Symbolic Computation of Conservation Laws of Nonlinear Partial Differential Equations in Multiple Space Dimensions

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1 Symbolic Computation of Conservation Laws of Nonlinear Partial Differential Equations in Multiple Space Dimensions Willy Hereman and Douglas Poole Department of Mathematical and Computer Sciences Colorado School of Mines Golden, Colorado AMS 2010 Central Sectional Spring Meeting Macalester College, St. Paul, Minnesota Saturday, April 10, 2010, 9:30a.m.

2 Acknowledgements Mark Hickman (Univ. of Canterbury, New Zealand Bernard Deconinck (Univ. of Washington, Seattle Several undergraduate and graduate students Research supported in part by NSF under Grant CCF This presentation was made in TeXpower

3 Outline Why compute conservation laws? The Korteweg-de Vries and Zakharov-Kuznetsov equations Demonstration of ConservationLawsMD.m Algorithm for computing conservation laws Tools from the calculus of variations Applicaton to the Zakharov-Kuznetsov equation Conclusions and future work

4 Additional examples Manakov-Santini system Camassa-Holm equation Khoklov-Zabolotskaya equation Shallow water wave model for atmosphere Kadomtsev-Petviashvili equation Potential Kadomtsev-Petviashvili equation Generalized Zakharov-Kuznetsov equation

5 Conservation Laws for Nonlinear PDEs System of evolution equations of order M u t = F(u (M (x with u = (u, v, w,... and x = (x, y, z. Conservation law in (1+1-dimensions D t ρ + D x J = 0 evaluated on the PDE. Conserved density ρ and flux J.

6 Conservation law in (2+1-dimensions D t ρ + J = D t ρ + D x J 1 + D y J 2 = 0 evaluated on the PDE. Conserved density ρ and flux J = (J 1, J 2. Conservation law in (3+1-dimensions D t ρ + J = D t ρ + D x J 1 + D y J 2 + D z J 3 = 0 evaluated on the PDE. Conserved density ρ and flux J = (J 1, J 2, J 3.

7 Reasons for Computing Conservation Laws Conservation of physical quantities (linear momentum, mass, energy, electric charge,... Testing of complete integrability and application of Inverse Scattering Transform Testing of numerical integrators Study of quantitative and qualitative properties of PDEs (Hamiltonian structure, recursion operators,... Verify the closure of a model

8 Examples of PDEs with Conservation Laws Korteweg-de Vries (KdV equation models shallow water waves, ion-acoustic waves in plasmas, etc. u t + u u x + 3 u x 3 = 0 or u t + uu x + u xxx = 0 Diederik Korteweg Gustav de Vries

9 First six (of infinitely many conservation laws: ( D t (u + D 12 x u2 + u xx = 0 ( D t u 2 ( + D 2 x 3 u3 u 2 x + 2uu xx = 0 D t (u 3 3u 2 x ( + D 3 x 4 u4 6uu 2 x + 3u2 u xx + 3u 2 xx 6u xu xxx = 0 ( D t (u 4 12uu 2 x u2 xx + D 4 x 5 u5 18uu 2 x + 4u 3 u xx +12u 2 xu xx uu2 xx 24uu x u xxx 36 5 u2 xxx u xxu 4x = 0

10 ( D t u 5 30 u 2 u 2 x + 36 uu 2 xx u2 xxx + D x ( 5 6 u6 40u 3 u 2 x u xxxu 5x = 0 D t (u 6 60 u 3 u 2 x 30 u 4 x u 2 u 2 xx u3 xx uu2 xxx u2 4x + D x ( 6 7 u7 75u 4 u 2 x u 4xu 6x = 0 Third conservation law: Gerald Whitham, 1965 Fourth and fifth: Norman Zabusky, Seventh (sixth thru tenth: Robert Miura, 1966

11 Conservation law explicitly dependent on t and x: D t (tu 2 2xu ( +D 2 x 3 tu3 xu 2 + 2u x tu 2 x + 2tuu xx 2xu xx = 0

12 First eleven densities: Control Data Computer CDC-6600 computer (2.2 seconds large integers problem! Control Data CDC-6600

13 Zakharov-Kuznetsov (ZK equation models ion-sound solitons in a low pressure uniform magnetized plasma u t + αuu x + β(u xx + u yy x = 0 Conservation laws: ( D t (u + D α2 x u2 + βu xx + D y (βu xy = 0 D t (u 2 ( + D 2α x 3 u3 β(u 2 x u2 y + 2βu(u xx + u yy ( D y 2βu x u y = 0

14 More conservation laws (ZK equation: ( ( D t u 3 3β α (u2 x + u2 y + D 3α x 4 u4 + 3βu 2 u xx 6βu(u 2 x + u2 y + 3β2 α (u2 xx u 2 yy 6β2 + D y ( 3βu 2 u xy + 6β2 α u xy(u xx + u yy α (u x(u xxx + u xyy + u y (u xxy + u yyy = 0 ( D t tu 2 2 α xu + D x (t( 2α 3 u3 β(u 2 x u 2 y + 2βu(u xx + u yy ( x(u 2 + 2β α u xx + 2β α u x D y 2β(tu x u y + 1 α xu xy = 0

15 Methods for Computing Conservation Laws Use Noether s theorem (Lagrangian formulation: connection between symmetries and conservation laws (Olver, and many others Integrating factor methods (Anderson, Bluman, Anco, Cheviakov, Wolf, etc. require solving ODEs (or PDEs

16 Proposed Algorithmic Method Density is linear combination of scaling invariant terms with undetermined coefficients Compute D t ρ with total derivative operator Use variational derivative (Euler operator to express exactness Solve a (parametrized linear system to find the undetermined coefficients Use the homotopy operator to compute the flux (invert D x or Div

17 Work with linearly independent pieces in finite dimensional spaces Use linear algebra, calculus, and variational calculus (algorithmic Implement the algorithm in Mathematica

18 Software Demonstration

19 Notation Computations on the Jet Space Independent variables x = (x, y, z Dependent variables u = (u (1, u (2,..., u (j,..., u (N In examples: u = (u, v, θ, h,... Partial derivatives u kx = k u x k, u kx ly = k+l u x k y l, etc. Examples: u xxxxx = u 5x = 5 u u xx yyyy = u 2x4y = 6 u x 2 y 4 u (M represents all components of u and all its x 5 partial derivatives up to order M. Differential functions Example: f = uvv x + x 2 u 3 x v x + u x v xx

20 Tools from the Calculus of Variations Definition: A differential function f is a exact iff f = DivF. Special case (1D: f = D x F. Question: How can one test that f = DivF? Theorem (exactness test: f = Div F iff L u (j (x f 0, j = 1, 2,..., N. N is the number of dependent variables. The Euler operator annihilates divergences

21 Euler operator in 1D (variable u(x: M L u(x = ( D x k k=0 = u D x u kx u x + D 2 x u xx D 3 x u xxx + Euler operator in 2D (variable u(x, y: L u(x,y = M x k=0 M y ( D x k ( D y l l=0 = u D x + D 2 x u x D y u xx +D x D y u y +D 2 y u xy u kx ly u yy D 3 x u xxx

22 Question: How can one compute F = Div 1 f? Theorem (integration by parts: In 1D: If f is exact then F = D 1 x f = f dx = H u(x f In 2D: If f is a divergence then F = Div 1 f = (H (x u(x,y f, H(y u(x,y f The homotopy operator inverts total derivatives and divergences!

23 Homotopy Operator in 1D (variable x: H u(x f = 1 0 N (I u (jf[λu] dλ λ j=1 with integrand I u (jf = M x (j k=1 k 1 i=0 ix ( D x k (i+1 u (j f u (j kx (I u (jf[λu] means that in I u (jf one replaces u λu, u x λu x, etc. More general: u λ(u u 0 + u 0 u x λ(u x u x 0 + u x 0 etc.

24 Application to Zakharov-Kuznetsov Equation u t + αuu x + β(u xx + u yy x = 0 Step 1: Compute the dilation invariance ZK equation is invariant under scaling symmetry (t, x, y, u ( t λ 3, x λ, y λ, λ2 u= ( t, x, ỹ, ũ λ is an arbitrary parameter. Assign weights to each variable W (u = 2, W (D t = 3, W (D x = 1, W (D y = 1. Rank of a monomial is the sum of the weights of the variables. Example: Rank(αuu x = 2W (u + W (D x = 5.

25 Key observation: A conservation law is invariant under the scaling symmetry of the PDE W (u = 2, W (D t = 3, W (D x = 1, W (D y = 1. For example, ( D t (u 3 3β α (u2 x +u 2 y + D 3α x 4 u4 +3βu 2 u xx 6βu(u 2 x + u 2 y + 3β2 α (u2 xx u 2 xy 6β2 α (u x(u xxx +u xyy + u y (u xxy +u yyy ( + D y 3βu 2 u xy + 6β2 α u xy(u xx + u yy = 0 Rank(ρ = 6, Rank(J = 8. Rank (conservation law = 9. Rank of the density needs to be selected!

26 Step 2: Construct the candidate density For example, construct a density of rank 6. Make a list of all terms with rank 6: {u 3, u 2 x, uu xx, u 2 y, uu yy, u x u y, uu xy, u 4x, u 3xy, u 2x2y, u x3y, u 4y } Remove divergences and divergence-equivalent terms. Candidate density of rank 6: ρ = c 1 u 3 + c 2 u 2 x + c 3 u 2 y + c 4 u x u y

27 Step 3: Compute the undetermined coefficients Compute D t ρ = ρ t + ρ (u[u t ] = ρ t + = M x k=0 M y l=0 ρ u kx ly D k x Dl y u t ( 3c 1 u 2 I + 2c 2 u x D x + 2c 3 u y D y + c 4 (u y D x + u x D y Substitute u t = (αuu x + β(u xx + u yy x. u t

28 E = D t ρ = 3c 1 u 2 (αuu x + β(u xx + u xy x + 2c 2 u x (αuu x + β(u xx + u yy x x + 2c 3 u y (αuu x + β(u xx + u yy x y + c 4 (u y (αuu x + β(u xx + u yy x x + u x (αuu x + β(u xx + u yy x y Apply the Euler operator (variational derivative M x M y L u(x,y E = ( D x k ( D y l E k=0 l=0 u kx ly ( = 2 (3c 1 β + c 3 αu x u yy + 2(3c 1 β + c 3 αu y u xy +2c 4 αu x u xy +c 4 αu y u xx +3(3c 1 β +c 2 αu x u xx 0

29 Solve a parameterized linear system for the c i : 3c 1 β + c 3 α = 0, c 4 α = 0, 3c 1 β + c 2 α = 0 Solution: c 1 = 1, c 2 = 3β α, c 3 = 3β α, c 4 = 0 Substitute the solution into the candidate density ρ = c 1 u 3 + c 2 u 2 x + c 3 u 2 y + c 4 u x u y Final density of rank 6: ρ = u 3 3β α (u2 x + u 2 y

30 Step 4: Compute the flux Use the homotopy operator to invert Div: ( J=Div 1 E = H (x u(x,y E, H(y u(x,y E where with I (x u E = H (x M x k=1 u(x,y E = 1 M y l=0 ( k 1 i=0 0 l j=0 (I u (x E[λu] dλ λ ( i+j i u ix jy ( D x k i 1 ( D y l j ( k+l i j 1 k i 1 ( k+l k E u kx ly Similar formulas for H (y u(x,y E and I(y u E.

31 Let A = αuu x + β(u xxx + u xyy so that E = 3u 2 A 6β α u xa x 6β α u ya y Then, ( J = H (x u(x,y E, H(y u(x,y E = ( 3 4 αu4 + βu 2 (3u xx + 2u yy βu(6u 2 x + 2u 2 y + 3β2 4α u(u 2x2y + u 4y β2 α u x( 7 2 u xyy + 6u xxx β2 α u y(4u xxy u yyy + β2 α (3u2 xx u2 xy u2 yy + 5β2 4α u xxu yy, βu 2 u xy 4βuu x u y 3β2 4α u(u x3y + u 3xy β2 4α u x(13u xxy + 3u yyy 5β2 4α u y(u xxx + 3u xyy + 9β2 4α u xy(u xx + u yy

32 However, Div 1 E is not unique. Indeed, J = J + K, where K = (D y θ, D x θ is a curl term. For example, θ =2βu 2 u y + β2 4α Shorter flux: ( 3u(u xxy +u yyy +10u x u xy +5u y (3u yy +u xx J = J K ( = 3α 4 u4 + 3βu 2 u xx 6βu(u 2 x + u 2 y + 3β2 α ( u 2 xx u 2 yy 6β2 α (u x(u xxx + u xyy + u y (u xxy + u yyy, 3βu 2 u xy + 6β2 α u xy(u xx + u yy

33 Additional Examples Manakov-Santini system u tx + u yy + (uu x x + v x u xy u xx v y = 0 v tx + v yy + uv xx + v x v xy v y v xx = 0 Conservation laws for Manakov-Santini system: ( D t fu x v x + D x (f(uu x v x u x v x v y u y v y ( f y(u t + uu x u x v y + D y f(u x v y + u y v x + u x vx 2 + f (u yu y yu x v x = 0 where f = f(t is arbitrary.

34 Conservation laws continued: ( D t (f(2u + vx 2 yu x v x + D x f(u 2 + uvx 2 + u y v vy 2 vxv 2 y y(uu x v x u x v x v y u y v y f y(v t + uv x v x v y (2fx f y 2 (u t + uu x u x v y D y (f(u x v 2v x v y vx 3 + y(u x vx 2 + u x v y + u y v x f (v y(2u + v y + vx 2 + (2fx f y 2 (u x v x + u y =0 where f = f(t is arbitrary. There are three additional conservation laws.

35 (2+1-dimensional Camassa-Holm equation (αu t + κu x u txx + 3βuu x 2u x u xx uu xxx x + u yy = 0 Interchange t with y (αu y + κu x u xxy + 3βuu x 2u x u xx uu xxx x + u tt = 0 Set v = u t to get u t = v v t = αu xy κu xx + u 3xy 3βu 2 x 3βuu xx + 2u 2 xx + 3u x u xxx + uu 4x

36 Conservation laws for the Camassa-Holm equation D t (fu + D x ( 1 α f(3 2 βu2 + κu 1 2 u2 x uu xx u tx ( 1 α fx 1 2 f y 2 (αu t + κu x + 3βuu x 2u x u xx uu xxx ( u txx D y u y ( 1 α fx 1 2 f y 2 + f yu = 0 ( D t (fyu + D 1 x α fy( 3 2 βu2 + κu 1 2 u2 x uu xx u tx y( 1 α fx 1 6 f y 2 (αu t + κu x + 3βuu x 2u x u xx uu xxx ( u txx D y yu y ( 1 α fx 1 6 f y 2 u( 1 α fx 1 2 f y 2 = 0 where f = f(t is an arbitrary function.

37 Khoklov-Zabolotskaya equation describes e.g. sound waves in nonlinear media (u t uu x x u yy u zz = 0 Conservation law: ( D t (fu D 1 x 2 fu2 + (fx + g(u t uu x ( D y (f y x + g y u (fx + gu y ( D z (f z x + g z u (fx + gu z = 0 under the constraints f = 0 and g = f t where f = f(t, y, z and g = g(t, y, z.

38 Shallow water wave model (atmosphere u t + (u u + 2 Ω u + (θh 1 2 h θ = 0 θ t + u ( θ = 0 h t + (uh = 0 where u(x, y, t, θ(x, y, t and h(x, y, t. In components: u t + uu x + vu y 2 Ωv hθ x + θh x = 0 v t + uv x + vv y + 2 Ωu hθ y + θh y = 0 θ t + uθ x + vθ y = 0 h t + hu x + uh x + hv y + vh y = 0

39 First few conservation laws of SWW model: ρ (1 = h ρ (2 = h θ J (1 = h J (2 = h θ ρ (3 = h θ 2 J (3 = h θ 2 ρ (4 = h (u 2 + v 2 + hθ ρ (5 = θ (2Ω + v x u y J (5 = 1 2 θ J (4 = h u v 4Ωu 2uu y + 2uv x hθ y 4Ωv + 2vv x 2vu y + hθ x u v u v u (u2 + v 2 + 2hθ v (v 2 + u 2 + 2hθ

40 More general conservation laws for SWW model: ( ( D t (f(θh + D x f(θhu + D y f(θhv = 0 D t (g(θ(2ω + v x u x ( + D 1 x 2 g(θ(4ωu 2uu y + 2uv x hθ y ( + D 1 y 2 g(θ(4ωv 2u yv + 2vv x + hθ x = 0 for any functions f(θ and g(θ.

41 Kadomtsev-Petviashvili (KP Equation (u t + αuu x + u xxx x + σ 2 u yy = 0 parameter α IR and σ 2 = ±1. Equation be written as a conservation law D t (u x + D x (αuu x + u xxx + D y (σ 2 u y = 0. Exchange y and t and set u t = v u t = v v t = 1 σ (u 2 xy + αu 2 x + αuu xx + u xxxx

42 Examples of conservation laws for KP equation (explicitly dependent on t, x, and y ( ( ( D t xu x +D x 3u 2 u xx 6xuu x +xu xxx +D y αxu y =0 ( ( D t yu x +D x (y(αuu x + u xxx +D y σ 2 (yu y u =0 ( tu ( D t + D α x 2 tu 2 + tu xx + σ2 y 2 4 t u t + σ2 y 2 4 t u xxx + ασ2 y 2 4 t uu x x tu t αx tuu x x tu xxx +D y ( x tu y + y2 u y 4 t yu 2 t = 0

43 More general conservation laws for KP equation: ( D t (fu + D x f( α 2 u2 + u xx +( σ2 2 f y 2 fx(u t + αuu x + u 3x +D y (( 1 2 f y 2 σ 2 fxu y f yu = 0 ( D t (fyu + D x fy( α 2 u2 + u xx +y( σ2 6 f y 2 fx(u t + αuu x + u 3x +D y (y( 1 6 f y 2 σ 2 fxu y +(σ 2 fx 1 2 f y 2 u =0 where f(t is arbitrary function.

44 Potential KP equation Replace u by u x and integrate with respect to x. u xt + αu x u xx + u xxxx + σ 2 u yy = 0 Examples of conservation laws (not explicitly dependent on x, y, t: ( ( ( D t u x + D 1 x 2 αu2 x + u xxx + D y σ 2 u y = 0 ( D t (u 2 x + D 23 x αu3 x u 2 xx + 2u x u xxx σ 2 u yy ( +D y 2σ 2 u x u y = 0

45 Conservation laws for pkp equation continued: ( ( D t u x u y + Dx αu 2 xu y + u t u y + 2u xxx u y 2u xx u xy +D y (σ 2 u 2 y 1 3 u3 x u t u x + u 2 xx = 0 ( D t (2αuu x u xx + 3uu 4x 3σ 2 u 2 y + D x 2αu t u 2 x + 3u 2 t 2αuu x u tx 3u tx u xx + 3u t u xxx + 3u x u t xx 3uu t xxx +D y ( 6σ 2 u t u y = 0 Various generalizations exist.

46 Generalized Zakharov-Kuznetsov equation u t + αu n u x + β(u xx + u yy x = 0 where n is rational, n 0. Conservation laws: ( D t (u + D α x n+1 un+1 + βu xx + D y ( βu xy = 0 ( D t u 2 ( + D 2α x n+2 un+2 β(u 2 x u 2 y + 2βu(u xx + u yy ( D y 2βu x u y = 0

47 Third conservation law for gzk equation: ( D t u n+2 (n+1(n+2β (u 2 2α x + u 2 y + D x ( (n+2α 2(n+1 u2(n+1 + (n + 2βu n+1 u xx (n + 1(n + 2βu n (u 2 x + u 2 y + (n+1(n+2β2 2α (n+1(n+2β2 α (u x (u xxx + u xyy + u y (u xxy + u yyy (u 2 xx u 2 yy + D y ( (n + 2βu n+1 u xy + (n+1(n+2β2 α u xy (u xx + u yy = 0.

48 Conclusions and Future Work The power of Euler and homotopy operators: Testing exactness Integration by parts: D 1 x and Div 1 Integration of non-exact expressions Example: f = u x v + uv x + u 2 u xx fdx = uv + u 2 u xx dx Use other homotopy formulas (prevent curl terms

49 Broader class of PDEs (beyond evolution type Example: short pulse equation (nonlinear optics u xt = u + (u 3 xx = u + 6uu 2 x + 3u 2 u xx with non-polynomial conservation law ( D t 1 + 6u 2 x D x (3u u 2 x = 0 Continue the implementation in Mathematica Software: whereman

50 Thank You

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