# Since the solution of the Legendre’s equations are the Legendre polynomials $P_n(\mu)$, write the general solution for $u(r, \theta)$.

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!

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

• 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. – Wyuw Sep 13 '18 at 23:17
• 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.. – воитель Sep 13 '18 at 23:27
• Hmm, but how does that make sense? After all, we cannot divide by $0$? – Wyuw Sep 13 '18 at 23:30
• 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. – воитель Sep 13 '18 at 23:32
• 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/… – воитель Sep 14 '18 at 0:15