How to solve the following Poisson equation? [closed]

Solve the following Poisson equation $$u_{xx}+u_{yy}=x^2+4xy \text{ for } x^2+y^2<4\\ u(x,y)=-2 \text{ for } x^2+y^2=4$$

I know how to solve homogenous Laplace equation but I have no idea how to solve can you help

Let $x = r \cos \varphi, y = r \sin \varphi$ for $r \in [0, 2]$ and $\phi \in [0, 2\pi)$. Then for $v(r, \varphi) = u(r \cos \varphi, r \sin \varphi)$: $$u_{xx} + u_{yy} = \frac{1}{r^2} \left(r^2 v_{rr} + r v_r + v_{\varphi \varphi} \right) = r^2 \left( \cos^2 \varphi + 4 \cos \varphi \sin \varphi\right) \Leftrightarrow \\ \Leftrightarrow r^2 v_{rr} + r v_r + v_{\varphi \varphi} = r^4 \left( \frac{1}{2} + \frac{1}{2}\cos2\varphi + 4 \cos \varphi \sin \varphi\right) $$ and $v(2, \varphi) = -2$. Let's look for a solution in form of $v(r, \varphi) = v_0 (r, \varphi) + v_1 (r, \varphi) + v_2 (r, \varphi)$, where $$ \begin{cases} r^2 (v_0)_{rr} + r (v_0)_r + (v_0)_{\varphi \varphi} = 0 \\ v_0(2, \varphi) = -2 \\ r^2 (v_1)_{rr} + r (v_1)_r + (v_1)_{\varphi \varphi} = \frac{1}{2}r^4 \\ v_1(2, \varphi) = 0 \\ r^2 (v_2)_{rr} + r (v_2)_r + (v_2)_{\varphi \varphi} = r^4 \cdot \left(\frac{1}{2}\cos2\varphi + 4 \cos \varphi \sin \varphi \right)\\ v_2(2, \varphi) = 0 \end{cases} $$

Let's make an ansatz that $v_1(r, \varphi) = f_1(r)$ and $v_2(r, \varphi) = f_2(r) \cdot \left(\frac{1}{2}\cos2\varphi + 4 \cos \varphi \sin \varphi \right)$. Then $$ \begin{cases} r^2 f_1''(r) + r f_1'(r) = \frac{1}{2} r^4 \\ f(2) = 0 \\ \end{cases} \Leftrightarrow \begin{cases} r f_1''(r) + f_1'(r) = \frac{1}{2}r^3 \\ f_1(2) = 0 \end{cases} \Leftrightarrow \begin{cases} \left(r f_1'(r) \right)' = \frac{1}{2}r^3 \\ f_1(2) = 0 \end{cases} \Rightarrow \\ \Rightarrow \begin{cases} f_1(r) = \frac{1}{32}r^4 + c_1 \ln r + c_2 \\ f_1(2) = 0 \end{cases} $$ As $f_1$ must be defined for $r=0$, $c_1 = 0$ and from $f_1(2) = 0$ $c_2 = -\frac{1}{2}$, so $v_1(r, \varphi) = f_1(r) = \frac{1}{32}r^4 - \frac{1}{2} $. Now let's take a look at $v_2$. As $$(v_2)_{\varphi \varphi} = f_2(r) \cdot \left(\frac{1}{2}\cos2\varphi + 4 \cos \varphi \sin \varphi\right)_{\varphi \varphi} = f_2(r) \cdot (-4) \cdot \left(\frac{1}{2}\cos2\varphi + 4 \cos \varphi \sin \varphi\right)$$ we get $$ \begin{cases} r^2 f_2''(r) + r f_2'(r) - 4 f_2(r) = r^4 \\ f_2(2) = 0 \end{cases} \Rightarrow \begin{cases} f_2(r) = \frac{1}{12}r^4 + \frac{c_1}{r^2} + c_2 r^2 \\ f_2(2) = 0 \end{cases} $$ Again, as $f_2$ must be defined for $r=0$, $c_1 = 0$ and from $f_2(2) = 0$ we get $c_2 = -\frac{1}{3}$, so $f_2(r) = \frac{1}{12}r^4 - \frac{1}{3}r^2$.

Finally, $v_0$ is easy to find with standard methods (seperation of variables). Letting $v_0(r, \varphi) = f(r) \cdot g(\varphi)$ leads to $$ v_0(r, \varphi) = \sum_{n=0}^{\infty} r^n \left( C_n \cos n\varphi + D_n \sin n\varphi \right) $$ As $v_0(2, \varphi) = -2 = \sum_{n=0}^{\infty} 2^n \left( C_n \cos n\varphi + D_n \sin n\varphi \right)$, we conclude that $C_0 = -2, D_0 = 0$, $C_n = D_n = 0, n \geq 1$, so $v_0(r, \varphi) = -2$. Finally, $$ v(r, \varphi) = -2 + \left(\frac{1}{32}r^4 - \frac{1}{2}\right) + \left( \frac{1}{12}r^4 - \frac{1}{3}r^2 \right) \cdot \left(\frac{1}{2}\cos2\varphi + 4 \cos \varphi \sin \varphi\right) = \\ = -\frac{5}{2} + \frac{1}{32}r^4 + \frac{1}{24}r^4 \cos2\varphi + \frac{1}{3}r^4 \cos \varphi \sin \varphi - \frac{1}{6}r^2 \cos2\varphi - \frac{4}{3}r^2 \cos \varphi \sin \varphi = \\ = - \frac{5}{2} + \frac{1}{12}r^4 \left(\frac{3}{8} + \frac{1}{2}\cos 2\varphi\right) + \frac{1}{3}r^4 \cos \varphi \sin \varphi - \frac{1}{3} r^2 \left(\frac{1}{2} + \frac{1}{2}\cos2\varphi\right) + \frac{1}{6}r^2 - \frac{4}{3}r^2 \cos \varphi \sin \varphi = \\ = - \frac{5}{2} + \frac{1}{12} r^4 \left(\cos^2 \varphi - \frac{1}{8}\right) + \frac{1}{3}r^4 \cos \varphi \sin \varphi - \frac{1}{3}r^2\cos^2\varphi + \frac{1}{6}r^2 - \frac{4}{3}r^2 \cos \varphi \sin \varphi $$ Remembering that $x = r \cos \varphi, y = r \sin \varphi$, we finally get $$ u(x, y) = -\frac{5}{2} - \frac{1}{96}(x^2 + y^2)^2 + \frac{1}{12} (x^2+y^2)x^2 + \frac{1}{3} (x^2+y^2)xy - \frac{1}{3}x^2 + \frac{1}{6}(x^2+y^2) -\frac{4}{3}xy $$

It is beneficial to adopt the plane polar coordinates for this problem. One has \begin{align} \frac{1}{r}\frac{\partial}{\partial{r}}\Bigg(r\frac{\partial}{\partial{r}}\psi\Bigg) +\frac{1}{r^{2}}\frac{\partial^{2}}{\partial{\varphi}^{2}}\psi=r^{2}[\cos^{2}\varphi+2\sin2\varphi], \quad \psi(2, \varphi)=-2. \end{align} The solution to the homogenous equation obeys \begin{align} \frac{1}{r}\frac{\partial}{\partial{r}}\Bigg(r\frac{\partial}{\partial{r}}\psi_{h}\Bigg) +\frac{1}{r^{2}}\frac{\partial^{2}}{\partial{\varphi}^{2}}\psi_{h}=0 \end{align} We let \begin{align} \psi_{h}(r, \varphi)=\frac{1}{\sqrt{2\pi}}\sum_{n\in\mathbb{Z}}\hat\psi_{h}^{(n)}(r)e^{in\varphi}, \end{align} so that \begin{align} r^{2}\frac{\partial^{2}}{\partial{r}^{2}}\hat\psi_{h}^{(n)}+r\frac{\partial}{\partial{r}}\hat\psi_{h}^{(n)}-n^{2}\hat\psi_{h}^{(n)}=0. \end{align} This is a standard Euler equation which has the solution \begin{align} \hat\psi_{h}^{(n)}(r)=C_{h, +}^{(n)}r^{n}+C_{h, -}^{(n)}r^{-n}. \end{align} Hence \begin{align} \psi_{h}(r, \varphi)=\sum_{n\in\mathbb{Z}}[C_{h, +}^{(n)}r^{n}+C_{h, -}^{(n)}r^{-n}]\frac{e^{in\varphi}}{\sqrt{2\pi}}. \end{align} Now we consider the inhomogeneous part, i.e. the particular solution \begin{align} \frac{1}{r}\frac{\partial}{\partial{r}}\Bigg(r\frac{\partial}{\partial{r}}\psi_{p}\Bigg) +\frac{1}{r^{2}}\frac{\partial^{2}}{\partial{\varphi}^{2}}\psi_{p}=r^{2}[\cos^{2}\varphi+2\sin2\varphi], \quad \psi(2, \varphi)=-2. \end{align} Again we Fourier expand \begin{align} \frac{1}{r}\frac{\partial}{\partial{r}}\Bigg(r\frac{\partial}{\partial{r}}\psi_{p}^{(n)}\Bigg) -\frac{n^{2}}{r^{2}}\psi_{p}^{(n)}=r^{2}\int_{-\pi}^{\pi}d\varphi[\cos^{2}\varphi+2\sin2\varphi]\frac{e^{-in\varphi}}{\sqrt{2\pi}}. \end{align} We have \begin{align} \int_{-\pi}^{\pi}d\varphi[\cos^{2}\varphi+2\sin2\varphi]\frac{e^{-in\varphi}}{\sqrt{2\pi}}=\frac{\sqrt{2\pi}}{4}[2\delta_{n, 0}+[1-4i]\delta_{n, 2}+[1+4i]\delta_{n, -2}]. \end{align} Hence the "non-trivial" modes satisfy \begin{align} \frac{\partial^{2}}{\partial{r}^{2}}\psi_{p}^{(0)}+\frac{1}{r}\frac{\partial}{\partial{r}}\psi_{p}^{(0)}=&r^{2}\frac{\sqrt{2\pi}}{2},\\ \frac{\partial^{2}}{\partial{r}^{2}}\psi_{p}^{(2)}+\frac{1}{r}\frac{\partial}{\partial{r}}\psi_{p}^{(2)} -\frac{4}{r^{2}}\psi_{p}^{(2)}=&r^{2}\frac{[1-4i]\sqrt{2\pi}}{4},\\ \frac{\partial^{2}}{\partial{r}^{2}}\psi_{p}^{(-2)}+\frac{1}{r}\frac{\partial}{\partial{r}}\psi_{p}^{(-2)} -\frac{4}{r^{2}}\psi_{p}^{(-2)}=&r^{2}\frac{[1+4i]\sqrt{2\pi}}{4}. \end{align} We let \begin{align} \psi_{p}^{(0)}=&\frac{\sqrt{2\pi}}{32}r^{4},\\ \psi_{p}^{(2)}=&\frac{[1-4i]\sqrt{2\pi}}{48}r^{4},\\ \psi_{p}^{(-2)}=&\frac{[1+4i]\sqrt{2\pi}}{48}r^{4}. \end{align} It follows \begin{align} \psi_{p}(r, \varphi)=&\frac{r^{4}}{8}\Bigg[\frac{1}{4}+\frac{2}{3}\cos(2\varphi)+\frac{8}{3}\sin(2\varphi)\Bigg]. \end{align} The general solution is thus given by \begin{align} \psi(r, \varphi)=\sum_{n\in\mathbb{Z}}[C_{h, +}^{(n)}r^{n}+C_{h, -}^{(n)}r^{-n}]\frac{e^{in\varphi}}{\sqrt{2\pi}}+\psi_{p}(r, \varphi). \end{align} The boundary conditions dictate \begin{align} \sum_{n\in\mathbb{Z}}[C_{h, +}^{(n)}2^{n}+C_{h, -}^{(n)}2^{-n}]\frac{e^{in\varphi}}{\sqrt{2\pi}}+\psi_{p}(2, \varphi)=-2. \end{align} We multiply by $\frac{e^{-in\varphi}}{\sqrt{2\pi}}$ an integrate over $2\pi$, so that \begin{align} C_{h, +}^{(n)}2^{n}+C_{h, -}^{(n)}2^{-n}+\psi_{p}^{(0)}(2)\delta_{n, 0}+\psi_{p}^{(2)}(2, \varphi)\delta_{n, 2}+\psi_{p}^{(-2)}(2)\delta_{n, -2}=-2\sqrt{2\pi}\delta_{n, 0}. \end{align} We have in particular \begin{align} C_{h, +}^{(0)}+C_{h, -}^{(0)}=&-\frac{5}{2}\sqrt{2\pi},\\ C_{h, +}^{(2)}4+C_{h, -}^{(2)}\frac{1}{4}=&-\frac{[1-4i]\sqrt{2\pi}}{3},\\ C_{h, +}^{(-2)}\frac{1}{4}+C_{h, -}^{(2)}4=&-\frac{[1+4i]\sqrt{2\pi}}{3},\\ C_{h, +}^{(n)}=-C_{h, -}^{(n)}2^{-2n},\quad n\neq0, \ \pm2. \end{align} This completes the solution.

The other answers are good. I'd like to give my own thoughts.

You can approach this problem in a similar way as I do here, by breaking the problem into simpler parts. Your problem is $$\begin{cases} (\Delta u)( x,y) =x^{2} +4xy & ( x,y) \in \mathbb{B}( 0,4)\\ u( x,y) =-2 & ( x,y) \in \partial \mathbb{B}( 0,4) \end{cases}\tag{0}$$ Where $\Delta=\partial_x^2+\partial_y^2$.

I would suggest first finding a solution (call it $u_1$) to $$\begin{cases} (\Delta u)( x,y) =x^{2} +4xy & ( x,y) \in \mathbb{B}( 0,4)\\ u( x,y) =0 & ( x,y) \in \partial \mathbb{B}( 0,4) \end{cases}\tag{1}$$ And then find a solution (call it $u_2$) to $$\begin{cases} (\Delta u)( x,y) =0 & ( x,y) \in \mathbb{B}( 0,4)\\ u( x,y) =-2 & ( x,y) \in \partial \mathbb{B}( 0,4) \end{cases}\tag{2}$$ And then by the linearity of the Laplacian, $u_1+u_2$ will be a solution to $\boldsymbol{(0)}$. What is nice about this approach is that the solution to $\boldsymbol{(2)}$ is already well known. What we can do is make a change of coordinates $$x'=x/4~~;~~y'=y/4$$ To move the problem to the unit disk, and then use the Poisson Kernel. Switching to polar coordinates, given the BVP on the unit disk $$\begin{cases} (\Delta u)(r,\theta) =0 & r<1\\ u(1,\theta)=h(\theta)& \\ (r,\theta)\in[0,1]\times(-\pi,\pi] \end{cases}\tag{0}$$ The solution is given by $$u(r,\theta)=\frac{1}{2\pi}\int_{-\pi}^{\pi}h(\phi)\frac{1-r^2}{1+r^2+2r\cos(\theta-\phi)}\mathrm d\phi$$ See pages 160-166 in Peter J Olver's Introduction to Partial Differential Equations for a derivation of this formula.

So in your case you can pick $h=-2$. Unfortunately I don't think $\boldsymbol{(1)}$ is quite this easy, since $x^2+4xy$ doesn't transform all that nicely under polar coordinates. So perhaps a closed form for this part is not possible. I'm sure the other answers will be plenty useful for this bit.

You can just set up a discretized linear equation system.

Make ansatz of truncated power series expansion $$u(x,y) = \sum_{\forall i,j \in\{0,\cdots,N\}} c_{ij}x^iy^j$$

Vectorize $c_{ij}$ lexicographically. Implement $\bf D$ as $\{0,1,\cdots N-1\}$ 1 step off-diagonal matrix. Example for 6th degree polynomial :

$${\bf D} = \left[\begin{array}{lllllll}&1&&&&&\\&&2&&&&\\&&&3&&&\\&&&&4&&\\&&&&&5&\\&&&&&&6\\&&&&&&\end{array}\right]$$

Now we can construct 2D partial differentiators like so:

$${\bf D_x = ({\bf I}\otimes {\bf D})}\\{\bf D_y = (D\otimes I)}$$

Now set up boundary conditions. We can sample uniformly $$x+yi = 2\exp(2\pi i k/N), \forall k \in\{0,\cdots,N-1\}$$

Remains to do is to build equation system and solve it. Here is what a solution can look like (the blue circles are the linearly spaced points used to ensure boundary conditions are followed) :

Just for verification purposes now that we have an exact solution by @Yalikesifulei here is a truncated representation of the numerical solution (any number abs val $<10^{-10}$ has been set to $0$) :

$$\left[\begin{array}{lllll}-2.500&0&0.166667&0&-0.01041666\\0&-1.3333&0&0.333333&0\\-0.166667&0&0.06249999&0&0\\0&0.33333&0&0&0\\0.07291666&0&0&0&0\end{array}\right]$$

The three trickiest coefficients which we may not see straight from ocular inspection are the same are

for $y^4$ :

$1/12-1/96 = 0.07916666...$

and $x^4$ :

$-1/96 = -0.010416666...$

and for $x^2y^2$ :

$1/12-2/96 = 0.0625$