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Math 444 — Spring 2001 — Test Solutions
Green’s Formulas: ∫
Ω
[v∆u + ∇v · ∇u] dx =
∂Ω
v ∂u ∂ν dS ∫
Ω
[v∆u − u∆v] dx =
∂Ω
[
v ∂u ∂ν
− u ∂v ∂ν
]
dS
1: (25 points) Assume G is a smooth function and u(x, y) is a weak solution of the conservation law G(u)x + uy = 0. (a) Show that if u is in fact a smooth solution except for a jump across the C^1 -curve x = ξ(y), then the direction of the curve is related to the jump by
ξ′(y) = G(u) − G(ur) u − ur
(b) State the entropy condition, say what it means physically, and illustrate with a diagram. (If you like, you can assume in part (b) that G(z) = 12 z^2 , i.e. Burgers’ equation.)
Solution: (a) Write u= u(ξ(y)−, y), ur = u(ξ(y)+, y). Then u 6 = ur because of the jump. Observe that u is a weak solution
=⇒ 0 = G(u(b, y)) − G(u(a, y)) + d dy
{∫ (^) ξ(y)
a
u(x, y) dx +
∫ (^) b
ξ(y)
u(x, y) dx
whenever a < ξ(y) < b
=⇒ 0 = G(u(b, y)) − G(u(a, y)) +
∫ (^) ξ(y)
a
uy(x, y) dx + u`ξ′(y) +
∫ (^) b
ξ(y)
uy(x, y) dx − urξ′(y)
by differentiating through the integral, and also using the fundamental theorem to differentiate the limit of integration, ξ(y)
=⇒ 0 = −G(u) + G(ur) + ξ′(y) [u − ur]
by substituting uy = −G(u)x in the two integrals (which is valid classically away from x = ξ(y))
=⇒ ξ′(y) = G(u) − G(ur) u − ur (b) The entropy condition is: G′(u`) > ξ′(y) > G′(ur). Physically it means that two characteristics can “come together” or collide at a shock, but cannot emanate from
the shock and then subsequently move apart. We drew some diagrams of this in class.
2: (25 points) Let f (x, t) be smooth. (a) State Duhamel’s Principle for solving the nonhomogeneous wave equation ztt − c^2 ∆z = f in Rn^ (with zero initial conditions) in terms of the solutions to certain homogeneous problems. (You are not required to solve these homogeneous problems.) (b) Prove Duhamel’s Principle. That is, show that your formula for z(x, t) really does solve ztt − c^2 ∆z = f.
Solution: (a) Duhamel’s Principle states that if Z(x, t; s) solves the homogeneous wave equation Ztt − c^2 ∆Z = 0 with initial conditions Z(x, 0; s) = 0 and Zt(x, 0; s) = f (x, s), then
z(x, t) =
∫ (^) t
0
Z(x, t − s; s) ds
solves the nonhomogeneous wave equation ztt − c^2 ∆z = f in Rn^ with zero initial conditions. (b) The proof of Duhamel’s Principle for dimension one was given in class, and can be found on page 81 of MCOWEN. For higher dimensions, just replace Zxx with ∆Z, and so on.
4: (25 points) Consider the nonhomogeneous wave equation utt − c^2 ∆u = f. Write
E(t) =
u^2 t + c^2 |∇u|^2
dx.
(Here the integral is over Rn. We assume u and f are smooth and have compact support, at each t). (a) Show E′(t) =
utf dx. (b) Explain why this formula is physically plausible. [For part (b) you can work in one dimension, so that f represents an external force on the string.]
Solution: (a) E′(t)
= d dt
u^2 t + c^2 ∇u · ∇u
dx
=
ututt + c^2 ∇u · ∇ut
dx (differentiation through the integral is valid since ut and ∇u have compact support)
=
ut
[
c^2 ∆u + f
]
dx by the PDE
=
c^2 ∇ · [ut∇u] + utf
dx
=
utf dx by the divergence theorem, since ∇u ≡ 0 near infinity.
[Note: Another way to organize this calculation is to just multiply the PDE by ut and integrate with respect to x.] (b) The formula E′(t) =
utf dx is physically plausible as follows. For example, if f > 0 and ut > 0 then the force and velocity (of the string) are both upwards, so that the force will increase the velocity (and hence the kinetic energy) beyond what it would have been otherwise. Since the total energy is conserved in the absence of any external force, we see that the presence of f leads to an increase in the total energy. This is confirmed by the formula: if f > 0 and ut > 0 then E′(t) =
utf dx > 0.
