Analytical Solution of a PDE I need to solve a PDE which seems to be quite simple and to have an analytical solution. I tried the method of separation of variables, but could not complete the solution. Could you please let me know whether this PDE is analytically solvable and if yes how...?
$$\frac{\partial F(z,t)}{\partial t} = \alpha \frac{\partial F(z,t)}{\partial z} + \beta F(z,t) + \gamma$$
$$F(z,0) = f(z),~~~~~~ \frac{\partial F(0,t)}{\partial t} = c.$$
 A: $$\frac{\partial F(z,t)}{\partial t} = \alpha \frac{\partial F(z,t)}{\partial z} + \beta F(z,t) + \gamma$$
Search for the general solution :
Let $F(z,t)=\varphi(z,t)-\frac{\gamma}{\beta}$
$$\frac{\partial \varphi(z,t)}{\partial t} = \alpha \frac{\partial \varphi(z,t)}{\partial z} + \beta \varphi(z,t) $$
Let $\varphi(z,t)=e^{G(z,t)}$
$$\frac{\partial G(z,t)}{\partial t} = \alpha \frac{\partial G(z,t)}{\partial z} + \beta $$
Let $G(z,t)=H(z,t)+\beta t$
$$\frac{\partial H(z,t)}{\partial t} = \alpha \frac{\partial H(z,t)}{\partial z} $$
The wellknown general solution is :
$$H(z,t)=\Phi(z+\alpha t)$$
where $\Phi$ is any differentiable function.
Bringing it back into $G$ , $\varphi$ and then $F$ leads to :
$$F(z,t)=e^{\Phi(z+\alpha t)+\beta t}-\frac{\gamma}{\beta}$$
which can be written as :
$$F(z,t)=e^{\beta t}\Psi(z+\alpha t)-\frac{\gamma}{\beta}$$
where $\Psi$ is any derivable function.
Then,taking account of the boundary condition $F(z,0)=f(z)$
$F(z,0)=\Psi(z)-\frac{\gamma}{\beta}=f(z)$ implies $\Psi(z)=f(z)+\frac{\gamma}{\beta}$
$$F(z,t)=e^{\beta t}\left(f(z+\alpha t)+\frac{\gamma}{\beta}\right)-\frac{\gamma}{\beta}$$
$\frac{\partial F(z,t)}{\partial t} = \beta e^{\beta t}\left(f(z+\alpha t)+\frac{\gamma}{\beta}\right)+\beta e^{\beta t}\frac{\partial f(z+\alpha t)}{\partial t}$
$\frac{\partial F(0,t)}{\partial t} = \beta e^{\beta t}\left(f(\alpha t)+\frac{\gamma}{\beta}\right)+\beta e^{\beta t}\frac{df(\alpha t)}{dt}$
The boundary condition $\frac{\partial F(0,t)}{\partial t} = c$ implies:
$ \beta f(\alpha t)+\beta \frac{df(\alpha t)}{dt}= c e^{-\beta t}-\gamma$
So, the problem has no solution for any function $f(z)$, but has a solution only in some particular cases of particular function $f(z)$ which satisfies the next ODE:
$ \beta f(z)+\alpha \beta \frac{df(z)}{dz}= c e^{-\frac{\beta}{\alpha} z}-\gamma$
Solving the ODE leads to 
$$f(z)=\frac{c}{\beta(1-\beta)}e^{-\frac{\beta}{\alpha}z}-\frac{\gamma}{\beta}+C e^{-\frac{z}{\alpha}}$$
If the function $f(z)$ is not on this form above, the PDE with the given boundary conditions has no solution. If $f(z)$ is on the form above, the solution is :
$$F(z,t)=e^{\beta t}\left(f(z+\alpha t)+\frac{\gamma}{\beta}\right)-\frac{\gamma}{\beta}$$
A: Consider
$$
F(z,t) = \frac{Z(z) T(t) - \gamma}{\beta}, \alpha' = - \alpha
$$
then we have
$$
Z(z)T'(t) + \alpha' T(t) Z'(z) = Z(z) T(t) \\
\implies \frac{T'(t)}{T(t)} + \alpha' \frac{Z'(z)}{Z(z)} = 1
$$
Now let's assume that each of the terms are constant (since they each only depend on $t,z$ respectively) but add up to be $1$ (something not dependent on $t$ or $z$). Now put
$$
\frac{T'(t)}{T(t)} = k \implies \frac{Z'(z)}{Z(z)} = \frac{1 - k}{\alpha'}
$$
so we have two ordinary differential equations:
$$
T'(t) = k T(t) \to \frac{\mathrm{d} T}{\mathrm{d} t} = k T \to \ln T = kt + C_1 \to T(t) = C_1'e^{kt} \\
Z'(z) = \frac{1-k}{\alpha'} Z(z) \to \frac{\mathrm{d} Z}{\mathrm{d} z} = \frac{1-k}{\alpha'} Z \to \ln Z = \frac{1-k}{\alpha'}z + C_2 \to Z(z) = C_2'e^{(1-k)z/\alpha'}
$$
Now putting this together we get that
$$
F(z,t) = \frac{C_2'e^{(1-k)z/\alpha'} C_1'e^{kt} - \gamma }{ \beta } = C_2'' e^{(1-k)z/\alpha'} C_1''e^{kt} - \frac{\gamma}{\beta}
$$
Let's plug this back in, first we have
$$
\frac{\partial F}{\partial t} = k C_2'' e^{(1-k)z/\alpha'} C_1''e^{kt} \\
\frac{\partial F}{\partial z} = \frac{1-k}{\alpha'}C_2'' e^{(1-k)z/\alpha'} C_1''e^{kt}
$$
and we get
$$
k C_2'' e^{(1-k)z/\alpha'} C_1''e^{kt} = (k-1)C_2'' e^{(1-k)z/\alpha'} C_1''e^{kt} + C_2'' e^{(1-k)z/\alpha'} C_1''e^{kt} - \gamma + \gamma
$$
which checks out.
