Shocks are discontinuities that can arise in the solution of
nonlinear PDEs.
For Example... consider the nonlinear PDE
ut + uux = 0
Preliminaries What path would we have to travel in x and t to stay at a constant value of u? Introduce a parameter s and let x = x(s) and t = t(s). Then u(x,t) = u(x(s),t(s)) and
du du dx du dt
-- = -- -- + -- --
ds dx ds dt ds
Since we get to pick
x(
s) and
t(
s), set
dx dt
-- = u and -- = 1.
ds ds
Then
du/
ds simplifies to the original PDE:
du du du
-- = -- u + -- = uu + u = 0.
ds dx dt x t
Therefore,
u is constant along the
curve defined by (
x(
s),
t(
s)). Integrate
dx/
ds and
dt/
ds to get
x = us + c1
t = s + c2
Letting c2 = 0 for convenience,
x = ut + c1 ==&rt; x - ut = c
We have found that
u is constant along the
characteristic lines
x -
ut =
c.
Initial Conditions Notice that the lines along which u is constant are dependent on u itself! This is something that happens in nonlinear PDEs that doesn't happen in linear PDEs--the "behavior" is dependent on initial conditions. Suppose u at time zero looks like this:
t
^
|
|
|
-----2-_
| \
| \
| \
1 -___________
|
|
-----0-------------------------&rt;x
|
|
|
Then the lines of constant
u will be something like this:
t
^ / / / /
| / | / |
| / |/ |
| / // /
| / * |
| / /| |
| / // /
| / / | |
|/ / | |
+ / / /
/| / | |
u=2 | u=2 u=1 u=1
/ | / / /
/ | / | |
/ |/ | |
-----+--------------------------&rt;x
|
|
|
But wait, the lines where
u=2 and
u=1 intersect! In other words, at the location
*,
u is multivalued?
Shocks In reality, whatever is represented by u probably cannot be multivalued. Therefore we introduce the "weak solution" which has a shock, or discontinuity, in it. The weak solution satisfies the PDE everywhere except at the shock.
t
^
|
|
|
-----2---------+
| |
| |
| |
1 +_______
|
|
-----0-------------------------&rt;x
| ^
| |
| shock
To find the speed of the shock, integrate the PDE with resepct to
x from
a to
b to get
2 2
d a u (a,t) - u (b,t)
-- ∫ u dx = -----------------
dt b 2
Let
a → f
- and
b → f
+, where
f(
t) is the location of the shock. We get
2 2
df - + (u-) - (u+)
-- (u - u ) + 0 = ------------
dt 2
or,
df (u-) + (u+)
-- = -----------
dt 2
The
shock wave travels at the average of the speed on the left and the speed on the right.