Consider the equation in the form $X + \alpha X = 0$, with Dirichlet boundary conditions at $x = 0$ and $x = π$.
a) Multiply the equation by X and integrate from 0 to π, then integrate the first term, $XX$, by parts to obtain the identity
$$\alpha \int_0^{\pi}X^2dx=-[X'X]\biggr|_0^{\pi} + \int_0^{\pi} (X')^2dx$$
b) Apply the Boundary Conditions to conclude that either $\alpha >0$ or ($\alpha = 0$ and X is constant)
I have obtained part a) and when applying the boundary conditions $X(0)=0=X(\pi )$ I get the following equation
$$\alpha \int_0^{\pi}X^2dx = \int_0^{\pi}(X')^2dx$$
It is clear to me why ($\alpha = 0$ and X constant) satisfy this equation but I'm having trouble explicitly showing that this equation is also true for $\alpha > 0$. Could someone walk me through this conclusion? Thanks!
Given the equation
$X'' + \alpha X = 0, \tag 1$
with the boundary conditions
$X(0) = X(\pi) = 0, \tag 2$
we may, as pointed out by our OP meff11 in the text of the question itself, multiply through by $X$ and obtain
$XX'' + \alpha X^2 = 0, \tag 3$
which may be written
$XX'' = -\alpha X^2, \tag 4$
and then integrated over the interval $[0, \pi]$ to yield
$\displaystyle \int_0^\pi XX'' \; dx = -\alpha \int_0^\pi X^2 \; dx; \tag 5$
observing that
$(XX')' = (X')^2 + XX'', \tag 6$
we write
$\displaystyle \int_0^\pi (XX')' \; dx - \int_0^\pi (X')^2 \; dx = \int_0^\pi XX'' \; dx, \tag 7$
and
$\displaystyle \int_0^\pi (XX')' \; dx = (XX'\vert_0^\pi = X(\pi)X'(\pi) - X(0)X'(0) = 0, \tag 8$
by virtue of the boundary conditions (2); (7) then becomes
$-\displaystyle \int_0^\pi (X')^2 \; dx = \int_0^\pi XX'' \; dx, \tag 9$
and thus (5):
$-\displaystyle \int_0^\pi (X')^2 \; dx = -\alpha \int_0^\pi X^2 \; dx, \tag{10}$
whence
$\displaystyle \int_0^\pi (X')^2 \; dx = \alpha \int_0^\pi X^2 \; dx, \tag{11}$
the desired result. $OE\Delta$.
(11) holds for any function $X$ and real scalar $\alpha$ satisfying equation (1) with boundary conditions (2), and may be exploited to investigate the relationship 'twixt values of $\alpha$ and properties of the function $X$. To this end we make some elementary observations about the integrals occurring in (11): first,
$\displaystyle \int_0^\pi X^2 \; dx \ge 0, \tag{12}$
and takes the value $0$ if and only if
$X(x) = 0, \; 0 \le x \le \pi \tag{13}$
(note we are assuming
$X(x) \in C^2[0, \pi] \tag{14}$
throughout this discussion); second,
$\displaystyle \int_0^\pi (X')^2 \; dx \ge 0, \tag{15}$
and this integral vanishes precisely when
$X'(x) = 0, \; 0 \le x \le \pi, \tag{16}$
i.e., when $X(x)$ is constant.
Now suppose that
$\alpha < 0 \tag{17}$
for some
$X \ne 0; \tag{18}$
then the right-hand side of (11) becomes negative, but the left-hand side remains non-negative, a contradiction. Therefore $\alpha$ cannot be negative for non-trivial (i.e. non-vanishing) $X$; (17) and (18) cannot simultaneously hold.
If we relax condition (17) and replace it with
$\alpha \ge 0, \tag{19}$
we still have two cases to consider, according to whether $\alpha$ is zero or not. When
$\alpha = 0, \tag{20}$
the left-hand integral vanishes, and as we have seen above, ca. (15)-(16), this forces $X' = 0$, that is, $X$ is constant; then in accord with (2) we see that
$X(x) = 0, \; 0 \le x \le \pi, \tag{21}$
i.e., (20) forces $X$ to vanish as well.
The preceding discussion in fact shows that
$\alpha \le 0 \tag{22}$
implies (21); (1)-(2) has only the trivial solution (21) in this case.
For
$\alpha > 0, \tag{23}$
there are in fact non-trivial solutions to (1), (2); however, they only exist for certain, isolated $\alpha$. This is perhaps most easily seen via explicit presentation of all solutions to (1); it is well-known, and easily seen by direct computation, that functions of the form
$X(x) = A\cos \sqrt \alpha x + B\sin\sqrt \alpha x \tag{24}$
satisfy (1); furthermore,
$A = A\cos \sqrt \alpha \cdot 0 + B\sin\sqrt \alpha \cdot 0 = X(0) = 0, \tag{25}$
and thus (23) becomes
$X(x) = B\sin\sqrt \alpha x, \tag{26}$
whence, from (2),
$B\sin \sqrt \alpha \cdot \pi = X(\pi) = 0; \tag{27}$
if $X(x)$ is non-trivial we must have
$B \ne 0, \tag{28}$
whence
$\sin \sqrt \alpha \cdot \pi = 0, \tag{29}$
only possible when
$\sqrt \alpha \in \Bbb Z; \tag{30}$
thus,
$\alpha = (\sqrt \alpha)^2 \in \Bbb Z_{\geq 0}, \tag{31}$
the set of non-negative integers. We may ensure $X(x)$ as in (26)-(28) is non-trivial by taking
$\alpha = (\sqrt \alpha)^2 \in \Bbb Z_{> 0}, \tag{32}$
i.e., $\alpha$ must be a positive square.
We may in fact proceed further in a slightly different direction and multiply (1) by $X'$ instead of $X$:
$X'X'' + \alpha XX' = 0; \tag{12}$
we then use the identities
$X'X'' = \dfrac{1}{2}((X')^2)', \tag{13}$
$XX' = \dfrac{1}{2}(X^2)'; \tag{14}$
(12) thus becomes
$\dfrac{1}{2}((X')^2)' + \dfrac{1}{2}\alpha(X^2)' = 0, \tag{15}$
or
$((X')^2)' + \alpha(X^2)' = 0, \tag{16}$
or
$((X')^2 + \alpha X^2)' = 0; \tag{17}$
now for any $y \in [0, \pi]$ we have
$(X'(y))^2 + \alpha X^2(y) - ((X'(0))^2 + \alpha X^2(0))$ $= \displaystyle \int_0^y ((X'(x))^2 + \alpha X^2(x))' \; dx = \int_0^y 0 \; dy = 0; \tag{18}$
that is,
$(X'(y))^2 + \alpha X^2(y) = (X'(0))^2 + \alpha X^2(0) = C, \; \text{a constant}; \tag{19}$
integrating this equation 'twixt $0$ and $\pi$ yields
$\displaystyle \int_0^\pi (X'(x))^2 + \alpha X^2(x)) \; dx = C\pi, \tag{20}$
which in the light of our previous result (11) allows us to conclude that
$\displaystyle \int_0^\pi (X'(x))^2 \; dx = \alpha \int_0^\pi \alpha X^2(x)) \; dx = \dfrac{C\pi}{2}. \tag{21}$