4 Example 3

Final example is when the ode has an arbitrary value (the eigenvalue) in the ode itself and the boundary condition also has an arbitrary value.

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

Let us solve this using normal method again. Since this is constant coefficient ode, then the characteristic equation is

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

Hence the general solution is

\begin{align} y & =c_{1}e^{\sqrt {\lambda }ix}+c_{2}e^{-\sqrt {\lambda }ix}\nonumber \\ & =c_{1}\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \tag {3}\end{align}

At \(y\left ( 0\right ) =0\) the above gives

\[ 0=c_{1}\]

And (3) becomes

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

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

\[ 0=c_{2}\sin \left ( \sqrt {\lambda }L\right ) \]

And since \(L\) can not be zero, we conclude that \(c_{2}=0\) which gives the trivial solution

\[ y=0 \]

Now the problem is solved using eigenvalue/eigenvector approach. Starting from \(r=\pm \sqrt {-\lambda }\) and since we do not know what \(\lambda \) is, we have to check each possible case. We assumed \(\lambda \) is real value in all of this.

case \(\lambda <0\) Hence \(-\lambda >0\) and we let \(\sqrt {-\lambda }=\mu \) where \(\mu >0\). The roots then are \(r=\pm \mu \) and the general solution is

\[ y=c_{1}\cos \left ( \mu x\right ) +c_{2}\sinh \left ( \mu x\right ) \]

At \(y\left ( 0\right ) =0\) the above gives

\[ 0=c_{1}\]

So the solution becomes

\[ y=c_{2}\sinh \left ( \mu x\right ) \]

At \(y\left ( L\right ) =0\) the above becomes

\[ 0=c_{2}\sinh \left ( \mu L\right ) \]

But \(\sinh \left ( \mu \right ) \) is only zero when \(\mu L=0\) or \(\lambda =0\) (since \(L\) can not be zero), which is not the case here. This means only other option is \(c_{2}=0\) which leads to trivial solution. Hence \(\lambda <0\) is not valid assumption.

case \(\lambda =0\) The roots now are \(r=0\) double root. In other words, the ode is \(y^{\prime \prime }=0\). Hence the general solution is

\[ y=c_{1}+c_{2}x \]

At \(y\left ( 0\right ) =0\) this gives \(c_{1}=0\). Hence \(y=c_{2}x\). At \(y\left ( L\right ) =0\) this gives \(0=Lc_{2}\). Which means \(c_{2}=0\) since \(L\) can not be zero. Hence the trivial solution \(y=0\). So \(\lambda =0\) is again not a valid assumption.

case \(\lambda >0\) The roots now are \(r=\pm i\sqrt {\lambda }\) which means the general solution is

\[ y=c_{1}\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \]

At \(y\left ( 0\right ) =0\) the above gives \(0=c_{1}\) and the solution now becomes

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

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

\[ 0=c_{2}\sin \left ( \sqrt {\lambda }L\right ) \]

For non-trivial solution we want \(\sin \left ( \sqrt {\lambda }L\right ) =0\). This means \(\sqrt {\lambda }L=\pi ,2\pi ,\cdots \) or

\begin{align*} \sqrt {\lambda } & =\frac {n\pi }{L}\qquad n=\mathbb {Z} ,n>0\\ \lambda _{n} & =\frac {n^{2}\pi ^{2}}{L^{2}}\end{align*}

Hence the general solution now becomes

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

In the above \(\lambda _{n}\) are called the eigenvalues and \(\Phi _{n}=\sin \left ( \sqrt {\lambda _{n}}x\right ) ,n=1,2,3,\cdots \) are called the eigenfunctions. This is basically the same solution as second example, with the difference is that the length is now unknown \(L\) and not a specific value \(1\) as before. That is why the length \(L\) shows in the eigenvalues.