Draw the characteristics for $u_x+u_y=0$ with data on line

Question

I know the concept and solution of method of charecteristic but I need to visualize how the characteristics look in $x$-$y$ Plane. I am stuck in this equation $u_x+u_y=0$ for Cauchy data on curve $\Gamma $ ; $U(\alpha y,y)=\exp(-y^2)$ .

$x(s, t) = s + αt ,$

$ y(s, t) = s + t $,

$z(s,t) =\exp(-t^2)$

then we get

$u(x,y)=\exp(-(\frac{y-x}{\alpha-1})^2)$

So how will we draw the curve gamma and the characteristic curve in $x$-$y$ Plane, as we do for advection equation?

Answer 1

The PDE is the linear advection equation with unit speed. The characteristic curves are straight lines parallel to the graph of $y = x$, along which $u$ is constant. The boundary data is located on the curve $\Gamma$ which graph is the line $y = x/\alpha$. If $\alpha = 1$, then the information located on the boundary can't propagate inside the domain: the boundary-value problem has no solution.

Let's do the analytics. The method of characteristics gives

• $x'(s) = 1$, letting $x(0) = \alpha t$ gives $x(s) = s + \alpha t $.
• $y'(s) = 1$, letting $y(0) = t$ gives $y(s) = s + t$.
• $u'(s) = 0$, letting $u(0) = e^{-t^2}$ gives $u(s) = e^{-t^2}$.

To plot the characteristics, plot the curves $s\mapsto (x(s), y(s))$ for several values of $t$. One notes that these curves satisfy $y - x = (1-\alpha)\, t$. Thus, for various fixed values of $t$, they are indeed parallel to the line $y=x$. Moreover, we get $$ u(x,y) = \exp \left(- \big(\tfrac{y-x}{1-\alpha}\big)^2 \right) $$ if $\alpha \neq 1$.