Solve the boundary value problem (C.S.I.R) Consider the boundary value problem
$$ -u''(x) = \pi^2 u(x) , \ \ x \in (0,1) \ \ and $$
$$ u(0) = u(1) = 0$$
if $u \ \ and \ \ u'$ are continous on $[0,1]$, then


*

*$\int_0^1 u^3(x) dx = 0$

*$u'^2(x) + \pi^2 u^2(x) = u'^2(0)$

*$u'^2(x) + \pi^2 u^2(x) = u'^2(1)$

*$\int_0^1 u^2(x) dx = \frac{1}{\pi^2} \int_0^1 u'^2(x) dx$
Solve the given equation, we get $u(x) = c_1 cos(\pi x) + c_2 sin(\pi x)$ and use boundary condition we get $u(x) = c_2 sin (\pi x)$ 
Thus $u'(x) = \pi c_2 cos(\pi x)$
$u'^2(x) + \pi^2 u^2(x) = \pi^2 c_2^2 = u'^2(0) = u'^2(1) $, So (2) and (3) are true
Please help me tosolve (1) and (4).
Thank you
 A: For (1) we see that $(\cos(x)+i\sin(x))^3=\cos^3(x)+3i\cos^2(x)\sin (x)-3\cos(x)\sin^2(x)-i\sin^3(x)$
$=\cos(3x)+i\sin(3x)$
Comparing imaginary parts gives us:
$3(1-\sin^2(x))\sin(x)-\sin^3(x)=\sin(3x)$, then:
$\sin^3(x)=(3/4)\sin(x)-(1/4)\sin(3x)$, so
$c_2^3\sin^3(\pi x)=(3/4)\sin(\pi x)-(1/4)\sin(3\pi x)$
$\int_0^1c_2^3\sin^3(\pi x)dx=\int_0^1(3/4)\sin(\pi x)-(1/4)\sin(3\pi x)dx=((1/12\pi)\cos(3\pi x)-(3/4\pi)\cos (\pi x))|^1_0=-(1/6\pi)+(3/2\pi)\ne 0$
(4) this one is interesting, rearranging we see that:
$\pi^2=\frac{\int_0^1(u')^2dx}{\int_0^1u^2dx}$
This quotient is called rayleighs quotient, and it is minimised by the smallest eigenvalue of the dirichlet eigenvalue problem, which is the equation you have solved :)
Lets try to prove it, multiply through by $u$:
$-u''u=\pi^2u^2$
$-\int_0^1u''udx=\pi^2\int_0^1u^2dx$
$-(u'u)|^1_0+\int_0^1(u')^2dx=\pi^2\int_0^1u^2dx$ (integration by parts)
$0+\int_0^1(u')^2dx=\pi^2\int_0^1u^2dx$
$\int_0^1(u')^2dx=\pi^2\int_0^1u^2dx$
A: for (4) It may help that integrating from 0 to $\pi$ means integrating over the whole period of $\sin^2(x)$. Since $\cos$ is only an phase-shiftet $\sin$ $\int_0^1\cos(\pi x)\mathrm dx= \int_0^1\cos(\pi x)\mathrm dx$. (1) seems to be false because $\sin(\pi x) > 0~  \forall x\in [0,1]$ and so the integral would be $>0$ too. Maybe the integral boundarys are $-1,1$ ?
A: Additional Information: For (2) and (3), without solving, notice that
$$\{u'^2\}'=2u'u''=2u'.-\pi u=-2\pi uu'$$
and
$$\{u^2\}'=2uu'$$
and both together gives
$$\{u'^2\}'=-\pi\{u^2\}'$$
Integrating from $0$ to $x$ and $x$ to $1$ gives 2,3.
