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I am working on the problem

Consider the steady-state of the heat equation in a ball of radius a centred at the origin. In spherical coordinates, the ball occupied the region $0 \le r \le a$, $0 \le \theta \le \pi$ and $0 \le \phi < 2\pi$. It has a given temperature $g(\theta)$ imposed along its boundary, which is the sphere of radius $a$. Since the boundary condition is independent of $\phi$, we can assume that the temperature at the point $(r, \theta, \phi)$ in the ball is given as $u(r, \theta)$, which is given by the solution of the following boundary value problem,

$$\dfrac{1}{r^2} \dfrac{\partial}{\partial{r}} \left( r^2 \dfrac{\partial{u}}{\partial{r}} \right) + \dfrac{1}{r^2 \sin(\theta)} \dfrac{\partial}{\partial{\theta}} \left( \sin(\theta) \dfrac{\partial{u}}{\partial{\theta}} \right) = 0,$$

subject to boundary conditions

$u(a, \theta) = g(\theta)$ for $0 ≤ \theta ≤ \pi$.

(i) Show that the separation of variables $u(r, \theta) = R(r)S(\theta)$ leads to the equations

$$\dfrac{1}{\sin(\theta)} \dfrac{d}{d \theta} \left( \sin(\theta) \dfrac{dS}{d \theta} \right) + \lambda S = 0$$

and

$$(r^2 R')' - \lambda R = 0$$

(ii) Now let $\lambda = n(n + 1)$ for $n = 0,1,2,3, \dots$ and let $\mu = \cos(\theta)$, transform the ODE for $S(\theta)$ to the following Legendre’s equation:

$$(1 - \mu^2) \dfrac{\partial^2{S}}{\partial{\mu}^2} - 2\mu \dfrac{dS}{d \mu} + n(n + 1)S = 0$$

(iii) Solve the differential equation for $R$ for each eigenvalue $\lambda n = n(n + 1)$. (Hint: Try $R = Ar^m$.)

(iv) Given the solution of the Legendre’s equations are the Legendre polynomials $P_n(\mu) = P_n(\cos(θ))$, write the general solution for $u(r, \theta)$ as an infinite series.

I'm stuck on (iv) and just don't understand how to do this. I don't have very much experience with Legendre polynomials, so this is probably why. My textbook also doesn't have any solutions, so I am totally stuck. I would be very thankful if someone could please take the time to explain what (iv) is asking and show how (iv) is done. Thank you very much for your help!

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You have slightly different notation than my book. I am just going to write here. The steady state of the heat equation is Laplace's equation.

We should get this equation.

$$ \nabla^{2}u = 0 \tag{1} $$

with boundary conditions

$$ u(a,\theta, \phi) = F(\theta,\phi) \tag{2} $$

which corresponds to what you were saying. You should get an infinite series like this

$$ u(r,\theta, \phi) =\sum_{m=0}^{\infty} \sum_{n=m}^{\infty} \rho^{n}\big[ A_{mn} \cos(m\theta) + B_{mn} \sin(m\theta) \big] P_{n}^{m}(\cos(\phi)) \tag{3}$$

the non-homogeneous boundary condition gives

$$ F(\theta,\phi) =\sum_{m=0}^{\infty} \sum_{n=m}^{\infty} a^{n}\big[ A_{mn} \cos(m\theta) + B_{mn} \sin(m\theta) \big] P_{n}^{m}(\cos(\phi)) \tag{4}$$

in order to find the coefficients, we use orthogonality.

$$ a^{n}B_{mn} = \frac{\iint F(\theta,\phi)\sin(m\theta) P_{n}^{m}(\cos\phi) \sin(\phi)d \phi d\theta }{\iint \sin^{2}(m\theta)[P_{n}^{m}(\cos(\phi))]^{2} \sin(\phi) d\phi d\theta } \tag{5} $$

by the same method we find $A_{mn}$

$$ a^{n}A_{mn} = \frac{\iint F(\theta,\phi)\cos(m\theta) P_{n}^{m}(\cos\phi) \cos(\phi)d \phi d\theta }{\iint \cos^{2}(m\theta)[P_{n}^{m}(\cos(\phi))]^{2} \cos(\phi) d\phi d\theta } \tag{6} $$

in your case, you would simply have

$$ u(a,\theta ) = g(\theta) \tag{7} $$

Note that $g(\theta)$ isn't a function of $\phi$ so $$ g(\theta) =\sum_{m=0}^{\infty} \sum_{n=m}^{\infty} a^{n}\big[ A_{mn} \cos(m\theta) + B_{mn} \sin(m\theta) \big] \tag{8}$$

So when we solve for $A_{mn}, B_{mn}$

$$ a^{n}B_{mn} = \frac{\int_{0}^{\pi} g(\theta)\sin(m\theta) d\theta }{\int_{0}^{\pi} \sin^{2}(m\theta) d\theta } \tag{9} $$

$$ a^{n}A_{mn} = \frac{\int_{0}^{\pi} g(\theta)\cos(m\theta) d\theta }{\int_{0}^{\pi} \cos^{2}(m\theta) d\theta } \tag{10} $$

There should have been boundary conditions on the integrals to figure out the normalization. I.e

$$ B_{00} = \frac{\int_{0}^{\pi} g(\theta) \cdot 0 d\theta }{\int_{0}^{\pi} 0 d\theta } \tag{11} $$

Which means $ B_{00} $ can be anything. However, this shouldn't matter for $A_{00}$ Test it. Ok. Now you should be actually able to get a real normalization part since you have definite integral boundaries.

$$ A_{00} = \frac{\int_{0}^{\pi} g(\theta) \cdot 1 d\theta }{\int_{0}^{\pi} \cdot 1 d\theta } \tag{12} $$

I am attempting to say this coefficient on the bottom is a function of $m$ and at $m=0$ it equals $\pi$

$$ A_{00} = \frac{\int_{0}^{\pi} g(\theta) \cdot 1d\theta } {\pi } \tag{13} $$

It is "normalizing" the top integral

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  • $\begingroup$ Thanks for the help again! :) Just to be sure, since we have $$\dfrac{1}{r^2} \dfrac{\partial}{\partial{r}} \left( r^2 \dfrac{\partial{u}}{\partial{r}} \right) + \dfrac{1}{r^2 \sin(\theta)} \dfrac{\partial}{\partial{\theta}} \left( \sin(\theta) \dfrac{\partial{u}}{\partial{\theta}} \right) = 0,$$ does this mean that we have $\theta \not= n \pi$, $n \in \mathbb{Z}$, despite the boundary condition $0 ≤ \theta ≤ \pi$? Just want to check that I am understanding this properly. $\endgroup$ – Wyuw Sep 13 '18 at 23:17
  • $\begingroup$ umm...well $\theta$ never takes on values however part of the ansatz is the infinite series so $0 \leq \theta \leq \pi $ so technically it is values of all $ m \pi , n \pi $ etc it is only taking them inside of the trig functions because it is inside this ball right.. $\endgroup$ – воитель Sep 13 '18 at 23:27
  • $\begingroup$ Hmm, but how does that make sense? After all, we cannot divide by $0$? $\endgroup$ – Wyuw Sep 13 '18 at 23:30
  • $\begingroup$ Oh, I see what you're saying.. where they're defining the coefficients. I forgot about that. I did this really quickly. Let me look in my notes. $\endgroup$ – воитель Sep 13 '18 at 23:32
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    $\begingroup$ I see what you're saying now.. It actually notes that end points are singular. eaton.math.rpi.edu/faculty/Wahle/Courses/2400-F05/notes/… $\endgroup$ – воитель Sep 14 '18 at 0:15

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