UMD PDE Qualifying Exams/EvansCh2

Solution
Consider characteristics $$(x(s),t(s)) = (x_0+bs,s)\in\mathbb R^n$$. Also, for any $$x\in \mathbb R^n, t\in \mathbb R$$, consider $$z(s)=u(x+sb,t+s)$$. Then taking a derivative gives

$$\dot{z}(s) = Du(x+sb,t+s) \cdot b + u_t(x+sb,t+s) = -c z(s)$$

where the last inequality is a result of the original PDE. The above ODE can be solved and we get $$z(s) = z(0) e^{-cs}$$

Finally, any point $$(x,t)$$ is connected to the characteristic curve $$(x_0,0)$$ where $$x_0=x-tb$$ and hence

$$u(x,t) = g(x-tb)e^{-ct}$$.