# Find all eigenvalues and eigenfunctions for the Boundary Value Problem

$u'' + 4u' + \lambda u = 0$, $\lambda > 0$

Subject to the Boundary Conditions: $u(0) = u'(0)$, $u(1) = 1$

Using the characteristic equation I got the general solution to be: $u(x) = c_1 e^{(-2 + \sqrt{4 - \lambda})x} + c_2 e^{(-2 - \sqrt{4 - \lambda})x}$

Then, applying the boundary conditions I was left with this system:

$c_1 (3 - \sqrt{4 - \lambda}) + c_2 (3 + \sqrt{4 - \lambda}) = 0$

$c_1 e^{-2 + \sqrt{4 - \lambda}} + c_2 e^{-2 - \sqrt{4 - \lambda}} = 1$

Now, since we are looking for a nontrivial solution to this system apparently we need to assume that the determinant equals zero. I do not understand why we do this, could somebody explain?

Anyways, assuming the determinant equals zero:

$e^{-2 - \sqrt{4 - \lambda}}(3 - \sqrt{4 - \lambda}) = e^{-2 + \sqrt{4 - \lambda}}(3 + \sqrt{4 - \lambda})$

$\implies \lambda = 4$.

Thus, the only eigenvalue for this problem is $\lambda = 4$ and the corresponding eigenfunction is $u(x) = c_1 e^{-2x} + c_2 e ^{-2x}$. Is this the correct procedure?

It seems you want the eigenfunctions with negative eigenvalues $-\lambda$ of the operator $$D = \frac{d^2}{dx^2} + 4\frac{d}{dx}$$ These are just the solutions you've found. Don't force the determinant to be $0$, solve $$\left(\begin{matrix}3 - \sqrt{4 - \lambda} & 3 + \sqrt{4 - \lambda} \\ e^{-2 + \sqrt{4 - \lambda}} & e^{-2 - \sqrt{4 - \lambda}}\end{matrix}\right)\left(\begin{matrix}c_1 \\ c_2\end{matrix}\right) = \left(\begin{matrix}0 \\ 1\end{matrix}\right)$$ by inversion or however you want to get the constant factors for an eigenfunction for that eigenvalue of $-\lambda$. Distinguish the case $\lambda = 4$ which is resonant and has an additional solution of the form $xe^{kx}$.
• I'm sorry, I do not understand what you mean by negative eigenvalues of the operator $D$. Could you explain? The problem started by stating that $\lambda > 0$. Also what is meant by the case $\lambda = 4$ being resonant? I can see why the eigenvalue $\lambda = 4$ has the solution $xe^{kx}$ – Javier Oct 27 '15 at 20:20