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I'm asked to solve

$$u_{xx} + u_{yy} = 1$$

on a circle of radius $a$, where $u = 0$ when $r = a$. This naturally leads to a transformation to polar coordinates, where the equation becomes

$$u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta} = 1$$

There's a similar example in the book where, since the source (1) and boundary condition are both independent of $\theta$, the author assumes $u$ will only depend on $r$. This reduces the equation to and ODE in $r$:

$$u_{rr} + \frac{1}{r}u_r = 1$$

I cannot find a solution to this ODE if the Earth depended on it. I've tried factoring out the partial operation into

$$\frac{1}{r} \frac{\partial}{\partial r}\left[r \cdot \frac{\partial u}{\partial r}\right] = 1$$

and solving via direct integration, but it doesn't seem to work out. Is there some assumption or trick that I'm missing here, or is the method flawed altogether? The following problem is almost identical, but in spherical coordinates, where the boundaries of two concentric spheres are zero, so I assume there's something I'm missing.

Joe Pigott
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  • A side remark is that this problem can be solved more or less by inspection, if you notice that $v(x,y) = u(x,y) - (x^2+y^2)/4$ satisfies $v_{xx} + v_{yy} = 0$ with $v = -a^2/4$ on the boundary circle, which (by uniqueness) means that $v$ must be constant. – Hans Lundmark Feb 05 '23 at 21:10

1 Answers1

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You are on the right track. You wrote

$$\frac1r \frac{\partial }{\partial r}\left(r\frac{\partial u}{\partial r}\right)=1\tag1$$

Note that $(1)$ is equivalent to

$$ \frac{\partial }{\partial r}\left(r\frac{\partial u}{\partial r}\right)=r\tag2$$

Integrating $(2)$ and dividing by $r$ reveals that

$$ \frac{\partial u}{\partial r}=\frac12r+A\frac1r\tag3$$

Finally integrating both sides of $(3)$ yields

$$u=\frac14r^2+A\log(r)+B\tag4$$

We enforce the condition that $u$ is bounded on $r\le 1$. This requires that $A=0$. Next, we enforce the condition that $u(a)=0$ to find that $B=-\frac14 a^2$. Therefore, the soulution is

$$u=\frac14 (r^2-a^2)$$

Mark Viola
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  • Oh man, that is so much simpler than I thought. This class gets so difficult that sometimes I forget the basics. Just so I fully understand, why can you assume that $A$ is zero? $u$ is bounded on $r \leq a$, not 1, no? – Joe Pigott Feb 05 '23 at 20:47
  • Yes, we want $u$ to be $C^2$, which implies bounded. – Mark Viola Feb 05 '23 at 21:25