# Method of Characteristics for Linear Equations

Nonlinear PDE's Course
Current Topic Method of Characteristics for Linear Equations
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We present here a brief account of the method of characteristic for linear waves.

## Introduction

The method of characteristics is an important method for hyperbolic PDE's which applies to both linear and nonlinear equations.

We begin with the simplest wave equation

$\displaystyle{ \partial_t u + \partial_x u = 0,\,\,-\infty \lt x \lt \infty,\,\,t\gt 0, }$

subject to the initial conditions

$\displaystyle{ \left. u \right|_{t=0} = f(x) }$

We consider the solution along the curve $\displaystyle{ (x,t) = (X(t),t) }$. We then have

$\displaystyle{ \frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - 1 \right) }$

Therefore along the curve $\displaystyle{ \frac{\mathrm{d} X}{\mathrm{d}t} = 1 }$ $\displaystyle{ u(x,t) }$ must be a constant. These are nothing but the straight lines $\displaystyle{ x = t+c }$ This means that we have

$\displaystyle{ u(x,t) = u(t+c,t) = u(c,0) = f(c) = f(x-t)\, }$

Therefore the solution is $\displaystyle{ u(x,t) = f(x-t)\, }$.

## General Form

If we consider the equation

$\displaystyle{ \partial_t u + a(x,t)\partial_x u = 0,\,\,-\infty \lt x \lt \infty,\,\,t\gt 0, }$

then we can apply the method of characteristics. We consider the solution along the curve $\displaystyle{ (x,t) = (X(t),t) }$. We then have

$\displaystyle{ \frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - a(x,t) \right). }$

This gives us the following o.d.e. for the characteristic curves (along which the solution is a constant)

$\displaystyle{ \frac{\mathrm{d} X}{\mathrm{d} t} = a(x,t) . }$

## Example 1

Consider the equation

$\displaystyle{ \partial_t u + x \partial_x u = 0,\,\,-\infty \lt x \lt \infty,\,\,t\gt 0, }$

subject to the initial conditions

$\displaystyle{ \left. u \right|_{t=0} = f(x) }$

We consider the solution along the curve $\displaystyle{ (x,t) = (X(t),t) }$. We then have

$\displaystyle{ \frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d}X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - x \right) }$

Therefore along the curve $\displaystyle{ \frac{\mathrm{d} X}{\mathrm{d}t} = x }$ $\displaystyle{ u(x,t) }$ must be a constant. These are the curves $\displaystyle{ x = ce^t }$ This means that we have

$\displaystyle{ u(x,t) = u(ce^t,t) = u(c,0) = f(c) = f(xe^{-t})\, }$

Therefore the solution is given $\displaystyle{ u(x,t) = f(xe^{-t})\, }$. Solution for Example 1 with $\displaystyle{ f(x) = e^{-x^2} }$

## Example 2

Consider the equation

$\displaystyle{ \partial_t u + t \partial_x u = 0,\,\,-\infty \lt x \lt \infty,\,\,t\gt 0, }$

subject to the initial conditions

$\displaystyle{ \left. u \right|_{t=0} = f(x) }$

We consider the solution along the curve $\displaystyle{ (x,t) = (X(t),t) }$. We then have

$\displaystyle{ \frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - t \right) }$

Therefore along the curve $\displaystyle{ \frac{\mathrm{d} X}{\mathrm{d}t} = t }$ $\displaystyle{ u(x,t) }$ must be a constant. These are the curves $\displaystyle{ x = t^2/2+c }$ This means that we have

$\displaystyle{ u(x,t) = u(t^2/2 + c,t) = u(c,0) = f(c) = f(x - t^2/2)\, }$

Therefore the solution is given $\displaystyle{ u(x,t) = f(x - t^2/2)\, }$.

## Non-homogeneous Example

We can also use the method of characteristics in the non-homogeneous case. We show this through an example Consider the equation

$\displaystyle{ \partial_t u + t \partial_x u = xt,\,\,-\infty \lt x \lt \infty,\,\,t\gt 0, }$

subject to the initial conditions

$\displaystyle{ \left. u \right|_{t=0} = f(x) }$

We consider the solution along the curve $\displaystyle{ (x,t) = (X(t),t) }$. We then have

$\displaystyle{ \frac{\mathrm{d} u}{\mathrm{d} t} = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d}X}{\mathrm{d}t} - t \right) + xt }$

Therefore along the curve $\displaystyle{ \frac{\mathrm{d} X}{\mathrm{d}t} = t }$ which are the curves $\displaystyle{ x = t^2/2+c }$

$\displaystyle{ \frac{\mathrm{d}}{\mathrm{d}t}u(x,t) = xt = t^3/2 + c t }$

Therefore

$\displaystyle{ u(t^2/2+c,t) = t^4/8 + c t^2/2 + c_2\, }$

Now

$\displaystyle{ u(c,0) = c_2 = f(c) }$

Therefore the solution is given $\displaystyle{ u(x,t) = t^4/8 + (x -t^2/2) t^2/2 + f(x-t^2/2)\, }$ or $\displaystyle{ u(x,t) = -t^4/8 + x t^2/2 + f(x-t^2/2)\, }$