3.4 Poisson PDE, Cartesian coordinates, source term depends on y

Solve Poisson PDE \[ \frac{\partial ^{2}f}{\partial x^{2}}+\frac{\partial ^{2}f}{\partial y^{2}}=6y \]

The solution is \begin{equation} f=f_{h}+f_{p} \tag{1} \end{equation} Where \(f_{h}\) is the homogenous solution to Laplace \(\frac{\partial ^{2}f}{\partial x^{2}}+\frac{\partial ^{2}f}{\partial y^{2}}=0\) and \(f_{p}\) is a particular solution. The homogeneous solution is easily found as follows. Let \[ f_{h}=F\left ( mx+y\right ) \] Then \(\frac{\partial f_{h}}{\partial x}=mF^{\prime }\) and \(\frac{\partial ^{2}f_{h}}{\partial x^{2}}=m^{2}F^{\prime \prime }\) and \(\frac{\partial f_{h}}{\partial y}=F^{\prime }\) and \(\frac{\partial ^{2}f_{h}}{\partial y^{2}}=F^{\prime \prime }\). Substituting these into \(\frac{\partial ^{2}f_{h}}{\partial x^{2}}+\frac{\partial ^{2}f_{h}}{\partial y^{2}}=0\) gives\begin{align*} m^{2}F^{\prime \prime }+F^{\prime \prime } & =0\\ m^{2}+1 & =0\\ m & =\pm i \end{align*}

Since we assumed \(f_{h}=F\left ( mx+y\right ) \), then the homogeneous solution is sum of two arbitrary functions (one for each root of \(m\))\[ f_{h}=F_{1}\left ( ix+y\right ) +F_{2}\left ( -ix+y\right ) \] To find particular solution, let \begin{equation} f_{p}=Ax^{n}y^{m} \tag{2} \end{equation} Hence \(\frac{\partial f_{p}}{\partial x}=Anx^{n-1}y^{m}\) and \(\frac{\partial ^{2}f_{p}}{\partial x^{2}}=An\left ( n-1\right ) x^{n-2}y^{m}\) and \(\frac{\partial f_{p}}{\partial y}=Ax^{n}my^{m-1}\) and \(\frac{\partial ^{2}f_{p}}{\partial y^{2}}=Ax^{n}m\left ( m-1\right ) y^{m-2}\). Substituting these into \(\frac{\partial ^{2}f_{p}}{\partial x^{2}}+\frac{\partial ^{2}f_{p}}{\partial y^{2}}=6y\) gives\begin{align*} An\left ( n-1\right ) x^{n-2}y^{m}+Ax^{n}m\left ( m-1\right ) y^{m-2} & =6y\\ y^{m}\left ( An\left ( n-1\right ) x^{n-2}+Ax^{n}m\left ( m-1\right ) y^{-2}\right ) & =6y \end{align*}

Comparing terms, then \(m=1\) and \[ An\left ( n-1\right ) x^{n-2}+Ax^{n}m\left ( m-1\right ) y^{-2}=6 \] Since \(m=1\) then the above simplifies to\[ An\left ( n-1\right ) x^{n-2}=6 \] Since there is no \(x\) in RHS, then \(n-2=0\) or \(n=2\) and the above becomes\begin{align*} 2A\left ( 2-1\right ) & =6\\ A & =3 \end{align*}

Hence (2) becomes\[ f_{p}=Ax^{n}y^{m}=3x^{2}y \] And the complete solution (1) is\begin{align*} f\left ( x,y\right ) & =f_{h}+f_{p}\\ & =F_{1}\left ( ix+y\right ) +F_{2}\left ( -ix+y\right ) +3x^{2}y \end{align*}