x′′ + λx = 0, x(0) + x′(0) = 0, x(1) = 0
I'm given this boundary value problem, and I'm supposed to look at the 3 cases of when λ > 0, λ = 0, and λ < 0. I can find the characteristic equations of these three cases with no problem, but the boundary conditions are giving me issue. Normally when I do these problems, I don't have x and x' in the same condition. Instead I'll have, for example, something like x(0) = 0 and x'(1) = 0. How can I solve this given these strange conditions.
My professor gave me the following hint: For this problem finding all the values of λ and the solutions will be tricky. So check whether λ = 0 works, and if so, find the corresponding solution. Also, find an approximate value for λ1, the nonzero value with the smallest absolute value, along with the corresponding function.
For $\lambda=0$, we have that $x(t) = c_1 t + c_2$. Applying the boundary conditions gives $x(0) + x'(0) = c_2 + c_1 = 0$. Similarly, $x(1) = c_1 + c_2 = 0$. Thus, these are linearly dependent and we deduce that the solution for $\lambda=0$ is $c(x-1)$ for all $c \in \mathbb{R}$.
Case $\lambda<0$.
Then our solution is $$c_1 e^{\sqrt{-\lambda} t} + c_2 e^{-\sqrt{-\lambda} t}$$ Applying boundary conditions, $$x(0) + x'(0) = \sqrt{-\lambda}(c_1 - c_2) + c_1 + c_2 = 0 \implies (\sqrt{-\lambda} +1)c_1 + (1-\sqrt{-\lambda})c_2 = 0$$ With the second condition, $c_1 e^{\sqrt{-\lambda} } + c_2 e^{-\sqrt{-\lambda} } = 0$. Observing that this is a homogeneous system, the only time we will get a nontrivial solution is when the associated determinant vanishes, which can be calculated as $(\sqrt{-\lambda} +1)e^{-\sqrt{-\lambda} } - (1-\sqrt{-\lambda}) e^{\sqrt{-\lambda} }$. Assuming this is zero, we must approximate the solution of $e^{2\sqrt{-\lambda}} = \frac{(\sqrt{-\lambda} +1)}{(1-\sqrt{-\lambda})}$. This $\textit{does}$ have an approximate solution of $\approx -1.439 \dots$ (I do not believe that there is an analytic form). Also, please check for errors in the above.
Now, to solve for the constants, it is obvious that for the above value of $\lambda$, $c_2 = -c_1 e^{2\sqrt{-\lambda}}$. Plugging in, we see that $$x(t) = c (e^{\sqrt{-\lambda} t} -e^{-\sqrt{-\lambda} t + 2\sqrt{-\lambda}})$$
For all constants $c \in \mathbb{R}$ where $\lambda \approx 1.439 \dots$..
The case for $\lambda>0$ is handled similarly. I will not type out the full details but setting the associated determinant equal to $0$ will yield $\lambda = \tan (\lambda)$. Thus we actually have an infinite number of eigenvalues $\lambda$ for which there exists a solution. The associated solution (eigenvector) corresponding to any $\lambda$ is of course simple to find.