PDEMOC15

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Consider $uu_x+u_t=0\,$ with the IC $u(x,0)=h(x)=\begin{cases} 0, & x<0 \\ u_0(x-1), & x>0 \end{cases}\,$

Find the weak solution.

When $y=0\,$ , if $x>0\,$ then $u=u_0(x-1)\,$ but the characteristic line (see PDEMOC14) is $x=ut+x_0\,$ . Therefore $u=\frac{u_0(x-1)}{u_0t+1}\,$ . Now find the line where the shock occurs.

The jump condition is defined as in PDEMOC14.

$\xi'(t) = \frac{1}{2}\frac{u_0(x-1)}{u_0t+1}\,$

To integrate with respect to $t\,$ , treat everything but $u_0\,$ as a constant and you'll see that it's a log:

$\xi(t) = \frac{1}{2}(x-1)\ln(u_0t+1)\,$

The weak solution is $u=\begin{cases} 0, & x<\xi(t) \\ \frac{u_0(x_0-1)}{u_0t+1}, & x>\xi(t) \end{cases}\,$

Main Page : Partial Differential Equations : Characteristics

PDEMOC15

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