Warm-Up Problem

Problem: Find the exact roots of

x2 + ✏x 1=0 and then perform a series expansion in ✏.

Solution: The roots are

1 1 2 1 4 1 ✏2 1 2 ✏ + 8 ✏ 128 ✏ + x = ✏ 1+ = ··· 2 ± r 4 ( 1 1 ✏ 1 ✏2 + 1 ✏4 + 2 8 128 ··· converging for ✏ < 2. | |

Asymptotics, APPM 5480 Approximation Accuracy

0 1 10

0.8 −5 10 0.6 + x −10 0.4 10 first order 0.2 Exact error in approximation second order −15 Approx 10 fourth order 0 −2 0 −2 0 10 10 10 10 ϵ ϵ

Asymptotics, APPM 5480 My name is

• Mark Hoefer • Course webpage: https://goo.gl/DmCKIb (amath->course pages) • Office: ECOT 325 • Email: [email protected] • My Office Hours: M 10am-11am, W 12pm-1pm or by appt • TAs: Michelle Maiden, Patrick Sprenger (emails: fi[email protected]) • Office Hours in APPM Conf Room ECOT 226: M 12pm-1pm

Asymptotics, APPM 5480 Applications of Asymptoics

• Approximate solutions: algebraic/ transcendental/diﬀerential equations • Approximate equations: simplify ODEs/PDEs • Dynamical stability of solutions to ODEs/PDEs • Numerical methods: approximation theory, convergence, ... • Pure math: prime number theorem, Riemann hypothesis, ...

Asymptotics, APPM 5480 Asymptotics: Physical Example

dyed, diluted corn syrup

light corn syrup computer

piston pump

nozzle

Asymptotics, APPM 5480 Interfacial Waves

• Solitary waves • Dispersive shock waves (DSWs) • How to model? ★ Solitary waves are long ★ DSWs are longer

Asymptotics, APPM 5480 Conduit Modeling

Model u(i,e) =0 r · @u(i,e) Two-ﬂuid incompressible Re(i,e) +(u(i,e) )u(i,e) = p(i,e) + 2u(i,e) @t · r r r Navier-Stokes equations ✓ ◆ (REALLY HARD) + free surface boundary conditions + far ﬁeld boundary conditions

u(i,e) =0 r · ⇢(i) Two-ﬂuid incompressible P (i) + z = 2u(i) r ⇢ r Stokes equations ✓ ◆ ⇢(e) µ(i) (HARD) Re(i,e) 0 P (e) + z = 2u(e) ! r ⇢ µ(e) r ✓ ◆ + free surface boundary conditions + far ﬁeld boundary conditions

Non-dimensional equations Asymptotics, APPM 5480 Conduit Modeling (cont)

Model 2 2 1 A˜ +(A ) (A (A A˜) ) =0 Long wave conduit equation t z˜ t z˜ z˜ (not so hard) buoyancy viscous response nonlinear self-steepening nonlinear dispersion Cross-sectional| {z } | conduit{z area: A}(˜z,t˜) Concepts: scaling, maximal 2014 April Defense, Ph.D. balance

µ(i) = 1 µ(e) (L/✏1/2)

O 8 Rescale vertical length:z ˜ = 1/2z 1/2 ˜

Rescale time: t = t

modulation scale modulation

wave scale wave experiments

(L)

O Asymptotics, APPM 5480

How does a wave train evolve? train wave a does How

How do long wavelength perturbations propagate? perturbations wavelength long do How

condition along free interface free along condition

Velocity continuity, stress matching and kinematic kinematic and matching stress continuity, Velocity

- -

Modeling Approach - Multiple Scales Multiple - Approach Modeling

Nonlinear Dispersive Conduit Waves Conduit Dispersive Nonlinear

- True physical scale: scale: physical True -

- Macroscale description: description: Macroscale -

- Mesoscale description: description: Mesoscale -

- “Exact” microscale mathematical description: mathematical microscale “Exact” - Conduit Modeling (cont)

Model Whitham modulation equations: slowly varying 2 2 periodic waves (A)T + A 2k!A✓ =0 Z 1/2 (not so hard) (L/(✏ ✏˜)) 1 2 2 ⇣2 ⌘ A + k A A✓ 2ln A =0 O T Z ⇣ ⌘ Concepts: singular kT + !Z =0 perturbation, kZ !T Periodic traveling wave: A = A (✓) , ✓ = solvability condition ✏˜ k = k(Z, T),a= a(Z, T), A = A(Z, T), ! = !(k, a, A)

Rescale vertical length: Z =˜z˜ Rescale time: T =˜t˜

(L) O Asymptotics, APPM 5480 Warm-Up Problem

Problem: Find approximate expansions for the two roots of

✏x2 + x 1=0.

Solution:

2 3 1 p1+4✏ 1 ✏ +2✏ 5✏ + x = ± = ··· ± 2✏ ( 1 1+✏ 2✏2 +5✏3 + ✏ ···

Asymptotics, APPM 5480 My name is

Asymptotics, APPM 5480 “Asymptotology” - Martin Kruskal • M. Kruskal, Math Models in Phys. Sci., ed. Drobot, Prentice Hall, pp. 17-42, 1962 • Asymptotics: science of the evaluation of integrals, equations, etc. in various limiting cases • Asymptotology: art (quasi-science) of dealing with applied mathematical systems in limiting cases 1. Simpliﬁcation (in spite of resulting complications) 2. Recursion: e.g., iteration 3. Interpretation: e.g., change of variables 4. Wild Behavior: solutions behave “wildly” in the limit 5. Annihilation: complete set of persistent annihilators 6. Maximal (Dominant) Balance: keep both terms for ﬂexibility, ... 7. Mathematical Nonsense: if result doesn’t make sense, try again

Asymptotics, APPM 5480 Warm-Up Problem

Problem: Given the eigenvalue-eigenvector pair (0, v0) for a square matrix A (Av0 = 0v0), formulate an ansatz (guess) to approximate an eigenvalue- eigenvector pair (, v) for the perturbed matrix C = A + B (Cv = v), assuming that is small?

Solution: Try

= + + 2 + , v = v + v + 2v + . 0 1 2 ··· 0 1 2 ··· Then,

Cv = v (A + B)(v + v + )=( + + )(v + v + ) 0 1 ··· 0 1 ··· 0 1 ··· Av + (Av + Bv )+ = v + ( v + v )+ 0 1 0 ··· 0 0 0 1 1 0 ···

Equating like powers of , we already have Av0 = 0v0 so we require

(A I)v = (B I)v . 0 1 1 0

But we need both v1 and 1!?!

Asymptotics, APPM 5480 Warm-Up Problem

T Problem: Given the eigenvalue-eigenvector pair (1, 10) for the matrix 11 11 A = , estimate the eigenvalues for the matrix C= for small. 01 1

Solution: Because A is degenerate, expect non-standard asymptotic expansion

1 =1+p + 2p + , v = + pv + 2pv + . 1 2 ··· 0 1 2 ··· Evaluating like powers of , we already satisfy 0 equation. If p = 1, then

01 1 :(A I)v = v = (B I)v = 1 0 1 00 1 1 0 1 NOT SOLVABLE! (P7: Mathematical nonsense) Need p<1, then

1/p 01 1 0 :(A 0I)v1 = v1 = 1v0 = v1 = 00 0 1

Asymptotics, APPM 5480 Warm-Up Problem

Solution (continued): Proceeding to the next order 2/p and assuming 2/p < 1 01 0 2/p :(A I)v = v = v + v = + 2 0 2 00 2 1 1 2 0 2 0 1 NOT SOLVABLE! Need 2/p =1(p =1/2) so that the matrix perturbation enters at this order

01 0 2/p = 1 :(A I)v = v = v Bv + v = + 2 0 2 00 2 1 1 0 2 0 2 1 0 1 2 ( ) Therefore, we must have 1 = 1 or 1± = 1 so that the approximate eigenvalues are =1 1/2. Actually, these are± the exact eigenvalues! ± ± More generally, if geom(0) = 1, alg(0) = 2, then

1/2 w0Bv0 = 0 ± ± w0u0 where u is a generalized eigenvector (A I)u = v and (A I)w = 0. 0 0 0 0 0 0

Asymptotics, APPM 5480 Warm-Up Problem

Problem: Find an asymptotic series approximation to

x 1 e f(!)=! dx, ! . ! + x !1 Z0 Hint: consider the Taylor expansion of 1 for z < 1. 1+z | | Solution: We have ! 1 x x2 x3 = =1 + + , converging for x < !. ! + x 1+x/! ! !2 !3 ··· | |

Then, formally

n n 1 ( 1) 1 n x 1 ( 1) n! 1 2! 3! f(!)= x e dx = =1 + + !n !n ! !2 !3 ··· n=0 0 n=0 X Z X 1 n x because 0 x e dx = n!. This series does not converge. Consider the ratio test R a (n + 1)! !n n +1 n+1 = = ,n . a !n+1 n! ! !1 !1 n Asymptotics, APPM 5480 Warm-Up Problem

Problem: Find the ﬁrst two terms in an asymptotic series approximation to the elliptic integral of the ﬁrst kind

/2 1 K(m)= d 2 0 1 m sin as m 0+.

Solution: Taylor expand the integrand in m 1 m 1+ sin2 ,m 0+, 1 m sin2 2 uniformly valid for [0, /2]. Then /2 m /2 m K(m) 1d + sin2 d = 1+ . 0 2 0 2 8

Asymptotics, APPM 5480 Warm-Up Problem Problem: Evaluate

1 ax2+bx+c e dx, a > 0. Z1 This integral arises in many asymptotic problems. Solution: Complete the square

2 2 1 ax2+bx+c c 1 b b e dx = e exp a x + dx. 2a 4a Z1 Z1 " ✓ ◆ # Let y = pa(x b/2a), then

2 1 ax2+bx+c 1 c+ b 1 y2 e dx = e 4a e dy pa Z1 Z1 2 2 1 c+ b 1 y2 1 y2 1 c+ b x 2 = e 4a e dy e dy = e 4a e| | dx pa s pa s 2 Z1 Z1 ZR

2 2⇡ 2 1 c+ b 1 r2 1 c+ b 1 r2 = e 4a e r d dr = e 4a 2⇡ e r dr pa s pa Z0 Z0 s Z0 2 ⇡ c+ b = e 4a . a r Asymptotics, APPM 5480 Warm-Up Problem

Problem: Find the leading order behavior of

1 I(x)= sin t exp x sinh4 t dt, x + . ! 1 Z0

Solution: Dominant contribution comes endpoint t = arg max( sinh4 t) = 0. t [0,1] 2 Expand sin t and sinh t

✏ xt4 1 xt4 I(x) te dt te dt. ⇠ ⇠ Z0 Z0 Let s = xt4 to obtain

1 1 1/2 s (1/2) 1 ⇡ I(x) s e ds = = ,x + . ⇠ 4px 4px 4 x ! 1 Z0 r So long as f(t) goes to zero algebraically at the critical point, the Laplace method applies.

Asymptotics, APPM 5480 Warm-Up Problem Problem: Solve the following initial value problem using the Fourier transform

ut + ux + uxxx =0,u(x, 0) = u0(x),xR. 2

Assume that u0(x) is smooth and decays fast enough so that the following technical requirements hold:

ikx ikx u0(x)= uˆ0(k)e dk, uˆ0(k)= u0(x)e dx, ZR ZR 3 1 exist and k uˆ0(k) L (R). 2 Solution: Take the Fourier transform (in space) of the PDE to obtain the ODE

3 ikx uˆ + ikuˆ ik uˆ =0, uˆ(k, t) u(x, t)e dx, t ⌘ ZR which has solution

i(k3 k)t i(k3 k)t uˆ(k, t)=ˆu(k, 0)e =ˆu0(k)e .

Then the solution to the PDE is

i[kx (k k3)t] i[kx !(k)t] 3 u(x, t)= uˆ (k)e dk = uˆ (k)e dk, !(k)=k k . 0 0 ZR ZR Asymptotics, APPM 5480 Cancellation Eﬀects

(k) b cos[(k)t] to approximate for large t: f(k)ei(k)t dk sin[(k)t] Za

t = 2 t = 20

k k

Asymptotics, APPM 5480 Warm-Up Problem

Problem: Determine the leading order behavior of the solution to

it + xx =0, (x, 0) = f(x),xR, for t , assuming sucient smoothness and decay on f(x).

Solution: The (exact) solution is obtained by use of the Fourier transform

1 i(kx k2t) (x, t)= fˆ(k)e dk, 2 R 2 ikx where (k)=k is the dispersion relation and fˆ(k)= f(x)e dx. R We can apply the stationary phase method to evaluate this integral for large t 1 x (x, t)= fˆ(k)eit(k;x/t) dk, (k; x/t)=k k2. 2 t R

Stationary points k satisfy (k ; x/t)=x/t 2k = 0 of which there is only 0 0 0 one k = x/(2t). Since (k ; x/t)= 2, the stationary phase formula yields 0 0 fˆ x x 2 x (x, t) 2t exp i t i , = (1),t . 4t 2t 4 t O Asymptotics, APPM 5480 Free Schrodinger and Stationary Phase

iΨt + Ψxx =0, Ψ(x, 0) = sech(x) 1

0.8

) 0.6 u

i + =0 Re( t xx 0.4

0.2

0 (x, 0) = sech(x) 50 40 30 20 10 0 10 20 30 40 50 t =0.5 1.2 exact 1 stationary phase k 0.8 ) ˆ(k, 0) = sech u 0.6 2 Re( 0.4 0.2 0

0.2 50 40 30 20 10 0 10 20 30 40 50

t =5 (x, t) 0.5 exact 0.4 stationary phase 2 x x 0.3 )

sech exp i t i u 0.2

4t 4t 2t 4 Re( 0.1 0 t 0.1

0.2 50 40 30 20 10 0 10 20 30 40 50 x

Asymptotics, APPM 5480 Asymptotic Error Decay Rate

Error in Stationary Phase Approximation 0 10

1 10

2 10

3 10

error 3 2 4 t − 10 0 1 2 10 10 10 t

Asymptotics, APPM 5480 Warm-Up Problem

Problem: Given the surface z = f(x, y)

1. What vectors are orthogonal to the level curve f(x, y)=f(x0,y0)? 2. What is the steepest descent direction at any point on the surface?

x(t) 3. The steepest descent method can be applied to I(x)= C h(t)e dt, x .If(t)=(u + iv)=(u, v)+i(u, v) and d/dt(t0) = 0, why are paths of steepest descent curves of constant ?

Solution: 1. a f(x ,y ), a =0 0 0 2. f/ f | | 3. (u, v)=(u0,v0) is a level curve such that (u, v) is orthogonal to it and (u, v) is tangential to it (by Cauchy-Riemann equations). Generally, points in the direction of steepest descent for the surface z = (u, v). Combining these two facts, we have the result.

Asymptotics, APPM 5480 Warm-Up Problem

Problem: Find the general solution to the second order, linear, constant coef- ﬁcient ODE

y¨ + y = ✏y, t > 0, ✏ > 0.

Now, assume ✏ 1 and determine a regular perturbation expansion for y(t). For what times ⌧t is this expansion valid?

Solution: Seek an exponential solution y(t)=ert leading to the characteristic equation

r2 +(1 ✏)=0 r = ip1 ✏. ) ± Then the general solution is

ip1 ✏t ip1 ✏t y(t)=Ae + Be .

Regular perturbation expansion: y(t)=y (t)+✏y (t)+ . 0 1 ··· it it (1) :y ¨ + y =0 y (t)=Ae + Be O 0 0 ) 0 it it i it it y¨ + y = y = Ae + Be y (t)= t Ae + Be . 1 1 0 ) 1 2 Then y(t)=y (t)+✏y (t)+ is an asymptotic series so long as ✏t 1. 0 1 ··· ⌧ Asymptotics, APPM 5480 Warm-Up Problem

Problem: Consider the nonlinear ODE

y¨ + y y3 =0,t>0. What ansatz and methods can be used to asymptotically analyze weakly nonlinear solutions?

Solution: There is no small parameter in the equation so we have to introduce it ourselves. Weakly nonlinear means small but ﬁnite amplitude. A natural expansion that achieves this is

y = y + 2y + , 0, 0 1 ··· | | where corresponds to the amplitude. Possible methods: The frequency shift method: let = t and expand =1+ +2 + . • 1 2 ··· What is 1? Multiple scales: let y = y(t, T ; )whereT = 2t is the slow timescale. •

Asymptotics, APPM 5480 Warm-Up Problem Problem: Consider the equation for the linear pendulum with slowly varying length l

d2 l(✏t)y + y =0,t>0, 0 < ✏ 1. dt2 ⌧ h i Suppose you use the multiple scales expansion

y(t; ✏)=y (t, T )+✏y (t, T )+ . 0 1 ···

What is the form of y0(t, T )? Is it secular?

Solution: The (1) equation is O

l(T )y0,tt + y0 =0.

Since T is considered frozen with respect to t, the general (real) solution is

it/pl(T ) y0(t, T )=A(T )e + c.c..

This approximation is not uniformly valid because

A(T )l0(T ) it/pl(T ) y = A0(T ) it e + c.c. 0,T 2l(T )3/2  Asymptotics, APPM 5480 Warm-Up Problem

Asymptotics, APPM 5480 Warm-Up Problem

Problem: Use the WKB method to ﬁnd an approximate solution to Schr¨odinger’s equation

2 ✏ y00 = Q(x)y, Q(x) =0, 0 < ✏ 1. 6 ⌧ Solution: Use the ansatz (x; ✏) y(x)=exp , (x; ✏)= (x)+✏ (x)+ . ✏ 0 1 ··· ✓ ◆ 2 2 /✏ Since y00 =(00/✏ + 0 /✏ )e ,weﬁnd

2 (0 + ✏10 + ) + ✏ (000 + ✏100 + ) Q(x)=0, ··· x ··· 2 (1) : 0 = Q(x) (x)= Q(z)dz, O 0 ) 0 ± Z px 000(z) Q0(z) 1 (✏): 20 0 + 00 =0 (x)= dz = dz = ln Q(x) O 0 1 0 ) 1 2 (z) 4Q(z) 4 Z 0 Zx Then an approximate solution is

x x 1/4 1 1/4 1 y(x) c Q (x)exp Q(z)dz + c Q (x)exp Q(z)dz , ✏ 0. ⇠ 1 ✏ 2 ✏ ! ✓ Zx0 ◆ ✓ Zx0 ◆ p p Asymptotics, APPM 5480 Nonlinear Pendulum with Slowly Varying Length

⇡

y(t) 0

⇡ 0 100 t 200

1 l(✏x)=1+ sin(✏x) 2

Asymptotics, APPM 5480 Comparison w/ Linear Solution

Asymptotics, APPM 5480 Warm-Up Problem Problem: Find the two-term outer expansion for the initial value problem

1 2 y0 + + ✏x y =0,x>1,y(1) = e, 0 < ✏ 1. x2 ⌧ ✓ ◆ For what values of x does the expansion become nonuniform? Solution: Assume the regular expansion

y(x; ✏)=y (x)+✏y (x)+ . 0 1 ··· Then

1 2 y0 + ✏y0 + + + ✏x (y + ✏y + )=0 0 1 ··· x2 0 1 ··· ✓ ◆ 1 1/x (1) : y0 + y =0,y(1) = e y (x)=e , O 0 x2 0 0 ) 0 1 2 1/x (✏): y0 + y = x e ,y(1) = 0. O 1 x2 1 1

1/x Use variation of parameters y1(x)=u(x)e

1 1 1/x 1/x 1 1/x 1/x y0 + y = e u + e u0 + e u = e u0 1 x2 1 x2 x2 1 1 = x2e1/x u(x)= x3 + C y (x)= (1 x3)e1/x. ) 3 ) 1 3

1/x ✏ 3 The two-term outer expansion is y(x; ✏)=e 1+ 3 (1 x )+ , ✏ 0. 1/3 ··· ! This expansion breaks down when x = (✏ ). O ⇥ ⇤ Asymptotics, APPM 5480 Some prominentAsymptotologists mathematicians in the area of Asymptotic Analysis

J. H. Poincaré 1854-1912 N. H. Abel 1802-1829 F.W. Bessel 1784-1846 H.E. Padé 1863-1953 France Norway Germany France

G.N. Watson 1886-1965 M. van Dyke 1922-2001 Julian Cole 1925-1999 George Green 1793-1841 England USA USA England

Asymptotics, APPM 5480

Brook Taylor 1685-1731 Colin MacLaurin 1698-1746 Leonhard Euler 1707-1783 Pierre-Simon Laplace 1749-1827 England Scotland Switzerland / Russia France

George B. Airy 1801-1892 George G. Stokes 1819-1903 Ludwig Prandtl 1875-1953 Paco Lagerstrom 1914-1989 England England Germany Sweden / USA Some prominent mathematicians in the area of Asymptotic Analysis

J. H. Poincaré 1854-1912 N. H. Abel 1802-1829 F.W. Bessel 1784-1846 H.E. Padé 1863-1953 France Norway Germany France

G.N. Watson 1886-1965 M. van Dyke 1922-2001 Julian Cole 1925-1999 George Green 1793-1841 England Asymptotologists USA USA England

Brook Taylor 1685-1731 Colin MacLaurin 1698-1746 Leonhard Euler 1707-1783 Pierre-Simon Laplace 1749-1827 England Scotland Switzerland / Russia France

George B. Airy 1801-1892 George G. Stokes 1819-1903 Ludwig Prandtl 1875-1953 Paco Lagerstrom 1914-1989 England England Germany Sweden / USA

Asymptotics, APPM 5480 Boundary Layer Theory Problem

1 2 y + + x y =0,x>1,y(1) = e, 0 < 1. x2 (o) 1/x 3 2 Outer= solution: y (x; )=e 1+ (1 x )+ ( ) �� 3 O ��[{��[(��[� / �]*(� + � / � *(� - ��))) - ��[� / �]*��[� *(� - ��)/�]]� 3 [ [- (i)* 1/3/ ]*( +x//3)- 1[ / ]*2/3 [ *( - )/ ]] * [ > ] Inner solution:�� y� (�� x�; )=�e� � 1+��x �+ �( ��) � � �� [(��[� / �]+(��[-� * ��/ �]-�)*(O� + �)/�)-��[� / �]*��[� *(� - ��)/�]]}� {�� }� ��(c) → {1��/x x ��3/3 ��x+1 }� Composite solution: y (x; ) e + e 1 �� → {{�� }� {��(-�)��}}] x 1

10-2 Outer

�� Inner 10-4 Uniform absolute error

10-6

1 10 100 1000 x Asymptotics, APPM 5480 PDE Boundary Layer Problem

Problem: Find an approximate solution to the initial/boundary value problem

ut + uux = uxxt,xR,t>0, x u(x, 0) = A tanh ,xR, 0 < 1 Downloaded from http://rspa.royalsocietypublishing.org/ on May 31, 2016 lim u(x, t)= A. x ± ± 1 t =0 t = 60 4 rspa.royalsocietypublishing.org u 0 0.5 ...... 0 =0.1,A=1 –0.5 –1 –0.5 0 0.5

1 numerical t = 20 t = 80 u 0 0.5 0.5 solution 0 0 –0.5 –0.5 –1 –0.5 0 0.5 –0.5 0 0.5 rc .Sc A Soc. R. Proc. approximate 1 t = 40 t = 100

solution u 0 0.5 0.5

0 0 472 –0.5 –0.5 –0.5 0 0.5 –0.5 0 0.5 : –1 20160141 –50 0 50 –50 0 50 xx

From El, Hoefer, Shearer, Expansion ShockFigure Waves1. Numerical in Regularized (solid, blue) Shallow-Water and asymptotic Theory, (dashed, Proceedings red) solutions of of the initialRoyal value Society problem A 472 for 201660141 equation (1.1 )with(2016). initial data equation (3.1)whereε 0.1 and A 1. (Online versionincolour.) = = Asymptotics, APPM 5480

some approximation to it. However, we ﬁnd that the dispersive regularization resulting from the BBM equation sustains solutions in which a smoothed stationary shock persists but decays algebraically, as shown in ﬁgure 1.Moreprecisely,westudytheinitialvalueproblemwithinitial data being a smoothed stationary shock, with width ε > 0.

3. The expansion shock To see the effect of dispersion on a stationary shock, we pose initial data x u(x,0) A tanh , < x < , (3.1) = ε −∞ ∞ with amplitude A > 0fortheBBMequation(1.1).Thus,asε 0, the initial data converge to a → jump from u A to u A,representingastationaryexpansionshocksolutiontotheinviscid = − = Burgers equation (1.5). The numerical solution of (1.1) and (3.1) is shown in ﬁgure 1. We observe the development of a rarefaction wave on either side of a stationary but decaying shock. We analyse the solution by matched asymptotics using ε as the small parameter. First note that the initial function u(x,0) is an odd function, and the solution u(x, t)shouldthereforebeanodd function of x for each t > 0.

(a) The inner solution To capture the inner solution, we introduce into equation (1.1) the short space ξ x/ε and long = time T εt scalings of the independent variables x and t,yielding = 1 1 εu uuξ uξξ . (3.2) T + ε = ε T Expanding the dependent variable

u u (ξ, T) εu (ξ, T) (3.3) = 0 + 1 +··· Downloaded from http://rspa.royalsocietypublishing.org/ on May 31, 2016

–1 10 7 rspa.royalsocietypublishing.org ......

10–2 abs error 10–3 Expansion Shock Wave 10–4 –100 –50 0 50 100 Characteristics:x curves along A Soc. R. Proc. Figure 2. Pointwise error between the uniform asymptotic expansion and numerical solution of fgure 1 at t 100 with Downloaded from http://rspa.royalsocietypublishing.org/ on May 31, 2016 which solution is constant Absolute errorε 0.1 and A 1. = = = Compressive Expansive 472

in composite solution : 20160141 10–1 shock shock (a)(100 7 b) rspa.royalsocietypublishing.org ...... 75 10–2

t 50 abs error 10–3 25

10–4 0 –100 –50 0 50 100 –100 –50 0 50 100 –100 –50 0 50 100 x x A Soc. R. Proc. x u(x, 0) u(x, 0) Figure 2. Pointwise error between the uniformt = asymptotic 100, expansion =0.1 and,A numericalFigure=13 solution. Characteristics of fgure for:1 at (at ) the100 compressivewith shock solution of equation (1.5) and (b) the expansion shock solution of ε 0.1 and A 1. equation (1.1). (Online version in1 colour.)= 1

= = 472 : 20160141 (a)(100 b) the stationary shock from left to right. However, the characteristics with speed λ leave the shock 1 1 − on both sides if u < h and u > h . This is the case for the choices of h , u in ﬁgure 4. − − + + ± ± 75 These choices also satisfy! the RH conditions! for a stationary shock (3.11). To see that similar data with u < 0canmakeastationaryshockexpansiveintheλAsymptotics,characteristic APPM 5480 family, note that the + system (3.9) is unchanged under the transformation x x, u u. t 50 →− →− We remark that the analytical treatment of the Boussinesq expansion shock appears to be more challenging than it was for BBM. For example, there is no clear means to separate variables in an 25 inner solution owing to the non-zero mean values of h and u.

0 –100 –50 0 50 100 –100 –50 0 50 100 x (e) Discussionx The expansion shock solutions we have discovered here do decay slowly in time, but their Figure 3. Characteristics for: (a) the compressive shock solution of equation (1.5) and (b) the expansion shock solution of persistence in the face of the usual rules of causality is a surprise. For the BBM expansion shock, equation (1.1). (Online version in colour.) we can identify further robustness to perturbation by considering the asymmetric initial condition passing through zero 1 x u(x,0) (u u )tanh u u , the stationary shock from left to right. However, the characteristics with speed λ leave the shock= + − − ε + + + − − 2 on both sides if u < h and u > h . This is the case for the choices of h , u in ﬁgure 4. " " # # − − + + where u < 0 < u± .Thenumericalsimulationofequation(1.1)withthisasymmetricdatais± These choices also satisfy the RH conditions for a stationary shock (3.11).− To see+ that similar data ! ! shown in ﬁgure 5.Ast increases, the solution quickly develops a stationary expansion shock with with u < 0canmakeastationaryshockexpansiveintheλ characteristic family, note that the + system (3.9) is unchanged under the transformation x x, u u. →− →− We remark that the analytical treatment of the Boussinesq expansion shock appears to be more challenging than it was for BBM. For example, there is no clear means to separate variables in an inner solution owing to the non-zero mean values of h and u.

(e) Discussion The expansion shock solutions we have discovered here do decay slowly in time, but their persistence in the face of the usual rules of causality is a surprise. For the BBM expansion shock, we can identify further robustness to perturbation by considering the asymmetric initial condition passing through zero 1 x u(x,0) (u u )tanh u u , = 2 + − − ε + + + − " " # # where u < 0 < u .Thenumericalsimulationofequation(1.1)withthisasymmetricdatais − + shown in ﬁgure 5.Ast increases, the solution quickly develops a stationary expansion shock with