Eigenvaluse and eigenvectors of differential equation

Question

Vector space $C_2$ consists of every function that's second derivative is continuous in [0,1].
Tensor $A$ on the space $C_2$ is defined as $Ay=\frac {d^2 y}{dt^2}$ where $y(t)$ is a vector on space $C_2$.
When eigenvalues of the tensor $A$ that satisfy the boundary condition $y(0)=y(1)=0$ are $\lambda_n =-n^2 \pi^2$,$n=1,2,\cdots$, how can you show that the corresponding eigenvector $z_n (t)$ is
$z_n (t)=sin n\pi t$ ?
[$Af=\lambda f, so \frac{d^2 y}{dt}-\lambda y=0$] Any help would be much appreciated!!

Answer 1

Well, if

$z_n(t) =\sin n \pi t, \tag{1}$

then

$\dfrac{dz_n(t)}{dt} = n \pi \cos n \pi t, \tag{2}$

and

$\dfrac {d^2 z_n(t)}{dt^2} = - n^2 \pi^2 \sin n \pi t = -n^2 \pi^2 z_n(t), \tag{3}$

which, since

$z_n(0) = \sin(0) = 0 = \sin (n\pi) = z_n(1), \tag{4}$

shows that $z_n(t)$ is an eigenfunction (eigenvector in suitable function space) of

$A = \dfrac{d^2}{dt^2} \tag{5}$

with eigenvalue $\lambda_n = -n^2 \pi^2$ in the subspace of those $y(t) \in C_2$ with $y(0) = y(1) = 0$. That this is the only possibility may be seen by noting that the equation

$\dfrac{d^2y}{dt^2} + n^2 \pi^2 y = 0 \tag{6}$

must have a solution of the form

$y(t) = B_s\sin n \pi t + B_c\cos n \pi t; \tag{7}$

but then

$0 = y(0) = B_c\cos 0 = B_c, \tag{8}$

which rules out the $\cos n \pi t$ term in (7).

And that's how you can show it!

Note added Saturday 7 June 2014 1:17 PM PST: Many thanks to user347612 for his "acceptance" of this answer! This is a question of some note for me, for my answer put me over 11,100 reputation points: reckoning in my personal calendar system based upon the MSE YEAR, in which

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Answer 2

So lets solve the ode $y''-\lambda y=0$, the characteristic equation is $m^2-\lambda=0$, so $m=\sqrt{\lambda}$, with practice of these equations, you will see with the boundary conditions, we consider $\lambda\lt 0$, so we let $\lambda=-\alpha^2$.

Here $m=\pm\alpha i$.

So $y=A\cos\alpha x+ B\sin\alpha x$

$y(0)=A=0$

$y(1)=B\sin\alpha=0$, to avoid a non trivial solution we pick $\alpha_n=n\pi$

Thus $y=B_n\sin n\pi$, and $\lambda_n=-n^2\pi^2$