temperature in a hemisphere

[logistic_guy]

The steady-state temperature in a hemisphere of radius \(\displaystyle r = c\) is determined from

\(\displaystyle \frac{\partial^2 u}{\partial r^2} + \frac{2}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} + \frac{\cot \theta}{r^2}\frac{\partial u}{\partial \theta} = 0\)

\(\displaystyle 0 < r < c, \ \ \ \ 0 < \theta < \frac{\pi}{2}\)

\(\displaystyle u\left(r,\frac{\pi}{2}\right) = 0, \ \ \ \ 0 < r < c\)

\(\displaystyle u(c,\theta) = f(\theta), \ \ \ \ 0 < \theta < \frac{\pi}{2}\)

Solve for \(\displaystyle u(r,\theta)\). [Hint: \(\displaystyle P_{n}(0) = 0\) only if \(\displaystyle n\) is odd.]

 

[khansaheb]

The steady-state temperature in a hemisphere of radius \(\displaystyle r = c\) is determined from

\(\displaystyle \frac{\partial^2 u}{\partial r^2} + \frac{2}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} + \frac{\cot \theta}{r^2}\frac{\partial u}{\partial \theta} = 0\)

\(\displaystyle 0 < r < c, \ \ \ \ 0 < \theta < \frac{\pi}{2}\)

\(\displaystyle u\left(r,\frac{\pi}{2}\right) = 0, \ \ \ \ 0 < r < c\)

\(\displaystyle u(c,\theta) = f(\theta), \ \ \ \ 0 < \theta < \frac{\pi}{2}\)

Solve for \(\displaystyle u(r,\theta)\). [Hint: \(\displaystyle P_{n}(0) = 0\) only if \(\displaystyle n\) is odd.]

show us your effort/s to solve this problem.

 

[logistic_guy]

show us your effort/s to solve this problem.

Let me try to solve this PDE by separation of variables.

\(\displaystyle u(r,\theta) = R(r) \ \Theta(\theta)\)

\(\displaystyle \Theta\frac{\partial^2 R}{\partial r^2} + \Theta\frac{2}{r}\frac{\partial R}{\partial r} + R\frac{1}{r^2}\frac{\partial^2 \Theta}{\partial \theta^2} + R\frac{\cot \theta}{r^2}\frac{\partial \Theta}{\partial \theta} = 0\)

\(\displaystyle R\frac{1}{r^2}\frac{\partial^2 \Theta}{\partial \theta^2} + R\frac{\cot \theta}{r^2}\frac{\partial \Theta}{\partial \theta} = -\Theta\frac{\partial^2 R}{\partial r^2} - \Theta\frac{2}{r}\frac{\partial R}{\partial r}\)

\(\displaystyle \frac{1}{\Theta}\frac{\partial^2 \Theta}{\partial \theta^2} + \frac{\cot \theta}{\Theta}\frac{\partial \Theta}{\partial \theta} = -r^2\frac{\partial^2 R}{R\partial r^2} - \frac{2r}{R}\frac{\partial R}{\partial r}\)

\(\displaystyle \frac{1}{\Theta}\frac{\partial^2 \Theta}{\partial \theta^2} + \frac{\cot \theta}{\Theta}\frac{\partial \Theta}{\partial \theta} = -r^2\frac{\partial^2 R}{R\partial r^2} - \frac{2r}{R}\frac{\partial R}{\partial r} = -\lambda\)

With this substitution, I will have two equations:

\(\displaystyle \frac{\partial^2 \Theta}{\partial \theta^2} + \cot \theta \frac{\partial \Theta}{\partial \theta} + \lambda \Theta = 0\)

\(\displaystyle r^2\frac{\partial^2 R}{\partial r^2} + 2r\frac{\partial R}{\partial r} - \lambda R = 0\)

I will continue in the next post.

 

[logistic_guy]

Let me work on the first differential equation.

\(\displaystyle \frac{\partial^2 \Theta}{\partial \theta^2} + \cot \theta \frac{\partial \Theta}{\partial \theta} + \lambda \Theta = 0\)

This is an ordinary differential equation, so we can change the partial operator to ordinary.

\(\displaystyle \frac{d^2 \Theta}{d \theta^2} + \cot \theta \frac{d \Theta}{d \theta} + \lambda \Theta = 0\)

At this point, we don't know how to solve this ordinary differential equationso we will try to play around to change its form. May be the new form will be known to solve. We know that \(\displaystyle \cot\theta = \frac{\cos \theta}{\sin \theta}\), so the equation will become:

\(\displaystyle \frac{d^2 \Theta}{d \theta^2} + \frac{\cos \theta}{\sin \theta} \frac{d \Theta}{d \theta} + \lambda \Theta = 0\)

Multiply each term by \(\displaystyle \sin \theta\).

\(\displaystyle \sin \theta\frac{d^2 \Theta}{d \theta^2} + \cos \theta \frac{d \Theta}{d \theta} + \sin \theta \lambda \Theta = 0\)

By the product rule, we can combine the first and the second derivatives together.

\(\displaystyle \frac{d}{d\theta}\left(\sin \theta \frac{d \Theta}{d \theta}\right) + \sin \theta \lambda \Theta = 0\)

Now let us focus on the inside of the brackets. Let \(\displaystyle x = \cos \theta\) and \(\displaystyle v(x) = \Theta(\theta)\).
From the chain rule we get:

\(\displaystyle \sin \theta \frac{d \Theta}{d \theta} = \sin \theta \frac{dv}{d\theta} = \sin \theta \frac{dv}{dx}\frac{dx}{d\theta} = -\sin^2 \theta \frac{dv}{dx}\)

\(\displaystyle = -(1 - \cos^2 \theta)\frac{dv}{dx} = -(1 - x^2)\frac{dv}{dx}\)

This means that:

\(\displaystyle \frac{d}{d\theta}\left(\sin \theta \frac{d \Theta}{d \theta}\right) = \frac{d}{dx}\left(\sin \theta \frac{d \Theta}{d \theta}\right)\frac{dx}{d\theta} = -\sin \theta \frac{d}{dx}\left(\sin \theta \frac{d \Theta}{d \theta}\right)\)

\(\displaystyle = -\sin \theta \frac{d}{dx}\left(-(1 - x^2)\frac{dv}{dx}\right)\)

Back to the ordinary differential equation with this new substitution.

\(\displaystyle -\sin \theta \frac{d}{dx}\left(-(1 - x^2)\frac{dv}{dx}\right) + \sin \theta \lambda v = 0\)

Divide each term by \(\displaystyle \sin \theta\).

\(\displaystyle -\frac{d}{dx}\left(-(1 - x^2)\frac{dv}{dx}\right) + \lambda v = 0\)

Take the derivative of the brackets and simplify.

\(\displaystyle (1 - x^2)\frac{d^2 v}{dx^2} - 2x\frac{dv}{dx} + \lambda v = 0\)

This is just the Legendre's equationand it has the solution:
\(\displaystyle v(x) = P_n(x)\) only when \(\displaystyle \lambda = n(n + 1), \ \ n = 0, 1, 2, \cdots \ \ \) in the interval \(\displaystyle -1 < x < 1\).

Then

\(\displaystyle \Theta(\theta) = v(x) = P_n(x) = P_n(\cos \theta)\)

And we have solved the first ordinary differential equation. I will continue in the next post.

 

[logistic_guy]

Let me try to solve the second differential equation.

\(\displaystyle r^2\frac{\partial^2 R}{\partial r^2} + 2r\frac{\partial R}{\partial r} - \lambda R = 0\)

This is also an ordinary differential equation, so we can change the partial operator to ordinary.

\(\displaystyle r^2\frac{d^2R}{dr^2} + 2r\frac{dR}{dr} - \lambda R = 0\)

I will let \(\displaystyle r = e^x \rightarrow x = \ln r\) and I will let \(\displaystyle y(x) = R(r)\).

From the chain rule, we get:

\(\displaystyle \frac{dR}{dr} = \frac{dy}{dr} = \frac{dy}{dx}\frac{dx}{dr} = \frac{dy}{dx}\frac{1}{r} = \frac{dy}{dx}\frac{1}{e^x}\)

\(\displaystyle \frac{d^2R}{dr^2} = \frac{d}{dr}\left(\frac{dR}{dr}\right) = \frac{d}{dr}\left(\frac{dy}{dx}\frac{1}{e^x}\right) = \frac{d}{dx}\left(\frac{dy}{dx}\frac{1}{e^x}\right)\frac{dx}{dr}\)

\(\displaystyle = \frac{d}{dx}\left(\frac{dy}{dx}\frac{1}{e^x}\right)\frac{1}{r} = \frac{d}{dx}\left(\frac{dy}{dx}\frac{1}{e^x}\right)\frac{1}{e^{x}}\)

\(\displaystyle = \left(\frac{d^2y}{dx^2}\frac{1}{e^x} - \frac{dy}{dx}\frac{1}{e^x}\right)\frac{1}{e^x} = \frac{d^2y}{dx^2}\frac{1}{e^{2x}} - \frac{dy}{dx}\frac{1}{e^{2x}}\)

Back to our ordinary differential equation with this new substitution.

\(\displaystyle r^2\left(\frac{d^2y}{dx^2}\frac{1}{e^{2x}} - \frac{dy}{dx}\frac{1}{e^{2x}}\right) + 2r\frac{dy}{dx}\frac{1}{e^x} - \lambda y = 0\)

\(\displaystyle \left(\frac{d^2y}{dx^2} - \frac{dy}{dx}\right) + 2\frac{dy}{dx} - \lambda y = 0\)

\(\displaystyle \frac{d^2y}{dx^2} + \frac{dy}{dx} - \lambda y = 0\)

This is one of the basic differential equations and it has the solution:

\(\displaystyle y(x) = c_1e^{k_1x} + c_2e^{k_2x}\)

where \(\displaystyle k_1 = \frac{-1 + \sqrt{1 + 4\lambda}}{2}\) and \(\displaystyle k_2 = \frac{-1 - \sqrt{1 + 4\lambda}}{2}\)

We already know that \(\displaystyle x = \ln r\) and \(\displaystyle y(x) = R(r)\), so the solution will become:

\(\displaystyle R(r) = c_1r^{k_1} + c_2r^{k_2}\)

We also know that \(\displaystyle \lambda = n(n+1)\), so \(\displaystyle 1 + 4\lambda = 1 + 4n(n + 1) = 4n^2 + 4n + 1 = (2n + 1)(2n + 1) = (2n + 1)^2\)

Therefore,

\(\displaystyle k_1 = \frac{-1 + \sqrt{(2n + 1)^2}}{2} = \frac{-1 + 2n + 1}{2} = n\)

And

\(\displaystyle k_2 = \frac{-1 - \sqrt{(2n + 1)^2}}{2} = \frac{-1 - 2n - 1}{2} = -n - 1\)

And the solution becomes:

\(\displaystyle R(r) = c_1r^{n} + c_2r^{-n - 1} = c_1r^{n} + c_2\frac{1}{r^{n + 1}}\)

We are only interested in bounded solutions. When \(\displaystyle r = 0\), the second term becomes \(\displaystyle \frac{1}{0} = \infty\). So we get rid of this unbounded solution and the final solution is:

\(\displaystyle R(r) = c_1r^{n}\)

Letting \(\displaystyle A = c_1\) and combining the two solutions that we found, we get:

\(\displaystyle u_n(r,\theta) = A_nR_n(r)\Theta_n(\theta) =A_nr^{n}P_n(\cos \theta)\)

I will continue in the next post.

 

[logistic_guy]

If we apply the superposition principle, we get:

\(\displaystyle u(r,\theta) = \sum_{n=0}^{\infty} A_nr^nP_n(\cos \theta)\)

Let us apply the first condition.

\(\displaystyle u\left(r,\frac{\pi}{2}\right) = \sum_{n=0}^{\infty} A_nr^nP_n\left(\cos \frac{\pi}{2}\right) = \sum_{n=0}^{\infty} A_nr^nP_n(0) = 0\)

We know that \(\displaystyle A_n \neq 0, r^n \neq 0\), then it must be \(\displaystyle P_n(0) = 0\)

According to the hint \(\displaystyle P_n(0) = 0\) only when \(\displaystyle n = 1, 3, 5, 7, \cdots\)

Or \(\displaystyle 2n + 1, \ \ n = 0, 1, 2, 3, \cdots\)

Then, the solution becomes:

\(\displaystyle u(r,\theta) = \sum_{n=0}^{\infty} A_{2n + 1}r^{2n + 1}P_{2n + 1}(\cos \theta)\)

Now, the second condition will allow us to find the constant \(\displaystyle A_{2n + 1}\). Let us apply it.

\(\displaystyle u(c, \theta) = f(\theta) = \sum_{n=0}^{\infty} A_{2n + 1}c^{2n + 1}P_{2n + 1}(\cos \theta)\)

I can apply the integral to both sides and apply the orthogonality property for Legendre Polynomials since the infinite series is a Fourier Legendre series. Then, I can obtain the coefficients inside the series.

\(\displaystyle \int_{0}^{\pi} f(\theta) P_{2n + 1}(\cos \theta) \sin \theta \ d\theta = \sum_{n=0}^{\infty} A_{2n + 1}c^{2n + 1}\int_{0}^{\pi} P_{2n + 1}(\cos \theta)P_{2n + 1}(\cos \theta) \sin \theta \ d\theta\)

\(\displaystyle \int_{0}^{\pi} f(\theta) P_{2n + 1}(\cos \theta) \sin \theta \ d\theta = \sum_{n=0}^{\infty} A_{2n + 1}c^{2n + 1}\int_{0}^{\pi} P^2_{2n + 1}(\cos \theta) \sin \theta \ d\theta\)

\(\displaystyle \int_{0}^{\pi} f(\theta) P_{2n + 1}(\cos \theta) \sin \theta \ d\theta = \sum_{n=0}^{\infty} A_{2n + 1}c^{2n + 1}\left(\frac{2}{2(2n + 1) + 1}\right)\)

This gives:

\(\displaystyle A_{2n + 1} = \frac{4n + 3}{2c^{2n + 1}}\int_{0}^{\pi} f(\theta) P_{2n + 1}(\cos \theta) \sin \theta \ d\theta\)

So, the final solution to the problem in the OP is:

\(\displaystyle u(r, \theta) = \sum_{n=0}^{\infty} A_{2n + 1}r^{2n + 1}P_{2n + 1}(\cos \theta)\)

where \(\displaystyle A_{2n + 1}\) as mentioned above.


Definition and properties of Fourier Legendre series are worth to remember.

Definition: The Fourier-Legendre series of a function \(\displaystyle f\) defined on the interval \(\displaystyle (-1, 1)\) is given by

\(\displaystyle f(x) = \sum_{n=0}^{\infty}c_nP_n(x)\)

where \(\displaystyle c_n = \frac{2n + 1}{2}\int_{-1}^{1}f(x)P_n(x) \ dx\)

Orthogonality property:

\(\displaystyle \parallel P_n(x)\parallel^2 = \int_{-1}^{1}P^2_n(x) \ dx = \frac{2}{2n + 1}\)