Web page maintained by Evans M. Harrell, II, harrell@math.gatech.edu.
We need to handle the problem which occurred in Example 4.2 and again in Example 5.4. In both these examples, the characteristic curves crossed. Here's an understanding for why this crossing of the characteristics causes a problem.
Suppose the characteristic curves cross at {a,b}. We had found the value of u(a,b) by tracing back along the characteristic to the line where the initial value of u was specified and determined how u evolved by following above that characteristic curve. The problem is that we find there might be two characteristic curves going through {a,b}. Tracing back to the initial values may put us at different initial values and u may evolve differently as it moves along the different characteristic curves. Thus, when we come to {a,b} along the two curves, we may have different values as candidates for the values of u(a,b).
This induces a shock curve. It is developing one understanding of the shock curve that we consider in this section. Two examples will be used to illustrate the problem again. Both use the same partial differential equation, with different initial values. We find that one has crossing characteristics and the other does not.
Example 7.1 (Burger's Equation)
The partial differential equation is
u_t + u u_x = 0. (7.1)
We take the initial value to by
u(0,x) = f(x).
The reader who has worked through these notes to this point can derive the solution for the characteristic curves:
x = f([[xi]]) t + [[xi]].
At first, one might think this is a good situation; the characteristic curves are straight lines. Alas, they don't have the same slope. The nature of the initial value f is the important key. For this equation, if the values of f never decrease, the characteristics will not cross. In problem (7.2), you are asked to examine the characteristics for two different initial distributions f:
f1(x) = BLC{(A(1 if x < 0, 1-x if 0 < x < 1, 0 if 1 < x))
and
f2(x) = BLC{(A(0 if x < 0, x if 0 < x < 1, 1 if 1 < x)).
For one of these the characteristic curves cross, for the other, they do not.
Warning: This property of the initial distribution being non-decreasing implying the characteristics do not cross is not necessarily true for equations other than the Burger's equation. Recall, for example, equation (5.1), example 5.2, and figure 5.3.
Having this illustration of how the characteristic curves might cross for a partial differential equation with one set of initial conditions and not cross with another, we address how to determine the value of u in this region where it might be multivalued. This can be resolved with some tools from the calculus.
There is information from the integral calculus which we should recall. Some of the important ideas in that topic involve comparisons of integrals over areas with integrals around contours.
The following is a fundamental idea in that comparison.
STOKES THEOREM IN 2D. Suppose that D is a bounded region in the plane with a piecewise smooth boundary denoted [[partialdiff]]D. Suppose also that {P(x,y), Q(x,y)} has continuous second partial derivatives and is a function from R^2 to R^2. Then
integral [[partialdiff]]D Pdx + Qdy = integral D ([[partialdiff]]Q/[[partialdiff]]x - [[partialdiff]]P/[[partialdiff]]y) dx dy.
This theorem will be used to determine the direction the shock curve travels if characteristic lines cross and it appears that u should be multivalued.
(Click here if you prefer the divergence theorem of Gauss to the Stokes theorem - comments by Evans Harrell.)
In the previous sections we found it was reasonable to have solutions of a partial differential equation that were not differentiable at all points in their domain. (Re-look at Figure 5.4.) A wider notion of what a "solution" is seems appropriate.
To formulate this larger notion of solution, we want to develop the notion of the adjoint of the linear mapping L defined by
L(u) = au_x +bu_y + cu.
First, note the calculus:
v L(u) = v (au_x + bu_y + cu)
= (vau)x + (vbu)y - u(av)x - u(bv)y + vcu.
This implies
integral D v(au_x + bu_y + cu) dx dy + integral D [(va)x + (vb)y - vc]u dx dy
= integral D [(avu)x + (bvu)y] dx dy = integral [[partialdiff]]D [-bvu dx + avu dy].
Use Stokes theorem in 2D and the fact that v = 0 on the boundary of D to get that this last integral is zero. It follows that
integral D v(au_x + bu_y + cu) dx dy = integral D [-(va)x - (vb)y + vc]u dx dy.
We define
L*(v) to be -(va)x - (vb)y + vc
and say that L* is the adjoint of L.
We see that if u is differentiable and satisfies L(u) = 0, then also
integral D L*(v) u dx dy = 0
for all v in the domain of L*.
Thus, an alternate definition of a solution for L(u) = 0 is to say that
u is a "weak" solution of L(u) = 0 provided
integral D L*(v) u dx dy = 0.
for all v that are differentiable in D and 0 on the boundary
of D.
The above makes the notion of a function u being a solution for a partial differential equation when u is not even differentiable a little more comfortable.
The question of what is the direction that the "shock" travels is next. The sets D, D^+, and D^- are suggested in Figure 7.3. We will establish each of the following steps:
1. G(u) and H(u) are defined so that
F([[partialdiff]]G(u),[[partialdiff]]x) = a u_x and F([[partialdiff]]H(u),[[partialdiff]]y) = b u_y .
We assume that
0 = a u_x + b u_y = F([[partialdiff]]G(u),[[partialdiff]]x) + F([[partialdiff]]H(u),[[partialdiff]]y) .
2. If v = 0 on [[partialdiff]]D then
0 = integral D B(G(u) vx + H(u) vy) dx dy.
3. integral D+ B(G(u) vx + H(u) vy) dx dy =
integral L v B(G(u^+) dy - H(u^+)dx)
4. integral D- B(G(u) vx + H(u) vy) dx dy =
integral L v B(-G(u^-) dy + H(u^-)dx)
5. Add 3 and 4 above to get 2 so that
0 = integral L v B((G(u^+)-G(u^-)) dy - (H(u^+) - H(u^-))dx).
Conclude: F(dy,dx) = F(H(u^+) - H(u^-),G(u^+) - G(u^-)) .
Proof of 2:
F([[partialdiff]][G(u) v],[[partialdiff]]x) = F([[partialdiff]]G(u),[[partialdiff]]x) v + G(u) F([[partialdiff]]v,[[partialdiff]]x) and F([[partialdiff]][H(u) v],[[partialdiff]]y) = F([[partialdiff]]H(u),[[partialdiff]]y) v + H(u) F([[partialdiff]]v,[[partialdiff]]y) .
Hence if v = 0 on [[partialdiff]]D then
0 = integral [[partialdiff]]D [-H(u) v dx + G(u) v dy ]
= integral D [ F([[partialdiff]][G(u) v],[[partialdiff]]x) + F([[partialdiff]][H(u) v],[[partialdiff]]y) ]dxdy
= integral D [G(u) F([[partialdiff]]v,[[partialdiff]]x) + H(u) F([[partialdiff]]v,[[partialdiff]]y) ]dx dy + integral D [ F([[partialdiff]]G(u),[[partialdiff]]x) v + F([[partialdiff]]H(u),[[partialdiff]]y) v ] dx dy.
The last of these two integrals is 0 because of (1) above.
Proof of 3:
Figure 7.3
The curve is given by y = s(x) and we suppose that s^-1(y) = x. Suppose that v vanishes on the boundary of the rectangle. Then
integral D+ G(u) F([[partialdiff]]v,[[partialdiff]]x) dx dy = I(y1,y2, ) I(x1, s^-1(y), ) (G(u) v)x - G(u)x v)dx )dy
= I(y1,y2, ) (I(x1, s^-1(y), ) (G(u)v)x dx) dy - integral D+ G(u)x v dx dy
= I(y1,y2, ) G(u)+ v dy - integral D+ G(u)x v dx dy.
Also,
integral D+ H(u) F([[partialdiff]]v,[[partialdiff]]y) dx dy
= I(x1,x2, ) I(s(x), y2, ) (H(u) v)y dy dx -integral D+ H(u)y v dx dy
= - I(x1,x2, ) H(u)+ v dy - integral D+ H(u)y v dx dy
Thus,
integral D+ G(u) F([[partialdiff]]v,[[partialdiff]]x) dx dy + integral D+ H(u) F([[partialdiff]]v,[[partialdiff]]y) dx dy = integral s v ( G(u^+)dy - H(u^+) dx)
Proof of 4:
In a similar manner, 4 can be established.
^
^Example: Here is a graphical representation of a solution for the equation
u_x + 2 u u_y = 0
with u(0,y) given by the continuous function
u(0,y) = 1 if y < 0, 1- y if 0 < y < 1, and 0 if 1 < y.
The extension past the shock point used the Rankine-Hagonoit jump condition. The shock associated with the solution can be seen in the graph.
* u:=proc(x,y)
if x <=1/2 then
if y < 2*x then 1
elif 2*x<= y and y < 1 then (1-y)/(1-2*x)
elif 1 <= y then 0
fi
elif 1/2 < x then
if y < x+1/2 then 1
else 0
fi fi
end:
* plot3d(u,0..2,-2..2,orientation=[-160,50],numpoints=900);
Solution with shock
Exercise
7.1 Application of Stokes Theorem in the plane: Integrate < curl F, [[eta]] > over D.
a. F(x,y) = {x,y}, D = D1(0) (= the unit disk).
b. F(x,y) = {-y,x}, D = D1(0),
c. F(x,y) = {3y, 5x}, D = D1(0), ans:2[[pi]]
d. F(x,y) = {0,x^2}, D is the rectangle with vertices at {0,0}, {a,0}, {a,b}, and {0,b}. ans: a^2b
e. F(x,y) = {3xy + y^2, 2xy + 5x^2}, D= D1({1,2}) (= the disk with radius 1 and center {1,2} ). ans:7[[pi]].
7.2 Examine the characteristics for Burger's equation (7.1) with two initial distributions f:
f1(x) = BLC{(A(1 if x < 0, 1-x if 0 < x < 1, 0 if 1 < x))
and
f2(x) = BLC{(A(0 if x < 0, x if 0 < x < 1, 1 if 1 < x)).
7.3. Solve Burger's equation (2.1) with both initial condition given in Exericse 7.2 above.
7.4. Solve the equation
u_t + 2 u u_x = 0
with both initial condition given in Exericse 7.2 above.
7.5. Draw the solution for Burger's equation with
u(0,x) = BLC{(A( 1-x^2 if |x| <= 1, 0 if |x| > 1)).