Hwk6-sol

MATH220/CME303: PDEs of Applied Mathematics

Solution to Homework #6

1. Problem 1: Solve using characteristics (a) x1

∂u ∂u + x2 = 2u, ∂x1 ∂x2

u(x1 , 1) = g(x1 )

(b) u

∂u ∂u + = 1, ∂x1 ∂x2

u(x1 , x1 ) =

x1 2

Solution: (a) For the Cauchy problem x1 ux1 + x2 ux2 = 2u,

u(x1 , 1) = g(x1 )

the surface Γ ⊂ R2 is defined by Γ = {(y1 , y2 ) : y2 = 1}. The characteristic equations become ξ1′ (s) ξ2′ (s) U ′ (s)

= ξ1 (s), = ξ2 (s), = 2U (s),

ξ1 (0) = y1 , ξ2 (0) = 1, U (0) = g(y1 )

This leads to the characteristics ξ1 (s) = ξ2 (s) =

y1 es , es ,

U (s) =

g(y1 ) e2s

Consider the values for s such that a characteristic crosses the point (x1 , x2 ) for x2 ≥ 1. We obtain ( ) x1 ξ1 (s) = x1 ⇒ s = log , y1 x1 ξ2 (s) = x2 ⇒ x2 = y1 Therefore y1 = x1 /x2 and s = log x2 . Consequently u(x1 , x2 ) = U (s) = g

(

x1 x2

) x22

The solution is undefined for x2 ≤ 0. Note that this is a consequence of the fact that ξ2 (s) > 0 and therefore the characteristics never cross the x1 -axis. The solution only can be extended over {(x1 , x2 ) : x2 > 0}. 1

(b) For the Cauchy problem uux1 + ux2 = 1,

u(x1 , x1 ) =

1 x1 2

the surface Γ ⊂ R2 is defined by Γ = {(y1 , y2 ) : y1 = y2 }. The characteristic equations become ξ1′ (s) ξ2′ (s)

= U (s), ξ1 (0) = y1 , = 1, ξ2 (0) = y1 , 1 = 1, U (0) = y1 2

U ′ (s)

The equation for ξ1 can be rewritten as ξ1′′ (s) = 1,

ξ1′ (0) =

ξ1 (0) = y1 ,

1 y1 2

We obtain ξ1 (s)

=

ξ2 (s) = U (s)

=

1 2 1 s + y1 s + y1 , 2 2 s + y1 , 1 s + y1 2

Consider the values s for which a characteristic crosses the point (x1 , x2 ). We have ξ2 (s) = x2

⇒

ξ1 (s) = x1

⇒

s = x2 − y1 , 1 1 x1 = (x2 − y1 )2 + y1 (x2 − y1 ) + y1 2 2

From the second equation we deduce x1 =

1 2 1 1 1 x − x2 y1 + y12 + y1 x2 − y12 + y1 = 2 2 2 2 2

( ) 1 1 1 − x2 y1 + x22 2 2

⇒

y1 =

x1 − 12 x22 1 − 12 x2

We see that for x2 = 2, the value of y1 is undefined. This is due to the fact that, for x2 = 2 we have s = 2 − y1 and ξ1′ (2 − y1 )

=

2,

ξ2′ (2

=

2

− y1 )

from which we deduce that the corresponding characteristic curve has to be tangent to Γ. In other words, the characteristic that crosses (x1 , 2) is precisely a characteristic that is tangent to Γ. This is the main problem of the solution over these points. We conclude that, for x2 ̸= 2 x22 + 2x1 − 4x2 1 1 x1 − 12 x22 = u(x1 , x2 ) = U (s) = x2 − y1 + y1 = x2 − 2 2 1 − 21 x2 2(x2 − 2)

2

2. Problem 2: Show that

{ u(t, x) =

√ ( ) − 23 t + 3x + t2 0

if 4x + t2 > 0 otherwise

is an entropy solution of the Burgers’ equation ut + uux = 0 Solution: Let F (u) = 12 u2 . We first check that the proposed function u defined as { √ ( ) − 23 t + 3x + t2 , 4x + t2 > 0, u(t, x) = 0, o.w. is a pointwise solution of Burguers’ equation away from its discontinuities. This is trivially true for 4x + t2 < 0. For 4x + t2 > 0 we have 1 −√ , 3x + t2 2 2 t ut (t, x) = − − √ 3 3 3x + t2 2 2 = − + tux (t, x), 3 3 2 2√ u(t, x)ux (t, x) = − tux (t, x) − 3x + t2 ux (t, x) 3 3 ux (t, x) =

Therefore

2 2 2 2√ ut + uux = − + tux − tux (t, x) − 3x + t2 ux (t, x) = 0 3 3 3 3

We need to verify Rankine-Hugoniot’s condition for the discontinuity points of u. It satisfies 4x(t)+t2 = 0, from which we deduce x(t) ˙ = − 12 t. Remark that u+ (t) = = = u− (t) = =

lim u(t, x) ( ) √ 2 3 2 2 − t+ − t +t 3 4 x↓x(t)

−t, lim u(t, x) x↑x(t)

0

so that Rankine-Hugoniot’s condition in our case becomes x˙ =

1 2 t −0 1 F (u+ (t)) − F (u− (t)) = 2 =− t u+ (t) − u− (t) −t − 0 2

So u is indeed a solution of Burgers’ equation. To prove it is an entropy solution, it suffices to verify that u+ < ul , which is clear from our previous calculations. 3

3. Problem 3: Compute explicitly the entropy solution for the Burger’s equation with the initial data  1 if x < −1    0 if −1 < x < 0 g(x) =  2 if 0 < x < 1    0 if 1 < x and explain what happens at various times t > 0. Plot the graph of u(t, x) at various times when it behaves differently. Solution: We denote as usual F (u) = 12 u2 . Qualitatively the leftmost wave will propagate as a shockwave. We denote this discontinuity as x1 . The second wave will develop a rarefaction wave on its left and a shockwave on its right. We denote the rarefaction front as xf and its shockwave as x2 . We construct each stage of the solution analyzing the point at each one of them begins to evolve. Stage 1 (begins at t = 0): The Rankine-Hugoniot condition for the shockwaves is x˙ 1

x˙ 2

F (1) − F (0) 1−0 1 = , 2 F (2) − F (0) = 2−0 = 1

=

The rarefaction wave is of the form x/t and therefore xr (t) = 2t. It is not difficult to see that   1, x < −1 + 21 t,    1   0, −1 + 2 t < x < 0, x u1 (t, x) = t , 0 < x < 2t,    2, 2t < x < 1 + t,    0, 1 + t < x This expression shows that the left shockwave will hit the rarefaction wave at t such that −1 + 21 t = 0, i.e. t = 2 whereas the rarefaction front will hit the right shockwave at t such that 2t = 1 + t, i.e. t = 1. Therefore, the next step is to analyze the evolution of the solution for t > 1. Stage 2 (begins at t = 1): In this case the rarefaction front hits the right shockwave. The result is a rarefaction wave with shockwave x2 of height α2 = x2 /t. The Rankine-Hugoniot condition for this case is x˙ 2

= =

F (α2 ) − F (0) α2 − 0 1 α2 2 4

1 x2 2

= Additionally, x2 (1) = 2. This implies

√ x2 (t) = 2 t

Therefore, the solution at the second stage is  1,    0, u2 (t, x) = x   t,  0,

x < −1 + 21 t, −1 + 21 t < x < 0, √ 0 < x < 2 t, √ 2 t<x

We remark that the left shockwave will hit the rarefaction wave at t = 2, which is the next stage of the solution. Stage 3 (begins at t = 2): In this case the left shockwave hits the rarefaction wave. The left shockwave will consequently change its propagation speed. The Rankine-Hugoniot condition is x˙ 1

F (1) − F (x1 ) 1 − x1 1 1 2 − 2 2t2 x1 1 − 1t x1 ( ) 1 1 1 + x1 2 t

= = =

Remark that

∫

t

2

and therefore

ds = log 2s

( )1/2 t 2

⇒

e

∫t

ds 2 2s

( )1/2 ∫t t = , e− 2 2

ds 2s

( )1/2 2 = t

( )1/2 ∫ t ( )1/2 ∫ t t 1 2 ds x1 (t) = ds = t1/2 1/2 2 s 2 2 2 2s

The solution is then x1 (t) = t −

√ 2t

and the solution in this stage is consequently  √  t − 2t, 1, x <√ √ u3 (t, x) = xt , t − 2t < x < 2 t,  √  0, 2 t < x Finally, note that both shockwaves will merge at t such that t − is the last stage of the solution.

5

√ √ √ 2t = 2 t, i.e. t = (2 + 2)2 . This

√ Stage 4 (begins t = (2 + 2)2 ): In this case, there is only a single shockwave that propagates. The Rankine-Hugoniot condition is simply x˙ 1

F (1) − F (0) 1−0 1 2

= =

The last stage of the solution is { √ 1, x < 12 (2 + 2)2 + 12 t, √ u4 (t, x) = 0, 21 (2 + 2)2 + 12 t < x In summary, the solution is (see attached diagram  u1 (t, x),    u (t, x), 2 u(t, x) =  u3 (t, x),    u4 (t, x),

at the end for a plot of the different stages) 0 < t < 1, 1 < t < 2, √ 2 < t < (2 + 2)2 , √ 2 t < (2 + 2)

4. Problem 4: Consider the viscous Burgers’ equation ut + uux = εuxx where ε > 0 is a constant. (a) Find all solutions of this equation that have the form u(t, x) = V (x − ct), where the function V (y) has the limits lim V (y) = 1, lim V (y) = 0 y→−∞

y→∞

In particular, find all c for which such solution exists. (b) Let ε → 0, and show that the limit u(t, x) from the previous item is an entropy solution of the Burgers’ equation. Solution: (a) We search a solution of the form u(t, x) = V (x − ct). Remark that ut (t, x) = −cV ′ (x − ct), ux (t, x) = V ′ (x − ct) and uxx (t, x) = V ′′ (x − ct). Let z = x − ct. Then the PDE becomes −cV ′ (z) + V V ′ (z) = εV ′′ (z) which is simply an ODE. We assume that V is regular at infinity and therefore that V ′ (z) → 0 as |z| → ∞. In this case we can integrate from z to ∞ to obtain 1 cV (z) − V 2 (z) = −εV ′ (z) 2 6

Using the regularity of V ′ at infinity and the condition V (z) → 1 as z → −∞ we obtain c − 12 = 0. Therefore, the only admissible speed of propagation of a stationary wave satisfying the imposed conditions is c = 21 . Finally, we obtain the nonlinear ODE V ′ (z) =

1 1 V (z)2 − V (z) 2ε 2ε

which is the Riccati equation (this is also a form of the Bernoulli equation). Let W be such that V (z) = −2εW ′ (z)W (z)−1 . Then ( ′ )2 W (z) W ′′ (z) 1 W ′′ (z) V ′ (z) = −2ε + 2ε = −2ε + V (z)2 W (z) W (z) W (z) 2ε from which we deduce −2ε

W ′′ (z) W ′ (z) 1 1 = V ′ (z) − V (z)2 = − V (z) = W (z) 2ε 2ε W (z)

Therefore W ′′ (z) +

1 ′ W (z) = 0 2ε

and, for constants A, B ∈ R, we deduce that W (z) = A e− 2ε z + B 1

so that

W ′ (z) A 1 = = 1 1 z W (z) A + B e 2ε 1 + C e 2ε z with C = A−1 B. Remark that V (z) → 0 as z → ∞ and V (z) → 1 as z → −∞ for any choice of constant C ̸= 0. If C < 0, V has a singularity at z = −2ε log(|C|) and therefore is not a solution of the viscous Burgers’ equation. Consequently for C > 0 the function V (z) defined is the desired solution. Note that, for these cases, C acts as a translation parameter. In fact, let z0 = log C. Then 1 V (z) = 1 1 + e 2ε z+z0 for any z0 ∈ R. V (z) = −2ε

(b) Define Vε (z) =

1 1

1 + C e 2ε z

We prove that Vε (z) → Θ(z) as ε ↓ 0, where Θ(z) = χ(−∞,0] . Fix z > 0 and let ε be small enough 1 so that C −1 e− 2ε z < 1/2. Then |Vε (z) − Θ(z)| ≤

1 −1 + C e

= C −1 e− 2ε z

1

1

1 2ε z

1 2ε z

1−

1 C −1 e− 2ε z

≤ 2C −1 e− 2ε z → 0

< 1/2 as ε → 0. For z < 0, let ε be such that C e 1 1 C e 2ε z 1 C e 2ε z |Vε (z) − Θ(z)| = ≤ ≤ 2C e 2ε z → 0 1 1 z 1 + C e 2ε z 1 − C e 2ε 7

1

as ε ↓ 0. Consequently Vε → Θ for z ̸= 0 pointwise. Therefore, define uε (t, x) = Vε (x − 12 t). We conclude ( ) 1 lim uε (t, x) = u(t, x) = Θ x − t ε↓0 2 We now prove that this is indeed an entropy solution for Burgers’ equation with initial condition u(0, x) = Θ(x). The function u(t, x) satisfies the initial conditions and it is a pointwise solution for x < 12 t and x > 12 t. Additionally, the discontinuity point propagates at speed c = 12 , which satisfies Rankine-Hugoniot’s condition. Finally, it is not difficult to see that the characteristics satisfy ξt (s)

=

ξx (s) =

s, { y + s, y,

y < 0, y>0

Therefore, for any (t, x), if x − 12 t < 0, we choose the characteristic ξx (s) = x − (t − s). Clearly ξx (s)− 21 ξt (s) = x− 21 t− 12 (t−s) < x− 12 t < 0, so this characteristic does not cross the discontinuity curve nor another characteristic. If x − 12 t > 0, then we simply choose ξx (s) = x. Therefore, u is an entropy solution of Burgers’ equation, as desired.

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