Spectrum of eigenvalues and eigenfunctions

Our O.D.Es professor had the "amazing" idea of heavily introducing advanced linear algebra material (which is not an official prerequisite for the course) along with boundary value problems. Not being trained in these types of exercises I am facing quite a few difficulties. If anybody would be willing to help, it would be most appreciated.

For the following sets of boundary conditions, consider the equation:

$$u''(x) + \lambda u(x) = 0, \ \ \ 0 < x < 1$$

and find the spectrum of eigenvalues, in essence the set of values $\lambda$ in the complex plane for which a nontrivial solution exists, and give the eigenfunction (or eigenfunctions) for each such eigenvalue.

1) $u(0) + u(1) = 0$ and $u'(0) + u'(1) = 0$ 2) $u(0) + u(1) = 0$ and $u'(0) - u'(1) = 0$

• Start by considering three cases: $\lambda > 0$, $\lambda = 0$ and $\lambda < 0$. What are the solutions in each case? Can they satisfy the given boundary conditions? – in_mathematica_we_trust Mar 12 '13 at 5:32
• I have an amazing idea! You should use Sturm-Liouville theory and the Rayleigh quotient! – AndrewG Mar 12 '13 at 5:46

The characteristic equation is $m^2 + \lambda = 0$.

We need to consider three cases, $(1) \lambda = 0$, $(2) \lambda \lt 0$ and $(3) \lambda \gt 0$.

For $\lambda = 0$:

The solution is $u(x) = a + bx$.

Applying the boundary condition $1)$, yields

$u(0) + u(1) = a + a + b = 0$ and $u'(0) + u'(1) = b + b = 0$, this yields $a = b = 0$, thus $u(x) = 0$.

Applying the boundary condition $2)$, yields

$u(0) + u(1) = a + a + b = 0$ and $u'(0) - u'(1) = b - b = 0$, this yields $b = -2a$, so choose $a = 0$, which gives $b = 0$ and thus $u(x) = 0$.

For $\lambda \lt 0$:

The solution is $\displaystyle u(x) = a e^{\sqrt{-\lambda}~x} + b e^{-\sqrt{-\lambda}~x}$, where $-\lambda$ and $\sqrt{-\lambda}$ are positive.

You'll have to do the two cases of IC's.

For $\lambda \gt 0$:

The solution is $\displaystyle u(x) = a \sin \sqrt{\lambda}~x + b \cos \sqrt{\lambda}~x$.

We still have to apply the boundary conditions.

After we find the $u(x)$, we need to string them together to figure out the eigenvalues and eigenfunctions.

Need to go to bed for now.

• Thank you very much for your hard work! Sweet dreams! – user44069 Mar 12 '13 at 6:35
• You are very welcome, but this is still not done, the ic's still have to be done for the last two, and then, you need to find thr general $\lambda$ – Amzoti Mar 12 '13 at 13:01
• Great details, and you received super feedback, too! – Namaste Apr 21 '13 at 0:26

See my solution at my website, http://people.cs.uchicago.edu/~lebovitz/Winter13/Slvdprblms

1. u(0) + u(1) = 0; u'(0) + u'(1) = 0

SOLUTION: With $\lambda$ = $\mu^2 \neq$ 0, any solution has the form

u = A cos(x) + B sin(x)

Applying the boundary conditions gives

(1 + cos$\mu$ )A + sin $\mu$B = 0

-$\mu$sin A + $\mu$(1 + cos $\mu$)B = 0

implying that 2$\mu$(1 + cos $\mu$) = 0. It is easy to see that $\mu$= 0 leads only to the solution u(x) = 0 so 0 is not an eigenvalue. The other values,

$\mu_k$ = (2k + 1)$\pi$ or $\lambda_k$ = $(2k + 1)^2\pi^2$

k = 0, 1, 2

are genuine eigenvalues. To each there corresponds a pair of eigenfunctions $u_k(x)$ = cos($\mu_k$x) and $v_k(x)$ = sin($\mu_k$x)

1. u(0) + u(1) = 0; u'(0)-u'(1) = 0

SOLUTION: A solution satisfying the first boundary condition is (with $\lambda = \mu^2 \neq$ 0)

$u_1$ = -sin($\mu$x) + sin ($\mu$(1-x)).

But this also easily seen to satisfy the second boundary condition, without any condition on $\mu$. If = 0 the function u0(x) = 1 2x satisfies the equation and the boundary conditions. The spectrum therefore consists of the entire complex plane and, for given $\lambda \neq 0$, $u_1$ is the eigenfunction (or $u_0$ if $\lambda$ = 0).

I don't like your tone of voice, young man. Please see me before the exam.

• It really helps to format your questions/answers using LaTex / MathJax and there are pointers to excellent resources in the FAQ. regards – Amzoti Mar 16 '13 at 21:12

See the simple harmonic oscillator, though I should mention this particular example will only give you "half" of the answers.