Author Topic: TT2--P3N  (Read 1537 times)

Victor Ivrii

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TT2--P3N
« on: March 23, 2018, 06:15:52 AM »
Using Fourier method find eigenvalues and eigenfunctions of Laplacian in the rectangle $\{0<x<a,\, 0<y<b\}$ with the boundary conditions:
\begin{align}
&u_{xx}+u_{yy}=-\lambda u\qquad 0<x<a,\ 0<y<b,\label{3-1}\\[3pt]
&u|_{x=0}=u_x|_{x=a}=u|_{y=0}=u_y|_{y=b}=0.\label{3-2}
\end{align}

Tristan Fraser

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Re: TT2--P3N
« Reply #1 on: March 23, 2018, 08:09:03 PM »
We start by taking the $ u = X(x)Y(y) $, then plugging in gives us:

$$ X'' Y + Y'' X  = -\lambda XY $$

$$ X(0)Y(y) = X'(a)Y(y) = 0  \ \ and \ \ X(x)Y(0) = X(x)Y(b)' = 0 $$

Dividing both of these expressions by $XY$ gives us

$$\frac{X''}{X}  + \frac{Y''}{Y} = -\lambda $$ 

$$ X(0) = X'(a) = 0  \ \ and \ \ Y(0) = Y(b)' = 0 $$

Now we know that both $\frac{Y''}{Y} $ and $\frac{X''}{X} $ are independent of each other, i.e. they should be equivalent to some constant. Introduce constants $\lambda_1, \lambda_2$ such that $ \lambda = \lambda_1 + \lambda_2$, thus  $\frac{Y''}{Y} = -\lambda_{1} $ and $\frac{X''}{X} = -\lambda_{2} $

Then, we can examine the different cases of $\lambda_{1,2}$.

i) If both $\lambda_{1} = 0 = \lambda_{2} $:

We get a simplified eigenvalue problem of:
 $$ X'' = 0 , Y'' = 0 $$

Meaning that:

$$ X = A_0 x + B_0 , Y = C_0y + D_0  $$

Running it through the boundary conditions, we can easily show that: $ B_0 = 0 , D_0 = 0 , A_0 = 0 , C_0 = 0 $
I.e. this leads to a trivial solution of the eigenvalue problem.

For $\lambda_{1}, \lambda_{2} >0 $

We will get eigenvalue problem of $$X'' + \lambda_2 X = 0 , Y'' + \lambda_1  Y = 0$$

This results in:

$$X(x) = Acos\sqrt{\lambda_2}x + Bsin\sqrt{\lambda_2}x$$
$$Y(y) = Ccos\sqrt{\lambda_1}y + Dsin\sqrt{\lambda_1}y$$

Apply the boundary conditions, and we get:
$ A = 0 , C= 0, 0 =  \sqrt{\lambda_2}Bcos\sqrt{\lambda_2}a, 0 = \sqrt{\lambda_1}Dcos\sqrt{\lambda_1}b$

We're in search of nontrivial solutions, which can be attained if $\sqrt{\lambda_{1,2}}b,a =  \frac{\pi(2n+1)}{2} $, thus we have eigenvalues and eigenfunctions of:

$$ \lambda_1 =  (\frac{\pi(2m+1)}{2b})^2 , Y_{m} = sin(\frac{\pi(2m+1)}{2b})y , \lambda_2 = (\frac{\pi(2n+1)}{2a})^2, X_{n} =  sin(\frac{\pi(2n+1)}{2a}x) $$

For the case of $\lambda_1 , \lambda_2 < 0 $ we solve the eigenvalue problem of:

$$X''  - \lambda_2 X = 0 , Y'' - \lambda_1 Y = 0 $$, which gives us, in turn:

$$X(x) =  Ae^{\sqrt{\lambda_2} x} + Be^{-\sqrt{\lambda_2} x} , Y(y) = Ce^{\sqrt{\lambda_1} y} + De^{-\sqrt{\lambda_1} y} $$

Apply the boundary conditions to get: $ A+ B = 0, C+D = 0$ , $ 0 =  \sqrt{\lambda_2} A (e^{\sqrt{\lambda_2}a} +e^{-\sqrt{\lambda_2}a}) = 2A\sqrt{\lambda_2}\cosh(\sqrt{\lambda_2}a) $

and $ 0 = \sqrt{\lambda_1}C(e^{\sqrt{\lambda_1}b} +e^{-\sqrt{\lambda_1}b})  = 2C\sqrt{\lambda_1}\cosh(\sqrt{\lambda_1}b) $

But since the $\cosh$ function never reaches 0, we can't have a nontrivial solution. Therefore there only exists a trivial solution in this case.

Note: updated solution to reflect feedback




« Last Edit: March 25, 2018, 07:13:02 PM by Tristan Fraser »

Victor Ivrii

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Re: TT2--P3N
« Reply #2 on: March 25, 2018, 04:40:50 AM »
Oh, no, there are no sinh or cosh , because only trivial solutions come out ...

Please correct ....

Andrew Hardy

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Re: TT2--P3N
« Reply #3 on: April 06, 2018, 03:07:29 PM »
Above is incomplete.
we begin with separation of variables $ U = X(x)Y(y) $
The equation simplifies to
$$ \frac{X"}{X} +  \frac{Y"}{Y}  = -\lambda $$

We know that these fractions must remain constant and so we have corresponding $  -\lambda_x + -\lambda_y = -\lambda $
We now have  Not enough letters? V.I. :D I'm sorry I don't follow
$$ X" + \lambda_xX = 0 $$
$$ Y" + \lambda_yY = 0 $$
Furthermore boundary conditions state that  $$ X(0) = X'(a) $$ and $$ Y(0) = Y'(b) $$ These are mixed boundary conditions (Dirichlet and Nuemann). They dictate the eigenvalues and corresponding eigenfunctions for the ODEs.
\begin{align*}
&\lambda_x =( \frac{\pi(2n+1)}{2a})^2 , &&X_n = \sin(\frac{\pi(2n+1)}{2a}) && n = 0,1,2,...\\
&\lambda_y =( \frac{\pi(2m+1)}{2b})^2  &&X_n = \sin(\frac{\pi(2m+1)}{2b}) &&m = 0,1,2,...
\end{align*}
We can and must then conclude that
$$
\lambda  =\pi^2( (\frac{(2n+1)}{2a})^2+ (\frac{(2m+1)}{2b})^2)$$
ERROR above. V.I. corrected
and our eigenfunctions are of the formula
$$ U_{n,m}(x,y) =  \sin(\frac{\pi(2n+1)}{2a})\sin(\frac{\pi(2m+1)}{2b}) $$

This is the complete answer. No need to go on a tangent.
« Last Edit: April 07, 2018, 10:12:47 AM by Andrew Hardy »

Jingxuan Zhang

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Re: TT2--P3N
« Reply #4 on: April 06, 2018, 04:54:41 PM »
Andrew,

Your square is misplaced at the place where it is pointed out.

Andrew Hardy

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Re: TT2--P3N
« Reply #5 on: April 07, 2018, 10:17:33 AM »
Just sloppy Latex'ing as I copied them. Thanks

Victor Ivrii

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Re: TT2--P3N
« Reply #6 on: April 07, 2018, 10:35:02 AM »
Just sloppy Latex'ing as I copied them. Thanks
It looks like you do not look how the result looks like––and one must do it, no matter how proficient he (or she) is