### Author Topic: FE-3  (Read 6738 times)

#### Victor Ivrii

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##### FE-3
« on: December 18, 2015, 07:42:21 PM »
\begin{align}
&u_{tt}-  9u_{xx}=0,\qquad 0<x<2, \; t>0,\label{3-1}\\[2pt]
& u (0,t)= u (2,t)=0,\label{3-2}\\[2pt]
& u(x,0)=f(x),\label{3-3}\\[2pt]
& u_t(x,0)=g(x)\label{3-4}\end{align}
with $f(x)=0$, $g(x)=x(2-x)$.  Write the answer in terms of  Fourier series.

#### Catch Cheng

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##### Re: FE-3
« Reply #1 on: December 18, 2015, 10:54:40 PM »
Please correct me if something is wrong, thank you.

#### Vivian Tan

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##### Re: FE-3
« Reply #2 on: December 18, 2015, 11:24:32 PM »
We separate variables. So $u(x,t) = T(t)X(x)$. Our equation is then $\frac{T''}{9T} = \frac{X''}{X} =- \lambda$.

So we solve the $X$ equation. We have $X'' = - \lambda X$. First consider $\lambda = \omega^2 > 0$. Then the solution is $X(x) = A\cos\omega x + B \sin \omega X$. Using the boundary conditions, we see that $A=0$, and $\omega 2 = n \pi \longrightarrow \omega = \frac{n \pi}{2}$.

Now consider negative eigenvalues. The solution will be $X(x) = A\cosh\omega x + B \sinh \omega X$. The boundary conditions clearly show that there is only a trivial solution, so we have no negative eigenvalues.

Likewise with the zero eigenvalue: the solution will be $X(x) = Ax + B$, and we can see that there is only a trivial solution. So no zero eigenvalues.

So then the eigenvalues are $\frac{n^2 \pi^2}{4}$ with eigenfunctions $X(x) = B \sin ( \frac{n \pi x}{2})$.

We can then solve the $T$ equation.

\frac{T''}{9T} = - \frac{n^2 \pi^2}{4} \longrightarrow T'' = - \frac{9 n^2 \pi^2 T}{4}

So the solution is

T(t) = C\cos(3n\pi t/2) + D\ sin(3n\pi t/2)

So overall the solution is:

u(x,t) = \sum_1^{\infty} \sin( \frac{n \pi x}{2}) \left( A_n\cos(3n\pi t/2) + B_n\sin(3n\pi t/2) \right)

When we use the initial condition at $u(x,0)$, we see right away that $A_n = 0$. So:

u(x,t) = \sum_1^{\infty} \sin( \frac{n \pi x}{2}) B_n\sin(3n\pi t/2)

u_t(x,t) = \sum_1^{\infty} \sin( \frac{n \pi x}{2}) B_n \frac{3 n \pi}{2} \cos(3n\pi t/2)

Then using the other condition:

u_t(x,0) = \sum_1^{\infty} \sin( \frac{n \pi x}{2}) \frac{3 n \pi}{2} B_n = x(2-x)

We then solve for the $B_n$'s:

B_n = \frac{2}{3 n \pi} \int_0^2 (2x-x^2) \sin( \frac{n \pi x}{2}) dx = -( \frac{2}{3 n \pi} ) \frac{8(\pi \sin(n \pi) + 2\cos(n \pi) - 2)}{\pi^3 n^3} =( \frac{2}{3 n \pi} ) \frac{16 ((-1)^n - 1)}{\pi^3 n^3}

If n is even we see that this is zero, and if n is odd we see that this is $\frac{-64}{3 \pi^4 n^4}$

So we see that

u(x,t) = -\frac{64}{3\pi^4}\sum_{n odd}^{\infty}\frac{1}{n^4}\sin(\frac{n \pi x}{2})\sin(3n\pi t/2)
« Last Edit: December 18, 2015, 11:34:41 PM by Vivian Tan »

#### Emily Deibert

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##### Re: FE-3
« Reply #3 on: December 18, 2015, 11:36:20 PM »
What do you mean by nodd?

#### Vivian Tan

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##### Re: FE-3
« Reply #4 on: December 18, 2015, 11:42:17 PM »
Sorry, I meant n, odd! As in all the odd terms, my bad!

#### Emily Deibert

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##### Re: FE-3
« Reply #5 on: December 18, 2015, 11:43:13 PM »
Thanks Vivian Tan! Great answer, as far as I recall I got the same!

#### Bruce Wu

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##### Re: FE-3
« Reply #6 on: December 18, 2015, 11:51:00 PM »
Great work everyone!

#### Victor Ivrii

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##### Re: FE-3
« Reply #7 on: December 20, 2015, 10:16:51 AM »

B_n = \frac{2}{3 n \pi} \int_0^2 (2x-x^2) \sin( \frac{n \pi x}{2}) dx = -( \frac{2}{3 n \pi} ) \frac{8(\pi \sin(n \pi) + 2\cos(n \pi) - 2)}{\pi^3 n^3} ={\color{red}{-}}( \frac{2}{3 n \pi} ) \frac{16 ((-1)^n - 1)}{\pi^3 n^3}\tag{11}

which leads to the error in the sign. Correct answer is
\begin{equation*}
u =\sum_{m=0}^\infty \frac{64}{3\pi ^4(2m+1)^4} \sin \bigl(\frac{3\pi (2m+1) t}{2}\bigr) \sin \bigl(\frac{\pi (2m+1) x}{2}\bigr).
\end{equation*}
which coincides with Vivian's except has an opposite sign.

PS To get $$\sum_{n\text{ odd}}$$ type \sum_{n\text{ odd}}  or \sum_{n\ \text{odd}}