Examples

BVP of sinusoids

BVP of sinusoids

Consider the ODE given by $y^{\prime \prime }+{\omega }^{2}y=0$. We study the BVP given by $y(0)=0$ and $y(x_1)=y_1$. The ‘boundaries’ in question are $x=0$ and $x=x_1$.

We have already found the complete family of solutions to the ODE:

$$y(x)=C_1\cos (\omega x)+C_2\sin (\omega x)=A\sin (\omega x+\varphi ).$$

The first boundary condition determines $C_1=0$, or (in the second formulation) $\varphi =\pi k$. If we assume $A>0$, then it will be enough to consider just the two options $\varphi =0$ and $\varphi =\pi$. (Why?)

The solutions compatible with $y(0)=0$ are therefore sinusoids with angular frequency $\omega$ that pass through the origin. At $x_1$ these solutions take the value $C_1\sin (\omega x_1)$. With fixed $\omega$, by varying $C_1$ we can control the values at $x_1$ to a certain extent, but we are not completely free.
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Here is a summary of what we can do:

• If $y_1=0$, then we must have $x_1=k\pi /\omega$, or else $C_1=0$ (hence $y=0$ everywhere)
• If $y_1\ne 0$, then we can solve for $C_1$ provided $x_1\ne k\pi /\omega$; but if $x_1=k\pi /\omega$ there is simply no solution.

That is a complete solution to the BVP in question.

Fixing boundary, varying parameter

Fixing boundary, varying parameter

We can instead consider the problem from another angle. Suppose we fix the boundary data, and ask instead for the possible parameters $\omega$ for which nonzero solutions exist.

Let us fix the second boundary point at $(x_1,0)$. Solutions (other than $y\equiv 0$) are only possible when $x_1=k\pi /\omega$. Treating $x_1$ as a constant, we see that nontrivial solutions exist if and only if $\omega =k\frac{\pi }{x_1}$.

In other words, the oscillation frequency must be such that the sinusoid crosses through $y=0$ at the boundary point $x_1$. This will occur precisely when $$\omega =\frac{\pi }{x_1},2\frac{\pi }{x_1},3\frac{\pi }{x_1},4\frac{\pi }{x_1},\dots .$$