### Author Topic: HA10 Problem 1  (Read 1470 times)

#### Victor Ivrii

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##### HA10 Problem 1
« on: March 26, 2015, 03:08:21 PM »
Consider variational problem under constrain:
\begin{align}
&\Phi(u):=\frac{1}{2}\int\_0^2 u\_{xx}^2\,dx,\label{eq-10.1}\\
&\Psi(u)= \frac{1}{2}\int\_0^2 u\_{x}^2\,dx =1,\label{eq-10.2}\\
&u(0)=u'(0)=u(2)=u'(2)=0.\label{eq-10.3}
\end{align}

a. Write down Eulerâ€“Lagrange equation $\delta (\Phi-\lambda \Psi)=0$.
b. Under boundary conditions (\ref{eq-10.3}) solve it and find out eigenvalues $\lambda$ for each solution exists.

Hint: it will be 4th order ODE.

Hint:  Changing interval to $(-1,1)$ and observing that ODE and boundary conditions are symmetric (survive $x\mapsto -x$) one can consider separately even and odd eigenfunctions and corresponding eigenvalues.

#### Jessica Chen

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##### Re: HA10 Problem 1
« Reply #1 on: April 02, 2015, 08:43:07 PM »
\begin{align*}
&\Phi(u):=\frac{1}{2}\int_0^2 u_{xx}^2\,dx,\\
&\Psi(u)= \frac{1}{2}\int_0^2 u_{x}^2\,dx =1,\\
&u(0)=u'(0)=u(2)=u'(2)=0.
\end{align*}

a. Write down Eulerâ€“Lagrange equation $\delta (\Phi-\lambda \Psi)=0$.
Let $L (x, u, u', u'') = u_{xx}^2$
then
\begin{align}
\delta \Phi&=
\iiint_\Omega\Bigl(\frac{\partial L}{\partial u} - \sum_{1\le j\le n} \frac{\partial\ }{\partial x_j} \frac{\partial L}{\partial u_{x_j}} \Bigr)\delta u \,dx \\
&=\frac{\partial^2}{\partial x^2}2u_{xx}\\
&=2u_{xxxx}
\end{align}
similarly,
\begin{align}
L_2(x,u,u') = u_x^2\\
\delta\Psi &=\frac{\partial L}{\partial u}-\frac{\partial}{\partial x}\frac{\partial L}{\partial u_x}\\
&=-\frac{\partial}{\partial x}2u_x\\
&=-2u_{xx}
\end{align}
Thus

\delta(\Phi-\lambda\Psi) = u_{xxxx}+\lambda u_{xx}=0

b. Under boundary conditions (\ref{eq-10.3}) solve it and find out eigenvalues $\lambda$ for each solution exists.
Characteristic Equation:

k^4+\lambda k^2 = 0\\
k = 0,\sqrt{\lambda}i, -\sqrt{\lambda}i

Let $\lambda = \omega ^2$
Then $u = A + Bx+C\cos(\omega x)+D\sin(\omega x)$
By hint, we have $u(-1)=u'(-1)=u(1)=u'(-1) = 0$
Even eigenfunction is:
\begin{align}
X &= A+C\cos(\omega x)\\
X(1) &= A+C\cos(\omega)=0\\
X'(1) &= -C\sin(\omega) = 0\\
\sin(\omega) &=0 \implies \omega = n\pi\\
\lambda_n &= \omega ^2=(n\pi)^2\\
A+C\cos(\omega) = 0\\
A = 1\\
C = (-1)^{n+1}\\
\end{align}
$X_n = 1+ (-1)^{n+1}cos(nx)$ is the even eigenfunction

Odd function
\begin{align}
X &= Bx + D \sin (\omega x)\\
X(1) &=B+D\cos(\omega ) = 0\\
X'(1) &=B+D\cos(\omega)\omega = 0\\

\cos(\omega)\omega &= \sin(\omega)\\
\omega&=\tan(\omega)\\
\end{align}
Then the graph $\omega$ and $\tan(\omega)$ intersections are the eigenvalues
« Last Edit: April 03, 2015, 12:17:50 PM by Victor Ivrii »