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!
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!
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.