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Quiz 8 (Notes, books, and calculators are not authorized)
Show all your work in the blank space you are given on the exam sheet. Always justify your answer. Answers with
no justification will not be graded.
The implicit representation of the solution to the equation ∂t v + ∂x q(v) = 0, v(x, 0) = v0 (x), is
X(s, t) = q 0 (v0 (s))t + s;
v(X(s, t), t) = v0 (s).
(1)
Question 1: Consider the following conservation equation
(
∂t ρ + ∂x (q(ρ)) = 0,
x ∈ (−∞, +∞), t > 0,
ρ(x, 0) = ρ0 (x) :=
1
2
1
if x < 0,
if x > 0,
where q(ρ) = 2ρ + cos(ρ2 ) (and ρ(x, t) is the conserved quantity). What is the wave speed for this problem?
The wave speed is the quantity q 0 (ρ) = 2 − 2ρ sin(ρ2 ).
Question 2: Consider the following conservation equation
(
∂t ρ + ∂x (q(ρ)) = 0,
where q(ρ) = 41 ρ2 −
1
8
x ∈ (−∞, +∞), t > 0,
ρ(x, 0) = ρ0 (x) :=
3π
4
π
4
if x < 1,
if x > 1,
cos(2ρ) (and ρ(x, t) is the conserved quantity). (i) What is the sign of q 00 (ρ)?
Note that q 0 (ρ) = 12 ρ + 14 sin(2ρ) and q 00 (ρ) = 12 + 12 cos(2ρ). This implies that q 00 (ρ) ≥
have q 00 (ρ) = cos2 (ρ), which clearly proves that q 00 (ρ) ≥ 0, i.e., q is convex.
1
2
−
1
2
≥ 0. Actually we also
(ii) Given that the given initial data produces a shock, give the speed of the shock.
Using ρ+ =
π
4
and π − =
3π
4 ,
the Rankin-Hugoniot relation gives the speed of this shock:
q(ρ+ ) − q(ρ− )
1 π 2 − 9π 2
8π 2
π
dxs
=
=
=
= .
+
−
dt
ρ −ρ
16 π − 3π
32π
4
(iii) Give the solution to the problem.
In conclusion we have dxs (t)/dt = π4 and xs (0) = 1. This means that the location of the shock is xs (t) =
the explicit representation of the solution is
3π
,
4
π
ρ= ,
4
ρ=
πt
+ 1 = xs (t),
4
πt
if x >
+ 1 = xs (t).
4
if x <
π
4t
+ 1 and
2
Quiz 8, April 15, 2014
Question 3: Solve the conservation equation ∂t ρ + ∂x q(ρ) = 0, x ∈ (∞, +∞), t > 0 with flux q(ρ) = ρ4 + 2ρ, and
with the initial condition ρ(x, 0) = 1, if x < 0, ρ(x, 0) = 2, if x > 0. Is it a shock or an expansion wave?
The solution is given by the implicit representation
ρ(X(s, t), t) = ρ0 (s),
X(s, t) = s + (4ρ0 (s)3 + 2)t.
We then have two cases depending whether s is positive or negative.
Case 1: s < 0, then ρ0 (s) = 1 and X(s, t) = (4 + 2)t + s = 6t + s. This means
ρ(x, t) = 1
if
x < 6t.
Case 2: s > 0, then ρ0 (s) = 2 and X(s, t) = (4 × 8 + 2)t + s = 34t + s. This means
ρ(x, t) = 2 if
x > 34t.
Case 3: We see that there are no characteristics in the region {34t > x > 6t}. This implies that there is an expansion
wave in this region. The characteristics are given by setting s = 0,
X = (4ρ30 + 2)t,
Then
1 ≤ ρ0 ≤ 2.
with
13
1 X
−2
.
ρ0 =
4 t
Note that the condition 1 ≤ ρ0 ≤ 2 implies that 6t ≤ X ≤ 34t. In conclusion
13
1 x
ρ(x, t) =
−2
,
4 t
if
6t ≤ X ≤ 34t.