Laplace equation - boundary value problem

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.

• 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.) – Ian Jan 23 '17 at 17:12
• 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 $\theta$ – Martin Jan 23 '17 at 17:24
• You are right about the name of the equation, but unfortunately it seems I cannot change name of the question now. I am sorry – Martin Jan 23 '17 at 17:34
• 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. – Chee Han Jan 23 '17 at 17:36
• 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) – Martin Jan 23 '17 at 17:56

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.
• actually $v$ is not zero on the boundary, because the boundary is given by the unknown function – Martin Jan 23 '17 at 17:38
• Ok, Ok, now I see you have $v_r=3v$ which doesn't die out – nikola Jan 23 '17 at 17:40