\partial G\left ( \vec{x},\vec{x}_{0}\right ) }{\partial y}\right ) _{y=0}\ dxdz\nonumber \end{equation} Rearranging\begin{equation} u\left ( \vec{x}_{0}\right ) ={\displaystyle \iiint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) dV+{\displaystyle \int \limits _{-\infty }^{\infty }}{\displaystyle \int \limits _{-\infty }^{\infty }} \left ( G\left ( \vec{x},\vec{x}_{0}\right ) \frac{\partial u\left ( \vec{x}\right ) }{\partial y}-u\left ( \vec{x}\right ) \frac{\partial G\left ( \vec{x},\vec{x}_{0}\right ) }{\partial y}\right ) _{y=0}\ dxdz\tag{2} \end{equation} But \(\left ( \frac{\partial u\left ( \vec{x}\right ) }{\partial y}\right ) _{y=0}=h\left ( x,z\right ) \) and we impose \(\left ( \frac{\partial G\left ( \vec{x},\vec{x}_{0}\right ) }{\partial y}\right ) _{y=0}=0\), therefore the above becomes\begin{equation} u\left ( \vec{x}_{0}\right ) ={\displaystyle \iiint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) dV+{\displaystyle \int \limits _{-\infty }^{\infty }}{\displaystyle \int \limits _{-\infty }^{\infty }} G\left ( \vec{x},\vec{x}_{0}\right ) _{y=0}h\left ( x,z\right ) \ dxdz\tag{3} \end{equation} Evaluating \(G\left ( \vec{x},\vec{x}_{0}\right ) _{y=0}\) gives\begin{align*} G\left ( \vec{x},\vec{x}_{0}\right ) _{y=0} & =\frac{1}{4\pi }\left ( -\frac{1}{\sqrt{\left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}+\left ( z-z_{0}\right ) ^{2}}}-\frac{1}{\sqrt{\left ( x-x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}+\left ( z-z_{0}\right ) ^{2}}}\right ) _{y=0}\\ & =\frac{1}{4\pi }\left ( \frac{-1}{\sqrt{\left ( x-x_{0}\right ) ^{2}+y_{0}^{2}+\left ( z-z_{0}\right ) ^{2}}}-\frac{1}{\sqrt{\left ( x-x_{0}\right ) ^{2}+y_{0}^{2}+\left ( z-z_{0}\right ) ^{2}}}\right ) \\ & =-\frac{1}{2\pi \sqrt{\left ( x-x_{0}\right ) ^{2}+y_{0}^{2}+\left ( z-z_{0}\right ) ^{2}}} \end{align*}
Using the above in (3) results in\[ u\left ( \vec{x}_{0}\right ) ={\displaystyle \iiint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) dV-\frac{1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }}{\displaystyle \int \limits _{-\infty }^{\infty }} \frac{1}{\sqrt{\left ( x-x_{0}\right ) ^{2}+y_{0}^{2}+\left ( z-z_{0}\right ) ^{2}}}h\left ( x,z\right ) \ dxdz \] And finally reversing the role of \(\vec{x}_{0},\vec{x}\) gives the final answer\[ u\left ( \vec{x}\right ) ={\displaystyle \iiint } G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}_{0}\right ) dV_{0}-\frac{1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }}{\displaystyle \int \limits _{-\infty }^{\infty }} \frac{1}{\sqrt{\left ( x_{0}-x\right ) ^{2}+y^{2}+\left ( z_{0}-z\right ) ^{2}}}h\left ( x_{0},z_{0}\right ) \ dx_{0}dz_{0}\]
Green function in 2D on infinite domain, which is the solution to \[ \nabla ^{2}G\left ( \vec{x},\vec{x}_{0}\right ) =\delta \left ( \vec{x}-\vec{x}_{0}\right ) \] Is given by\[ G_{\infty }\left ( \vec{x},\vec{x}_{0}\right ) =\frac{1}{2\pi }\ln \left ( r\right ) \] A negative impulse is placed \(\vec{x}_{1}=\left ( x_{0},-y_{0}\right ) \) and another negative impulse at \(\vec{x}_{2}=\left ( -x_{0},y_{0}\right ) \) and positive one at \(\vec{x}_{3}=\left ( -x_{0},-y_{0}\right ) \). The following is a diagram showing the placement of images.
The resulting Green function becomes\[ G\left ( \vec{x},\vec{x}_{0}\right ) =\frac{1}{2\pi }\ln \left ( r\right ) -\frac{1}{2\pi }\ln \left ( r_{1}\right ) -\frac{1}{2\pi }\ln \left ( r_{2}\right ) +\frac{1}{2\pi }\ln \left ( r_{3}\right ) \] Or\begin{align} G\left ( \vec{x},\vec{x}_{0}\right ) & =\frac{1}{4\pi }\ln \left ( \left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}\right ) -\frac{1}{4\pi }\ln \left ( \left ( x-x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}\right ) \nonumber \\ & -\frac{1}{4\pi }\ln \left ( \left ( x+x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}\right ) +\frac{1}{4\pi }\ln \left ( \left ( x+x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}\right ) \tag{1} \end{align}
The following is 3D plot of the above Green function, showing the image impulse and showing that \(G=0\) at the line \(y=0\) and also at line \(x=0\). (Lines marked as red and blue)
Now that the Green function is found, it is used to solve \(\nabla ^{2}u\left ( \vec{x}\right ) =f\left ( \vec{x}\right ) \), with \(u\left ( x,0\right ) =h\left ( x\right ) \), \(u\left ( 0,y\right ) =g\left ( y\right ) \,.\)Starting with Green formula for 2D\begin{align*}{\displaystyle \iint } u\left ( \vec{x}\right ) \nabla ^{2}G\left ( \vec{x},\vec{x}_{0}\right ) -G\left ( \vec{x},\vec{x}_{0}\right ) \nabla ^{2}u\left ( \vec{x}\right ) dA & ={\displaystyle \oint _{s_{1}}} \left ( u\left ( \vec{x}\right ) \nabla G\left ( \vec{x},\vec{x}_{0}\right ) -G\left ( \vec{x},\vec{x}_{0}\right ) \nabla u\left ( \vec{x}\right ) \right ) \cdot \hat{n}\ ds\\ & +{\displaystyle \oint _{s_{2}}} \left ( u\left ( \vec{x}\right ) \nabla G\left ( \vec{x},\vec{x}_{0}\right ) -G\left ( \vec{x},\vec{x}_{0}\right ) \nabla u\left ( \vec{x}\right ) \right ) \cdot \hat{n}\ ds \end{align*}
To simplify the notation, from now on, \(G\,\) is used of \(G\left ( \vec{x},\vec{x}_{0}\right ) \), and also \(u\) instead of \(u\left ( \vec{x}\right ) \). The line \(s_{1}\) in the above is the line \(x>0,y=0\) and \(s_{2}\) is the line \(x=0,y>0\). Therefore the above becomes\[{\displaystyle \iint } u\nabla ^{2}G-G\nabla ^{2}u\ dA={\displaystyle \oint _{s_{1}}} \left ( u\nabla G-G\nabla u\right ) \cdot \left ( -\hat{\jmath }\right ) \ ds+{\displaystyle \oint _{s_{2}}} \left ( u\nabla G-G\nabla u\right ) \cdot \left ( -\hat{\imath }\right ) \ ds \] Or\[{\displaystyle \iint } u\nabla ^{2}G\ dA-{\displaystyle \iint } G\nabla ^{2}u\ dA={\displaystyle \int \limits _{0}^{\infty }} \left ( G\frac{\partial u}{\partial y}-u\frac{\partial G}{\partial y}\right ) _{y=0}\ dx+{\displaystyle \int \limits _{0}^{\infty }} \left ( G\frac{\partial u}{\partial x}-u\frac{\partial G}{\partial x}\right ) _{x=0}\ dy \] But \(\nabla ^{2}G\left ( \vec{x},\vec{x}_{0}\right ) =\delta \left ( \vec{x},\vec{x}_{0}\right ) \) and \(\nabla ^{2}u\left ( \vec{x}\right ) =f\left ( \vec{x}\right ) \), hence the above reduces to\[{\displaystyle \iint } u\left ( \vec{x}\right ) \delta \left ( \vec{x},\vec{x}_{0}\right ) \ dA-{\displaystyle \iint } Gf\left ( \vec{x}\right ) \ dA={\displaystyle \int \limits _{0}^{\infty }} \left ( G\frac{\partial u}{\partial y}-u\frac{\partial G}{\partial y}\right ) _{y=0}\ dx+{\displaystyle \int \limits _{0}^{\infty }} \left ( G\frac{\partial u}{\partial x}-u\frac{\partial G}{\partial x}\right ) _{x=0}\ dy \] But \({\displaystyle \iint } u\left ( \vec{x}\right ) \delta \left ( \vec{x},\vec{x}_{0}\right ) dA=u\left ( \vec{x}_{0}\right ) \) therefore\[ u\left ( \vec{x}_{0}\right ) -{\displaystyle \iint } Gf\left ( \vec{x}\right ) \ dA={\displaystyle \int \limits _{0}^{\infty }} \left ( G\frac{\partial u}{\partial y}-u\frac{\partial G}{\partial y}\right ) _{y=0}\ dx+{\displaystyle \int \limits _{0}^{\infty }} \left ( G\frac{\partial u}{\partial x}-u\frac{\partial G}{\partial x}\right ) _{x=0}\ dy \] Or\[ u\left ( \vec{x}_{0}\right ) ={\displaystyle \iint } Gf\left ( \vec{x}\right ) \ dA+{\displaystyle \int \limits _{0}^{\infty }} \left ( G\frac{\partial u}{\partial y}-u\frac{\partial G}{\partial y}\right ) _{y=0}\ dx+{\displaystyle \int \limits _{0}^{\infty }} \left ( G\frac{\partial u}{\partial x}-u\frac{\partial G}{\partial x}\right ) _{x=0}\ dy \] Since \(G\left ( \vec{x},\vec{x}_{0}\right ) =0\) at \(y=0\), and \(G\left ( \vec{x},\vec{x}_{0}\right ) =0\) at \(x=0\), the above becomes\[ u\left ( \vec{x}_{0}\right ) ={\displaystyle \iint } Gf\left ( \vec{x}\right ) \ dA-{\displaystyle \int \limits _{0}^{\infty }} \left ( u\frac{\partial G}{\partial y}\right ) _{y=0}\ dx-{\displaystyle \int \limits _{0}^{\infty }} \left ( u\frac{\partial G}{\partial x}\right ) _{x=0}\ dy \] Since \(u\left ( \vec{x}\right ) =h\left ( x\right ) \) at \(y=0\) and \(u\left ( \vec{x}\right ) =g\left ( y\right ) \) at \(x=0\) then\begin{equation} u\left ( x_{0},y_{0}\right ) ={\displaystyle \iint } Gf\left ( \vec{x}\right ) \ dA-{\displaystyle \int \limits _{0}^{\infty }} h\left ( x\right ) \left ( \frac{\partial G}{\partial y}\right ) _{y=0}\ dx-{\displaystyle \int \limits _{0}^{\infty }} g\left ( y\right ) \left ( \frac{\partial G}{\partial x}\right ) _{x=0}\ dy\tag{2} \end{equation} \(\left ( \frac{dG}{dy}\right ) _{y=0}\) and \(\left ( \frac{\partial G}{\partial x}\right ) _{x=0}\) are now evaluated to complete the solution. Using \(G\left ( \vec{x},\vec{x}_{0}\right ) \) in equation (1) gives\begin{align*} \frac{\partial G}{\partial y} & =\frac{1}{4\pi }\left ( \frac{2\left ( y-y_{0}\right ) }{\left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}}\right ) -\frac{1}{4\pi }\left ( \frac{2\left ( y+y_{0}\right ) }{\left ( x-x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}}\right ) \\ & -\frac{1}{4\pi }\left ( \frac{2\left ( y-y_{0}\right ) }{\left ( x+x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}}\right ) +\frac{1}{4\pi }\left ( \frac{2\left ( y+y_{0}\right ) }{\left ( x+x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}}\right ) \end{align*}
Evaluating the above at \(y=0\) results in\begin{align*} \left ( \frac{\partial G}{\partial y}\right ) _{y=0} & =\frac{1}{4\pi }\left ( \frac{-2y_{0}}{\left ( x-x_{0}\right ) ^{2}+y_{0}^{2}}\right ) -\frac{1}{4\pi }\left ( \frac{2y_{0}}{\left ( x-x_{0}\right ) ^{2}+y_{0}^{2}}\right ) \\ & -\frac{1}{4\pi }\left ( \frac{-2y_{0}}{\left ( x+x_{0}\right ) ^{2}+y_{0}^{2}}\right ) +\frac{1}{4\pi }\left ( \frac{2y_{0}}{\left ( x+x_{0}\right ) ^{2}+y_{0}^{2}}\right ) \end{align*}
Or\begin{equation} \left ( \frac{\partial G}{\partial y}\right ) _{y=0}=\frac{y_{0}}{\pi }\left ( \frac{1}{\left ( x+x_{0}\right ) ^{2}+y_{0}^{2}}-\frac{1}{\left ( x-x_{0}\right ) ^{2}+y_{0}^{2}}\right ) \tag{3} \end{equation} Finding \(\frac{\partial G}{\partial x}\) gives\begin{align*} \frac{\partial G}{\partial x} & =\frac{1}{4\pi }\left ( \frac{2\left ( x-x_{0}\right ) }{\left ( x-x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}}\right ) -\frac{1}{4\pi }\left ( \frac{2\left ( x-x_{0}\right ) }{\left ( x-x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}}\right ) \\ & -\frac{1}{4\pi }\left ( \frac{2\left ( x+x_{0}\right ) }{\left ( x+x_{0}\right ) ^{2}+\left ( y-y_{0}\right ) ^{2}}\right ) +\frac{1}{4\pi }\left ( \frac{2\left ( x+x_{0}\right ) }{\left ( x+x_{0}\right ) ^{2}+\left ( y+y_{0}\right ) ^{2}}\right ) \end{align*}
Evaluating the above at \(x=0\) results in\begin{align*} \left ( \frac{\partial G}{\partial x}\right ) _{x=0} & =\frac{1}{4\pi }\left ( \frac{-2x_{0}}{x_{0}^{2}+\left ( y-y_{0}\right ) ^{2}}\right ) -\frac{1}{4\pi }\left ( \frac{-2x_{0}}{x_{0}^{2}+\left ( y+y_{0}\right ) ^{2}}\right ) \\ & -\frac{1}{4\pi }\left ( \frac{2x_{0}}{x_{0}^{2}+\left ( y-y_{0}\right ) ^{2}}\right ) +\frac{1}{4\pi }\left ( \frac{2x_{0}}{x_{0}^{2}+\left ( y+y_{0}\right ) ^{2}}\right ) \end{align*}
Or\begin{align} \left ( \frac{\partial G}{\partial x_{0}}\right ) _{x=0} & =\frac{1}{\pi }\left ( \frac{x_{0}}{x_{0}^{2}+\left ( y+y_{0}\right ) ^{2}}\right ) -\frac{1}{\pi }\left ( \frac{x_{0}}{x_{0}^{2}+\left ( y-y_{0}\right ) ^{2}}\right ) \nonumber \\ & =\frac{x_{0}}{\pi }\left ( \frac{1}{x_{0}^{2}+\left ( y+y_{0}\right ) ^{2}}-\frac{1}{x_{0}^{2}+\left ( y-y_{0}\right ) ^{2}}\right ) \tag{4} \end{align}
Substituting (3,4) into (2) gives the final answer\begin{align*} u\left ( x_{0},y_{0}\right ) & ={\displaystyle \int \limits _{0}^{\infty }}{\displaystyle \int \limits _{0}^{\infty }} G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( \vec{x}\right ) \ dxdy\\ & -\frac{y_{0}}{\pi }{\displaystyle \int \limits _{0}^{\infty }} h\left ( x\right ) \left ( \frac{1}{\left ( x+x_{0}\right ) ^{2}+y_{0}^{2}}-\frac{1}{\left ( x-x_{0}\right ) ^{2}+y_{0}^{2}}\right ) \ dx\\ & -\frac{x_{0}}{\pi }{\displaystyle \int \limits _{0}^{\infty }} g\left ( y\right ) \left ( \frac{1}{x_{0}^{2}+\left ( y+y_{0}\right ) ^{2}}-\frac{1}{x_{0}^{2}+\left ( y-y_{0}\right ) ^{2}}\right ) \ dy \end{align*}
Reversing the role of \(\vec{x},\vec{x}_{0}\) gives\begin{align*} u\left ( x,y\right ) & ={\displaystyle \int \limits _{0}^{\infty }}{\displaystyle \int \limits _{0}^{\infty }} G\left ( \vec{x},\vec{x}_{0}\right ) f\left ( x_{0},y_{0}\right ) \ dx_{0}dy_{0}\\ & -\frac{y}{\pi }{\displaystyle \int \limits _{0}^{\infty }} h\left ( x_{0}\right ) \left ( \frac{1}{\left ( x_{0}+x\right ) ^{2}+y^{2}}-\frac{1}{\left ( x_{0}-x\right ) ^{2}+y^{2}}\right ) \ dx_{0}\\ & -\frac{x}{\pi }{\displaystyle \int \limits _{0}^{\infty }} g\left ( y_{0}\right ) \left ( \frac{1}{x^{2}+\left ( y_{0}+y\right ) ^{2}}-\frac{1}{x^{2}+\left ( y_{0}-y\right ) ^{2}}\right ) \ dy_{0} \end{align*}
Where \(G\left ( \vec{x},\vec{x}_{0}\right ) \) is given by equation (1). This complete the solution.
The following is 3D plot of the solution (for small area is first quadrant) generated using Mathematica using \begin{align*} f\left ( x\right ) & =-20e^{-\left ( x-4\right ) ^{2}-\left ( y-5\right ) ^{2}}\\ g\left ( y\right ) & =10\sin \left ( 5y\right ) \\ h\left ( x\right ) & =5\cos \left ( 2x\right ) \end{align*}
This is a contour plot of the above solution
