Question

Sturm Louiville Problem: Eigenvalues and Eigenfunction

Answer 1

what I have so far:
if lambda>0,
\[y(x)=c_1\cos(\sqrt(\lambda)x)+c_2\sin(\sqrt(\lambda)x)\]
applied initial condition, c1 = 0.
Differentiate the equation,
\[y'(x) = \sqrt(\lambda)c_2\cos(\sqrt(\lambda)x\]
applied the initial condition again,
\[\sqrt(\lambda)c_2 \cos(4\sqrt(\lambda) = 0\]
and then I am lost. as I do not how to continue: I try assuming c2 equal to zero. but still after i am not sure what to do.

Answer 2

I prefer to use complex numbers, so the solution above can be rewritten as follows:
\[y\left( x \right) = {c_1}{e^{ - i\sqrt \lambda x}} + {c_2}{e^{i\sqrt \lambda x}}\]
now if we apply the initial conditions, we get:
\[ - i\sqrt \lambda c\left( {{e^{ - i4\sqrt \lambda }} + {e^{i4\sqrt \lambda }}} \right) = 1\]

Answer 3

wherein \(c_2=-c_1=-c\)

Answer 4

the corresponding solution of last equation is:
\[\begin{gathered}
1 + {e^{i8\sqrt \lambda }} = 0 \hfill \\
\hfill \\
{e^{i8\sqrt \lambda }} = - 1 = {e^{i\left( {2n + 1} \right)\pi }},\quad n \in \mathbb{Z} \hfill \\
\end{gathered} \]

Answer 5

so we get:
\[\large {\lambda _n} = \frac{{{\pi ^2}}}{{64}} \cdot {\left( {2n + 1} \right)^2},\quad n \in \mathbb{Z}\]

Answer 6

oops.. I made a typo:
\[y'\left( 4 \right) = - i\sqrt \lambda c\left( {{e^{ - i4\sqrt \lambda }} + {e^{i4\sqrt \lambda }}} \right) = 0\]

Answer 7

nevertheless the solutions are the same

Answer 8

So do you get lambda_n from the equation after applying the initial conditions?

Answer 9

@Michele_Laino i just don't quite understand how do you transform the equation to find lambda.

Answer 10

here are the steps:

\[\begin{gathered}
y'\left( 4 \right) = - i\sqrt \lambda c\left( {{e^{ - i4\sqrt \lambda }} + {e^{i4\sqrt \lambda }}} \right) = 0 \Rightarrow {e^{ - i4\sqrt \lambda }} + {e^{i4\sqrt \lambda }} = 0 \Rightarrow \hfill \\
\hfill \\
\Rightarrow \frac{1}{{{e^{i4\sqrt \lambda }}}} + {e^{i4\sqrt \lambda }} = 0 \Rightarrow \frac{{1 + {e^{i8\sqrt \lambda }}}}{{{e^{i4\sqrt \lambda }}}} = 0 \Rightarrow {e^{i8\sqrt \lambda }} = - 1 \hfill \\
\end{gathered} \]

Answer 11

furthermore, in the complex plane we can write this:
\[ - 1 = {e^{i\left( {2n + 1} \right)\pi }},\quad n \in \mathbb{Z}\]

Answer 12

@Michele_Laino ah... now i see. just basically now i just apply the lambda back to my equation y. which mean my eigenfucntion would be:
\[y(x) = \sin (\pi x/8)\]
is it?

Answer 13

I got this:
\[{y_n}\left( x \right) = {a_n}\sin \left( {\sqrt {{\lambda _n}} x} \right)\]