# 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!

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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.

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