Find the eigenvalues and eigenfunctions for
$y^{\prime \prime}+\lambda y=0$
with the boundary conditions
$y^{\prime} (0)=0$ , $y^{\prime} (1)=0$
then prove that the eigen functions are orthogonal, and find normalized eigenfunctions "I don't sure if that it's name in English, I don't speak English well:("
my try :
if $\lambda =0$ then $y=ax+b$ and $y^{\prime} (0)=y^{\prime} (1)=0$ if $a=0$. Hence the constant function is an eigenfunction corresponding to the eigenvalue $0$.an eigenvalue for this problem cannot be negative. if $\lambda >0$ then
$y(x)=c_1 cos(x \sqrt{\lambda}) + c_2 sin (x \sqrt{\lambda})$
$y'(x)=-c_1 \sqrt{\lambda} sin (x \sqrt{\lambda})+c_2 \sqrt{\lambda} cos(x \sqrt{\lambda})$
$y'(0)=0$ if $c_2=0$
then $y(x)=c_1 cos(x \sqrt{\lambda})$
and $y'(x)=-c_1 \sqrt{\lambda} sin (x \sqrt{\lambda})$
$y'(1)=-c_1 \sqrt{\lambda} sin ( \sqrt{\lambda})=0$ if $sin ( \sqrt{\lambda})=0$
$sin ( \sqrt{\lambda})=sin ( n\pi)$
so the eigenvalues $\lambda_n={(n\pi)}^2$
but what about eigenfunctions,
I think $y_n(x)=B_n cos (n\pi x)$ but I found this $y_n(x)=B_n sin (n\pi x)$ in internet ?
if eigenfunctions are $y_n(x)=B_n cos (n\pi x)$
then
$\int_{0}^{1} ({B_n cos (n\pi x)}{B_m cos (m\pi x)}) dx$
$={{B_n B_m}\over 2} \int_{0}^{1} (cos (n+m)x + cos (n-m)x) dx $
$={{B_n B_m}\over 2} ({sin ((m-n)\pi x) \over {m-n}} + {sin ((m+n)\pi x) \over {m+n}})_{0}^{1}$
$={{B_n B_m}\over 2} ({sin ((m-n)\pi ) \over {m-n}} + {sin ((m+n)\pi ) \over {m+n}})=0$
and for normalized eigenfunctions
$\int_{0}^{1} {B_n^2 cos^2 (n\pi x)} dx=1$
${{B_n^2}\over 2} \int_{0}^{1} (1+cos(2n \pi x) dx=1$
${{B_n^2}\over 2} ({x+{sin(2n \pi x)}\over {2n}})_{0}^{1}=1$
$B_n={\sqrt 2}$
so $y_n={\sqrt 2} cos (n \pi x)$
please help me 🙂
Thanks.
Best Answer
Finding $\lambda$. Once this is clear, I'll keep adding onto this to show how the rest of the problem works.
Claim 1: $\lambda <0$.
Then $y(x)=C_1e^{\sqrt{\lambda} x}+C_2e^{-\sqrt{\lambda} x}$ and $y'(x)=C_1\sqrt{\lambda}e^{\sqrt{\lambda} x}-C_2\sqrt{\lambda}e^{-\sqrt{\lambda} x}.$
Using the boundary conditions: $0=y'(0)=C_1-C_2$. Say $C_1=C_2$
Then for the other boundary condition:
$0=y'(1)=C_2\sqrt{\lambda}e^{\sqrt{\lambda}}-C_2\sqrt{\lambda}e^{-\sqrt{\lambda}}$
$C_2\sqrt{\lambda}e^{-\sqrt{\lambda}}=C_2\sqrt{\lambda}e^{\sqrt{\lambda}}$
$1=e^{2\sqrt{\lambda}}$
$0=2\sqrt{\lambda}$
$0=\lambda$
Claim 2: $\lambda=0$ This is the easy one.
Claim 3: $\lambda >0$
$y(x)=C_1\cos(\sqrt{\lambda}x)+C_2\sin(\sqrt{\lambda}x)$
$y'(x)=-C_1\sqrt{\lambda}\sin(\sqrt{\lambda}x)+C_2\sqrt{\lambda}\cos(\sqrt{\lambda}x)$
$0=y'(0)=C_2 \Rightarrow C_2=0$
$0=y'(1)=C_1\sqrt{\lambda}\sin(\sqrt{\lambda})$
The above shows that $\lambda_n=(n\pi)^2$ so $y_n=\cos\left(n\pi x\right)$