Spectrum of eigenvalues and eigenfunctions

Question

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 $u'(0) - u'(1) = 0$

Answer 1

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

Answer 2

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.

Answer 3

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.