Question

Find the eigenvalues and eigenfunctions for the given boundry-value problem. Consider only the case$\mathit{\lambda }\mathbf{=}{\mathit{\alpha }}^{\mathbf{4}}\mathbf{,}\mathit{\alpha }\mathbf{>}\mathbf{0}$.( hint: read (ii)in the remarks.)

${\mathit{y}}^{\left(4\right)}\mathbf{-}\mathit{\lambda }\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathit{\pi }\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathit{\pi }\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

$n=0,1,2,.....$The eigenvalues are${\lambda }_n={\left(n+\frac{1}{2}\right)}^4$, and the corresponding eigenfunctions are$y_n\left(x\right)=cos{\left(n+\frac{1}{2}\right)}^{4}x$, where$n=1,2,3,\cdots$

Step-by-step solution

given information

$y^{\left(4\right)}-\lambda y=0,y\text{'}\left(0\right)=0,y\text{'}\text{'}\text{'}\left(0\right)=0,y\left(\pi \right)=0,y\text{'}\text{'}\left(\pi \right)=0$

find differential equation

Fourth order homogeneous differential equation

$\frac{d^{4}y}{d{x}^4}-\lambda y=0$where $\lambda ={\alpha }^4$and $\alpha >0$

$\frac{d^{4}y}{d{x}^4}-{\alpha }^{4}y=0$

Boundary conditions

$y\text{'}\left(0\right)=y\text{'}\text{'}\text{'}\left(0\right)=0$and $y\left(\pi \right)=y\text{'}\text{'}\left(\pi \right)=0$

The characteristic equation of the differential equation is

$$\begin{gathered} m^4-{\alpha }^4=0 \\ \left(m^2-{\alpha }^2\right)\left(m^2+{\alpha }^2\right)=0 \\ \left(m^2-{\alpha }^2\right)\left(m^2-i^2{\alpha }^2\right)0 \end{gathered}$$

$\left(m-\alpha \right)\left(m+\alpha \right)\left(m-i\alpha \right)\left(m+i\alpha \right)=0$

$m=\pm \alpha$and$m=\pm i\alpha$

find first, second and third derivative and apply the first boundary conditions

The general solution, which has the form, has two unique real roots and two complex ones.

$y\left(x\right)=\underset{\text{corresponding the real roots}}{\underbrace{c_1\cosh \alpha x+c_2\sinh \alpha x}}+\underset{\text{corresponding the complex roots}}{\underbrace{c_3\sin \alpha x+c_4\cos \alpha x}}\dots \text{(1)}$

We find the first, the second, and the third derivative of (1)

$$\begin{gathered} \frac{dy}{dx}=\alpha c_{1}sinh\alpha x+\alpha c_{2}cosh\alpha x+\alpha c_{3}cos\alpha x-\alpha c_{4}sin\alpha x \\ \frac{d^{2}y}{d{x}^2}={\alpha }^2{c}_{1}cosh\alpha x+{\alpha }^2{c}_{2}sinh\alpha x-{\alpha }^2{c}_{3}sin\alpha x-{\alpha }^2{c}_{4}cos\alpha x...\left(2\right) \end{gathered}$$

$\frac{d^{3}y}{d{x}^3}={\alpha }^3{c}_{1}sinh\alpha x+{\alpha }^3{c}_{2}cosh\alpha x-{\alpha }^3{c}_{3}cos\alpha x+{\alpha }^3{c}_{4}sin\alpha x$

To apply the first boundary conditions,

$\alpha c_2+\alpha c_3=0$for$y\text{'}\left(0\right)=0$

${\alpha }^3{c}_2-{\alpha }^3{c}_3=0$for$y\text{'}\text{'}\text{'}\left(0\right)=0$

to find equation (1) and (2) reduced form and apply second boundary conditions

$c_2=c_3=0$

Reducing the equation (1) and (2)

$$\begin{gathered} \left(x\right)=c_{1}cosh\alpha x+c_{4}cos\alpha x \\ y\text{'}\text{'}\left(x\right)={\alpha }^2{c}_{1}cosh\alpha x-{\alpha }^2{c}_{4}cos\alpha x \end{gathered}$$

To apply second boundary conditions,

$c_{1}cosh\alpha \pi +c_{4}cos\alpha \pi =0...\left(3\right)$for$y\left(\pi \right)=0$

${\alpha }^2{c}_{1}cosh\alpha \pi -{\alpha }^2{c}_{4}cos\alpha \pi =0...\left(4\right)$for$y\text{'}\text{'}\left(\pi \right)=0$

to write two equations in a matrix form

$$\begin{gathered} \left[\cosh \mathrm{\alpha \pi }\cos \mathrm{\alpha \pi } \\ {\mathrm{\alpha }}^2\cosh \mathrm{\alpha \pi }-{\mathrm{\alpha }}^2\cos \mathrm{\alpha \pi }\right] \end{gathered}$$ $$\begin{gathered} \left[c_1 \\ {c}_4\right] \end{gathered}$$ =$$\begin{gathered} \left[0 \\ 0\right] \end{gathered}$$

Which can be written in the form

$AC=0$

If A is a matrix with a non-zero value, it must have an inverse.

$$\begin{gathered} C=A^{-1}0=0\Rightarrow \left[c_1 \\ {c}_4\right]=\left[0 \\ 0\right] \end{gathered}$$

The trivial solution, which is $y\left(x\right)=0$A is a zero matrix, we get the nontrivial solutions,

$$\begin{gathered} \left[\cosh \mathrm{\alpha \pi }\cos \mathrm{\alpha \pi } \\ {\mathrm{\alpha }}^2\cosh \mathrm{\alpha \pi }-{\mathrm{\alpha }}^2\cos \mathrm{\alpha }\mathrm{\pi }\right] \end{gathered}$$=0

$-{\alpha }^2\cosh \alpha \pi \cos \alpha \pi -{\alpha }^2\cosh \alpha \pi \cos \alpha \pi =0$

$$\begin{gathered} -2{\alpha }^2\cosh \alpha \pi \cos \alpha \pi =0 \\ \cosh \alpha \pi \cos \alpha \pi =0 \end{gathered}$$

$cosh\alpha \pi \ne 0$ for $\alpha =m$where m is a real number,

$cos\alpha \pi =0$at${\alpha }_n=n+\frac{1}{2}$ where$n=0,1,2,...$

find eigenvalues and eigenfunctions

$y\left(x\right)=c_{4}cos\alpha x$

The eigenvalues are

${\lambda }_n={\alpha }^4={\left(n+\frac{1}{2}\right)}^4$where $n=0,1,2...$

Eigenfunctions are

$y_n\left(x\right)=cos{\left(n+\frac{1}{2}\right)}^{4}x$

conclusion

the eigenvalues are${\lambda }_n={\left(n+\frac{1}{2}\right)}^4$ , and the corresponding eigenfunctions are , $y_n\left(x\right)=cos{\left(n+\frac{1}{2}\right)}^{4}x$where$n=0,1,2,.....$