Solution satisfies boundary condition but does not satisfy the PDE $\nabla^2 u(r,\theta) = f(r,\theta)$. Does anyone see the issue?

Question

As @Aruralreader pointed out, we could find this by understanding that our solution has angular symmetry and thus: $$(\partial_{rr} + \frac{1}{r}\partial_r) u(r,\theta) = u(r)_{rr} + \frac{1}{r}u(r)_r = 1$$

However, the desired goal here is to arrive to the solution using only Green's Function, so I'm asking for help with finding the solution using only Green's function.

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This is all in polar coordinates.

$$\nabla^2 u(r,\theta) = f(r,\theta); \cases{0\le\theta\le 2\pi \\ 0\le r < \infty \\ u(r,\theta) = u(r,\theta\pm2n\pi)\\ u(a,\theta) = 0 \\ f(r,\theta) = \cases{1 & $0\le r \le R \land 0\le\theta\le 2\pi$ \\ 0}}$$

Green's function (which we arrive to using method of images for a circle) for this equation (such that $G(r = a,\theta,r_0,\theta_0) = 0$): $$G(\textbf{r},\textbf{r}_0) = \frac{1}{4\pi}\ln\Bigg(\frac{a^2}{r_0^2}\frac{r^2+r_0^2-2rr_r\cos(\theta-\theta_0)}{r^2+a^2\Big(\frac{a}{r_0}\Big)^2-2r\frac{a^2}{r_0}\cos(\theta-\theta_0)}\Bigg)$$

And the solution obtained from Green's formula is then:

$$\tag{1} u(r,\theta) = \int_0^{2\pi} \int_0^R f(\textbf{r}_0)G(\textbf{r},\textbf{r}_0)r_0dr_0d\theta_0 - \int_0^{2\pi}u(a,\theta)\frac{\partial G(\textbf{r},\textbf{r}_0)}{\partial r_0}|_{|\textbf{r}| = r= a}d\theta_0 \\ = \frac{1}{4\pi}\int_0^{2\pi} \int_0^R \ln\Bigg(\frac{a^2}{r_0^2}\frac{r^2+r_0^2-2rr_r\cos(\theta-\theta_0)}{r^2+a^2\Big(\frac{a}{r_0}\Big)^2-2r\frac{a^2}{r_0}\cos(\theta-\theta_0)}\Bigg)r_0 dr_0d\theta_0 + 0$$

Computing the angular integral which we note the integrand was simplified from $\cos(\theta-\theta_0)$ to $\cos(\theta_0)$ as semiclassical alludes to here due to integrating a $2\pi$ periodic function over the period, after doing some factoring to rewrite the square roots in terms of magnitudes which I worked out at the end of my own answer here, we arrive to:

$$\tag{2}u(r,\theta) =-\frac{1}{2}\int_0^R\ln\Bigg(\frac{r_0^2}{a^2}\frac{r^2+\Big(\frac{a^2}{r_0}\Big)^2+|r^2-\Big(\frac{a^2}{r_0}\Big)^2|}{r^2+r_0^2+|r^2-(r_0)^2|}\Bigg)r_0dr_0; $$

From the magnitudes, we rewrite them into the following piecewise definitions:

$$|r^2-\Big(\frac{a^2}{r_0}\Big)^2| = \cases{r^2-\Big(\frac{a^2}{r_0}\Big)^2 & $r_0\ge\frac{a^2}{r}$ \\ \Big(\frac{a^2}{r_0}\Big)^2-r^2 & $r_0<\frac{a^2}{r}$} \\ |r^2-r_0^2| = \cases{r^2-r_0^2 & $r_0\le r$ \\ r_0^2-r^2 & $r_0>r$}$$

Which results in the following integrals:

$$-\frac{1}{2}\Bigg[\underbrace{2 \int_{0}^{\frac{a^2}{r}}r_0\ln\Big(\sqrt{2}a^2\Big)dr_0}_{(A)} + 2\underbrace{\int_\frac{a^2}{r}^{R}r_0\ln\Big(\sqrt2 rr_0\Big)dr_0}_{(B)} \\ -2\underbrace{\int_0^r r_0\ln(\sqrt{2}ar)dr_0}_{(C)} - 2\underbrace{\int_r^R r_0\ln(\sqrt{2}ar_0)dr_0}_{(D)}\Bigg]$$

$$(A) = -\frac{1}{2}\Big(\frac{a^2}{r}\Big)^2\ln(\sqrt{2}a^2)$$

For (B): $$u = \ln(\sqrt{2}rr_0) \ \ \ \ dv = r_0dr_0 \\ du = \frac{\sqrt{2}r}{\sqrt{2}rr_0}dr_0 \ \ \ \ v= r_0^2/2 \\ (B) = -\frac{1}{2}\Bigg[\Big(R^2\ln(\sqrt{2}rR)-\Big(\frac{a^2}{r}\Big)^2\ln(\sqrt{2}a^2)\Big) + \frac{1}{2}\bigg(\Big(\frac{a^2}{r}\Big)^2-R^2\Big)\bigg)\bigg]$$

$$(C) = +\frac{1}{2}r^2\ln(\sqrt{2}ar)$$

For (D):

$$u = \ln\Big(\sqrt{2}ar_0\Big) \ \ \ \ dv = r_0dr_0 \\ du = \frac{1}{\sqrt{2}ar_0}\sqrt{2}a dr_0 \ \ \ \ v= r_0^2/2 \\ (D) = +\frac{1}{2}\Bigg[R^2\ln(\sqrt{2}aR)-r^2\ln(\sqrt{2}ar) + \frac{1}{2}(r^2-R^2)\Bigg] $$

$$I(r) = (A)+(B)+(C)+(D) = -\frac{1}{2}R^2\ln(\sqrt{2}rR) + \frac{1}{4}\bigg(r^2-\Big(\frac{a^2}{r}\Big)\bigg) + \frac{1}{2}R^2\ln(\sqrt{2}aR) \\ u(r,\theta) = \frac{1}{4r^2}\Bigg[r^4 + R^2r^2\ln(a/r) - a^4\Bigg] $$

This solution satisfies the boundary condition: $u(r=a,\theta) = 0$

What it doesn't satisfy is the Poisson Equation (in polar coordinates):

$$\nabla^2 u(r,\theta) = \cases{1 & $0\le r\le R \land 0 \le \theta \le 2\pi$ \\ 0} \\ = \frac{\partial^2 u(r,\theta)}{\partial r^2} + \frac{1}{r}\frac{\partial u(r,\theta)}{\partial r} \\ =1-\frac{a^4}{r^4} \ne 1$$

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NOTE: $\frac{1}{4r^2}\Big(R^2r^2\ln(a/r)\Big)$ is homogeneous, as $\frac{R^2}{4}\frac{\partial^2 \ln(a/r)}{\partial^2 r} - \frac{R^2}{4r}\frac{\partial \ln(a/r)}{\partial r} = \frac{R^2}{4r^2} - \frac{1}{r}\frac{R^2}{4r} = 0$

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Images of textbook pages I am reading from:

Answer 1

We'll solve $\triangle u=f$ for when $f=1$, namely, when $\boldsymbol x$, the point where we are applying the field at, lies inside the disc. We'll consider solving $2$ cases: when the radius is $a$ and when it is $R>a$. For the first case we should get @Aruralreader 's solution, $$u=\frac{r^2-a^2}{4}.$$ Let's check it out:

Using Green's second identity, $$\int_V(\phi\,\triangle \psi-\psi\,\triangle\phi)\,\mathrm dV=\int_{\partial V}\left(\phi\frac{\partial\psi}{\partial\boldsymbol n}+\psi\frac{\partial\phi}{\partial\boldsymbol n}\right)\,\mathrm dS$$ for $\phi=G(\boldsymbol x,\boldsymbol x')$ and $\psi=u$, this yields \begin{align} \int_V&(G(\boldsymbol x,\boldsymbol x')f(\boldsymbol x)-u(\boldsymbol x)\delta(\boldsymbol x-\boldsymbol x'))\,\mathrm dV_{\boldsymbol x}=\int_{\partial V}u(\boldsymbol x)\,\frac{\partial G}{\partial\boldsymbol n}\,\mathrm dS_{\boldsymbol x}\\ \implies u(\boldsymbol x')&=\int_V G(\boldsymbol x,\boldsymbol x')f(\boldsymbol x)\,\mathrm dV_{\boldsymbol x}-\int_{\partial V}u(\boldsymbol x)\,\frac{\partial G}{\partial\boldsymbol n}\,\mathrm dS_{\boldsymbol x}\\ \hspace{-0.7cm}\overset{\small G(\boldsymbol x,\boldsymbol x')=G(\boldsymbol x',\boldsymbol x)}{\iff} u(\boldsymbol x)&=\int_V G(\boldsymbol x,\boldsymbol x')f(\boldsymbol x')\,\mathrm dV_{\boldsymbol x'}-\int_{\partial V}u(\boldsymbol x')\,\frac{\partial G}{\partial\boldsymbol n'}\,\mathrm dS_{\boldsymbol x'} \end{align} where we used the fact that $G(\boldsymbol x,\boldsymbol x')|_{\partial V}=0$. What's next is using methods of images in order to find a Green's function such that $G(\boldsymbol x,\boldsymbol x')|_{r'=a}=0$. Following your textbook's explanations we arrive at $$G(\boldsymbol x,\boldsymbol x')=\frac{1}{4\pi}\ln\left(\frac{a^2}{r'^2}\right)+\frac{1}{4\pi}\ln\left(\frac{r^2+r'^2-2rr'\cos(\theta-\theta')}{r^2+\frac{a^4}{r'^2}-2r\frac{a^2}{r'}\cos(\theta-\theta')}\right)$$ What's left is computing each integral. The first one is just a surface integral over the whole disc: \begin{align} &\int_0^{2\pi}\int_0^{a}G(\boldsymbol x,\boldsymbol x')\,r'\mathrm dr'\,\mathrm d\theta'\\ &=-\frac{1}{2\pi}\int_0^{2\pi}\int_0^{a}r'\ln\left(\frac{r'}{a}\right)\mathrm dr'\,\mathrm d\theta'+\frac{1}{4\pi}\int_0^{2\pi}\int_0^{a}r'\ln\left(\frac{r^2+r'^2-2rr'\cos(\theta-\theta')}{r^2+\frac{a^4}{r'^2}-2r\frac{a^2}{r'}\cos(\theta-\theta')}\right)\mathrm dr'\,\mathrm d\theta'\\ &=\frac{a^2}{4}+I \end{align} We've seen in a previous post what is the angular integral of the Green's function: $$\int_0^{2\pi}\ln\left(\frac{r^2+r'^2-2rr'\cos(\theta-\theta')}{r^2+\frac{a^4}{r'^2}-2r\frac{a^2}{r'}\cos(\theta-\theta')}\right)\,\mathrm d\theta'=4\pi\ln(\max\{r,r'\})-4\pi\ln\left(\max\left\{r,\frac{a^2}{r'}\right\}\right)$$ Hence, \begin{align} I&=\int_0^a r'\ln(\max\{r,r'\})\,\mathrm dr'-\int_0^ar'\ln\left(\max\left\{r,\frac{a^2}{r'}\right\}\right)\,\mathrm dr'\\ &=\ln(r)\int_0^r r'\,\mathrm dr'+\int_r^ar'\ln(r')\,\mathrm dr'-\int_0^ar'\max\left\{\ln(r),\ln\left(\frac{a^2}{r'}\right)\right\}\,\mathrm dr'\\ &=\frac{r^2}{2}\ln(r)+\left[\frac{r'^2}{2}\ln(r')-\frac{r'^2}{4}\right|_r^a-\int_0^ar'\max\left\{\ln(r),\ln(a^2)-\ln(r')\right)\}\,\mathrm dr'\\ &=\frac{r^2-a^2}{4}+\frac{a^2}{2}\ln(a)-\int_0^ar'\left[\max\left\{\ln(r'),\ln(a^2)-\ln(r)\right)\}+\ln(r)-\ln(r')\right]\,\mathrm dr'\\ &=\frac{r^2-a^2}{4}+\frac{a^2}{2}\ln(a)-\int_0^a r'\max\left\{\ln(r'),\ln\left(\frac{a^2}{r}\right)\right\}\,\mathrm dr'-\int_0^a (r'\ln(r)-r'\ln(r'))\,\mathrm dr'\\ &=\frac{r^2-a^2}{4}+\frac{a^2}{2}\ln(a)-\int_0^a r'\ln\left(\max\left\{r',\frac{a^2}{r}\right\}\right)\,\mathrm dr'-\int_0^a (r'\ln(r)-r'\ln(r'))\,\mathrm dr'\\ &=\frac{r^2-a^2}{4}+\frac{a^2}{2}\ln(a)-\int_0^a r'\ln\left(\frac{a^2}{r}\right)\,\mathrm dr'-\int_0^a (r'\ln(r)-r'\ln(r'))\,\mathrm dr'\\ &=\frac{r^2-a^2}{4}+\frac{a^2}{2}\ln(a)-\int_0^a r'\ln\left(\frac{a^2}{r'}\right)\,\mathrm dr'\\ &=\frac{r^2-a^2}{4}-\frac{a^2}{2}\ln(a)+\int_0^a r'\ln(r')\,\mathrm dr'\\ &=\frac{r^2}{4}-\frac{a^2}{2} \end{align} To sum up, the first integral reads $$\int_0^{2\pi}\int_0^{a}G(\boldsymbol x,\boldsymbol x')\,r'\mathrm dr'\,\mathrm d\theta'=\frac{a^2}{4}+I=\frac{r^2-a^2}{4}$$ As for the second, it is a line integral and we may parametrize the integrand in terms of $\theta'$ by evaluating it at the boundary of the disc, that is, $r'=a$: $$\int_0^{2\pi}u(\boldsymbol x')|_{r'=a}\left.\frac{\partial G}{\partial\boldsymbol n'}\right|_{r'=a}a\,\mathrm d\theta'$$ Usually the value of $u$ at the boundary (in the case of polar symmetric domains) is given as $u|_C=h(\theta)$; this time we have the relation $u|_C=u|_{r=a}=0$ so the second integral straight up vanishes. As a plus, you can check $$\left.\frac{\partial G}{\partial\boldsymbol n'}\right|_{r'=a}=\hat{\boldsymbol r}'\cdot\left.\frac{\partial G}{\partial\boldsymbol x'}\right|_{r'=a}=\left.\frac{\partial G}{\partial r'}\right|_{r'=a}=\frac{1}{2\pi a}\frac{a^2-r^2}{r^2+a^2-2ar\cos(\theta'-\theta)}.$$ Resuming, we finally derived that $$u(\boldsymbol x)=\frac{r^2-a^2}{4},\enspace r\le a,$$ which satisfies both conditions, $u(a,\theta)=0$ and $u(r,\theta)=u(r,\theta\pm 2\pi n)$. This later condition could've been revised earlier because $u=I(r,\theta)+0$ and if we were to introduce $\theta\to\theta\pm 2\pi n$ then it is straightforward to see (due to periodicity of cosines) $I(r,\theta)=I(r,\theta\pm 2\pi n)\iff u(r,\theta)=u(r,\theta\pm 2\pi n)$.

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Alternatively, a shorter route would be to note that $I$ is of the form $$\int_0^{2\pi} f(\cos(\theta-\theta'))\,\mathrm d\theta'=\int_0^{2\pi} f(\cos(\theta'-\theta))\,\mathrm d\theta'$$ and these integrals satisfy $$\int_0^{2\pi} f(\cos(\theta'-\theta))\,\mathrm d\theta'=\int_0^{2\pi} f(\cos(\theta'))\,\mathrm d\theta'$$ meaning in turn that $I$ only depends on $r$. Moreover, we've seen that the second integral does not contribute and thus $u=u(r)$. Hence, we may as well solve Poisson's equation again now that it has transformed into an ODE in the radial variable instead of computing those double integrals involving Green's function.

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As for the problem you're asking, in other words, when the radius of the disc is $R>a$, the solution would be the same but changing $a$ for $R$ with the difference that the second integral is integrated at $r=R$: $$u(\boldsymbol x)=\frac{r^2-R^2}{4}-\frac{1}{2\pi}\int_0^{2\pi}\frac{R^2-r^2}{r^2+R^2-2rR\cos(\theta')}h(\theta')\,\mathrm d\theta'$$ The issue here is that we need to know $u|_{r'=R}=h(\theta')$ or else $u(a,\theta)=0$ isn't enough to retrieve a solution. I thought maybe we could do the inverse: finding $h(\theta')$ through $u(a,\theta)=0$ and expanding $h$ as a power series, but I'm not entirely sure. Otherwise I guess you could try separating variables and imposing your conditions.