Question

Consider the boundary-value problem

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

(a) The type of boundary conditions specified are called periodic boundary conditions. Give a geometric interpretation of these conditions.

(b) Find the eigenvalues and eigenfunctions of the problem.

(c) Use a graphing utility to graph some of the eigenfunctions. Verify your geometric interpretation of the boundary conditions given in part (a).

Short answer

(a) A solution curve has the same coordinate at both ends of the interval$[-\pi ,\pi ]$ and the tangent lines at the endpoints of the interval are parallel.

(b) ${\lambda }_n={\alpha }^2=n^2,y_n\left(x\right)=cosnx$and $y_n=\sin nx,n=1,2,3,\cdots .$

(c) See the graph.

Step-by-step solution

To Find the geometric interpretation of these conditions.

(a) A solution curve has the same coordinate at both ends of the interval$[-\pi ,\pi ]$ and the tangent lines at the endpoints of the interval are parallel.

Find the eigenvalues and eigenfunctions of the problem 

(b) For $\lambda =0$the solution of $y^{\varphi \varphi }=0$is $y=c_{1}x+c_2.$From the first boundary condition we have $y(-\pi )=-c_1\pi +c_2=y\left(\pi \right)=c_1\pi +c_2$or $2c_1\pi =0$. Thus, $c_1=0$and $y=c_2$This constant solution is seen to satisfy the boundary-value problem. For $\lambda =-{\alpha }^2<0$we have $y=c_{1}cosh\alpha x+c_{2}sinh\alpha x.$In this case the first boundary condition gives

$$\begin{gathered} y(-\mathrm{\pi })={\mathrm{c}}_1\cosh (-\mathrm{\alpha \pi })+{\mathrm{c}}_2\sinh (-\mathrm{\alpha \pi }) \\ ={\mathrm{c}}_1\mathrm{cosh\alpha \pi }-{\mathrm{c}}_2\mathrm{sinh\alpha \pi } \\ =\mathrm{y}\left(\mathrm{\pi }\right)={\mathrm{c}}_1\mathrm{cosh\alpha \pi }+{\mathrm{c}}_2\mathrm{sinh\alpha \pi } \end{gathered}$$

or$2c_{2}sinh\alpha \pi =0.$ Thus$c_2=0$ and$y=c_{1}cosh\alpha x.$ The second boundary condition implies in a similar fashion that $c_1=0.$

Find the graph

C) See the graph

$y=2\sin \left(3x\right)$

$y=\sin 4x-2\cos 3x$

Find the boundary-value problem

Thus, for $\lambda <0,$ the only solution of the boundary-value problem is $y=0.$ $\lambda ={\alpha }^2>0$we have $y=c_{1}cos\alpha x+c_{2}sin\alpha x.$ The first boundary condition implies

$y(-\pi )=c_{1}cos(-\alpha \pi )+c_{2}sin(-\alpha \pi )$

$=c_{1}cos\alpha \pi -c_{2}sin\alpha \pi$$=\quad y(\pi)$

$=c_{1}cos\alpha \pi +c_{2}sin\alpha \pi$

or $2c_{2}sin\alpha \pi =0.$Similarly, the second boundary condition implies $2c_1\alpha sin\alpha \pi =0$. If $c_1=c_2=0$the solution is $c_1=c_2=0$ However, if $c_1\ne 0$or $c_2\ne 0,$then which implies that $\alpha =n.$Therefore, for $c_1$and $c_2$not both $0,y=c_{1}cosnx+c_{2}sinnx$ is a nontrivial solution of the boundary-value problem. Since $\cos (-nx)=\cos nx$ and $\sin (-nx)=-\sin nx,$ we may assume without loss of generality that the eigenvalues are ${\lambda }_n={\alpha }^2=n^2,$ for n a positive integer. The corresponding eigenfunctions are $y_n\left(x\right)=cosnx$ and $y_n=sinnx.$

Final proof

(a) A solution curve has the same y- coordinate at both ends of the interval $[-\pi ,\pi ]$ and the tangent lines at the endpoints of the interval are parallel.

(b) ${\lambda }_n={\alpha }^2=n^2,y_n\left(x\right)=cosnx$ and $y_n=\sin nx,n=1,2,3,\cdots .$

(c) See the graph.