Thread: 2nd order PDE- general solution

1. 2nd order PDE- general solution

Hi all,

I am trying to determine the general solution for the following equation: First of all i factorised it and determined the 1st order ODEs to be: For equation (1) there are no characterisitcs. While, for equation (2), i determined the characterisitics to be c_2 = x - t

The answer is.... the general solution,
phi(x,t) = F(t) + G(x -t)

Could some explain to me why there is F(t) in the general solution and why is it a function of t?

ArTiCk

2. Well, if $\displaystyle \phi(x,t)=F(t)$, then $\displaystyle \frac{\partial \phi}{\partial x}=0$.

$\displaystyle \frac{\partial \phi}{\partial x}$ kills of all functions of t.

3. Originally Posted by ArTiCK Hi all,

I am trying to determine the general solution for the following equation: First of all i factorised it and determined the 1st order ODEs to be: This is simply wrong. You cannot "factor" derivatives like that.
You can write it as $\displaystyle \frac{\partial}{\partial x}\left(\frac{\partial \phi}{\partial x}+ \frac{\partial \phi}{\partial t}\right)= 0$ which says that $\displaystyle \frac{\partial \phi}{\partial x}+ \frac{\partial \phi}{\partial t}$ is independent of x. That is, that $\displaystyle \frac{\partial \phi}{\partial x}+ \frac{\partial \phi}{\partial t}= F(t)$ where F(t) can be any function of t only.

For equation (1) there are no characterisitcs. While, for equation (2), i determined the characterisitics to be c_2 = x - t

The answer is.... the general solution,
phi(x,t) = F(t) + G(x -t)

Could some explain to me why there is F(t) in the general solution and why is it a function of t?

2nd, general, order, pde, solution 