I've been looking for the solution of the following diffusion equation and I haven't been able to find it.
Can anyone help me, please?
$$U_t=kU_{xx} \\ U(x,0)=f(x)\\ U(0,t)=0 ;U(L,t)=A\sin(\omega t+\phi)$$
Considering the answer underneath, to get the Jacobian could I do the following:
\begin{bmatrix} \frac{\partial U_s}{\partial x} \frac{\partial U_s}{\partial t} \\ \frac{-\partial U_s}{\partial t} \frac{\partial U_s}{\partial x} \end{bmatrix}
I got this from:
Complex function and Jacobian matrix
And once I do this, I proceed getting the determinant to find a $z$ from the equation:$$(U_s)_t=(U_s)_{xx}$$
Big thanks,
Emeric
Find $z\in{\mathbb C}$ such that \begin{equation} U_S(x, t) = \Im \left(\frac{e^{z x} - e^{-z x}}{e^{z L} - e^{-z L}}A e^{i(\omega t + \phi)}\right) \end{equation} satisfies the equation and the boudary conditions (but not the initial conditions). Then write the problem satisfied by \begin{equation} V(x, t) = U(x, t) - U_S(x, t) \end{equation} In order to get the value of $z$, let us compute \begin{equation} (U_S(x, t))_t = \Im \left(\frac{e^{z x} - e^{-z x}}{e^{z L} - e^{-z L}}A i\omega e^{i(\omega t + \phi)}\right) \end{equation} and \begin{equation} k(U_S(x, t))_{xx} = k \Im \left(z^2\frac{e^{z x} - e^{-z x}}{e^{z L} - e^{-z L}}A e^{i(\omega t + \phi)}\right) \end{equation} Hence if one chooses $i \omega = k z^2$, the equation $(U_S)_t = k(U_S)_{xx}$ is satisfied. Assuming $\omega$ and $k$ are positive, it follows that \begin{equation} z = \pm \sqrt{\frac{\omega}{2 k}}(1 + i) \end{equation}