3
$\begingroup$

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...?

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

$\endgroup$
  • $\begingroup$ Do we have $\alpha, \beta, \gamma$ are just constants? $\endgroup$ – DanZimm Jun 23 '14 at 9:49
  • $\begingroup$ Just constants! $\endgroup$ – Reza Jun 23 '14 at 9:52
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ And then for your boundary conditions consider fourier series of these solutions. $\endgroup$ – DanZimm Jun 23 '14 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 '14 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 '14 at 8:46
  • $\begingroup$ @RezaHashemi how do you figure? I divide everything through by $\beta$ thus canceling that out $\endgroup$ – DanZimm Jun 25 '14 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 '14 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.