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I am trying to solve Laplace equation on a sphere of radius $1$. $$\Delta u =0, B(0,1) \subset \mathbb{R}^3 $$ $$u|_{S(0,1)}=z^2$$ Of course, I deduced through separation of variables and some substitions the general form of the solition for interior problems: $$\tilde{u}=\sum_{l=0}^{\infty} \left(\dfrac{r}{a}\right)^l\sum_{m=-l}^{l} a_{lm} Y_l^m(\theta, \phi)$$ Now, imposing initial conditions we get: $$z^2=\cos^2\theta=\sum_{l=0}^{\infty}\sum_{m=-l}^{l}a_{lm}Y_l^m(\theta, \phi)$$ I know the more explicit form of spherical harmonics in terms of associated legendre functions and complex exponential. I tried to find the coefficients through identification (I've computed a few spherical harmonics) I also tried using fourier expansion, but in the integral I don't know how to get rid of legendre and also I obtained that all of the coefficients should be 0. Please any hints would be much appreciated.

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For $m = 0$, the spherical harmonics reduce to the Legendre polynomials in $\cos \theta$. In particular, $$ Y_0^0(\theta, \phi) = \sqrt{\frac 1 {4\pi}}, \ \ \ Y_2^0(\theta, \phi)=\sqrt{\frac 5 {16\pi}}(3\cos^2 \theta - 1),$$ and from here, you should be able to solve for $\cos^2 \theta$ in terms of $Y_0^0$ and $Y_2^0$.

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  • $\begingroup$ This is what I was looking for. It was just beneath my nose. Thanks a lot. $\endgroup$ – asd11 May 30 '18 at 14:02
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The easiest way to guess: here, you know that $\Delta z^2 = 2$, so all you need is a harmonic function $v$ that has $\Delta v=-2$ and vanishes on $r=1$. An obvious candidate is $a(1-r^2)$ for some $a$: taking the derivatives shows that $a=1/3$, so the answer is $$ \frac{1}{3}(1-r^2) + z^2 = \frac{1}{3} (1+2z^2-x^2-y^2). $$


Okay, what if you don't trust your guessing capability? We can simplify the problem immediately by noting that the boundary conditions are axisymmetric, which means that the solution is very likely to also be axisymmetric (indeed, I have a feeling this is the case for any linear differential equation, although I've probably forgotten something). This reduces the expansion to one in terms of Legendre polynomials. As you note, writing it in terms of the angle gives $\cos^2{\theta}$ as the boundary condition. Since the Legendre polynomials up to degree $n$ are an orthogonal basis of the polynomials of degree at most $n$, you only need to consider $n \leq 2$. Moreover, $\cos^2{\theta}$ is even, so you actually only need to look at $P_0(X)$ and $P_2(X)$, and find a linear combination that gives $X^2$.

The moral of this story is that in the case of certain simple boundary conditions (in this case, polynomials), you can completely avoid actually calculating using the orthogonality relations, for the same reason that you can calculate the Fourier series of, e.g. $\sin^3{\theta}$ without actually integrating anything.

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  • $\begingroup$ Thank you for your answer. I'll try educated guesses from now on! $\endgroup$ – asd11 May 30 '18 at 14:03
  • $\begingroup$ This is really nice answer, Richard! It's nice to bump into you on the internet, and I hope everything's well at Trinity/DAMTP. $\endgroup$ – Kenny Wong May 30 '18 at 14:06

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