Question

find the eigenvalues and eigenfunctions of the regular Sturm Liouville System:
d^(2)y/dx^2 + (lambda)y=0,
y(0)+y'(0)=0,
y(1)+y'(1)=0

Answer 1

There's generally a lot of work that goes into solving problems like this step-by-step, so I'll split up how to work this out into three separate comments (for the three special cases to consider for \(\lambda\)).

This ODE has characteristic equation
\[\chi(r)=r^2+\lambda=0\]which will yield different solution for \(y(x)\) depending on whether \(\lambda\) is positive, zero, or negative. We also require that \(y\) satisfy the boundary values of
\[\begin{cases}y(0)+y'(0)=0&&(1)\\[1ex]y(1)+y'(1)=0&&(2)\end{cases}\]

Answer 2

Suppose \(\lambda>0\). Then the \(\chi(r)\) has roots at \(r=\pm i\sqrt\lambda\), and so the solution is
\[y=C_1\cos\sqrt\lambda x+C_2\sin\sqrt\lambda x\]which satisfies
\[\begin{cases}
C_1+\sqrt\lambda C_2=0&&(1)_{\lambda>0}\\[1ex]
C_1\cos\sqrt\lambda+C_2\sin\sqrt\lambda-\sqrt\lambda C_1\sin\sqrt\lambda+\sqrt\lambda C_2\cos\sqrt\lambda=0&&(2)_{\lambda>0}
\end{cases}\]Rewrite \((2)_{\lambda>0}\) to get
\[\begin{cases}
C_1+\sqrt\lambda C_2=0&&(1)_{\lambda>0}\\[1ex]
(\sqrt\lambda C_2+C_1)\cos\sqrt\lambda=(\sqrt\lambda C_1-C_2)\sin\sqrt\lambda&&(2)_{\lambda>0}
\end{cases}\]From the first condition, you have
\[(\sqrt\lambda C_1-C_2)\sin\sqrt\lambda=0\]The left-hand side will disappear whenever \(\sqrt\lambda C_1-C_2=0\), so the solution in that case would satisfy
\[\begin{cases}
C_1+\sqrt\lambda C_2=0&&(1)_{\lambda>0}\\[1ex]
\sqrt\lambda C_1-C_2=0&&(2)_{\lambda>0}
\end{cases}\]but the only solutions for the constants here are \(C_1=C_2=0\), so we ignore these.

With \(\lambda>0\), \(\sin\sqrt\lambda\) will disappear for any \(\lambda_{n_1}=4{n_1}^2\pi^2\) or \(\lambda_{n_2}=\pi^2+4n_2\pi^2+4{n_2}^2\pi^2\) where \(n_1\) is any positive integer and \(n_2\) is any non-negative integer. For any of these possible \(\lambda\), the second equation reduces to \(0=0\) and leaves us with the lone
\[C_1+\sqrt\lambda C_2=0\]but we have infinitely many choices for these constants, so there are infinitely many solutions in the case of \(\lambda>0\).

Answer 3

Now suppose \(\lambda=0\). Then \(\chi(r)\) has one root \(r=0\) with multiplicity \(2\), so you have two independent solutions
\[y=C_1+C_2x\]which the boundary conditions require that this satisfy
\[\begin{cases}C_1+C_2=0\\[1ex]
C_1+2C_2=0\end{cases}\]This only has the trivial solution \(C_1=C_2=0\), so we don't have a zero eigenvalue.

Answer 4

Finally, suppose \(\lambda<0\). \(\chi(r)\) will have two real roots \(r=\pm\sqrt{-\lambda}\), and the ODE's general solution is
\[y=C_1\cosh\sqrt{-\lambda}x+C_2\sinh\sqrt{-\lambda}x\]This satisfies
\[\begin{cases}
C_1+\sqrt{-\lambda}C_2=0\\[1ex]
(C_1+\sqrt{-\lambda}C_2)\cosh\sqrt{-\lambda}+(\sqrt{-\lambda}C_1+C_2)\sinh\sqrt{-\lambda}=0
\end{cases}\]which reduces to
\[\begin{cases}
C_1+\sqrt{-\lambda}C_2=0\\[1ex]
(\sqrt{-\lambda}C_1+C_2)\sinh\sqrt{-\lambda}=0
\end{cases}\]\(\sinh\sqrt{-\lambda}\) has only one real solution at \(\lambda=0\), so we ignore this scenario. The resulting system
\[\begin{cases}
C_1+\sqrt{-\lambda}C_2=0\\[1ex]
\sqrt{-\lambda}C_1+C_2=0
\end{cases}\]again forces \(C_1=C_2=0\), so there are no non-trivial solutions for \(y\).

Answer 5

\[\boxed{
\newcommand \ps [1] {\left(#1\right)} % dynamic
\newcommand \den [3] {\frac{\mathrm d^#3 #1}{\mathrm d#2^#3}} % n-th d.
}\]
\[\begin{align}
\den yx2 + \lambda y
&= 0
& y(0)+y'(0)=0\\
&& y(1)+y'(1)=0\\
m^2+\lambda
&= 0\\[2ex]
\textbf{Case I}\\
\lambda &= 0\\
m &= 0,0\\
y &= A+Bx\\
y' &= x\\
y+y'&= A+(B+1)x\\
&& y(0)+y'(0)&= A = 0\\
&& y(1)+y'(1)&= B+1 = 0\\
&& & B = -1\\
&& & \mu_0 = 0\qquad n=0\\
y_0(x) &= -x && \lambda_0=0
\end{align}\]
\[\begin{align}
\textbf{Case II}\\
\lambda &= -\mu^2 < 0\\
m &= \pm\mu\\
y &= Ce^{\mu x}+De^{-\mu x}\\
y' &= \ps{Ce^{\mu x}-De^{-\mu x}}\mu\\
y+y'&= Ce^{\mu x}+De^{-\mu x}
+\ps{Ce^{\mu x}-De^{-\mu x}}\mu\hspace{-4cm}\\[2ex]
&&y(0)+y'(0)&= C+D+\ps{C-D}\mu\\
&& &\quad C+D=0, \qquad C-D=0\\
&& &\qquad\quad C=0, \qquad\qquad D=0
\end{align}\]
\[\begin{align}
\textbf{Case III}\\
\lambda &= \mu^2 > 0\\
m &= \pm i\mu\\
y &= F\cos\mu x+G\sin\mu x\\
y' &= \ps{-F\sin\mu x+G\cos\mu x}\mu\hspace{-2cm}\\
y+y'&= F\cos\mu x+G\sin\mu x
+\ps{-F\sin\mu x+G\cos\mu x}\mu\hspace{-5cm}\\
&= (F+G\mu)\cos\mu x+(G-F\mu)\sin\mu x\hspace{-5cm}\\[2ex]
&& y(0)+y'(0) &= F+G\mu = 0\\
&& & G = -F/\mu\\
y+y'&= -F\ps{\tfrac1\mu+\mu}\sin\mu x\\
&& y(1)+y'(1) &= -F\ps{\tfrac1\mu+\mu}\sin\mu = 0\\
&& & \mu_n = n\pi\qquad n=1,2,3,\dots\\
y_n(x)
&= \cos n\pi x-\tfrac1{n\pi}\sin n\pi x
& & \lambda_n = n^2\pi^2\\
\end{align}\]

Answer 6

In \(\textbf{Case I}\), shouldn't \(y'=B\) ? In which case you have the inconsistent system
\[\begin{cases}A+B=0\\[1ex]A+2B=0\end{cases}\]

Answer 7

oh dear

Answer 8

\[\begin{align}
\textbf{Case I}\\
\lambda &= 0\\
m &= 0,0\\
y &= A+Bx\\
y' &= B\\
y+y'&= A+B(1+x)\\
&& y(0)+y'(0)&= A+B = 0\\
&& & B=-A\\
y+y'&= A-Ax+A\\
&=-Ax\\
&& y(1)+y'(1)&= -A = 0\\
&& & A = 0\\
\end{align}\]

Answer 9

At first glance, your \(\textbf{Case III}\) is an improvement on mine, I think. Not sure what I was thinking...

Answer 10

I'm not used to using \(n_1,n_2\),

Answer 11

Actually, something doesn't sit right with me. You could remove the sine term in the solution for that case, since they all disappear, but the remaining cosine term doesn't satisfy the conditions. If \(y=\cos n\pi x\), then:
\[\begin{cases}y(0)+y'(0)=1\\[1ex]
y(1)+y'(1)=\cos n\pi
\end{cases}\]Am I missing something obvious?

Answer 12

they don't disappear because they are ...sin (nπx) 's not ...sin(nπ) 's

Answer 13

Oh, there it is -_-
Thanks!

Answer 14

; )