Add Me!Close Menu Navigation

Solving first order PDEs with Characteristics

Use the method of characteristics to solve the PDE system

     $$\begin{cases} u_x + xu_y = 0 \\ u(0,y) = f(y) \end{cases}$$

We begin by parameterizing u,x,y as functions of t such that

    \begin{align*} \dfrac{d}{dt} u(x,y) &= u_x x' + u_y y' \\ \intertext{observe that if we $x' = 1$ and $y' = x$ gives us the PDE} &= u_x + x u_y = 0 \end{align*}

This gives the system of ODE

    \[\begin{cases} x' = 1 \\ y' = x \\ x(0) = 0 \\ y(0) = y_0 \\ u(x,y) = f(y(0)) \\ \end{cases}\]

Unfortunately, we don’t know what y_0 is; but we do have the parameterization parameter t, so we can solve the system backwards in time.

    \begin{align*} v' &= -1 \\ w' &= -v \\ v(0) &= x \\ w(0) &= y \end{align*}

which is equivalent to the second order ODE

    \[w'' - 1 = 0\]

which has the solution

    \begin{align*} w(t) &= \frac{t^2}{2} - c_1 t + c_0 \\ v(t) &= -t + c_1 \\ v(0) &= c_1 = x \\ w(0) &= c_0 = y \end{align*}

Now, we know that x_0 = 0, so we want to find the time t such that v(t) = 0, or x - t = 0 \implies t = x, then

    \[y_0 = w(x) = \frac{x^2}{2} - x^2 + y\]

and u(x,y) = f\pa{\frac{x^2}{2} - x^2 + y} = f\pa{y-\frac{x^2}{2}}

Let’s quickly verify that this is indeed the solution

    \begin{align*} u_x + xu_y &= \dfrac{d[y-\frac{x^2}{2}]}{dx}f' + x\dfrac{d[y-\frac{x^2}{2}]}{dy}f' \\ &= -xf' + xf' \\ &= 0 \\ u(0,y) &= f\pa{y-\frac{0^2}{2}} \\ &= f(y) \end{align*}

Since this is fully linear, the solution is unique and the characteristics intersect.

Posted By Lee

Woosh