2 Example 1

Let us solve the second order BVP

\begin{align*} y^{\prime \prime }+3y & =0\\ y\left ( 0\right ) & =0\\ y\left ( L\right ) & =0 \end{align*}

If the boundary condition do not have an arbitrary \(L\) in them, i.e. if \(y\left ( L\right ) =0\) happened to be for example \(y\left ( 1\right ) =0\) then this can be solved using normal methods giving \(y=0\) as solution and there is nothing more to say.

But if we try to solve this using normal methods with \(y\left ( L\right ) =0\), where \(L\) is now a symbol and not a number, we will see that the result will be also be \(y\left ( 0\right ) =0\) which is the trivial solution again.

But there is a non-trivial solution to this when \(L\) is unknown. We have to use the eigenvalue/eigenfunction method instead of normal methods to find this non-trivial solution. This note shows this difference.

First we will solve

\begin{align*} y^{\prime \prime }+3y & =0\\ y\left ( 0\right ) & =0\\ y\left ( L\right ) & =0 \end{align*}

Using normal methods. Since this is constant coefficient ode, then the characteristic equation is

\begin{align*} r^{2}+3 & =0\\ r & =\pm \sqrt {3}i \end{align*}

This means the basis solutions are \(y_{1}=e^{\sqrt {3}ix},y_{2}=e^{-\sqrt {3}ix}\) and the general solution is

\[ y=c_{1}e^{\sqrt {3}ix}+c_{2}e^{-\sqrt {3}ix}\]

Which can be written using Euler formula as

\begin{equation} y=c_{1}\cos \left ( \sqrt {3}x\right ) +c_{2}\sin \left ( \sqrt {3}x\right ) \tag {1}\end{equation}

Boundary conditions are now applied to find \(c_{1},c_{2}\). When \(y\left ( 0\right ) =0\) the above becomes

\[ 0=c_{1}\]

Hence (1) now becomes

\[ y=c_{2}\sin \left ( \sqrt {3}x\right ) \]

And when \(y\left ( L\right ) =0\) the above gives

\begin{equation} 0=c_{2}\sin \left ( \sqrt {3}L\right ) \tag {2}\end{equation}

So we conclude here that \(c_{2}=0\) since \(L\) is not known and can not be zero, which results in trivial solution

\[ y=0 \]

The difference between the above method and the eigenvalue/eigenfunction method to solving BVP, is that in (2), we do not choose \(c_{2}=0\), but instead we find \(L\) values for \(L\) that makes \(\sin \left ( 3L\right ) =0\) instead of the constant \(c_{2}\) and thus avoiding trivial solution.

We know that \(\sin \left ( \sqrt {3}L\right ) =0\) when \(\sqrt {3}L=0,\pm \pi ,\pm 2\pi ,\cdots \) but since \(L>0\) this becomes \(\sqrt {3}L=\pi ,2\pi ,3\pi ,\cdots \)

\[ \sqrt {3}L=n\pi \qquad n=\mathbb {Z} ,n>0 \]

Hence

\[ L=\frac {n\pi }{\sqrt {3}}\qquad n=\mathbb {Z} ,n>0 \]

The \(L\) values above are called the eigenvalues of the ode. These are the values that gives non trivial solution. The general solution now becomes

\[ y=\left \{ \begin {array} [c]{ccc}c_{2}\sin \left ( \sqrt {3}x\right ) \qquad & & L=\frac {n\pi }{\sqrt {3}},n=\mathbb {Z} ,n>0\\ 0 & & \text {otherwise}\end {array} \right . \]

\(\sin \left ( \sqrt {3}x\right ) \) above is the eigenfunction. There is only one eigenfuction in this example. The above says that the solution is trivial only when \(L\) does not satisfy \(L=\frac {n\pi }{\sqrt {3}}\).

Mathematica DSolve command handles this automatically and gives both trivial and non-trivial and the conditions on \(L\).