superposition in the solution of pdes I have a single pde in the following form:
$$\frac{\partial u}{\partial t} = -\alpha\frac{\partial u}{\partial z}+f(z,t).$$
where $f(z,t)$ is a non-linear function and the initial condition is: $u(z,0)=u_0$.
My question is: Is it correct to solve this pde in the following two steps:
$$\frac{\partial u_1}{\partial t} = -\alpha\frac{\partial u_1}{\partial z}$$ and $$\frac{\partial u_2}{\partial t} = f(z,t).$$
And then $u = u_1 + u_2$ ...?
If yes, how the initial conditions must be adopted...?
thanks,
Reza
 A: No. If $v=u_1+u_2$ then
$$
\frac{\partial v}{\partial t}=\frac{\partial u_1}{\partial t}+\frac{\partial u_2}{\partial t}=-\alpha\frac{\partial u_1}{\partial z}+f(z,t)=-\alpha\frac{\partial v}{\partial z}+f(z,t)+\alpha\frac{\partial u_2}{\partial z}.
$$
$v$ is not a solution unless $\dfrac{\partial u_2}{\partial z}=0$.
What you can do is:


*

*Find the solution $u_h$ of the homogeneous equation $\dfrac{\partial u}{\partial t} = -\alpha\,\dfrac{\partial u}{\partial z}$.

*Find a particular solution $u_p$ of the complete equation $\dfrac{\partial u}{\partial t} = -\alpha\,\dfrac{\partial u}{\partial z}+f(z,t)$.

*Then $u=u_h+u_p$.


In step (2) you need just one solution. The form of $f$ can suggest sometimes a simple way of finding $u_p$. For instance, if $f(z,t)=f(t)$ depends only on $t$, you can find $u_p$ as a solution of $\dfrac{\partial u}{\partial t} =f(t)$. Similarly if $f$ depends only on $z$.
A: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=1$ , we have $t=s$
$\dfrac{dz}{ds}=\alpha$ , letting $z(0)=z_0$ , we have $z=\alpha s+z_0=\alpha t+z_0$
$\dfrac{du}{ds}=f(z,t)=f(\alpha s+z_0,s)$ , letting $u(0)=F(z_0)$ , we have $u(z,t)=F(z_0)+\int_0^sf(\alpha\tau+z_0,\tau)~d\tau=F(z-\alpha t)+\int_0^tf(z+\alpha(\tau-t),\tau)~d\tau$
$u(z,0)=u_0$ :
$F(z)=u_0$
$\therefore u(z,t)=u_0+\int_0^tf(z+\alpha(\tau-t),\tau)~d\tau$
A: The first solution inserted in the PDE would lead to the equation 
$$
0 = f(z,t)
$$
and the solution $u$ thus would only be valid at these points $(z,t)$.
So it is not a method to get a general solution that works for the points where $f$ is non-zero.
