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The Method of Characteristics

\[ a(x,t)\,u_x + b(x,t)\,u_t + c(x,t)\,u = 0 \]

The method of characteristics solves first-order PDEs by finding curves—characteristics—along which the PDE reduces to an ODE. We solve the resulting ODEs and then map the solution back to \((x,t)\).

Theory Behind It

Consider the first-order equation (without the zeroth-order term for clarity)

\[ a(x,t)\,u_x + b(x,t)\,u_t = 1. \]

Parameterize \(x\) and \(t\) by a parameter \(s\): \(x=x(s)\), \(t=t(s)\). The chain rule gives

\[ \frac{du}{ds} \;=\; u_x\,\frac{dx}{ds} \;+\; u_t\,\frac{dt}{ds}. \]

If we choose the characteristic equations

\[ \frac{dx}{ds} = a\!\big(x(s),t(s)\big), \qquad \frac{dt}{ds} = b\!\big(x(s),t(s)\big), \]

then along the curve \((x(s),t(s))\),

\[ \frac{du}{ds} \;=\; a\,u_x + b\,u_t \;=\; 1, \]

which is an ODE with solution \(u(x(s),t(s)) = s + C\).

We will also introduce a parameter \(\tau\) to label where characteristics start on the \(x\)-axis at \(t=0\). Thus, characteristics are indexed by \(\tau\) and advanced by \(s\).

Example Problem

\[ 2x\,u_x + u_t = 1, \qquad u(0,x) = \phi(x). \]

Characteristic ODEs:

\[ \frac{dx}{ds} = 2x, \qquad \frac{dt}{ds} = 1. \]

Integrate:

\[ \int \frac{1}{x}\,dx = \int 2\,ds \;\Rightarrow\; x(s)=C_1 e^{2s}, \qquad t(s)=s+C_2. \]

Along characteristics, by the chain rule,

\[ \frac{du}{ds} = 1 \;\;\Rightarrow\;\; u(s)=s+C_3. \]

Apply Initial Data and Label Curves

Enforce that characteristics start at \(t=0\) when \(s=0\): \(\;t(0)=0 \Rightarrow C_2=0 \Rightarrow t(s)=s.\) Label their \(x\)-start by \(\tau\): \(x(0)=\tau \Rightarrow C_1=\tau \Rightarrow x(s)=\tau e^{2s}.\)

At \(s=0\) we have \(t=0\) and \(x=\tau\), so the initial condition gives

\[ u(0,\tau) \;=\; \phi(\tau) \;=\; C_3. \]

Therefore the solution in \((s,\tau)\) is

\[ u(s,\tau) \;=\; s + \phi(\tau). \]

Return to \((x,t)\)

We have \(t=s\). From \(x(s)=\tau e^{2s}\), solve for \(\tau\):

\[ \tau \;=\; \frac{x}{e^{2t}}. \]

Substitute into \(u(s,\tau)\):

\[ u(x,t) \;=\; t \;+\; \phi\!\left(\frac{x}{e^{2t}}\right). \]

This is the (classical) solution to the PDE with arbitrary initial condition \(\phi\).