How prove this inequality with $\int_{0}^{1}(f(x))^2dx\le\frac{1}{\theta^4}\int_{0}^{1}(f''(x))^2dx$ let $f(x)$ is smooth function，and such $$f(0)=f'(0)=f(1)=f'(1)=0$$  show that
$$\int_{0}^{1}(f(x))^2dx\le\dfrac{1}{\theta^4}\int_{0}^{1}(f''(x))^2dx$$
where the $\theta $ is  smallest positive root
$$\cos{\theta}\cosh{\theta}=1$$,and the $\dfrac{1}{\theta^4}$ is best coefficient.
I know that this form of inequality is commonly called Wirtinger's inequality：links,but this  my question conditions are quite many, and it doesn’t seem feasible to use a proof method like that Wirtinger's  inequality,
 A: Using functional analysis, consider the expression
$$S[f]=\int_0^1 {\rm d}t \left\{ f''(t)^2 - \theta^4 f(t)^2 \right\}$$
and we want to show $S[f]\geq 0$ given the stated conditions. For now we assume $\theta$ to be arbitrary, and we will see that it has to satisfy certain criteria. We first find some $f$ that extremizes $S[f]$, by calculating the variation of $S[f]$, i.e. $$\delta S[f]=2\int_0^1 {\rm d}t \left\{f''(t) \delta f''(t) - \theta^4 f(t) \delta f(t) \right\} \\
\stackrel{\text{P.I. in 1. term}}{=} 2\int_0^1 {\rm d}t \left\{f''''(t) - \theta^4 f(t) \right\} \delta f(t) = 0 \, .$$
Boundary terms vanished upon the condition $\delta f(0)=\delta f(1)=\delta f'(0)=\delta f'(1)=0$.
For arbitrary variations $\delta f$ we therefore have the equation
$$f''''(t)=\theta^4 f(t) \, ,$$
and hence
$$f(t)=c_1 \cosh(\theta t) + c_2 \sinh(\theta t) + c_3 \cos(\theta t) + c_4 \sin(\theta t) \, .$$
Now imposing the conditions $f(0)=f(1)=f'(0)=f'(1)=0$ (without going through the algebra, the first 3 conditions eliminate a constant each and the final condition fixes $\theta$), we obtain
$$f(t) = c_4 \left( \frac{\sinh(\theta)-\sin(\theta)}{\cosh(\theta)-\cos(\theta)} \left[\cosh(\theta t)-\cos(\theta t)\right] - \left[\sinh(\theta t)-\sin(\theta t)\right] \right) \tag{1} \\
\Rightarrow \quad f'(1)= 2 c_4 \theta \, \frac{\cosh(\theta)\cos(\theta)-1}{\cosh(\theta)-\cos(\theta)} = 0 \, .$$
For non-trivial solutions ($c_4\neq 0$) it therefore follows $$\cosh(\theta)\cos(\theta)=1 \tag{2}$$
for $\theta >0$.
Finally, calculating $S[f]$ using (1) gives
$$S[f]=4c_4^2 \theta^3 \, \frac{\left(\cosh(\theta)\cos(\theta)-1\right)\left(\cos(\theta)\sinh(\theta)-\cosh(\theta)\sin(\theta)\right)}{\left( \cosh(\theta)-\cos(\theta) \right)^2} \, ,$$
which vanishes if $\theta$ is a solution to (2).
Since $f(t)$ as in (1) with a solution to (2) is a stationary point of $S[f]$, with extremal value $0$, we need to check under which conditions $S[f]=0$ is actually a minimum. In analogy to the definiteness of the hessian, we need to determine the second variation of $S[f]$ (which btw "approximates" $S[f]$ exactly, since it is quadratic in $f$),
$$\delta^2 S[f] = 2\int_0^1 {\rm d}t \, \delta f(t) \left( \frac{{\rm d}^4}{{\rm d}t^4} - \theta^4 \right) \delta f(t) \, .$$
If the eigenvalues $E$ of the operator in the middle are all $\geq 0$, we have a minimum. Therefore it is necessary to solve the eigenvalue equation
$$\left( \frac{{\rm d}^4}{{\rm d}t^4} - \theta^4 \right) y(t)=E y(t) \\
\Rightarrow \quad y''''(t) = (E+\theta^4)y(t) \equiv \lambda^4 y(t) \, .$$
The solution to this equation is the same as before, with $\theta$ replaced by $\lambda$ and the boundary conditions give rise to the allowed values for $\lambda$ as in Equation (2), $$\lambda_k=\pm 4.7300... , \pm 7.8532... , ... \, .$$
We can then solve for the eigenvalues $$E_k = \lambda_k^4 - \theta^4 \geq 0 \, .$$
This inequality can only be satisfied if the chosen $\theta$ is minimal, i.e. $\theta=\theta_1=4.7300...$, since in this case the smallest eigenvalue is $0$ and the others positive. If we would have chosen a larger $\theta$, the definiteness of the operator would be unclear, as there would be at least one negative eigenvalue.
Hence, the smallest solution $\theta>0$ to Equation (2) gives rise to a minimum for $S[f]$ and $S[f] \geq 0$ for any $f$ follows.

In the special case $$\int_0^1 f(t) \, {\rm d}t=0 \, ,$$ you can find a better inequality by the same procedure that lead to Wirtinger's inequality, i.e. by expanding $f(t)$ as a Fourier-Series on $[0,1]$. Imposing the conditions you arrive at
$$f(t)=\sum_{k=1}^\infty a_k \left(1-\cos(2\pi kt)\right) \\
f''(t)=\sum_{k=1}^\infty a_k (2\pi k)^2 \cos(2\pi kt) \, ,$$
and so
$$\int_0^1 (f''(t))^2 \, {\rm d}t = \sum_{k=1}^\infty \frac{a_k^2}{2} \, (2\pi k)^4 \\
\Rightarrow \quad \int_0^1 f(t)^2 \, {\rm d}t = \underbrace{\left(\sum_{k=1}^\infty a_k\right)^2}_{\left(\int_0^1 f(t) \, {\rm d}t\right)^2} + \sum_{k=1}^\infty \frac{a_k^2}{2} \leq \left(\int_0^1 f(t) \, {\rm d}t\right)^2 + \frac{1}{(2\pi)^4} \int_0^1 (f''(t))^2 \, {\rm d}t \, .$$
So in the case $\int_0^1 f(t) \, {\rm d}t =0$ this inequality is better, since $\theta = 4.7300... < 2\pi = 6.2831...$.
