# Critical Thresholds in Euler-Poisson Equations. Shlomo Engelberg Jerusalem College of Technology Machon Lev

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1 Critical Thresholds in Euler-Poisson Equations Shlomo Engelberg Jerusalem College of Technology Machon Lev 1

2 Publication Information This work was performed with Hailiang Liu & Eitan Tadmor. These results appeared in the Indiana University Math Journal, Vol. 50, No. 1 (2001). 2

3 Shock Formation a Summary t = 0 t = 0 t = 1/2 t = 1 t = 3/2 t = 2 speed speed speed speed speed x x x x x 3

4 Conservation of Mass Letρ(x,t)bethedensityat(x,t)and letu(x,t)bethevelocityoftheparticles at position x at time t. d b dt a ρ(x,t)dx = ρ(a,t)u(a,t) ρ(b,t)u(b,t) b a { t ρ(x,t) + (ρ(x,t)u(x,t))}dx = 0. x If ρ and u are continuously differentiable, then we find that: t ρ+ (ρu) = 0. x 4

5 Newton s Second Law and Its Implications In general, we know that F = ma. Letx(t,x 0 )bethepositionoftheparticle that was at position x 0 when t = 0. Then: d dt x(t,x 0) = u(x(t,x 0 ),t). As dx/dt = u, we find that: F = m d2 dt 2x(t,x 0) d = m dt u(x(t,x 0),t) = m t u+ d dt x x u = m t u+u x u 5

6 Newton s Second Law and Its Implications Continued We find: ρ(u t +uu x ) = F. If there are no interactions, then F = 0, and u t +uu x = 0. If the particles are electrons and there is no fixed background charge, then the forceincludesatermoftheformρe. E satisfies the equation: x E(x,t) = ρ(x). 6

7 Newton s Second Law and Its Implications Continued If the particles are electrons, and there is a fixed positive background charge, then the force includes a term of the form ρe. E satisfies the equation: x E(x,t) = ρ(x) 1. Finally, if the fluid is assumed to be viscous, then the force includes a term of the form: u xx /ρ. 7

8 Some Systems We Consider u t +uu x = 0. (1) ρu t +ρuu x = ρe (2) ρ t +(ρu) x = 0 (3) x E = ρ. (4) u t +uu x = E (5) ρ t +(ρu) x = 0 (6) x E = ρ 1. (7) u t +uu x = E + u x (8) x ρ ρ+uρ x = u x ρ (9) x E = ρ. (10) 8

9 Why Solutions of the Burgers Equation Loose Smoothness The Burgers equation describes the motion of particles that move without any forces acting on them. Thus, each particle continues moving at its initial speed. That means that if the particles in the rear are moving faster than those in front, then particles in the rear should catch up with the ones in the front. 9

10 Graphical Method t * x u u 2 u 1 x x m 2 u 1 u 2 2 m x 10

11 Analytical Method u t +uu x = 0. Letz(t) = x u(x(t,x 0),t). Thendz/dt = z 2. Thus, 1 z(x,t) = 1/z(0)+t. From the definition of z, we find that: x u(x(t,x 1 0),t) = 1/( x u(x 0,0)+t). Clearly,if x u(x 0,0) < 0foranyvalue ofx 0,then x u ast 1/ x u(x 0,0). 11

12 The Burgers -Poisson Equations I u t +uu x = E ρ t +(ρu) x = 0 x E = ρ. These are the Burgers Equation with a nonlocal term (the electric field) and conservation of mass added. They are a hyperbolic system with a nonlocal term added. 12

13 Conservation of Momentum On Physical Grounds, we require that the total momentum be conserved. Multiplying the first equation by ρ, the second by u and adding, we find that: (ρu) t +(ρu 2 ) x = KρE = E 2 2. Integrating with respect to x, we find that: ( ρudx ) t +(ρu2 ) x= = E2 2 x= This is the equation that describes the evolution of the total momentum of the system.. 13

14 Conservation of Momentum Effect on E If one starts with reasonable initial conditions (say ρ and u being of compact support) then one can require that: ρu 2 0 as x ±. For the total momentum to be conserved: E 2 ( ) = E 2 ( ). As E x = ρ > 0, we find that E is an increasing function of x. Thus, we find that E( ) = E( ). We find that the electric field is given by: E(x,t) = 1 2 ( x ρ(ξ,t)dξ x ρ(ξ,t)dξ ). 14

15 More about the Electric Field We have found that the electric field is, in this case, half the difference of the amountofchargetotherightoftheparticle and the amount of charge to the left of the particle. As long as particles do not collide as long as the solution remains smooth this number must be fixed along particle trajectories along characteristics. We see that: d u(x(t,α),t) = E(x(t,α),t). dt 15

16 The Characteristic Curves Let: We find that: E 0 E(α,0) u 0 u(α,0) ρ 0 ρ(α,0). u(x(t,α),t) = u 0 +E 0 t. Given u(x(t,α)) it is easy to see that: x(t,α) = α+u o t+e 0 t 2 /2. If and when these curves start intersecting one another, the solution must lose smoothness. 16

17 When Characteristics Collide... As the characteristic curves first collide precisely when: x(t,x o ) = 0 α for the first time, we examine the equation: x(t,x o ) = 1+u α 0t+ρt 2 /2 = 0. The zeros of this equation are: t = u 0± (u 0 )2 2ρ 0 ρ 0. 17

18 When Characteristics Collide... We find that if u 0 < 0 at some point, and at that point: u 0 > 2ρ 0 then the solution loses smoothness in finitetime. Ifthesolutiondoesnotbreak down, then ρ as t. 18

19 The Burgers -Poisson Equations II u t +uu x = E ρ t +(uρ) x = 0 x E = ρ 1. The energy associated with this equation is (ρu 2 /2+E 2 /2)dx. Forthisquantitytobefinite,E(±,t) = 0andρu 2 (±,t) = 0. Forchargeneutrality to be maintained, Therefore, (ρ(x) 1)dx = 0. E(x,t) = x (ρ(ξ,t) 1)dξ. 19

20 The Solution Thepositionofaparticleofour fluid, x(t,x 0 ), satisfies the equation: d dt x(t,x 0) = u(x(t,x 0 ),t). Newton s second law reads: d dt u(x(t,x 0),t) = E(x(t,x 0 ),t). Note that: d dt E(x(t,x 0),t) = d dt x(t,x 0)(ρ(x(t,x 0 ),t) 1) + x(t,x 0 ) ρ t (ξ,t)dξ = d dt x(t,x 0)(ρ(x(t,x 0 ),t) 1) x(t,x 0 ) (ρu(ξ,t)) x dξ = u(x(t,x 0 ),t). 20

21 The Solution Continued Combining these results, we find that: d 2 dt 2u(x(t,x 0),t) = u(x(t,x 0 ),t). Solving for all the terms of interest, we find that: u(x(t,x 0 ),t) = u 0 cos(t)+e 0 sin(t) x 0 u(x(t,x 0 ),t) = u x x x0 (t,x 0 ) = u 0cos(t)+E 0sin(t) u x (x(t,x 0 ),t) = u 0(x 0 )cos(t)+ρ 0 (x 0 )sin(t) D(x 0,t)( x x0 (t,x 0 )) ρ(x,t) = ρ 0(x 0 ) D(x 0,t). 21

22 When Characteristics Collide... We find that the denominator, D(x 0,t) is: D(x 0,t) = 1+u 0(x 0 )sin(t) ρ 0 (x 0 )(cos(t) 1). We find that there is a global in time solution if and only if: u (x 0 ) < 2ρ(x 0 )+1 is true for all x 0. 22

23 The Euler-Poisson Equations We previously showed that for the Euler- Poisson equations(which include a pressure term) it is possible to extend Lax s trick for 2x2 systems. In that case we were able to get conditions that guaranteed that the solution would remain smooth for all time. We also gave sufficient conditions for finite-time breakdown. Those(and other) results appeared in: Formation of Singularities in the Euler and Euler-Poisson Equations, Physica D. 98 (1996). 23

24 Spherical Symmetry Suppose that: r = x 2 +y 2 +z 2 ρ(x,y,z,t) = ρ(r,t) u(x,y,z,t) = u(r,t)ˆr. The divergence of a spherically symmetric function satisfies: (f(r)ˆr) = 1 r 2 r (r2 f(r)). 24

25 The Spherically Symmetric Burgers-Poisson Equations We find that conservation of mass can be written as: (r 2 ρ(r,t)) t +(r 2 ρ(r,t)u(r,t)) r = 0 Similarly, we find that the electric field satisfies the equation: (r 2 E(r,t)) t = r 2 ρ(r,t). Finally, by using the spherical symmetry one can also show that the equation for conservation of momentum can be written as: u(r,t) t +u(r,t)u(r,t) r = E. 25

26 Some Manipulations Let: e(r,t) r 2 E(r,t) ρ(r,t) r 2 ρ(r,t). With these definitions, we find that the equations that we must solve are: ρ t +( ρu) r = 0 u t +uu r = e r 2 e r = ρ(r,t). 26

27 The Electric Field is Constant along Characteristics Let x(t,α) be the (radial) position of a particle that is at position r = α when t = 0. As e r = ρ, we find that: e(x(t,α),t) = x(t,α) 0 ρ(ξ, t) dξ. As thisisjust the charge to the left of a fixed characteristic, this number must be constant until(and unless) characteristics collide. 27

28 Newton s Second Law We know that: d dt u(x(t,α),t) = e x 2 (t,α). We also know that: d x(t,α) = u(x(t,α),t). dt We find that in our case Newton s second law reads: d 2 dt 2x(t,α) = e x 2 (t,α). 28

29 An Appropriate Definition of Energy Let us consider the energy associated with the ODE: y = g(y). Picking E(y) = (y ) 2 + f(y), we find that to force the energy to be constant along solutions we must have: E (y) = 2y y +f (y)y = y (2g(y)+f (y)) = 0. We find that: f(y) f(y 0 ) = 2 y y0 g(y)dy. 29

30 The Correct Energy In our case: g(y) = e y 2. We find that f(y) must be: f(y) f(y 0 ) = 2 e y. Finally, we find that the energy is: E(y) = (y ) 2 +2 e y. 30

31 The Energy in Our Case Noting that: d x(t,α) = u(x(t,α),t), dt we find that the fact that the energy in our system is constant implies that: u 2 (x(t,α),t) = u 2 1 (α,0)+2e(α,0) α 1 x(t, α) 31

32 Consequences Asourparticleismovingawayfromthe originataspeedthatalwaysexceedsits initial speed, we know that in the fullness of time, x(t,α). Thus, the final speed of the particle that started at position α is just: Q(α) lim t u(x(t,α),t) Clearly if: = u 2 (α,0)+ 2e(α,0). α d dα Q(α) < 0 then the solution will lose smoothness in finite time. (The particles in the back will catch up with those in the front!) 32

33 The Final Form of the Condition The solution of the spherically symmetric equation will lose smoothness in finitetime if there exists an α for which: u(α,0) α < 1 u(α,0) [ρ(α,0)α E(α,0)]. One can also use the energy to find conditions for global in time existence though this is trickier. 33

34 How Shocks Form Consider the Burgers equations with the equation for conservation of mass adjoined to it. If we consider u x /ρ, it is easy to show that: d u x dt ρ = 0. As we know that u x as a shock forms, we find that: ρ as a shock forms. From the equations for conservation of mass it is clear that: d dt ρ = u xρ. This too implies that if mass is conserved, then as u steepens ρ must increase. 34

35 The Viscous Burgers Equation u t +uu x = u xx Thisequationisusedtomodelasetof particles that only interact with one another through viscosity. It is well known that the initial-value problem for this equation is well posed. 35

36 The Modified Viscous Burgers-Poisson Equations u t +uu x = E + ρ+uρ x = u x ρ x E = ρ. u x ρ x Burgers becauseofthelackofapressure term. Modified viscous because the viscous term is not of the form u xx /ρ. Poisson because of the inclusion of the electric field. Note that as u s profiles steepens, we know that ρ increases and the effect of the viscous term decreases. The viscosity has its smallest effect when we need it most! 36

37 An Interesting Parabolic Equation Let w = u x. Differentiating the first equation with respect to x gives us w t +uw x = w 2 +ρ+(w/ρ) xx. Multiplying this equation by ρ and subtracting from the resulting equation the second equation multiplied by w, we find that: ρw t wρ t +uρw x uρ x w = ρ 2 +ρ(w/ρ) xx. If one divides this equation by ρ 2, one finds that: α t +uα x = 1+α xx /ρ where α = w/ρ. This equation is a nonlinear parabolic equations for α. 37

38 Goals We would like to make use of a maximum principle to claim that α must satisfy: inf x 0 α(x 0,0)+t α(x,t) sup x 0 α(x 0,0)+t. Unfortunatelyitisnotclearthatamaximum principle obtains as the diffusion coefficient, 1/ρ(x, t), generally tends to infinity at ±. Later we show that α does satisfy a maximum principle. The maximum principle shows that u x /ρ is bounded in a nice fashion. 38

39 Some Manipulations From conservation of mass, we find that: d dt ρ(x(t,x 0),t) = wρ. This implies that: ρ(x(t,x 0 ),t) = ρ(x 0,0)e t 0 w(x(ξ,x 0 ),ξ)dξ. One sees then that: α = w/ρ = we t 0 w(x(ξ,x 0 ),ξ)dξ /ρ(x 0,0) = d dt e t 0 w(x(ξ,x 0 ),ξ)dξ /ρ(x 0,0). 39

40 Results Making use of the previous equation and the maximum principle(whose proof we discuss shortly), we find that: ρ(x 0,0) ρ(x 0,0)(t 2 /2+sup x α(x,0)t)+1 ρ(x(t,x 0 ),t) ρ(x 0,0) ρ(x 0,0)(t 2 /2+inf x α(x,0)t)+1. and that as long as the numerators remain positive: ρ(x 0,0)(t+inf x α(x,0)) ρ(x 0,0)(t 2 /2+inf x α(x,0)t)+1 w(x(t,x 0 ),t) ρ(x 0,0)(t+sup x α(x,0)) ρ(x 0,0)(t 2 /2+sup x α(x,0)t)+1 Clearly, we can have finite-time blowup (or loss of niceness ) under the right conditions. 40

41 An Example Supposethatwetakeu(x,0) = 2tanh(x) and ρ(x,0) = sech 2 (x). Then we find that α(x,0) = 2, and we must have finite-time blowup or loss of niceness. The weakest ways that the solution can stop being nice lead to loss of smoothness or lead to u as x ±. It is easy to show that if the solution remains nice, then as t we find that w 2/t and ρ 2/t 2. This is precisely the behavior seen in the solutions of the equation without any viscosity. 41

42 The Maximum Principle Theorem 1.1 (Maximum Principle) Ifu t +f(x,t)u x = a(x,t)u xx, u(x,t) D(x)E(t) where D(x) is sub-linear in x and E(t) is exponential in time; u,u t,u x,u xx C, f(x,t) d; and 0 a(x,t); a(x,t) C, then the solution of the equation obeys the maximum principle: u(x,t) sup x u(x,0). We will explain later why the bound on f is truly needed. 42

43 The Elements of the Proof We prove this maximum principle by finding a function, z(x,t) that is positive, that grows exponentially in time and linearly in space, and that satisfies z t +f(x,t)z x a(x,t)z xx. The function z(x,t) is assumed to be of the form F(x)H(t). We find an F thatsatisfiesf = F/(2b)whereb(x) > a(x,t) and b(x) cx 2+ǫ +1. Because b(x) is a rapidly growing function, we know that the solution of the ODE behaves linearly at ±. Also, F is concave up. Finally, we find that H(t) = e t. Thus, z(x,t) = e t F(x). 43

44 The Condition on f(x,t) Letu t = (b(x)u x ) x = b (x)u x +b(x)u xx. Letususeseparationofvariablestofind asolutionofthisequations. Letu(x,t) = F(x)H(t). Then F(x) = (F (x)b(x)). ConsiderthesolutionofG (x) = G(x)/b(x). G(x) behaves linearly at infinity. Let F = G. Then F satisfies the equation. Moreover: lim x ± F(x) = c ±. H(t) = e t solves the equation for H. Thus, the function u(x,t) = e t F(x) solves the original equation. We note that the supremum and the infimum of this function are not bounded function of time. 44

45 Implications The original equation for u(x,t) is u t + uu x = E+(u x /ρ) x. Asthisisparabolic, one might hope for a maximum principal for u(x,t). Since ρ generally decays at ±, we find that the hyperbolic part has an unbounded coefficient. As our previous example showed, in such cases there will not generally be any maximum principle. 45

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