Solve the Dirichlet problem for the Laplace equation in $\Bbb{R}^2$

Question

Solve the Dirichlet problem for the Laplace equation in $\Bbb{R}^2$: $$\begin{cases}\Delta u=0&\text{ in } 1<|x|<2\\u=x_1&\text{ on }|x|=1\\u=1+x_1x_2&\text{ on } |x|=2\end{cases}$$

The hint says to use Laurent series.

Attempt: We know both boundary polynomials are harmonic ($\Delta x_1=\Delta( 1+x_1 x_2)=0$), so we write a series $u=\sum p_m$ where $p_m$ are harmonic polynomials. This series should go to $x_1$ as $|x|=1$ and $1+x_1 x_2$ for $|x|=2$. So $u=x_1+1+x_1 x_2+\sum p_m$, but I don't know what the rest of the $p_m$ should be.

Any hints? I would prefer hints to full answers.

Answer 1

Since you asked for hints:

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Hint

Let $v = \ln(|x|)$, then $\partial_{x_1} v$ is a harmonic function (why?) and equals $x_1$ on $|x| = 1$ (why?)

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Do something similar for $1 + x_1x_2$.

Another hint:

The $1$ and the $x_1 x_2$ are not on the same footing in the expression $1 + x_1 x_2$. (In terms of Taylor series at the origin, the first is the 0th order term and the second is the second order term.) Their "Laurent series" partners are therefore also not on the same footing.

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Third and final hint:

This one may give away the answer, so I am hiding it. Don't mouse-over until you are really stuck.

What we are getting at is the multipole expansion in 2 dimensions. The function $x_1$ is "singular at infinity", as is the function $x_1 x_2$. In fact, the same is true of all harmonic polynomials. They have partners which are "singular at zero" (here's where the Laurent series comes in). We choose zero and infinity as the two singular points since they are "equidistant" from the circular boundaries of your annulus.

Answer 2

This one has the full answer. Hidden from view unless you mouse-over

Step 1:

We can find a harmonic function that is equal to $1$ on $|x| = 2$ and $0$ on $|x| = 1$. To do so we use the ansatz $$ \phi(x) = a \ln |x| + b $$ and we see that $a = \frac{1}{\ln 2}$ and $b = 0$ works.

Step 2:

We can find a harmonic function that is equal to $x_1$ on $|x| = 1$ and $0$ on $|x| = 2$. To do so we use the ansatz $$ \phi(x) = a \frac{x_1}{|x|^2} + b x_1 = x_1 (\frac{a}{|x|^2} + b) $$ and we see that $a + b = 1$ and $a/4 + b = 0$.

Step 3:

We can find a harmonic function that is equal to $x_1 x_2$ on $|x| = 2$ and $0$ on $|x| = 1$. We proceed with the ansatz $$ \phi(x) = a \partial^2_{x_1x_2} \ln(|x|) + b x_1 x_2 $$ and solve for $a$ and $b$.

Explanation:

In terms of Laurent series, what we are doing is using that $1/z^n$ are holomorphic functions outside the origin, and so their real and imaginary parts are harmonic. In particular, you have that $\frac{1}{z^n} = \frac{\bar{z}^n}{|z|^{2n}}$ the denominator now taking constant values over circles centered at zero. Note also that $z^n$ also are harmonic functions. So with two circular boundaries you can take linear combinations between $z^{-n}$ and $z^n$ to get "polynomial-like" boundary values.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ The general solution of $\ds{2D}$-Laplace Equation is given by: \begin{align} \Phi\pars{\vec{r}} & = \pars{A + B\theta}\bracks{C\ln\pars{r} + D} + \sum_{n = 1}^{\infty}\bracks{A_{n}\sin\pars{n\theta + \phi_{n}}r^{n} + B_{n}\sin\pars{n\theta + \delta_{n}}{1 \over r^{n}}} \end{align} As the boundary conditions are 'quite simple' $\ds{\pars{~\cos\pars{\theta}\ \mbox{at}\ r = 1\ \mbox{and}\ 1 + 2\sin\pars{2\theta}\ \mbox{at}\ r = 2~}}$ it's sufficient to write: $$ \Phi\pars{\vec{r}} = {\ln\pars{r} \over \ln\pars{2}} + \mrm{b}\pars{r}\cos\pars{\theta} + 2\mrm{c}\pars{r}\sin\pars{2\theta}\,,\qquad \left\{\begin{array}{rcl} \ds{\mrm{b}\pars{1}} & \ds{=} & \ds{1\,,} & \ds{\mrm{b}\pars{2}} & \ds{=} & \ds{0} \\[1mm] \ds{\mrm{c}\pars{1}} & \ds{=} & \ds{0\,,} & \ds{\mrm{c}\pars{2}} & \ds{=} & \ds{1} \end{array}\right. $$

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$$ \nabla^{2}\phi\pars{\vec{r}} = \partiald[2]{\phi\pars{\vec{r}}}{r} + {1 \over r}\,\partiald{\phi\pars{\vec{r}}}{r} + {1 \over r^{2}}\,\partiald[2]{\phi\pars{\vec{r}}}{\theta} $$ Note that $\ds{\left.\nabla^{2}\ln\pars{r}\right\vert_{\ r\ >\ 0} = 0}$.

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$$ \left.\begin{array}{rcl} \ds{\partiald[2]{\mrm{b}\pars{\vec{r}}}{r} + {1 \over r}\,\partiald{\mrm{b}\pars{\vec{r}}}{r} - {\mrm{b}\pars{r} \over r^{2}}} & \ds{=} & \ds{0} \\[2mm] \ds{\partiald[2]{\mrm{c}\pars{\vec{r}}}{r} + {1 \over r}\,\partiald{\mrm{c}\pars{\vec{r}}}{r} - {4\mrm{c}\pars{r} \over r^{2}}} & \ds{=} & \ds{0} \end{array}\right\}\implies \left\{\begin{array}{rcl} \ds{\mrm{b}\pars{r}} & \ds{=} & \ds{-\,{1 \over 3}\pars{r - {4 \over r}}} \\[2mm] \ds{\mrm{c}\pars{r}} & \ds{=} & \ds{{4 \over 15}\pars{r^{2} - {1 \over r^{2}}}} \end{array}\right. $$ In the OP notation, the solution is given by: $$\bbox[#ffe,15px,border:1px dotted navy]{\ds{% \Phi\pars{\vec{r}} = {\ln\pars{x_{1}^{2} + x_{2}^{2}} \over 2\ln\pars{2}} - {1 \over 3}\pars{x_{1} - {4x_{1} \over x_{1}^{2} + x_{2}^{2}}} + {16 \over 15}\bracks{x_{1}x_{2} - {x_{1}x_{2} \over \pars{x_{1}^{2} + x_{2}^{2}}^{2}}}}} $$