I want to find a trigometric function for which:
$u(0)=u(1)=0$
and
$\frac{\partial u}{\partial y}(y=0)=\frac{\partial u}{\partial y}(y=1)=0$.
Lets look at e.g. $u=C1 cos(a y+\phi_0)+C2 sin(a y +\phi_1)$.
It is possible to choose a, $\phi_0, \phi_1$, C1 and C2 such that u(0)=u(1)=0, however I can only find the trivial solution in which u=0, to also solve the boundary condtions for the derivatives.
The function u itself should behave like a trigonometric function, but does not necessarily have to be one. So you can also modify the shown function itself. I feel like I am overlooking something very obvious right now, I would appreciate any kind of help.
What you believe to be a combination of trigonometric functions with four independent parameters only has two in fact. Because
$$c_c\cos(ay+\phi_c)+c_s\sin(ay+\phi_s)\\=(c_c\cos\phi_c+c_s\cos\phi_s)\cos(ay)+(-c_c\sin\phi_c+c_s\sin\phi_s)\sin(ay)\\=p\cos(ay)+q\sin(ay)=c\cos(ay+\phi).$$
This is a general sinusoid, which cannot enjoy the requested properties.
A possible answer is
$$u(y)=2\sin^2(\pi y)=1-\cos(2\pi y),$$ if that is "trigonometric" enough.