Heat equation with periodic BC

[Dennis_P]

I got the following PDE.

$${\frac {\partial }{\partial t}}u \left( y,t \right) = \nu \cdot {\frac {\partial ^{2}}{\partial {y}^{2}}}u \left( y,t \right)$$

With boundary conditions:
$$y=0: u(0,t)=0$$
$$y=h: u(h,t)=U_0 \cdot cos(\omega \cdot t)$$

Now I need to show by using substitution that,
$$\frac{u}{U_0} = \left( A\cos \left( \omega\,t \right) +B\sin \left( \omega\,t \right) \right) \sinh \left( {\frac {\alpha\,y}{h}} \right) \cos \left( {\frac {\alpha\,y}{h}} \right) + \left( C\cos \left( \omega\,t \right) + \left( D \right) \sin \left( \omega\,t \right) \right) \sin \left( {\frac {\alpha\,y}{h}} \right) \cosh \left( {\frac {\alpha \,y}{h}} \right)$$
, is a solution of the above equation.

Here:
$$\alpha = \sqrt{\frac{\Omega}{2}}$$ where $$\Omega = \frac{\omega \cdot h^2}{\nu}$$

Next to that I need to find the expressions to calculate the coefficients for a given $$\alpha$$.

Attempt at a solution:
Starting with the hint I substitute the expression back into the PDE.
$$\left( -A\sin \left( \omega\,t \right) \omega+B\cos \left( \omega\,t \right) \omega \right) \sinh \left( {\frac {\alpha\,y}{h}} \right) \cos \left( {\frac {\alpha\,y}{h}} \right) + \left( -C\sin \left( \omega\,t \right) \omega+ \left( D \right) \cos \left( \omega\,t \right) \omega \right) \sin \left( {\frac {\alpha\,y}{h}} \right) \cosh \left( {\frac {\alpha\,y}{h}} \right)$$
=
$$-2\, \left( A\cos \left( \omega\,t \right) +B\sin \left( \omega\,t \right) \right) \cosh \left( {\frac {\alpha\,y}{h}} \right) {\alpha}^{2}\sin \left( {\frac {\alpha\,y}{h}} \right) {h}^{-2}+2\, \left( C\cos \left( \omega\,t \right) + \left( D \right) \sin \left( \omega\,t \right) \right) \cos \left( {\frac {\alpha\,y}{h}} \right) {\alpha}^{2}\sinh \left( { \frac {\alpha\,y}{h}} \right) {h}^{-2}$$

This gives me that A = D and B = C. Now using this information I continue with the BC's. The first BC is automatically satisfied by the sine and hyperbolic sine term. Next I continue with the second BC.

$$\left( A\cos \left( \omega\,t \right) +B\sin \left( \omega\,t \right) \right) \sinh \left( {\frac {\alpha\,y}{h}} \right) \cos \left( {\frac {\alpha\,y}{h}} \right) + \left( B\cos \left( \omega\,t \right) + \left( A \right) \sin \left( \omega\,t \right) \right) \sin \left( {\frac {\alpha\,y}{h}} \right) \cosh \left( {\frac {\alpha \,y}{h}} \right) =\cos \left( \omega\,t \right)$$

Here I don't see how to continue. Maple gives a rather lengthy solution. Doing it by hand it seems I miss information, but my professor said this should be enough. I have been looking at this problem for a while and maybe I am overlooking things. So I need some assistance from someone with a fresh look.

 

[lippyka]

Solution:To solve this problem, we can first substitute the given expression for $u$ into the PDE and boundary conditions to determine the coefficients. From the boundary condition at $y=0$, we have $u(0,t)=0$. This implies that the coefficient of $\sinh\left(\frac{\alpha y}{h}\right) \cos\left(\frac{\alpha y}{h}\right)$ must be equal to the coefficient of $\sin\left(\frac{\alpha y}{h}\right) \cosh\left(\frac{\alpha y}{h}\right)$, i.e., $A = D$. We can also use the second boundary condition to find the other coefficients. Evaluating the expression for $u$ at $y=h$, we have:$$\left( A\cos \left( \omega\,t \right) +B\sin \left( \omega\,t \right) \right) \sinh \left( {\frac {\alpha\,h}{h}} \right) \cos \left( {\frac {\alpha\,h}{h}} \right) + \left( B\cos \left( \omega\,t \right) + \left( A \right) \sin \left( \omega\,t \right) \right) \sin \left( {\frac {\alpha\,h}{h}} \right) \cosh \left( {\frac {\alpha\,h}{h}} \right) = U_0\cos(\omega t).$$Simplifying the above equation yields $2AB = U_0$, which gives us the relationship between $A$ and $B$. We can now solve for each coefficient individually. For example, solving for $A$, we have $A = \frac{U_0}{2B}$. We can similarly solve for the other coefficients $B$, $C$, and $D$. The final expressions are:$$A = \frac{U_0}{2B}, \quad B = \frac{U_0}{2A}, \quad C = \frac{U_0}{2A}, \quad D = \frac{U_0}{2B}.$$