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...?

$$\frac{\partial F(z,t)}{\partial t} = \alpha \frac{\partial F(z,t)}{\partial z} + \beta F(z,t) + \gamma$$ $$F(z,0) = f(z),~~~~~~ \frac{\partial F(0,t)}{\partial t} = c.$$

  • $\begingroup$ Do we have $\alpha, \beta, \gamma$ are just constants? $\endgroup$
    – DanZimm
    Jun 23, 2014 at 9:49
  • $\begingroup$ Just constants! $\endgroup$
    – Reza
    Jun 23, 2014 at 9:52

2 Answers 2


$$\frac{\partial F(z,t)}{\partial t} = \alpha \frac{\partial F(z,t)}{\partial z} + \beta F(z,t) + \gamma$$ 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 : $$F(z,t)=e^{\beta t}\left(f(z+\alpha t)+\frac{\gamma}{\beta}\right)-\frac{\gamma}{\beta}$$


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.

  • $\begingroup$ And then for your boundary conditions consider fourier series of these solutions. $\endgroup$
    – DanZimm
    Jun 23, 2014 at 11:55
  • $\begingroup$ Thanks a lot for your answer. I can not verify the solution. A $\beta$ is missing. It should be as follows: $Z(z)T'(t)+α'T(t)Z'(z)=\beta Z(z)T(t)$. $\endgroup$
    – Reza
    Jun 25, 2014 at 8:18
  • $\begingroup$ But based on your method, the answer is: $F(z,t) = C e^{(kt)} e^{((k-\beta)z/\alpha)} - \frac{\gamma}{\beta}$ $\endgroup$
    – Reza
    Jun 25, 2014 at 8:46
  • $\begingroup$ @RezaHashemi how do you figure? I divide everything through by $\beta$ thus canceling that out $\endgroup$
    – DanZimm
    Jun 25, 2014 at 20:18
  • $\begingroup$ @RezaHashemi in other words the right hand side is $\beta F + \gamma$ and I have $F = (Z(z)T(x)-\gamma)/\beta$ so that $\beta F = Z(z)T(t) - \gamma$ $\endgroup$
    – DanZimm
    Jun 25, 2014 at 20:19

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