3 Example 2

Another variation of this problem is when the ode has the arbitrary value in the ode itself, instead of in the boundary condition as in the above example. Now the boundary conditions do not have an arbitrary symbol in them. An example is

\begin{align*} y^{\prime \prime }+\lambda y & =0\\ y\left ( 0\right ) & =0\\ y\left ( 1\right ) & =0 \end{align*}

Let us solve this using normal method again. Since this is a constant coefficient ode, 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 \(y\left ( 1\right ) =0\) the above gives

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

And we conclude that \(c_{2}=0\) which gives the trivial solution

\[ y=0 \]

Now the ode 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 assume \(\lambda \) 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 \(y\left ( 1\right ) =0\) the above becomes

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

But \(\sinh \left ( \mu \right ) \) is only zero when \(\mu =0\) or \(\lambda =0\) 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 \(y\left ( 1\right ) =0\) this gives \(0=c_{2}\). Which means 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 \(y\left ( 1\right ) =0\) the above gives

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

For non-trivial solution we want \(\sin \left ( \sqrt {\lambda }\right ) =0\). This means \(\sqrt {\lambda }=\pi ,2\pi ,3\pi ,\cdots \) or

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

Notice that \(n>0\) because we assumed \(\lambda >0\) we can’t pick \(-\pi ,-2\pi ,\cdots \) values. We also can not pick \(n=0\). The general solution hence becomes

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

In the above \(\lambda _{n}\) are called the eigenvalues and \(\Phi _{n}=\sin \left ( \sqrt {\lambda _{n}}x\right ) ,n=1,2,\cdots \) are called the eigenfunctions. In this example, there is one eigenfunction \(\Phi _{n}\) associated with each eigenvalue \(\lambda _{n}\).