Initial Value Method of Characteristics Question Help me to solve the following Partial differential equation:
$$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=-ku, \;\;u(x,0)=2, \;\; k>0 \;\text{is a constant}\;\; \text{and} \;\; x \;\text{is real}$$
Thanks!
 A: Let $\gamma_y(t)$ be the curve that solves the ordinary differential equation 
$$ \frac{\mathrm{d}}{\mathrm{d}t}\gamma_y(t) = u(t,\gamma_y(t)) $$
with initial condition
$$ \gamma_y(0) = y$$
We have that
$$ \frac{\mathrm{d}}{\mathrm{d}t} u(t,\gamma_y(t)) = \frac{\partial}{\partial t} u(t,\gamma_y(t)) + u(t,\gamma_y(t)) \frac{\partial}{\partial x} u(t,\gamma_y(t)) = -k u(t,\gamma_y(t)) $$
from the equation. So this means that
$$ u(t,\gamma_y(t)) = u(0,\gamma_y(0)) e^{-k t} $$
Plug this back into the equation for $\gamma$ we have
$$ \frac{\mathrm{d}}{\mathrm{d}t} \gamma_y(t) = u(0,\gamma_y(0)) e^{-kt} = u(0,y) e^{-k t} $$
so that
$$ \gamma_y(t) = \gamma_y(0) - \frac{1}{k} u(0,y) e^{-k t} = y - \frac{1}{k} u(0,y) e^{-k t}$$
So given a point $(t,x)$ we first solve for $y$ using the equation 
$$ x = y - \frac{1}{k} u(0,y) e^{-k t} $$
then we apply the equation that
$$ u(t,x) = u(t,\gamma_y(t)) = u(0,y) e^{-k t} $$
For our specific case, using that $u(0,y) = 2$ by assumption, we have that
$$ x = y - \frac{2}{k} e^{-kt}$$
or that
$$ y = x + \frac{2}{k} e^{-kt}$$
Then we have that
$$ u(t,x) = u(t,\gamma_y(t)) = u(0,y)e^{-kt} = 2 e^{-kt} $$

Remark: note that by the symmetry of the equation, given that the initial data does not depend on $x$, the solution also does not depend on $x$. So your PDE actually can be reduced to the ODE $\frac{\mathrm{d}}{\mathrm{d}t} u = - k u$ and the solution read off accordingly. But since you mentioned "method of characteristics" in the question title, I showed you how to solve the problem for any initial condition using said method. 
