I would like to ask for your help with an IBVP. We have the following partial differential equation:
$u_{t} = u_{xxxx}, x \in (0, \pi), t > 0$
with boundary conditions:
$u(0,t) = u(\pi,t) = 0, t > 0$
$u_{x}(0,t) = e^{-4t}, t > 0$
$u_{x}(\pi,t) = -\cosh(\pi) \cdot e^{-4t}, t > 0$
and initial condition:
$u(x,0) = \sin(x)\cosh(x), x \in (0, \pi)$
Furthermore, we know that $u$ is bounded as $t \to \infty$.
I applied the Laplace transform and found:
$U(x,s) = c_1\cdot e^{s^{\frac{1}{4}}x} + c_2\cdot e^{-s^{\frac{1}{4}}x} + c_3\cdot \cos(s^{\frac{1}{4}}x) + c_4\cdot \sin(s^{\frac{1}{4}}x)$
After that, I obtained a system of equations for $c_1$, $c_2$, $c_3$, and $c_4$ from the given conditions. However, my solution is very lengthy and I cannot apply the inverse transform.
The laplace transformed IBVP with initial condition is $$s U(x,s)-\frac{\partial ^4U(x,s)}{\partial x^4}-u(x,0)=s U(x,s)-\frac{\partial ^4U(x,s)}{\partial x^4}-sin(x)cosh(x)$$
Now we solve the IBVP with the boundary conditions $U(0,s)=0,U(\pi ,s)=0,U_{x}(0,s)=\frac{1}{s+4},U_{x}(\pi ,s)=-\frac{\cosh (\pi )}{s+4}$
This results in
$$U(x,s)\to\frac{\sin (x) \cosh (x)}{s+4}$$
and back in time domain
$$u(x,t)\to e^{-4 t} \sin (x) \cosh (x)$$