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Lecture 4: Math 553
Weak Solutions: Rarefactions
Last time we talked about introducing discontinuities into the solution of
a first order quasilinear equation. This becomes necessary when one wants to
maintain single-valuedness of the solution. There is another type of discontinuity
that one might need to introduce, called a “rarefaction fan”.
We’ll start with the inviscid Burger’s equation, which is given by
ut + uux = 0
with initial data u(x, 0) = H(−x), with H the Heaviside function. The characteristics look like this:
So there is a wedge which is not covered by characteristics. So what should
we do?
Well, we have the special solution to the Burger’s equation given by
u(x, t) =
x − x0
t − t0
which is constant along lines (x − x0 ) = α(t − t0 ).
This is a one-parameter family of lines through the point (x0 , t0 ) parameterized by the slope, α. This suggests the following: we could just “stick in” a fan
of such characteristics. In the case above we need to fill in the region between
the lines x = 0 and x = t. It is easy to see in the above case that this works:
the solution is at least continuous across the lines x = t and x = 0, although
not differentiable. However given that the initial data is not even continuous
this is progress.
Now let’s look at a more complicated problem
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Example 1. Find the solution to the initial value problem
ut + uux = 0
u(x, 0) = χ[0,1] (x)
Let’s start by drawing a picture: The characteristics look like
There is a region which is doubly covered by the characteristics and a region
which has no characteristics through it. We know how to fix the region which is
doubly covered: we introduce a discontinuity into the region. By the RankineHugoniot condition we know that the shock propagates with speed equal to the
average of the values from the left and the right. In region with no characteristics
we create some by putting in a rarefaction fan. Thus we have the solution:
However something goes wrong at time t = 2. What is it?
After this time the location of the shock propagates somewhat differently. If
the shock is located at η(t) then the Rankine-Hugoniot condition becomes
η0 =
η
t
2
This can be integrated up to find that
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η(t) = At 2
At time t = 2 the shock is located at η = 2 (This is when the shock first hits the
fan) and thus the location is given by
√
η(t) = 2t
Thus for t > 2 the solution to the problem is given by
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In the case of a general first order quasilinear equation the idea is similar: one looks for a scale-invariant type solution to “patch” the area where no
characteristics propagate. If we have the conservation law
ut + (G(u))x = 0
we can look for a solution of the form u = f (x/t). In this case ut = − xt ux , so
the above equation becomes
(−
x
+ g(u))ux = 0
t
If the function g(u) is invertible then the above equation is equivalent to u =
g −1 ( xt ) where, of course, the notation g −1 denotes the inverse function, not the
reciprocal. We have a similar construction in this case to the one given in the
case of the Burger’s equation. Let’s assume that the function g is a monotone
increasing function.
Example 2. Find the weak solution to the quasi-linear equation
ut + u2 ux = 0
u(x, 0) = 1 + χ[0,1] (x)
The characteristics are given by
x=t+α
x = 2t + α
x=t+α
α<0
α ∈ [0, 1]
α>1
(1)
(2)
(3)
which look like
Obviously we need to incorporate a shock and a rarefaction fan. As far as
the shock goes the Rankine-Hugoniot condition is given by
η0 =
u3L − u3R
u2 + uR uL + u2R
= L
3(ul − ur )
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so the shock propagates with speed 76 .
As far as the rarefaction fan goes the function g(u) = u2 , so the inverse
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function is g −1 ( xt ) =
px
t.
Thus for short times the solution is given by
u=1
r
x
u=
t
x<t
u=2
4t < x <
u=1
t<
(4)
t < x < 4t
7
t+1
6
7
t+1
6
(5)
(6)
(7)
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The shock hits the fan when 4t = 67 t + 1 or t = 17
. After the shock and the
fan collide the Rankine-Hugoniot condition changes. The value of the function
√
to the left of the shock is xt and to the right is 1. Thus the R-H condition
becomes
r
1 x
x
0
η =
+
+1
3 t
t
pη
This equation is actually solvable via the change of variables w =
t , which
gives a separable equation that can be solved via partial fractions (try it!)
General Nonlinear Case
Consider the general nonlinear equation
F (x, y, z = u, p = ux , q = uy ) = 0
This equation is also solvable via the method of characteristics, but the procedure is somewhat complicated. The quasi-linear equation has a fairly straightforward geometric interpretation - in essence the quasilinear equation defines
the tangent plane to the surface at each point x, y, z. For the more general case,
however, things are not quite so straightforward. Essentially we have to propagate the tangent planes along with the characteristics. This gives us a larger
set of ordinary differential equations to solve.
Consider some point (x0 , y0 , z0 ) in R3 . The equation for the tangent plane
is
(z − z0 ) = p(x − x0 ) + q(y − y0 )
Unfortunately neither p nor q are known apriori - they are things that we need
to solve for. Using the defining equation we assume that q can be solved for as
a function of p as well as (x0 , y0 , z0 )). Thus we have a one-parameter family of
tangent planes
(z − z0 ) = p(x − x0 ) + q(p)(y − y0 )
This is a one-parameterfamily of tangent planes. We can take the envelope of
this family. This defines a ruled surface called the Monge cone. This is not a
circular cone but some ruled surface that comes to a point at (x0 , y0 , z0 ). The
Monge cone is the simultaneous solution to
(z − z0 ) = p(x − x0 ) + q(p)(y − y0 )
dq
0 = (x − x0 ) + (y − y0 )
dp
(8)
(9)
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In differential form we can write the above as
dz = pdx + q(p)dy
dq
0 = dx + dy
dp
(10)
(11)
Now we’d like to find a surface which is tangent to the Monge cone. Letting
the characteristic be parameterized by t the above becomes
dz
dx
dy
=p
+q
dt
dt
dt
By the implicit function theorem we have
dq
Fp
=−
dp
Fq
thus
dx
= Fp
dt
dy
= Fq
dt
(12)
(13)
At this point, as a reality check, let’s confirm that this reproduces the usual
characteristic equations:
We now need two more conditions: Differentiating the defining equation
with respect to x and y gives
Fx + pFz + px Fp + qx Fq = 0
(14)
Fy + qFz + py Fp + qy Fq = 0
(15)
Using the fact that qx = py the first is equivalent to
dp
= −(Fx + pFz )
dt
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and similarly
dq
= −(Fy + qFz )
dt
This gives a system of five equations in five unknowns. Now we need to impose
the initial conditions. Assume that the surface is parameterized by α. x(0) =
f (α), y(0) = g(α), z(0) = h(α). This has to be supplemented by two equations.
The first is
F (f (α), g(α), h(α), φ(α), ψ(α)) = 0
The second is
h0 (α) = φ(α)f 0 (α) + ψ(α)g 0 (α)
These two equations define (ψ(α), φ(α)) along the initial surface.
In summary we have the following five equations
dx
dt
dy
dt
dp
dt
dq
dt
dq
dt
∂F
∂p
∂F
=
∂q
∂F
∂F
= −(
+p
)
∂x
∂z
∂F
∂F
= −(
+q
)
∂y
∂z
∂F
∂F
=p
+q
∂p
∂q
=
(16)
(17)
(18)
(19)
(20)
In the case in which the equation is quasi-linear notice that the equations for
x, y, z decouple from those for p, q. In this case one can solve for these without
bothering to solve for p, q.
Example 3.
u2x + u2y − u = 0
The characteristics are
u(x, 0) = f (x)
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dx
dt
dy
dt
dz
dt
dp
dt
dq
dt
= 2p
(21)
= 2q
(22)
= 2p2 + 2q 2 = 2
(23)
= −p
(24)
= −q
(25)
(26)
which have solutions
z = 2t + f (α)
(27)
−t
p = p0 (α)e
(28)
−t
q = q0 (α)e
(29)
−t
x = 2p0 (α) e
−t
y = 2q0 (α) e
−1 +α
−1
We still have to solve for p0 (α), q0 (α). One equation is given by
p20 (α) + q02 (α) = f (α)
while the second is given by
f 0 (α) = p0 (α)
(30)
(31)