Question

In Problem 31 suppose . If the shaft is fixed at both ends then the boundary conditions are

$\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}^{\mathbf{\backslash }\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}^{\mathbf{\backslash }\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}$

(a) Show that the eigenvalues ${\mathit{\lambda }}_{\mathbf{n}}\mathbf{=}{\mathit{\alpha }}_{\mathbf{n}}^{\mathbf{4}}$ are defined by the positive roots of $\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{\alpha }\mathbf{=}\mathbf{1}$. [Hint: See the instructions to Problems 21 and 22.]

(b) Show that the eigenfunctions are

${\mathit{y}}_{\mathbf{n}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\left(-sin{\alpha }_n+sinh{\alpha }_n\right)\left(cos{\alpha }_{n}x-cosh{\alpha }_{n}x\right)\mathbf{+}\left(cos{\alpha }_n-cosh{\alpha }_n\right)\left(sin{\alpha }_{n}x-sinh{\alpha }_{n}x\right)$

Short answer

$\left(a\right)cosh\alpha cos\alpha =1$

(b)$y_n\left(x\right)=\left(-sin{\alpha }_n+sinh{\alpha }_n\right)\left(cos{\alpha }_{n}x-cosh{\alpha }_{n}x\right)+\left(cos{\alpha }_n-cosh{\alpha }_n\right)\left(sin{\alpha }_{n}x-sinh{\alpha }_{n}x\right)$

Step-by-step solution

To Find the nontrivial solution

(a) Recall Problem 31, the general solution is

$y\left(x\right)=c_{1}cosh\alpha x+c_{2}sinh\alpha x+c_{3}cos\alpha x+c_{4}sin\alpha x.$

The initial conditions $y\left(0\right)=y^{\backslash \mathrm{cent}}\left(0\right)=0$gives $c_1+c_3=0$and $c_2+c_4=0$so

$y\left(x\right)=c_3(cos\alpha x-cosh\alpha x)+c_4(sin\alpha x-sinh\alpha x)$$y\left(x\right)=c_3(cos\alpha x-cosh\alpha x)+c_4(sin\alpha x-sinh\alpha x)$

The conditions $y\left(1\right)=y^{\backslash \mathrm{cent}}\left(1\right)=0$gives

$c_3(cos\alpha -cosh\alpha )+c_4(sin\alpha -sinh\alpha )=0$and

$c_3(-sin\alpha -sinh\alpha )+c_4(cos\alpha -cosh\alpha )=0$

The system has a nontrivial solution if

$$\begin{gathered} \left|\cos \alpha -\cos h\alpha \sin \alpha -\sin h\alpha \\ -(\sin \alpha +\sin h\alpha )\cos \alpha -\cos h\alpha \right|=0 \end{gathered}$$and so

$(cos\alpha -cosh\alpha {)}^2+(sin\alpha +sinh\alpha )(sin\alpha -sinh\alpha )=0$

Therefore,

$\left(co{s}^2\alpha +si{n}^2\alpha \right)+\left(cos{h}^2\alpha -sin{h}^2\alpha \right)=2cos\alpha cosh\alpha$and so

$cosh\alpha cos\alpha =1$

Find the eigenfunctions

(b) Since

$c_3(cos\alpha -cosh\alpha )+c_4(sin\alpha -sinh\alpha )=0$

Hence,

$c_3=c_4\frac{(-sin\alpha +sinh\alpha )}{(cos\alpha -cosh\alpha )}$Thus the eigenfunctions are given by

$y_n\left(x\right)=c_4\frac{(-sin\alpha +sinh\alpha )}{(cos\alpha -cosh\alpha )}(cos\alpha x-cosh\alpha x)+c_4(sin\alpha x-sinh\alpha x)$since the solutions are linearly independent, so we can multiply by the nonzero value to$(cos\alpha -cosh\alpha )/c_4$ get

(b)$y_n\left(x\right)=\left(-sin{\alpha }_n+sinh{\alpha }_n\right)\left(cos{\alpha }_{n}x-cosh{\alpha }_{n}x\right)+\left(cos{\alpha }_n-cosh{\alpha }_n\right)\left(sin{\alpha }_{n}x-sinh{\alpha }_{n}x\right)$

Final proof 

$\left(a\right)cosh\alpha cos\alpha =1$

(b)$y_n\left(x\right)=\left(-sin{\alpha }_n+sinh{\alpha }_n\right)\left(cos{\alpha }_{n}x-cosh{\alpha }_{n}x\right)+\left(cos{\alpha }_n-cosh{\alpha }_n\right)\left(sin{\alpha }_{n}x-sinh{\alpha }_{n}x\right)$