IBV

• Mar 16th 2011, 05:47 PM
dwsmith
IBV
$\text{D.E.}: \ u_t=ku_{xx}, \ 00$
$\displaystyle\text{B.C.}=\begin{cases}u(0,t)=0\\u_ x(L,t)=0\end{cases}$
$\text{I.C.}: \ u(x,0)=f(x) \ \ 0

$u(x,t)=\varphi(x)T(t)$

$\displaystyle\varphi(x)T'(t)=k\varphi''(x)T(t)\Rig htarrow\frac{\varphi''}{\varphi}=\frac{T'}{Tk}=-\lambda$

$\varphi''+\lambda\varphi=0\Rightarrow \varphi=A\cos(x\sqrt{\lambda})+B\sin(x\sqrt{\lambd a})$

$\displaystyle T=Ce^{-\lambda kt}$

$\varphi_1(0)=1 \ \varphi_2(0)=0$
$\varphi_1'(0)=0 \ \varphi_2'(0)=1$

$\varphi_1: \ A=1 \ \varphi_2: \ A=0$
$\displaystyle\varphi_1': \ B=0 \ \varphi_2': \ B=\frac{1}{\sqrt{\lambda}}$

$\displaystyle\varphi=A_2\cos(x\sqrt{\lambda})+B_2\ frac{\sin(x\sqrt{\lambda})}{\sqrt{\lambda}}$

$\displaystyle\varphi(0): \ A_2=0$

$\displaystyle\varphi': \ B_2\cos(x\sqrt{\lambda})=0$

The eigenvalues must satisfy:

$\displaystyle\cos(x\sqrt{\lambda})=0$

$\displaystyle\lambda_n=\pi^2\left(\frac{1+2n}{2L}\ right)^2, \ n\in\mathbb{Z}$

The eigenfunction is:

$\displaystyle\varphi_n(x)=\frac{2L}{\pi(1+2n)}\sin \left(\frac{\pi x(1+2n)}{2L}\right), \ n\in\mathbb{Z}^+$

Is this much correct?

Thanks.
• Mar 19th 2011, 02:54 PM
dwsmith
$\displaystyle u(x,t)=\sum_{n=1}^{\infty}b_n\frac{2L}{\pi(1+2n)}\ exp\left(-\frac{\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\rig ht)\sin\left(\frac{\pi x(1+2n)}{2L}\right)$

I have shown $\displaystyle\int_0^L\varphi_n(x)\varphi_m(x) \ dx=0, \ m\neq n$.

Before I solve for $b_n$ is u(x,t) correct?

Thanks.
• Mar 19th 2011, 03:31 PM
HallsofIvy
Quote:

Originally Posted by dwsmith
$\text{D.E.}: \ u_t=ku_{xx}, \ 00$
$\displaystyle\text{B.C.}=\begin{cases}u(0,t)=0\\u_ x(L,t)=0\end{cases}$
$\text{I.C.}: \ u(x,0)=f(x) \ \ 0

$u(x,t)=\varphi(x)T(t)$

$\displaystyle\varphi(x)T'(t)=k\varphi''(x)T(t)\Rig htarrow\frac{\varphi''}{\varphi}=\frac{T'}{Tk}=-\lambda$

$\varphi''+\lambda\varphi=0\Rightarrow \varphi=A\cos(x\sqrt{\lambda})+B\sin(x\sqrt{\lambd a})$

$\displaystyle T=Ce^{-\lambda kt}$

$\varphi_1(0)=1 \ \varphi_2(0)=0$
$\varphi_1'(0)=0 \ \varphi_2'(0)=1$

Where did $\varphi_1$ and $\varphi_2$ suddenly come from?
Are they $e^{\lambda kt}(Acos(x\sqrt{\lambda}))$ and $e^{\lambda kt}(B sin(x\sqrt{\lambda}))$?

If so, you cannot, generally, do that. But here it does not hurt.

Quote:

$\varphi_1: \ A=1 \ \varphi_2: \ A=0$
$\displaystyle\varphi_1': \ B=0 \ \varphi_2': \ B=\frac{1}{\sqrt{\lambda}}$

$\displaystyle\varphi=A_2\cos(x\sqrt{\lambda})+B_2\ frac{\sin(x\sqrt{\lambda})}{\sqrt{\lambda}}$
All you really need is that $\phi(0, t)= e^{-\lambda kt}(Acos(0)+ Bsin(0))= Ae^{-k\lambda kt}= 0$. Since the exponential is never 0, A= 0.

Then $\phi(L, t)= e^{-\lambda kt}B cos(L\sqrt{\lambda})= 0$. Again, the exponential is not 0 so we must have either B= 0, which would mean the solution was identically equal to 0, and so not an eigenfunction, or cos(L\sqrt{\lambda})= 0.

Quote:

$\displaystyle\varphi(0): \ A_2=0$

$\displaystyle\varphi': \ B_2\cos(x\sqrt{\lambda})=0$

The eigenvalues must satisfy:

$\displaystyle\cos(x\sqrt{\lambda})=0$
You mean $cos(L\sqrt{\lambda})= 0$

Quote:

$\displaystyle\lambda_n=\pi^2\left(\frac{1+2n}{2L}\ right)^2, \ n\in\mathbb{Z}$
The eigenfunction is:

$\displaystyle\varphi_n(x)=\frac{2L}{\pi(1+2n)}\sin \left(\frac{\pi x(1+2n)}{2L}\right), \ n\in\mathbb{Z}^+$

Is this much correct?

Thanks.
Yes, that is correct and your second post is correct.
• Mar 19th 2011, 03:33 PM
dwsmith
Quote:

Originally Posted by HallsofIvy
Where did $\varphi_1$ and $\varphi_2$ suddenly come from?
Are they $e^{\lambda kt}(Acos(x\sqrt{\lambda}))$ and $e^{\lambda kt}(B sin(x\sqrt{\ambda}))$?

If so, you cannot, generally, do that. But here it does not hurt.

My book solves all PDEs in the same manner. I don't know why it is done that way but it works in all cases.

Here is what the book says about $\varphi$ 1 and 2.

The theorem in the theory of ODE which explicitly states the proposition above is: if the coefficients $c_0(x), \ c_1(x), \ c_2(x)$ are continuous on the interval [a,b] and $c_0(x)$ is different from zero throughout the interval, then for any fixed $x_0$ in the interval [a,b] and for any constants $\alpha, \ \beta$. The DE has one and only one solution $\varphi(x)=\varphi(x,\lambda)$ satisfying at $x_0$ the initial conditions $\varphi(x_0)=\alpha, \ \varphi'(x_0)=\beta$.
Moreover, $\varphi(x,\lambda)$ and $\varphi'(x,\lambda)$ are continuous and continuously differentiable functions of both x for $a\leq x\leq b$ and all lambda.
It is convenient to single out the two special solutions $\varphi_1=\varphi_1(x,\lambda), \ \varphi_2=\varphi_2(x,\lambda)$ of which satisfy the DE
$\varphi_1(0)=1, \ \varphi_2(0)=0$
$\varphi'_1 (0)=0, \ \varphi'_2(0)=1$.

We call 1 and 2 the basic solutions at x nought.
• Mar 19th 2011, 07:32 PM
dwsmith
$\displaystyle \int_0^Lf(x)\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx=\int_0^Lb_m\sin^2\left(\frac{\pi x(1+2m)}{L}\right) \ dx$

$\displaystyle\int_0^Lf(x)\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx=b_m\frac{L}{2}\Rightarrow b_m=\frac{2}{L}\int_0^Lf(x)\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx$
• Mar 20th 2011, 11:42 AM
dwsmith
For the case $f(x)=1$

$\displaystyle b_m=\frac{2}{L}\int_0^L\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx=\frac{2}{L}\left[\frac{-2L}{\pi(1+2m)}\cos\left(\frac{\pi x(1+2m)}{2L}\right)\right]_0^L$

$\displaystyle b_m=\frac{4}{\pi(1+2m)}, \ \ m=n$

$\displaystyle u(x,t)=\frac{8}{\pi^2}\sum_{n=0}^{\infty}\frac{L}{ (1+2n)^2}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi x(1+2n)}{2L}\right)$

If everything above is correct, can I do this?

$\displaystyle\sum_0^{\infty}\frac{1}{(1+2m)^2}=\fr ac{\pi^2}{8}$

$\displaystyle u(x,t)=\frac{8}{\pi^2}\sum_{n=0}^{\infty}\frac{1}{ (1+2n)^2}\cdot L\sum_{n=0}^{\infty}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi x(1+2n)}{2L}\right)$

$\displaystyle\Rightarrow u(x,t)=L\sum_{n=0}^{\infty}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi x(1+2n)}{2L}\right)$
• Mar 21st 2011, 03:11 PM
dwsmith
Ok, I pretty sure the above answer is wrong.

I now have:

$\displaystyle u(x,t)=\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{1}{1+ 2n}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi x(1+2n)}{2L}\right)$

How do I show this is true for all x with the IC:

$\displaystyle u(x,0)=\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{1}{1+ 2n}\sin\left(\frac{\pi x(1+2n)}{2L}\right)=1$