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I need to find the eigenvalues and eigenfunctions for the basic strum-liouville problem below with boundary conditions given:

$$y''+\lambda y=0$$ $$y(0)=0, y(2)=0 $$ on[0,2]. I know this is a basic SL form and problem but I am still quite confuse with the idea. This is what I did so far: $$y''+\lambda y=0 $$ $$m^2+ \lambda=0$$ $$y(x)=acos(\sqrt{\lambda} x)+bsin(\sqrt{\lambda} x)$$ Now using our boundary conditions: $y(0)=0=> acos(0)+bsin(0)=> a*1+0$ so a=0 and $y(2)=0=> acos(\sqrt{\lambda} 2)+bsin(\sqrt{\lambda} 2)=>bsin(\sqrt{\lambda} 2)$ because a=0 so $b \neq 0$ thus $$sin(\sqrt{\lambda} 2)=0$$

Don't know what to do from here to find the eigenvalues and eigenfunctions.

geeking4math
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1 Answers1

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$\begin{array}{c} \sin \left( {\sqrt \lambda 2} \right) = 0 = \sin \left( {n\pi } \right)\\ \sqrt \lambda 2 = n\pi \\ \lambda = \frac{{{n^2}{\pi ^2}}}{4} \end{array}$

then, the eigenvalues are

${\lambda _n} = \frac{{{n^2}{\pi ^2}}}{4}$

now, the eigen functions are

${y_n} = \sin \left( {\frac{{n\pi }}{2}} \right)x$

I think this will be helpful for you.

Krishna Srivastav
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  • Hi, how do you know it is equal to $sin(n\pi )$ and how did you get the eigenfunction. A little bit of steps in between please just so I understand clearly. thanks. – geeking4math May 12 '16 at 08:15
  • nvm, got it why it is equal to $sin(n\pi )$ but I still don't understand how you got the eigenfunction – geeking4math May 12 '16 at 10:13
  • you have to substitute the value of lambda in y(x)=c2*sin(lambda)x, as the first constant c1 is vanished (0), – Krishna Srivastav May 12 '16 at 10:45
  • Thanks, quite obvious now haha. Okay, I am also working on how to find the orthogonal and orthonormal to this problem as well. I will post it as another question but hopefully, if you're willing, do you mine helping me again?? – geeking4math May 12 '16 at 10:49