Chapter 11.RP, Problem 1RP

Find all the real eigen-values and eigen-functions for the given eigen value problem.

a. $\begin{array}{l}{y}^{″}+6y^{\prime }+\lambda y=0;\\ y\left(0\right)=0,y\left(1\right)=0\end{array}$

b. $\begin{array}{l}{y}^{″}+\lambda y=0;\\ y\left(0\right)=0,y\left(\pi \right)+2y^{\prime }\left(\pi \right)=0\end{array}$

To determine

(a)

To find:

All the real eigenvalues and eigenfunctions for the eigenvalue problem $y^{″}+6y^{\prime }+\lambda y=0$; $y(0)=0$, $y(1)=0$.

Answer to Problem 1RP

Solution:

The real eigenvalues are ${\lambda }_n=9+n^2{\pi }^2$ and eigenfunctions are $y_n=C_n{e}^{-3x}\sin \left(n\pi x\right)$, here, $n=1,2,3,\mathrm{...}$.

Explanation of Solution

Calculation:

The given eigenvalue problem is,

$y^{″}+6y^{\prime }+\lambda y=0 ...(1)$

The boundary values are given as,

$y(0)=0$, $y(1)=0$.

The auxiliary equation of equation $(1)$ can be written as,

$r^2+6r+\lambda =0 ...(2)$

The nature of roots of equation $(2)$ will depends upon the value of $\lambda$.

Consider, $\lambda <0$

Let $\lambda =-{\mu }^2$

Equation $(2)$ become,

$r^2+6r-{\mu }^2=0$

Roots of the above equation will be given as,

$\begin{array}{rll}r& =& \frac{-6\pm \sqrt{36+4{\mu }^2}}{2}\\ & =& \frac{-6\pm 2\sqrt{9+{\mu }^2}}{2}\\ & =& -3\pm \sqrt{9+{\mu }^2}\end{array}$

The solution is given as follows.

$y=c_1{e}^{\left(-3+\sqrt{9+{\mu }^2}\right)x}+c_2{e}^{\left(-3-\sqrt{9+{\mu }^2}\right)x}$ …$(3)$

Since, $y(0)=0$, thus

$\begin{array}{rll}0& =& c_1{e}^{\left(-3+\sqrt{9+{\mu }^2}\right)0}+c_2{e}^{\left(-3-\sqrt{9+{\mu }^2}\right)0}\\ 0& =& c_1{e}^0+c_2{e}^0\\ 0& =& c_1+c_2\\ c_1& =& -c_2\end{array}$

Again, since $y(1)=0$ thus,

$\begin{array}{rll}0& =& c_1{e}^{\left(-3+\sqrt{9+{\mu }^2}\right)1}+c_2{e}^{\left(-3-\sqrt{9+{\mu }^2}\right)1}\\ 0& =& c_1{e}^{-3}{e}^{\sqrt{9+{\mu }^2}}-c_1{e}^{-3}{e}^{-\sqrt{9+{\mu }^2}}\\ 0& =& c_1{e}^{-3}\left(e^{\sqrt{9+{\mu }^2}}-e^{-\sqrt{9+{\mu }^2}}\right)\\ c_1& =& 0\end{array}$

Thus, $c_2=0$

This implies, there is no solution.

Now, consider, $\lambda =0$

Equation $(2)$ become,

$\begin{array}{l}{r}^2+6r=0\\ r\left(r+6\right)=0\end{array}$

Roots of the above equation will be given as,

$r=0,-6$

The solution is given as follows.

$y=c_1{e}^{-6x}+c_2$ …$(4)$

Since, $y(0)=0$, thus

$\begin{array}{rll}y& =& c_1{e}^{-6\times 0}+c_2\\ 0& =& c_1{e}^0+c_2\\ 0& =& c_1+c_2\\ c_1& =& -c_2\end{array}$

Again, since $y(1)=0$ thus,

$\begin{array}{rll}0& =& c_1{e}^{-6(1)}+c_2\\ 0& =& c_1{e}^{-6}-c_1\\ 0& =& c_1\left(e^{-6}-1\right)\\ c_1& =& 0\end{array}$

Thus, $c_2=0$

This implies, there is no solution.

Now, consider, $\lambda >0$

Let $\lambda ={\mu }^2$

Equation $(2)$ become,

$r^2+6r+{\mu }^2=0$

Roots of the above equation will be given as,

$\begin{array}{rll}r& =& \frac{-6\pm \sqrt{36-4{\mu }^2}}{2}\\ & =& \frac{-6\pm 2i\sqrt{{\mu }^2-9}}{2}\\ & =& -3\pm i\sqrt{{\mu }^2-9}\end{array}$

The solution is given as follows.

$\begin{array}{rll}y& =& c_1{e}^{\left(-3+i\sqrt{{\mu }^2-9}\right)x}+c_2{e}^{\left(-3-i\sqrt{{\mu }^2-9}\right)x}\\ & =& c_1{e}^{-3x}\left(\cos \sqrt{{\mu }^2-9}x+i\sin \sqrt{{\mu }^2-9}x\right)+c_2{e}^{-3x}\left(\cos \sqrt{{\mu }^2-9}x-i\sin \sqrt{{\mu }^2-9}x\right)\\ & =& \left(c_1+c_2\right)e^{-3x}\cos \sqrt{{\mu }^2-9}x+i\left(c_1-c_2\right)e^{-3x}\sin \sqrt{{\mu }^2-9}x\\ & =& C_1{e}^{-3x}\cos \sqrt{{\mu }^2-9}x+C_2{e}^{-3x}\sin \sqrt{{\mu }^2-9}x\end{array}$ …$(5)$

Since, $y(0)=0$, thus

$\begin{array}{rll}0& =& C_1{e}^{-3(0)}\cos \sqrt{{\mu }^2-9}(0)+C_2{e}^{-3(0)}\sin \sqrt{{\mu }^2-9}(0)\\ 0& =& C_1{e}^0\cos (0)+C_2{e}^0\sin (0)\\ C_1& =& 0\end{array}$

Again, since $y(1)=0$ thus,

$\begin{array}{rll}0& =& (0)e^{-3(1)}\cos \sqrt{{\mu }^2-9}(1)+C_2{e}^{-3(1)}\sin \sqrt{{\mu }^2-9}(1)\\ C_2{e}^{-3}\sin \sqrt{{\mu }^2-9}& =& 0\end{array}$

Since, $C_2\ne 0$.

Thus,

$\begin{array}{rll}\sin \sqrt{{\mu }^2-9}& =& 0\\ \sqrt{{\mu }^2-9}& =& n\pi \\ {\mu }^2-9& =& n^2{\pi }^2\\ {\mu }^2& =& 9+n^2{\pi }^2\end{array}$

Here, $n=1,2,3,\mathrm{...}$

The real eigenvalues are as follows,

$\begin{array}{rll}{\lambda }_n& =& {\mu }_n^2\\ & =& 9+n^2{\pi }^2\end{array}$

Substitute $C_1=0$ in equation $(5)$ as follows,

$\begin{array}{rll}y& =& (0)e^{-3x}\cos \sqrt{{\mu }^2-9}x+C_2{e}^{-3x}\sin \sqrt{{\mu }^2-9}x\\ & =& C_2{e}^{-3x}\sin \sqrt{{\mu }^2-9}x\end{array}$

The real eigenfunctions are given as follows,

$y_n=C_n{e}^{-3x}\sin \left(n\pi x\right)$

Here, $n=1,2,3,\mathrm{...}$

Therefore, the real eigenvalues are ${\lambda }_n=9+n^2{\pi }^2$ and eigenfunctions are $y_n=C_n{e}^{-3x}\sin \left(n\pi x\right)$, here, $n=1,2,3,\mathrm{...}$.

Conclusion:

Hence, the real eigenvalues are ${\lambda }_n=9+n^2{\pi }^2$ and eigenfunctions are $y_n=C_n{e}^{-3x}\sin \left(n\pi x\right)$, here, $n=1,2,3,\mathrm{...}$.

To determine

(b)

To find:

All the real eigenvalues and eigenfunctions for the eigenvalue problem $y^{″}+\lambda y=0$; $y(0)=0$, $y\left(\pi \right)+2y^{\prime }\left(\pi \right)=0$.

Answer to Problem 1RP

Solution:

The real eigenvalues are ${\lambda }_n={\mu }_n^2$, here, $\tan {\mu }_n\pi =-2{\mu }_n$, $n=1,2,3,\mathrm{...}$ and the eigenfunctions are $y_n=C_n\sin \left({\mu }_{n}x\right)$, $n=1,2,3,\mathrm{...}$.

Explanation of Solution

Calculation:

The given eigenvalue problem is,

$y^{″}+\lambda y=0$ (6)

The boundary values are given as,

$y(0)=0$, $y\left(\pi \right)+2y^{\prime }\left(\pi \right)=0$.

The auxiliary equation of equation $(6)$ can be written as,

$r^2+\lambda =0$ (7)

The nature of roots of equation $(7)$ will depends upon the value of $\lambda$.

Consider, $\lambda <0$

Let $\lambda =-{\mu }^2$

Equation $(7)$ become,

$r^2-{\mu }^2=0$

Roots of the above equation will be given as,

$r=\pm \mu$

The solution is given as follows.

$\begin{array}{rll}y& =& c_1{e}^{\mu x}+c_2{e}^{-\mu x}\\ & =& c_1\left(\cosh \mu x+\sinh \mu x\right)+c_2\left(\cosh \mu x-\sinh \mu x\right)\\ & =& \left(c_1+c_2\right)\cosh \mu x+\left(c_1-c_2\right)\sinh \mu x\\ & =& C_1\cosh \mu x+C_2\sinh \mu x& \text{…}(8)\end{array}$

Since, $y(0)=0$, thus

$\begin{array}{rll}0& =& C_1\cosh \mu (0)+C_2\sinh \mu (0)\\ 0& =& C_1(1)+C_2(0)\\ C_1& =& 0\end{array}$

Again, since $y\left(\pi \right)+2y^{\prime }\left(\pi \right)=0$ thus,

$\begin{array}{rll}{C}_1\cosh \left(\mu \pi \right)+C_2\sinh \left(\mu \pi \right)+2\left(C_1\mu \sinh \left(\mu \pi \right)+C_2\mu \cosh \left(\mu \pi \right)\right)& =& 0\\ (0)\cosh \left(\mu \pi \right)+C_2\sinh \left(\mu \pi \right)+2\left((0)\mu \sinh \left(\mu \pi \right)+C_2\mu \cosh \left(\mu \pi \right)\right)& =& 0\\ C_2\sinh \left(\mu \pi \right)+2C_2\mu \cosh \left(\mu \pi \right)& =& 0\\ C_2\left(\sinh \left(\mu \pi \right)+2\mu \cosh \left(\mu \pi \right)\right)& =& 0\end{array}$

Since, $C_2\ne 0$

Thus,

$\begin{array}{rll}\sinh \left(\mu \pi \right)+2\mu \cosh \left(\mu \pi \right)& =& 0\\ \tanh \left(\mu \pi \right)& =& -2\mu \end{array}$

This implies, there is only one solution, having positive value to this problem and it is denoted by ${\mu }_0$ and it has only one negative eigenvalue ${\lambda }_0=-{\mu }^2$.

Here, $\tanh \left({\mu }_0\pi \right)=-2{\mu }_0$

Substitute $0$ for $C_1$ in equation $(8)$ as follows.

$\begin{array}{rll}y& =& (0)\cosh \mu x+C_2\sinh \mu x\\ & =& C_2\sinh \mu x\end{array}$

Thus, corresponding eigenfunctions are,

$y_0=C_0\sinh {\mu }_{0}x$

Now, consider, $\lambda =0$

Equation $(7)$ become,

$r^2+0=0$

Roots of the above equation will be given as,

$r=0,0$

The solution is given as follows.

$y=c_{1}x+c_2$ …$(9)$

Since, $y(0)=0$, thus

$\begin{array}{rll}0& =& c_1(0)+c_2\\ c_2& =& 0\end{array}$

Again, since $y\left(\pi \right)+2y^{\prime }\left(\pi \right)=0$ thus,

$c_1\left(\pi \right)+c_2+2\left(c_1\right)=0$

since, $c_2=0$

$\begin{array}{rll}{c}_1\left(\pi \right)+(0)+2\left(c_1\right)& =& 0\\ c_1\left(2+\pi \right)& =& 0\\ c_1& =& 0\end{array}$

This implies, there is no solution.

Now, consider, $\lambda >0$

Let $\lambda ={\mu }^2$

Equation $(7)$ become,

$r^2+{\mu }^2=0$

Roots of the above equation will be given as,

$\begin{array}{rll}r& =& \sqrt{-{\mu }^2}\\ & =& \pm \mu i\end{array}$

The solution is given as follows.

$\begin{array}{rll}y& =& c_1{e}^{\mu ix}+c_2{e}^{-\mu ix}\\ & =& c_1\left(\cos \mu x+i\sin \mu x\right)+c_2\left(\cos \mu x-i\sin \mu x\right)\\ & =& \left(c_1+c_2\right)\cos \mu x+i\left(c_1-c_2\right)\sin \mu x\\ & =& C_1\cos \mu x+C_2\sin \mu x\end{array}$ …$\left(10\right)$

Since, $y(0)=0$, thus

$\begin{array}{rll}0& =& C_1\cos \mu (0)+C_2\sin \mu (0)\\ 0& =& C_1\cos (0)+0\\ C_1& =& 0\end{array}$

Again, since $y\left(\pi \right)+2y^{\prime }\left(\pi \right)=0$ thus,

$\begin{array}{rll}{C}_1\cos \mu \pi +C_2\sin \mu \pi +2\left(-C_1\mu \sin \mu \pi +\mu C_2\cos \mu \pi \right)& =& 0\\ (0)\cos \mu \pi +C_2\sin \mu \pi +2\left(-(0)\mu \sin \mu \pi +C_2\mu \cos \mu \pi \right)& =& 0\\ C_2\sin \mu \pi +2C_2\mu \cos \mu \pi & =& 0\\ C_2\left(\sin \mu \pi +2\mu \cos \mu \pi \right)& =& 0\end{array}$

Since, $C_2\ne 0$.

Thus,

$\begin{array}{rll}\sin \mu \pi +2\mu \cos \mu \pi & =& 0\\ \tan \mu \pi & =& -2\mu \end{array}$

The eigenvalues are as follows,

$\begin{array}{rll}{\lambda }_n& =& {\mu }_n^2\\ \tan {\mu }_n\pi & =& -2{\mu }_n\end{array}$

Here, $n=1,2,3,\mathrm{...}$

Substitute $0$ for $C_1$ in equation $\left(10\right)$ as follows,

$\begin{array}{rll}y& =& (0)\cos \mu x+C_2\sin \mu x\\ & =& C_2\sin \mu x\end{array}$

The real eigenfunctions are given as follows,

$y_n=C_n\sin \left({\mu }_{n}x\right)$

Here, $n=1,2,3,\mathrm{...}$

Therefore, the real eigenvalues are ${\lambda }_n={\mu }_n^2$, here, $\tan {\mu }_n\pi =-2{\mu }_n$, $n=1,2,3,\mathrm{...}$ and the eigenfunctions are $y_n=C_n\sin \left({\mu }_{n}x\right)$, $n=1,2,3,\mathrm{...}$.

Conclusion:

Hence, the real eigenvalues are ${\lambda }_n={\mu }_n^2$, here, $\tan {\mu }_n\pi =-2{\mu }_n$, $n=1,2,3,\mathrm{...}$ and the eigenfunctions are $y_n=C_n\sin \left({\mu }_{n}x\right)$, $n=1,2,3,\mathrm{...}$.