Solve second order PDE with cauchy boundary How do I solve the heat equation with cauchy conditions -
$$
u_t = D u_{xx} \\
u(x,t=0)=\phi(x) \\
u(x=0,t)=f(t) \\
u_x(x=0,t)=0
$$
I know how to solve this if this was just standard non-homogenous Dirichlet conditions ($u(x=L,t)=g(t)$) using eigenfunction expansion. But I am not sure how to solve these conditions.
 A: $$
u_t = D u_{xx} \\
u(x,t=0)=\phi(x) \\
u(x=0,t)=f(t) \\
u_x(x=0,t)=0
$$
First, we look for particular solutions (method of separation of variables) on the form $u(x,t)=X(x)T(t)$
$$XT'=DX''T\quad\implies\quad \frac{T'}{T}=D\frac{X''}{X}=\text{constant}$$
$$\begin{cases}
T(t)=e^{-\lambda\:t}\\
X(t)=\cos\left(\sqrt{\frac{\lambda}{D}}x\right)\quad\text{or}\quad \sin\left(\sqrt{\frac{\lambda}{D}}x\right)
\end{cases}$$
An infinity of particular solutions are found, each one according to $\lambda$ :
$$u_\lambda(x,t)=\left(A_\lambda \cos\left(\sqrt{\frac{\lambda}{D}}x\right)+B_\lambda \sin\left(\sqrt{\frac{\lambda}{D}}x\right)\right)e^{-\lambda\:t} $$
$A_\lambda$ and $B_\lambda$ are arbitrary constants.
Since the PDE is linear any linear combination of those particular solutions is solution of the PDE.
$$u(x,y)=\sum_{\forall\lambda}\left(A_\lambda \cos\left(\sqrt{\frac{\lambda}{D}}x\right)+B_\lambda \sin\left(\sqrt{\frac{\lambda}{D}}x\right) \right)e^{-\lambda\:t}$$
In order to satisfy the condition $u_x(x=0,t)=0\quad\implies\quad B_{\lambda}=0$ :
$$u(x,y)=\sum_{\forall\lambda}A_\lambda \cos\left(\sqrt{\frac{\lambda}{D}}x\right)e^{-\lambda\:t}$$
The condition $u(x,t=0)=\phi(x)$ implies :
$\phi(x)=\sum_{\forall\lambda}A_\lambda \cos\left(\sqrt{\frac{\lambda}{D}}x\right)$ .
The expansion of the known function $\phi(x)$ into Fourier series gives the values of the coefficients $A_{\lambda}$.
The condition $u(x=0,t)=f(t)$ implies :
$f(t)=\sum_{\forall\lambda}A_\lambda e^{-\lambda\:t}$ .
One would have to expand the known function $f(t)$ into series of exponential functions $e^{-\lambda\:t}$ leading to other coefficients $A_{\lambda}$ than the preceeding ones. This is not consistent.
So, the three conditions make the problem overdetermined.
