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1
Quiz 9 (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.
Question 1: Consider the equation ∂t u + ∂x (u4 ) = 0, where x ∈ (−∞, +∞), t > 0, with initial data u0 (x) = 0 if
2
x < 0, u0 (x) = x 3 if 0 < x < 1, and u0 (x) = 0 if 1 < x. (a) What will produce the discontinuity of u0 at x = 1.
Clearly justify your answer?
We know from class that since the flux is convex is can only be either a shock or an expansion. On the left of x + 1 the
equation of the characteristics is X(t, s) = s + 4tu0 (s)3 = s + 4ts2 . These are line with positive slope 4s12 , for s ∈ [1, 0).
On the right the characteristics are vertical, X(s, t) = s. Hence the characteristics cross, which implies that there is a
shock.
(b) What is the solution on the right of the shock?
On the right the characteristics are vertical and the solution is constant, i.e., U (X, t) = 0 on the right of the shock if
xs (t) < X
(c) What is the solution on the left of the shock?
Case 1 s < 0: We have X(x, t) = s, which implies u(X, t) = 0 if X < 0.
Case 2: 0 < s < 1: We have X(t, s) = s + 4tu0 (s)3 = s + 4ts2 . We need to solve for s ∈ (0, 1] in the equation
X = s + 4ts2 . This is a quadratic equation with two solutions
√
−1 ± 1 + 16Xt
s=
.
8t
The only positive solution is the one of interest s =
Hence
u(X, t) =
−1 +
√
√
−1+ 1+16Xt
.
8t
1 + 16Xt
+ 4t
8t
−1 +
√
1 + 16Xt
8t
2
,
if X < xs (t).
Question 2: Consider the equation ∂t u + ∂x (−u6 ) = 0, where x ∈ (−∞, +∞), t > 0, with initial data u0 (x) = 1
if x < 0, u0 (x) = 0 otherwise. (a) Show that u(x, t) = 1 − H(x + t) is a solution in the distribution sense.
Using the chain rule and denoting by δ the Dirac measure, we have
∂t u = −δ(x + t).
Moreover, using that (1 − H(x + t))6 = 1 − H(x + t), we also have
∂x (−u6 ) = +δ(x + t).
This implies that ∂t u + ∂x (−u6 ) = 0, i.e., u(x, t) = H(x + t) is a solution in the distribution sense.
2
Quiz 9, Dec 4, 2014
(b) What is the entropy flux associated with the entropy η(v) = v 2 ?
By definition, the entropy flux is
Z
F (v) =
v
0
Z
0
f (z)η (z)dz =
0
v
−6z 5 2zdz = −
0
12 7
v ,
7
7
i.e., F (v) = − 12
7 v .
(c) Is it the entropy solution? Justify clearly your answer.
Solution 1: By looking at the characteristics (X(t, s) = s − 6t for s < 0 and X(s, t) = s for s > 0) we observe that the
correct solution should be an expansion wave.
Rv
7
Solution 2: Consider the entropy η(v) = v 2 , then the entropy flux is F (v) = 0 −6z 5 2zdz = − 12
7 v . Then upon
12 7
12
2
observing that η(u) = u = 1 − H(x + t) and F (u) = − 7 u = − 7 (1 − H(x + t)), we infer that
∂t η(u) = −∂t H(x + t) = −δ(x + t)
and
∂t F (u) =
12
12
∂t H(x + t) =
δ(x + t).
7
7
In conclusion
5
12
)δ(x + t) = δ(x + t) > 0.
7
7
The entropy residual ∂t η(u) + ∂t F (u) is a positive measure, which implies that the 1 − H(x + t) is not the entropy
solution.
Question 3: Solve the conservation equation ∂t ρ + ∂x q(ρ) = 0, x ∈ (∞, +∞), t > 0 with flux q(ρ) = ρ2 + ρ, and
with the initial condition ρ(x, 0) = −1, if x < 0, ρ(x, 0) = 1, if x > 0. Do we have a shock or an expansion wave?
∂t η(u) + ∂t F (u) = (−1 +
The solution is given by the implicit representation
ρ(X(s, t), t) = ρ0 (s),
X(s, t) = s + (2ρ0 (s) + 1)t.
Case 1: s < 0. Then ρ0 (s) = −1 and X(s, t) = s + (−2 + 1)t. This means s = X + t. The solution is
ρ(x, t) = −1,
if x < t.
Case 2: s < 0. Then ρ0 (s) = 1 and X(s, t) = s + (2 + 1)t. This means s = X − 3t. The solution is
ρ(x, t) = 1,
if 3t < x.
We have a expansion wave. We need to consider the case ρ0 ∈ [−1, 1] at s = 0.
Case 3: s = 0 and ρ0 ∈ [−1, 1]. Then X(s, t) = s + (2ρ0 + 1)t = (2ρ0 + 1)t. This means ρ0 = (X/t − 1)2. In conclusion
ρ(x, t) =
1 x
−1 ,
2 t
if − t < x < 3t.