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I am trying to solve this boundary value problem in three dimensions:

$$L(u)= 2rcos(\theta)-1$$ inside of unit ball ($r<1$) $$\frac{\partial u}{\partial r}= 3u$$ on the boundary ($r=1$)

where L(u) is Laplace operator.

I should solve it using Legendre polynomials but I do not know how to aproach it. The second question is if there is only one bounded solution to this problem.

Thank you for your help.

Edit:

I use this spherical coordinates: $x=rsin(\theta)cos(\phi), y=rsin(\theta)sin(\phi), z=rcos(\theta)$.

So the problem is formulated in these coordinates. I want to solve this using Legendre polynomials to somehow prove it has only one unique bounded solution or it has more than one solution that is bounded.

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  • $\begingroup$ What is $\theta$, the polar $\theta$ or the spherical $\theta$? From other contextual clues I guess $r$ is the spherical $r$, so probably $\theta$ is the spherical $\theta$. Still warrants clarification. (Also, this is a Poisson equation, not a Laplace equation.) $\endgroup$ – Ian Jan 23 '17 at 17:12
  • $\begingroup$ I looked it up on the internet and it says that polar is for 2D and spherical is for 3D so in that case it is spherical $\theta$ $\endgroup$ – Martin Jan 23 '17 at 17:24
  • $\begingroup$ You are right about the name of the equation, but unfortunately it seems I cannot change name of the question now. I am sorry $\endgroup$ – Martin Jan 23 '17 at 17:34
  • $\begingroup$ You should do separation of variables, which will ultimately lead you to Legendre polynomials as solutions. I guess this is just an exercise of applying the boundary condition to find the constants. $\endgroup$ – Chee Han Jan 23 '17 at 17:36
  • $\begingroup$ Well thanks to the right side of the equation, if I separate variables it will not lead to something useful (I do not know what to do with it) $\endgroup$ – Martin Jan 23 '17 at 17:56
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you first need to clear your mind, first you have the equation of Laplacian equals something using polar coordinates I guess. So, the problem is badly formulated, it should be $\Delta u = 2z-1$. A problem like this we solve by trying to guess "particular solution" (one which kills 2z-1 on the other side) and you should try with something as $u(x,y,z) = A(x^2+y^2+z^2)z+ B(x^2+y^2+z^2)$ with appropriate $A$ and $B$, so you have $u_{xx} = 2Az+2B,\ u_{yy} = 2Az+2B,\ u_{zz} = 6Az + 2B$, so $\delta u = 10Az + 6B = 2z-1$ leading to $A = \frac{1}{5}$, $B=\frac{-1}{6}$.

Now,consider a shift $u(x,y) = v(x,y) + \frac{1}{5}z(x^2+y^2+z^2)-\frac{1}{6}(x^2+y^2+z^2)$, now at a boundary you have $x^2+y^2+z^2=1$, so you have $u(x,y,z) = \frac{1}{5}z - \frac{1}{6}+v(x,y,z)$, so $u(\phi, \theta, r) = \frac{1}{5}r\cos(\theta) - \frac{1}{6}+v(\phi, \theta, r)$, so $u_r = \frac{1}{5}\cos(\theta)+v_r(\phi,\theta, r)$ so $\frac{1}{5}\cos(\theta)+v_r(\phi,\theta, r)=\frac{3}{5} \cos(\theta) + 3v(\phi,\theta, r)$ from which we see $v_r=3v+\frac{2}{5}\cos(\theta)$ on the boundary.

So now you have $\delta v = 0$ and $v_r=3v+\frac{2}{5}\cos(\theta)$. What you now need is a Laplacian in polar coordinates in three dimensions and then you look for solution in the shape $v(x,y,z) = R(x)T(y)C(z)$ and discuss what those three functions are, this would be a five-pages novel, I think you can solve it no problems.

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  • $\begingroup$ actually $v$ is not zero on the boundary, because the boundary is given by the unknown function $\endgroup$ – Martin Jan 23 '17 at 17:38
  • $\begingroup$ Ok, Ok, now I see you have $v_r=3v$ which doesn't die out $\endgroup$ – nikola Jan 23 '17 at 17:40
  • $\begingroup$ The other thing is, that I know how to solve this by guessing, the real problem is to solve this using Legendre polynomials and somehow (using Legendre polynomials) prove that it has unique solution (or it has more than one solution). I will make some edits to the question $\endgroup$ – Martin Jan 23 '17 at 17:44

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