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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.

  • 1
    You'll have cosines, not sines, because you have the derivatives vanishing at the endpoints. When the function itself vanishes at the endpoints you get sines. As for your integral calculations, notice that you forgot the $\pi$! With the $\pi$ present the numerators vanish, while for $m \neq n$ the denominators don't vanish. – 2017-01-19
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    oh yeah you are right, Thank you so much, now is all my answer true? – 2017-01-19
  • 1
    Seems reasonable. You have another thing to check: are there any eigenfunctions with $\lambda<0$ satisfying these boundary conditions? – 2017-01-19

1 Answers 1

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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 $y_n=\cos\left(n\pi x\right)$