Question

Find the eigenvalues and eigenfunctions for the given boundary-value problem.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{\lambda }\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathit{L}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

Therefore, the solution is${\lambda }_n={\left(n+\frac{1}{2}\right)}^2{\pi }^2/L^2,y_n=\sin \left(\left(n+\frac{1}{2}\right)\pi x/L\right),n=0,1,2,\cdots .$

Step-by-step solution

Given information

The given value is:

$y^{\text{'}\text{'}}+\lambda y=0,y^{\text{'}}\left(0\right)=0,y\left(L\right)=0$

Three cases will be considered

$\lambda =0,\lambda <0,$and $\lambda >0$

Case I: For $\lambda =0$the solution of $y^{\text{'}\text{'}}=0$is $y=c_{1}x+c_2.$The conditions $y\left(0\right)=0$makes ${\text{c}}_2\text{= 0}$and $y^{\text{'}}\left(L\right)=0$makes ${\text{c}}_1\text{= 0}$and so $y\left(x\right)=0$which is trivial.

Case II: For $\lambda <0,$say$\lambda =-{\alpha }^2,$for some positive number $\alpha .$So that the characteristic equation is$m^2-{\alpha }^2=0$whose roots are$\pm \alpha .$Thus the general solution is

$y=c_1{e}^{\alpha x}+c_2{e}^{-\alpha x}$

The condition $y\left(0\right)$gives $c_1+c_2=0$and $y^{\text{'}}\left(L\right)=0$gives $c_1{e}^{\alpha L}-c_2{e}^{-\alpha L}=0$. Solving the previous equations simultaneously implies $c_1=c_2=0$to get the trivial solution $y\left(x\right)=0.$

Case III: For $\lambda >0$we write$\lambda ={\alpha }^2,$where $\alpha$is a positive number. The characteristic equation is $m^2+{\alpha }^2=0$which has complex roots $\pm \alpha i.$The general solution is

$y\left(x\right)=c_1\cos \alpha x+c_2\sin \alpha x.$

As before, $y\left(0\right)=0$yields $c_1=0$and so

$y\left(x\right)=c_2\sin \alpha x.$

Now the last condition$y\text{'}\left(L\right)=0$ and so

$c_2\alpha \cos \alpha L=0$

Choosing $c_2\ne 0,$we get $\cos \alpha L=0$and hence $\alpha L=\left(n+\frac{1}{2}\right)\pi$for every integer n. So that $\alpha =\left(n+\frac{1}{2}\right)\pi /L$and ${\lambda }_n={\left(n+\frac{1}{2}\right)}^2{\pi }^2/L^2.$Therefore for any real nonzero

$c_2,y_n\left(x\right)=c_2\sin \left(\left(n+\frac{1}{2}\right)\pi x/L\right.)$is a solution of the problem for each nonnegative integer n.

Because the differential equation is homogeneous, any constant multiple of a solution is also a solution, so we may, if desired, simply take $c_2=1.$

Thus the nontrivial eigenvalues are

${\lambda }_n={\left(n+\frac{1}{2}\right)}^2{\pi }^2/L^2$

with corresponding eigenfunctions

$y_n=\sin \left(\left(n+\frac{1}{2}\right)\pi x/L\right)$

for every$n=1,2,\cdots .$

Therefore, the solution is${\lambda }_n={\left(n+\frac{1}{2}\right)}^2{\pi }^2/L^2,y_n=\sin \left(\left(n+\frac{1}{2}\right)\pi x/L\right),n=0,1,2,\cdots .$