APM346-2018S > Term Test 2
TT2--P3N
Victor Ivrii:
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:
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
Victor Ivrii:
Oh, no, there are no sinh or cosh , because only trivial solutions come out ...
Please correct ....
Andrew Hardy:
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.
Jingxuan Zhang:
Andrew,
Your square is misplaced at the place where it is pointed out.
Navigation
[0] Message Index
[#] Next page
Go to full version