# Solve this PDE using the characteristic form

Solve this PDE using the characteristic form

$$$$\frac{\partial w}{\partial t} -\frac{\partial w}{\partial x}=-w \\ w(0,t)=4e^{-3t}$$$$

My attempt

We know $$w(t)=w(x(t),t)$$

Then by chain rule

$$\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\times \frac{\partial x}{\partial t} + \frac{\partial w}{\partial t}$$

This implies

$$$$\frac{\partial x}{\partial t}=-1 \\ \frac{\partial w}{\partial t}=-w$$$$

Here, i'm stuck. Can someone help me?

## 1 Answer

You are not applying the method correctly. You are using $$t$$ to denote a coordinate in the plane and again for the parameter.

Write the equation in the form $$1 \cdot w_x + (-1) \cdot w_t + (-1) \cdot w = 0$$.

The characteristic curve $$(x(s),t(s))$$ then satisfies $$x'(s) = 1$$ and $$t'(s) = -1$$.

The solution $$z(s)$$ satisfies $$z'(s) = z(s)$$.

Solve these ODE with initial data $$x(0) = 0$$, $$t(0) = t_0$$, and $$z(0) = w(0,t_0)$$ to get \begin{align*} x(s) &= s \\ t(s) &= -s + t_0 \\ z(s) &= w(0,t_0)e^{s} = 4e^{-3t_0}e^{s}. \end{align*}

You can add the first two of these equations to find that the characteristic curve containing a point $$(x,t)$$ satisfies $$t_0 = x+t$$ and $$s = x$$. Thus $$\boxed{w(x,t) = z(s) = 4e^{-3(x+t)}e^{x} = 4e^{-2x - 3t}.}$$

• Thanks for your answer! I have a couple of question. First here$1 \cdot w_x + (-1) \cdot w_t + 1 \cdot w = 0$ i think should be $(-1) \cdot w_x + (1) \cdot w_t + 1 \cdot w = 0$ Oct 18, 2018 at 16:32
• Second, i don't see very clear this step: $z'(s) = -z(s)$ where $z(s)$ is the solution... (I'm very newbie in this) , and this $x'(s) = 1$ , $t'(s) = -1$ come from the chain rule. and comparing with the original form of the equation no? Oct 18, 2018 at 16:38
• And this last step: $z(s) = w(0,t_0)e^{-s} = 4e^{-3t_0}e^{-s}.$ Why you get $z(s)=w(0,t_0))e^{-s}$? Thanks for all!!! Oct 18, 2018 at 16:40