Hi, awl-ga:
Thanks for posting this problem alongside the other one on Partial
Differential Equations, as it allows me to give a fuller explanation
of the method of characteristics which I used there in the solution of
one part.
The fully worked solution of this current problem:
U_t + U U_x = 0
with the piecewise-linear continuous initial conditions:
/ 1 for x <= 0
U(x,0) = { (1-x) for 0 < x <= 1
\ 0 for 1 < x
may be found in a PostScript document authored by Agnes Tourin at:
[Systems of Conservation laws and vanishing viscosity]
(see page 5 of 12)
http://www.math.toronto.edu/~atourin/chapter10.ps
We first describe the solution, and then briefly sketch how it is
found by the method of characteristics.
The homogeneous differential equation gives rise to solutions which
are constant along lines (characteristics) whose slopes depend upon
the initial condition. Up to time t=1 these characteristic lines do
not intersect, and a continuous closed-form solution may be given as
follows:
For 0 < t < 1:
/ 1 for x <= t
U(x,t) = { (1-x)/(1-t) for t < x <= 1
\ 0 for 1 < x
The resulting limit as t tends to 1 produces a discontinuity, which
corresponds to the fact that at t = 1 and above there are
intersections between characteristic lines. The following is a
solution for t >= 1 which satisfies the Rankine-Hugoniot condition:
/ 1 for x < (1+t)/2
U(x,t) = {
\ 0 for x > (1+t)/2
The jump discontinuity at x = (1+t)/2 is often colorfully described as
a "shock wave" because of the historical association of Burger's
equation with fluid mechanics.
The solution is said to be "physically correct" in the sense that
following any point backward in time along a characteristic line, we
do not not encounter another such line. However the same is not true
going forward in time, as all characteristic lines eventually
intersect the line of discontinuity or "shock wave".
What the Rankine-Hugoniot condition does is to specify the "speed of
propagation" of the jump discontinuity. How it does so in this case
may be summarized as follows.
Recall that we can express Burger's equation in "conservation form"
thusly:
U_t + (F(U))_x = 0
where scalar function F(u) = u^2/2.
The Rankine-Hugoniot condition requires that the speed of propagation
v of the jump discontinuity must agree with the difference quotient:
v = (F(U_+) - F(U_-))/( U_+ - U_- )
where U_+ and U_- correspond to the limiting values of U on the upper
and lower sides of the jump discontinuity, respectively. Here we have
a solution that "jumps" from 1 (on the lower side) to 0 (on the upper
sides), so the computation of speed gives:
v = ( 0 - 1/2 )/( 0 - 1) = 1/2
In this way the discontinuity's boundary line x = (1+t)/2 is fully
determined by having slope v = 1/2 and passing through the point (x,t)
= (1,1) where we know the discontinuity initally arises.
Let us say a few more words about the method of characteristics,
particularly to point out why the characteristic curves are straight
lines in the current problem but were parabolic curves in part (a) of
your other posting.
What is common to both analyses is the use of a characteristic curve x
= x(t) satisfying:
dx/dt = U
so that the total derivative wrt time agrees with the left-hand side
of the PDE:
dU(x(t),t)/dt = U_t + U U_x
In the current case the homogeneous PDE gives us that:
dU/dt = 0
along a characteristic curve, and therefore that U is constant along
such a path.
But then dx/dt = U is also constant along such paths, and therefore
the characteristic curves are simply straight lines. Their slopes,
however, will depend on the points at which the various characteristic
lines hit the x-axis and the intial value of U(x,0) prescribed there
(since the value of U is constant along this path).
One therefore has characteristic lines of slope 1 which issue forth
from the negative side of the x-axis (where dx/dt = U(x,0) = 1 by the
initial condition), and lines of slope 0 (considering x as a function
of t) issuing from the portion of the x-axis where x > 1 (with
characteristic lines of intermediate slopes in between). In
particular the characteristic lines issuing from x in [0,1] all
intersect at (x,t) = (1,1), and hence the discontinuity arises
"naturally" at that point.
We did not dwell on the point in our construction of the solution to
part (a) of the other problem, but the characteristics curves there do
not intersect. This was implicit, however, in our computation of a
unique "issue" point (x(0),0) corresponding to any given point (x,t)
in the upper half plane t > 0.
Search Strategy:
Keywords: "Burger's equation" "method of characteristics"
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regards, mathtalk-ga |
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