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Solve this PDE using the characteristic form

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

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

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

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

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1 Answer 1

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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 $$\boxed{w(x,t) = z(s) = 4e^{-3(x+t)}e^{x} = 4e^{-2x - 3t}.}$$

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  • $\begingroup$ 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 $(-1) \cdot w_x + (1) \cdot w_t + 1 \cdot w = 0$ $\endgroup$
    – rcoder
    Oct 18, 2018 at 16:32
  • $\begingroup$ 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? $\endgroup$
    – rcoder
    Oct 18, 2018 at 16:38
  • $\begingroup$ 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!!! $\endgroup$
    – rcoder
    Oct 18, 2018 at 16:40

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