# Characterizing all possible solutions for a simple differential equation

Consider the differential equation$\frac{d^2 X(x)}{d^2 x}= C_0 X(x)$. Here $C_0$ is a complex number, but $X(x)$ is real.

Suppose we impose the boundary conditions $X(0)=X_0, X(a)=X_a$. How can we characterize all possible solutions? Since the differential equation is second order, there should be two independent solutions, correct? Don't the two boundary conditions specify the two coefficients of the linear combination of the two independent solutions, leading to one unique solution? Or, am I missing some more subtleties?

For instance if $X_0=X_a=0$ then I see at least two different solutions, a sine and a cosine, contradicting the above statement.

• 1) If $X(x)$ is real then $X''(x)$ should be real as well, so $C_0$ can't be complex (unless $X(x) \equiv 0$). 2) If $X(0) = 0$ then $X(x) = \cos(\sqrt{C_0}x)$ is not a solution. – Abstraction Mar 3 '17 at 14:49

Boundary problems and initial value problems are different.

Picard's existence and uniqueness theorem is for initial value problems, where the conditions are imposed on the same value of the independent variable. In your example, there would be two conditions; $x(x_0)$ and $X'(x_0)$.

In boundary value problems the conditions are imposed on different values of the independent variable. The existence and uniqueness theorem does not apply. There is a certain type of boundary value problem that appears often: the eigenvalue problem. An example is $X''=\lambda\,X$ with homogeneous boundary conditions, that can be of different types:

1. Dirichlet: $X(a)=X(b)=0$
2. Neumann: $X'(a)=X'(b)=0$
3. Robin: $X(a)+h\,X'(a)=0$, $X(b)+h\,X'(b)=0$
4. Mixed (different type in $a$ and $b$)
5. Periodis: $X(a)=X(b)$, $X'(a)=X'(b)$
6. ...

In this cases, $X\equiv0$ is always a solution. The eigenvalue problem consists in finding values of $\lambda$ such that a non-zero solution exists.

Differentials equations of this type can be solved by the characteristic polynomial. That is, the characteristic polynomial of $X''=CX$ is $\lambda^2-C=0$. Hence $\lambda=\pm\sqrt{C}$. The general theory yields that $e^{\sqrt{C}}x$ and $e^{-\sqrt{C}}$ are independent solutions.

Suppose for simplicity that $C<0$, then $e^{-i\sqrt{C}}$ and $e^{+\sqrt{C}}$ are solutions, taking appropriate linear combinations we get that $\sin(\sqrt{C}x)$ and $\cos(\sqrt{C}x)$ are independent solutions. A general solution is then a linear combination of these ones.

Using the boundary conditions, we see that $\cos(\sqrt{C}x)$ is not a solution. Hence the solution is a proper multiple of $\sin(\sqrt{C}x)$.

Can you figure out what happens if $C=0$?