Question

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

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

Short answer

The solution is $\sin \left(nx\right),\lambda =n^2$$n\in \mathrm{\mathbb{N}}$

Step-by-step solution

Given information

The given value is:

For the boundary value problem, we're looking for eigenvalues and eigenfunctions.

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

Auxiliary equation to solve

$e^{mx}\left(m^2+\lambda \right)=0$

$m^2+\lambda =0$

$m^2=-\lambda$

$\sqrt{m^2}=\sqrt{-\lambda }$

$m=\pm \sqrt{-\lambda }$ it is trivial.

Three cases will be considered

We now have three options, all of which are dependent on the value of$\lambda .$The discriminant is zero and the auxiliary equation has a repeating root if$\lambda =0$

The discriminant is negative if and the auxiliary equation has two imaginary roots.

The discriminant is positive and the auxiliary equation has two unique real roots if$\lambda <0,$is positive.

During our search for eigenvalues and functions, we'll have to put all three to the test.

Case 1: $\lambda =0\to 2$duplicate roots

Case 2:$\lambda >0\to 2$ imaginary roots

Case 3:$\lambda <0\to 2$ distinct real roots

The complimentary solution is being evaluated for case I

We solve for each constant by evaluating the complimentary solutions at the two places. We discover that for $\lambda =0,$ the complementary solution is easy, thus we go on. In this boundary, we're looking for nonzero values that are dependent on lambda.

$\lambda =0$

$y_c=c_1+x{c}_2,y\left(0\right)=0,y\left(\pi \right)=0$

$y_c\left(0\right)=c_1+\left(0\right)c_2=0$

$c_1=0$

$y_c\left(\pi \right)=\left(0\right)+\left(\pi \right)c_2=0$

$\pi \ne 0,$so $c_2=0$

$y_c=\left(0\right)+\left(0\right)=0$

$y_c=0$

The complimentary solution is being evaluated for case II

$\lambda <0$

$y_c=c_1{e}^{x\sqrt{-\lambda }}+c_2{e}^{-x\sqrt{-\lambda }},y\left(0\right)=0$

$y\left(\pi \right)=0$

$y_c\left(0\right)=c_1{e}^{\left(0\right)\sqrt{-\lambda }}+c_2{e}^{-\left(0\right)\sqrt{-\lambda }}=0$

$c_1{e}^0+c_2{e}^0=0$

$c_1+c_2=0$

$y_c\left(\pi \right)=c_1{e}^{\left(\pi \right)\sqrt{-\lambda }}-c_1{e}^{-\left(\pi \right)\sqrt{-\lambda }}=0$

$c_1\left(e^{\pi \sqrt{-\lambda }}-e^{-\pi \sqrt{-\lambda }}\right)=0$

$c_1=0$

or

$$\begin{gathered} {}^{\pi \sqrt{-\lambda }}-e^{-\pi \sqrt{-\lambda }}=0(I.) \\ {e}^{\pi \sqrt{-\lambda }}=e^{-\pi \sqrt{-\lambda }} \\ \lambda =0 \end{gathered}$$

• We assume the form of a complementary solution with two real and distinct roots. We plug in the two boundary points.
• Notice on the second point that we run into a contradiction. Because of the zero product property, one of these values has to equal zero $c_1$, or or both, when the expression equals zero.
• The only way that (I.) can be zero is if$\lambda =0$ which would yield $1-1=0.$This is a contradiction because we have assumed that$\lambda <0$ and assuming otherwise at this point would be illogical. We cannot proceed with this case, so we move on.

The complimentary solution is being evaluated for case III 

$\lambda >0$

$\sqrt{-\lambda }=\sqrt{(-1)\lambda }=\pm i\sqrt{\lambda }$

$y_c=c_1\sin \left(x\sqrt{\lambda }\right)+c_2\cos \left(x\sqrt{\lambda }\right)$

$y\left(0\right)=0,y\left(\pi \right)=0$

$y\left(0\right)=0,y\left(\pi \right)=0y_c\left(0\right)=c_1\sin \left(\right(0\left)\sqrt{\lambda }\right)+c_2\cos \left(\right(0\left)\sqrt{\lambda }\right)=0$

$c_1\sin \left(0\right)+c_2\cos \left(0\right)=0$

$y_c\left(\pi \right)=c_1\sin \left(\right(\pi \left)\sqrt{\lambda }\right)+\left(0\right)\cos \left(\right(\pi \left)\sqrt{\lambda }\right)=0$

$c_1\sin \left(\right(\pi \left)\sqrt{\lambda }\right)=0$

$c_1=0$

or

$\sin \left(\pi \sqrt{\lambda }\right)=0(\Delta .)$

$\pi \sqrt{\lambda }=\pi n,n\in \mathrm{\mathbb{N}}(*)$

$\sqrt{\lambda }=n(**)$$\lambda =n^2$

$\sin \left(x\sqrt{\lambda }\right)$

$\sin \left(nx\right),from(**)$

$\lambda =n^2$

• We assume the form of a complementary solution involving complex numbers. We evaluate the complementary solution at the two points. Since,$C_2=0,$we are faced again with the zero product property which leads to us knowing that$c_1$can be zero,$(\Delta .)$can be zero, or both can be zero.
• $\sin \left(x\right)$is zero at integer multiples of pi, $\pi ,2\pi ,3\pi ,.$etc. This allows us to set up$(*)$and isolate$\lambda .$Notice that $n$ is not an integer, however. We will only be using the positive answer from$\pm \sqrt{\lambda }$ because we have assumed that$\lambda >0.$
• If the RHS of$(*)$is negative, this would imply that$\lambda <0$ which would contradict us. Also, the RHS cannot be zero because this would imply $\lambda =0:$that another contradiction. So, we assign$n\in \mathrm{\mathbb{N}},$ or,$n=1,2,3,4,\dots$
• We solve for $\lambda$and find the elusive eigenvalue and the function which depends on it.