Author Topic: FE-4  (Read 3733 times)

Victor Ivrii

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FE-4
« on: December 18, 2015, 07:44:29 PM »
Consider the Laplace equation in the ring
\begin{align}
&&&u_{xx} +  u_{yy} =0\qquad &&\text{in  }\ 1< r = \sqrt{x^2+y^2} < 2,
\label{4-1} \\
&\text{with the boundary conditions}\notag\\
&&& u =\sin(\theta)\qquad &&\text{for }\  r=1,\label{4-2}\\
&&& u= 3 \sin(\theta)\qquad &&\text{for }\  r=2.\label{4-3}\end{align}

(a)  Look for solutions $u$ in the form of  $u(r,\theta)= R(r) P(\theta)$  (in polar coordinates) and derive a set of  ordinary differential equations for $R$ and $P$. Write the correct  boundary conditions for $P$.

(b) Solve the eigenvalue problem for $P$ and find all eigenvalues.

(c)  Solve the differential equation  for $R$.

(d)  Find the solution $u$ of (\ref{4-1})--(\ref{4-3}).
« Last Edit: December 18, 2015, 07:49:48 PM by Victor Ivrii »

Vivian Tan

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Re: FE-4
« Reply #1 on: December 18, 2015, 10:23:03 PM »
(a) We solve this question using the two dimensional Laplacian in polar coordinates. So the equation becomes:

\begin{equation}
u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta \theta} = 0
\end{equation}

So the equation is

\begin{equation}
\frac{R''}{R} + \frac{1}{r}\frac{R'}{R} + \frac{1}{r^2}\frac{P''}{P} = 0
\end{equation}

Multiplying through by $r^2$, we see that the equation is:

\begin{equation}
r^2\frac{R''}{R} + r\frac{R'}{R} + \frac{P''}{P} = 0
\end{equation}

The equation is now separable. We have a set of ordinary differential equations:

\begin{gather}
r^2\frac{R''}{R} + r\frac{R'}{R} = \lambda \\
\frac{P''}{P} = - \lambda
\end{gather}

The boundary condition for P is that it is $2\pi$-periodic: $P(\theta) = P(2\pi + \theta)$.

Vivian Tan

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Re: FE-4
« Reply #2 on: December 18, 2015, 10:32:56 PM »
(b) The eigenvalue problem for $P$ (as seen in part a) is:

\begin{equation}
P'' = - \lambda P
\end{equation}

First let $\lambda = \omega^2 > 0$. Then the equation is $P'' = - \omega^2 P$, and the solution is $P(\theta) = A\cos(\omega \theta) + B\sin(\omega \theta)$.

Let $\lambda = - \omega^2 < 0$. Then the equation is $P'' = \omega^2 P$, and the solution is $P(\theta) = C\cosh(\omega \theta) + D\sinh(\omega \theta)$.

Then let $\lambda = 0$. Then the equation is $P'' = 0$, and the solution is $P(\theta) = E \theta + F$.

So the eigenvalues are $n^2$, and the eigenfunctions are $P(\theta) = A\cos(n \theta) + B\sin(n \theta)$.
« Last Edit: December 18, 2015, 10:35:48 PM by Vivian Tan »

Vivian Tan

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Re: FE-4
« Reply #3 on: December 18, 2015, 10:39:13 PM »
(c) The differential equation that we must solve for $R$ is $r^2\frac{R''}{R} + r\frac{R'}{R} = \lambda$, or $r^2R'' + rR' = \lambda R$, so we assume a solution $r^m$, and then we get:

\begin{equation}
r^2m(m-1)r^{m-2} + rmr^{m-1} - \lambda r^m = 0 \longrightarrow m^2 - m + m - \lambda = 0 \longrightarrow m^2 = n^2
\end{equation}

So we have $m = \pm n$, which means that the solution is $R(r) = r^n + r^{-n}$.
« Last Edit: December 18, 2015, 10:45:33 PM by Vivian Tan »

Vivian Tan

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Re: FE-4
« Reply #4 on: December 18, 2015, 10:55:16 PM »
(d) We then have a general solution $u(r, \theta) = \sum_1^{\infty}$, and we have to make use of the boundary conditions. So we have:

\begin{equation}
u(r, \theta) =  \sum_1^{\infty} \left[ r^n \left(A_n\cos n \theta + B_n \sin n \theta \right) + r^{-n} \left( C_n \cos n \theta + D_n \sin n \theta \right) \right]
\end{equation}

As $r=1$:

\begin{equation}
u(1, \theta) =  \sum_1^{\infty} \left[ \left(A_n\cos n \theta + B_n \sin n \theta \right) + \left( C_n \cos n \theta + D_n \sin n \theta \right) \right] = \sin\theta
\end{equation}

We see that this is only matched when all terms that aren't $n=1$ are zero. Also, $A_1$ and $C_1$ are zero. So we're left with:

\begin{equation}
B_1 + D_1 = 1
\end{equation}

Likewise for $r=2$:

\begin{equation}
u(2, \theta) =  \sum_1^{\infty} \left[ 2^n \left(A_n\cos n \theta + B_n \sin n \theta \right) + 2^{-n} \left( C_n \cos n \theta + D_n \sin n \theta \right) \right] = 3\sin\theta
\end{equation}

By the same logic as before, notice that we must have all coefficients zero, except for $B_1$ and $D_1$, where we have that:

\begin{equation}
2B_1 + \frac{1}{2}D_1 = 3
\end{equation}

We can solve these two equations for $B_1$ and $D_1$ to get that $D_1 = -\frac{2}{3}$ and $B_1 = \frac{5}{3}$. Our answer is then:

\begin{equation}
u(x,t) = u(r, \theta) = r \frac{5}{3} \sin n \theta - \frac{1}{r} \frac{2}{3} \sin n \theta
\end{equation}

Vivian Tan

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Re: FE-4
« Reply #5 on: December 19, 2015, 12:06:34 AM »
\begin{equation}
u(x,t) = u(r, \theta) = r \frac{5}{3} \sin n \theta - \frac{1}{r} \frac{2}{3} \sin n \theta
\end{equation}

Sorry, there aren't supposed to be n's, those are supposed to be just $\sin \theta$.

So the solution should be:
\begin{equation}
u(x,t) = u(r, \theta) = r \frac{5}{3} \sin \theta - \frac{1}{r} \frac{2}{3} \sin\theta
\end{equation}

Victor Ivrii

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Re: FE-4
« Reply #6 on: December 20, 2015, 04:17:24 PM »
Vivian's solution is correct.