Chapter 3.8, Problem 1P

Program Plan Intro

Program Description: Purpose of problem is to determine whether $\lambda =0$ is an Eigen value and then find the positive eigenvalues and associated Eigen functions.

Explanation of Solution

Given information:

The second order differential equation is,

  $y^{″}+\lambda y=0$

The value of first order differential equation at $0$ is,

  $y^{\prime }\left(0\right)=0$

The value of the function at $1$ is,

  $y\left(1\right)=0$

Explanation:

Consider $\lambda =0$ . Then, the given differential equation can be expressed as follows,

  $y^{″}=0$

The characteristic equation can be expressed as follows,

  $r^2=0$

The roots of the equation are as follows,

  $r_1=0\text{ or }{r}_2=0$

Therefore, the general solution of the differential equation can be expressed as,

  $y\left(x\right)=c_{2}x+c_1$

Apply the initial condition in equation $y\left(x\right)=c_{2}x+c_1$ .

  $c_1=0\text{ or }{c}_2=0$

Therefore, the value of $y\left(x\right)$ can be expressed as follows,

  $\begin{array}{c}y\left(x\right)=0\times x+0\\ =0\end{array}$

Since, $y\left(x\right)=0$ . Therefore, $\lambda =0$ is not an Eigen value.

Now,consider $\lambda <0$ .

Then, the given differential equation can be expressed as follows,

  $\lambda =-{\alpha }^2$

The characteristic equation can be expressed as follows,

  $r^2-{\alpha }^2=0$

The roots of the equation are as follows,

  $r_1=\alpha \text{ or }{r}_2=-\alpha$

Therefore, the general solution of the differential equation can be expressed as,

  $y\left(x\right)=c_1{e}^{\alpha x}+c_2{e}^{-\alpha x}$

Differentiate the above equation with respect to $x$ .

  $y^{\prime }\left(x\right)=\alpha c_1{e}^{\alpha x}-\alpha c_2{e}^{-\alpha x}$

Apply the initial condition $y^{\prime }\left(0\right)=0$ .

  $\begin{array}{c}{y}^{\prime }\left(0\right)=\alpha c_1-\alpha c_2\\ 0=\alpha c_1-\alpha c_2\\ \alpha c_1=\alpha c_2\\ c_1=c_2\end{array}$

Apply the initial condition $y\left(1\right)=0$ .

  $\begin{array}{c}y\left(1\right)=c_1+c_2\\ 0=c_1+c_2\\ 0=c_1+c_1\\ 0=2c_1\end{array}$

Therefore, the value of $c_1$ and $c_2$ is given below.

  $c_1=0\text{ or }{c}_2=0$

Then the value of $y\left(x\right)$ can be obtained as follows,

  $y\left(x\right)=0$

Since, $y\left(x\right)=0$ . Therefore, $\lambda <0$ is not an Eigen value.

Now,consider $\lambda >0$ .

Then, the given differential equation can be expressed as follows,

  $\lambda ={\alpha }^2$

The characteristic equation can be expressed as follows,

  $r^2+{\alpha }^2=0$

The roots of the equation are as follows,

  $r_1=\alpha i\text{ or }{r}_2=-\alpha i$

Therefore, the general solution of the differential equation can be expressed as,

  $y\left(x\right)=c_1\cos \alpha x+c_2\sin \alpha x$ ....... (1)

Differentiate the above equation with respect to $x$ .

  $y^{\prime }\left(x\right)=-\alpha c_1\sin \alpha x+\alpha c_2\cos \alpha x$ ....... (2)

Apply the initial condition $y^{\prime }\left(0\right)=0$ in equation (2) as,

  $\begin{array}{c}{y}^{\prime }\left(0\right)=-\alpha c_1\sin \left(\alpha \cdot 0\right)+c_2\alpha \cos \left(\alpha \cdot 0\right)\\ 0=\alpha c_2\\ 0=c_2\end{array}$

Apply the initial condition $y\left(1\right)=0$ in equation (1) as,

  $\begin{array}{c}y\left(1\right)=c_1\cos \left(\alpha \cdot 1\right)\\ 0=c_1\cos \left(\alpha \right)\end{array}$

Since, the value of $c_1$ cannot be zero. Therefore,

  $\begin{array}{c}0=\cos \alpha \\ \alpha =\frac{n\pi }{2}\\ {\alpha }^2=\frac{n^2{\pi }^2}{4}\end{array}$

For all odd value $n$ .

Hence, the eigenvalues for the given equation are as follows,

  ${\lambda }_n=\frac{n^2{\pi }^2}{4}\text{for }n=1,3,5,\dots$

Conclusion:

Thus, the Eigen functions corresponding to the eigenvalues are $y_n=\cos \left(\frac{n\pi }{2}\right)x$ .