PDEMOC15

From Exampleproblems

Jump to: navigation, search

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

Retrieved from "http://www.exampleproblems.com/wiki/index.php/PDEMOC15"
Argan Oil
Natural Skin Care
Organic Skin Care
visitor stats