Solutions of $u^{(4)}=f; u=u'$

Question

Let $$ \psi_1(x) = \sinh(x)-\sin(x) \quad \quad\psi_2(x)=\cosh(x)-\cos(x)$$

$$\phi_a(x)=\psi_1(x)\psi_2(a)-\psi_1(a)\psi_2(x)$$

Prove there exists a sequence $0 \lt a_1 \lt a_2 ....$ such that $\phi_a'(a)=0$ for $a=0, a=a_k, a=-a_k$ for some $k \in \mathbb N_+$

This is a subproblem to find a fundamental system of $$\begin{align} u^{(4)}=f && x \in(0,1) \\u=u'=0 && x=0,x=1 \end{align}$$

which is supposed to be $\lambda _k(x)=\phi_{a_k}(a_kx)$

What I tried: I calculated $\phi_a'(a)=1-\cos (a)\cosh(a)$

which only has the solution $a=0$

I think I am misinterpreting the question. Would appreciate if anyone, could clarify

Answer 1

You have $\phi_a'(a) = 2-2\cosh(a)\cos(a)$ which leads to the equation

$$ 1-\cosh(a)\cos(a) = 0\implies \cos(a) - \frac{1}{\cosh(a)} = 0$$

It's true that $a=0$ is a trivial solution, but it's not the only solution. Notice that as $a\to\infty$, $\frac{1}{\cosh(a)}\to 0$ while $\cos(a)$ remains oscillating. Hence, the solution asymptotically approaches the zeroes of $\cos(a)$, which is $a \approx (n+\frac12)\pi$

Below is a plot of $\cos(x)-\frac{1}{\cosh(x)}$ (in blue), with $x$ in units of $\pi$. See how quickly it approaches $\cos x$ (in red)

Answer 2

The formula for $\phi'_a(a)$ is indeed related to the above, $$ \phi'_a(a)=2-2\cos a\cosh a\ . $$ Let us plot this function:

And indeed, it is simple to see that $0$ is a solution $\phi'_a(a)=0$, the function is even, so with a solution $a_k$, its negative is also a solution, and that in each interval between $2k\pi$ and $2k\pi+\frac \pi2$ the function changes the sign, $k\ge 2$, so by continuity it has a vanishing.