I have the PDE
$$\frac{\partial{u}}{\partial{t}} + u \frac{\partial{u}}{\partial{x}} = −2u$$
Written explicitly, I thought the characteristic equations were $\frac{\partial{t}}{\partial{r}} = 1$, $\frac{\partial{x}}{\partial{r}} = u$, and $\frac{\partial{u}}{\partial{r}} = -2u$
But the textbook says that the characteristic equations are $\frac{\partial{x}}{\partial{r}} = 1$, $\frac{\partial{u}}{\partial{r}} = -2u$, which I believe is implicit form
Why is there this difference? Are these two forms equivalent? Is one correct and the other incorrect? If they are equivalent, then why does the textbook use the latter characteristic equations instead of the former?
I would greatly appreciate it if people could please take the time to clarify this.
$$\frac{\partial{u}}{\partial{t}} + u \frac{\partial{u}}{\partial{x}} = −2u$$
The three characteristic differential equations are :
$$\frac{dt}{dr} = 1 \quad;\quad \frac{dx}{dr} = u \quad;\quad\frac{du}{dr} = -2u$$ which can be written on condensed form : $$\frac{dt}{1} = \frac{dx}{u} = \frac{du}{-2u} = dr$$
This shows that the three equations are not independent. Any one of them is related to the two others. Equivalently they are related to any linear combination of the three equations.
SOLVING :
A first characteristic equation comes from $\frac{dx}{u} = \frac{du}{-2u}$ : $$u+2x=c_1$$ A second characteristic equation comes from $\frac{dt}{1} = \frac{du}{-2u}$ : $$ue^{2t}=c_2$$ The general solution of the PDE expressed on the form of implicite equation is : $$\Phi(u+2x\:,\:ue^{2t})=0$$ $\Phi$ is an arbitrary function of two variables, to be determined according to some boundary and/or initial conditions.