Question

38. Show that the eigenvalues and eigenfunctions of the boundary value problem

${\mathit{\lambda }}^{\mathbf{\text{'}}\mathbf{}\mathbf{\text{'}}}\mathbf{+}\mathit{\lambda }\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{+}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$

are${\mathit{\lambda }}_{\mathbf{n}}\mathbf{=}{\mathit{\alpha }}_{\mathbf{n}}^{\mathbf{2}}$ and${\mathit{y}}_{\mathbf{n}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}{\mathit{\alpha }}_{\mathbf{n}}\mathit{x}$ , respectively, where ${\mathit{\alpha }}_{\mathbf{n}}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{\dots }$are the consecutive positive roots of the equation $\mathit{t}\mathit{a}\mathit{n}\mathit{\alpha }\mathbf{=}\mathbf{-}\mathit{\alpha }\mathbf{.}$

Short answer

${\lambda }_n={\alpha }_n^2=x_n^2,n=1,2,3,...,$where the $n$ are the consecutive positive roots of$tan\alpha =-\alpha$

Step-by-step solution

To Find the eigenvalues and eigenfunctions

For $\lambda ={\alpha }^2>0$ the general solution is$y=c_{1}cos\alpha x+c_{2}sin\alpha x.$ Setting$y\left(0\right)=0$ we find $c_1=0$, so that$y=c_{2}sin\alpha x.$ The boundary condition $y\left(1\right)+y^{\text{'}}\left(1\right)=0$implies

$c_{2}sin\alpha +c_2\alpha cos\alpha =0$

Taking $c_2\ne 0,$this equation is equivalent to role="math" localid="1663856759003" $tan\alpha =-\alpha$. Thus, the eigenvalues are role="math" localid="1663856899267" ${\lambda }_n={\alpha }_n^2=x_{n^{\backslash \mathrm{cent}}}^{2}n=1,2,3,....,$where the n are the consecutive positive roots of $tan\alpha =-\alpha .$

Final proof

${\lambda }_n={\alpha }_n^2=x_n^2,n=1,2,3,...,$where the n are the consecutive positive roots of $tan\alpha =-\alpha$