Application of Euler and Homotopy Operators to Integrability Testing Willy Hereman Department of Applied Mathematics and Statistics Colorado School of Mines Golden, Colorado, U.S.A. http://www.mines.edu/fs home/whereman/ The Ninth IMACS International Conference on Nonlinear Evolution Equations and Wave Phenomena: Computation and Theory Athens, Georgia Thursday, April 2, 2015, 4:40p.m.
Acknowledgements Douglas Poole (Chadron State College, Nebraska) Ünal Göktaş (Turgut Özal University, Turkey) 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 No. CCF-0830783 This presentation was made in TeXpower
Outline Motivation of the Research Part I: Continuous Case Calculus-based formulas for the continuous homotopy operator (in multi-dimensions) See: Anco & Bluman (2002); Hereman et al. (2007); Olver (1986) Symbolic integration by parts and inversion of the total divergence operator Application: symbolic computation of conservation laws of nonlinear PDEs in multiple space dimensions
Part II: Discrete Case Simple formula for the homotopy operator See: Hereman et al. (2004); Hydon & Mansfield (2004); Mansfield & Hydon (2002) Symbolic summation by parts and inversion of the forward difference operator Analogy: continuous and discrete formulas Application: symbolic computation of conservation laws of nonlinear DDEs Conclusions and Future Work
Motivation of the Research 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 where = means evaluated on the PDE. Conserved density ρ and flux J.
Conservation law in (2+1)-dimensions D t ρ + J = D t ρ + D x J 1 + D y J 2 = 0 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 Conserved density ρ and flux J = (J 1, J 2, J 3 ).
Notation Computations on the Jet Space Independent variables: t (time), x = (x, y, z) Dependent variables u = (u (1), u (2),..., u (j),..., u (N) ) In examples: u = (u, v, w, 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 x 5 u xx yyyy = u 2x 4y = Differential functions 6 u x 2 y 4 Example: f = uvv x + x 2 u 3 xv x + u x v xx
Total derivatives: D t, D x, D y,... Example: Let f = uvv x + x 2 u 3 xv x + u x v xx Then D x f = f x + u f x u + u xx +v x f v + v xx f u x f + v xxx v x f v xx = 2xu 3 xv x + u x (vv x ) + u xx (3x 2 u 2 xv x + v xx ) +v x (uv x ) + v xx (uv + x 2 u 3 x) + v xxx (u x ) = 2xu 3 xv x + vu x v x + 3x 2 u 2 xv x u xx + u xx v xx +uv 2 x + uvv xx + x 2 u 3 xv xx + u x v xxx
Example 1: The Zakharov-Kuznetsov Equation u t + αuu x + β(u xx + u yy ) x = 0 models ion-sound solitons in a low pressure uniform magnetized plasma. 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
More conservation laws (ZK equation): ) ( D t (u 3 3β α (u2 x + u 2 y) + D x 3u 2 ( α 4 u2 + βu xx ) 6βu(u 2 x + u 2 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 u2 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
Example 2: Shallow Water Wave Equations P. Dellar, Phys. Fluids 15 (2003) 292-297 u t + (u )u + 2 Ω u + (θh) 1 2 h θ = 0 θ t + u ( θ) = 0 h t + (uh) = 0 In components: u t + uu x + vu y 2 Ωv + 1 2 hθ x + θh x = 0 v t + uv x + vv y + 2 Ωu + 1 2 hθ y + θh y = 0 θ t + uθ x + vθ y = 0 h t + hu x + uh x + hv y + vh y = 0 where u(x, y, t), θ(x, y, t) and h(x, y, t)
First few densities-flux pairs of SWW system: 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 u v u v u (u2 + v 2 + 2hθ) v (v 2 + u 2 + 2hθ) 4Ωu 2uu y + 2uv x hθ y 4Ωv + 2vv x 2vu y + hθ x
Generalizations: D t ( f(θ)h ) + Dx ( f(θ)hu ) + Dy ( f(θ)hv ) = 0 ( D t g(θ)(2ω + vx 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)
Conservation Laws for Nonlinear Differential-Difference Equations (DDEs) System of DDEs u n =F(, u n 1, u n, u n+1, ) Conservation law in (1 + 1) dimensions D t ρ n + J n = D t ρ n + J n+1 J n = 0 conserved density ρ n and flux J n Example: Toda lattice u n = v n 1 v n v n = v n (u n u n+1 )
First few densities-flux pairs for Toda lattice: ρ (0) n =ln(v n ) J n (0) =u n ρ (1) n =u n J n (1) =v n 1 ρ (2) n = 1 2 u2 n + v n J n (2) =u n v n 1 ρ (3) n = 1 3 u3 n + u n (v n 1 + v n ) J n (3) =u n 1 u n v n 1 + vn 1 2
PART I: CONTINUOUS CASE Problem Statement Continuous case in 1D: Example: For u(x) and v(x) f =8v x v xx u 3 x sin u +2u x u xx cos u 6vv x cos u +3u x v 2 sin u Question: Can the expression be integrated? If yes, find F = f dx (so, f = D x F ) Result (by hand): F = 4 vx 2 + u2 x cos u 3 v2 cos u
Continuous case in 2D or 3D: Example: For u(x, y) and v(x, y) f = u x v y u 2x v y u y v x + u xy v x Question: Is there an F so that f = Div F? If yes, find F. Result (by hand): F = (uv y u x v y, uv x + u x v x ) Can this be done without integration by parts? Can the computation be reduced to a single integral in one variable?
Tools from the Calculus of Variations Definition: A differential function f is a exact iff f = Div F. Special case (1D): f = D x F. Question: How can one test that f = Div F? 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
Euler operator in 1D (variable u(x)): L u(x) = M ( D x ) k k=0 = u D x u kx u x + D 2 x u xx D 3 x 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 + u xxx
Application: Testing Exactness Continous Case Example: f =8v x v xx u 3 x sin u +2u x u xx cos u 6vv x cos u +3u x v 2 sin u where u(x) and v(x) f is exact After integration by parts (by hand): F = f dx = 4 vx 2 + u 2 x cos u 3 v 2 cos u
Exactness test with Euler operator: f =8v x v xx u 3 x sin u +2u x u xx cos u 6vv x cos u +3u x v 2 sin u L u(x) f = f u D x L v(x) f = f v D x f u x + D 2 x f v x + D 2 x f u xx 0 f v xx 0
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!
Peter Olver s Book
Homotopy Concept in Olver s Book
Homotopy Formula in Olver s Book
Zoom into Homotopy Formula in Olver s Book
Homotopy Operator in 1D (variable x): H u(x) f = 1 0 N (I u (j)f)[λu] dλ λ j=1 with integrand I u (j)f = M x (j) k=1 k 1 i=0 ix ( D x) k (i+1) u (j) f u (j) kx (I u (j)f)[λu] means that in I u (j)f 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.
Homotopy Operator in 2D (variables x and y): H (x) u(x,y) f = 1 0 N j=1 (I (x) u (j) f)[λu] dλ λ H (y) u(x,y) f = 1 0 N j=1 (I (y) u (j) f)[λu] dλ λ where for dependent variable u(x, y) M x M y ( k 1 l i+j I u (x) f = i u ix jy k=1 l=0 i=0 j=0 ( D x ) k i 1 ( D y ) l j ) )( k+l i j 1 k i 1 ( k+l ) k f u kx ly )
Application of Homotopy Operator in 1D Example: f =8v x v xx u 3 x sin u +2u xu xx cos u 6vv x cos u +3u x v 2 sin u Goal: Find F = 4 v 2 x + u 2 x cos u 3 v 2 cos u Easy to verify: f = D x F Compute I u f = u f u x + (u x I ud x ) f u xx = uu 2 x sin u + 3uv 2 sin u + 2u 2 x cos u
Similarly, I v f = v f v x + (v x I vd x ) f v xx Finally, = 6v 2 cos u + 8v 2 x 1 F = H u(x) f = (I u f + I v f) [λu] dλ 0 λ 1 ( = 3λ 2 uv 2 sin(λu) λ 2 uu 2 x sin(λu) 0 +2λu 2 x cos(λu) 6λv2 cos(λu) + 8λv 2 x ) dλ = 4v 2 x + u2 x cos u 3v2 cos u
Application of Homotopy Operator in 2D Example: f = u x v y u xx v y u y v x + u xy v x By hand: F = (uv y u x v y, uv x + u x v x ) Easy to verify: f = Div F Compute I u (x) f = u f + (u x I ud x ) f u x ( ) + 1 f 2 u yi 1 2 ud y u xy u xx = uv y + 1 2 u yv x u x v y + 1 2 uv xy
Similarly, I v (x) f = v f = u y v + u xy v v x Hence, F 1 = H (x) u(x,y) f = 1 = 1 0 λ ( I (x) u ) f + I v (x) f [λu] dλ 0 λ ( uv y + 1 2 u yv x u x v y + 1 2 uv xy u y v + u xy v = 1 2 uv y + 1 4 u yv x 1 2 u xv y + 1 4 uv xy 1 2 u yv + 1 2 u xyv ) dλ
Analogously, F 2 = H (y) u(x,y) f = 1 = 1 0 ( λ ( ) I u (y) f + I v (y) f [λu] dλ 0 λ ) ) ( uv x 12 uv xx + 12 u xv x + λ (u x v u xx v) dλ = 1 2 uv x 1 4 uv xx + 1 4 u xv x + 1 2 u xv 1 2 u xxv So, F= 1 4 2uv y + u y v x 2u x v y + uv xy 2u y v + 2u xy v 2uv x uv xx + u x v x + 2u x v 2u xx v
Let K= F F then K= 1 4 2uv y u y v x 2u x v y uv xy + 2u y v 2u xy v 2uv x + uv xx + 3u x v x 2u x v + 2u xx v then Div K = 0 Also, K = (D y φ, D x φ) with φ = 1 4 (2uv uv x 2u x v) (curl in 2D) After removing the curl term K: F = F + K = (uv y u x v y, uv x + u x v x ) Avoid curl terms algorithmically!
Why does this work? Sketch of Derivation and Proof (in 1D with variable x, and for one component u) Definition: Degree operator M Mf = M i=0 u ix f u ix = u f u +u x f u x +u 2x f u 2x + +u Mx f u Mx f is of order M in x Example: f = u p u q xu r 3x (p, q, r non-negative integers) g = Mf = 3 i=0 u ix f u ix = (p + q + r) u p u q xu r 3x Application of M computes the total degree
Theorem (inverse operator) M 1 g(u) = 1 0 g[λu] dλ λ Proof: d M dλ g[λu] = i=0 g[λu] dλu ix λu ix dλ = 1 λ M i=0 Integrate both sides with respect to λ u ix g[λu] u ix = 1 λ Mg[λu] 1 0 d dλ g[λu] dλ = g[λu] λ=1 λ=0 = 1 0 = g(u) g(0) Mg[λu] dλ λ = M 1 0 g[λu] dλ λ Assuming g(0) = 0, M 1 g(u) = 1 0 g[λu] dλ λ
Example: If g(u) = (p + q + r) u p u q xu r 3x, then g[λu] = (p + q + r)λ p+q+r u p u q xu r 3x Hence, M 1 g = 1 0 (p + q + r) λ p+q+r 1 u p u q x ur 3x dλ = u p u q xu r 3x λ p+q+r λ=1 = λ=0 up u q xu r 3x
Theorem: If f is an exact differential function, then F = Dx 1 f = f dx = H u(x) f Proof: Multiply L u(x) f = M ( D x ) k k=0 f u kx by u to restore the degree. Split off u f u. Split off u x f u x. Lastly, split off u Mx Integrate by parts. Repeat the process. f u Mx.
ul u(x) f = u = u f u D x = u f u + u x =... M ( D x ) k k=0 u f u kx M ( D x ) k 1 k=1 f u x D x M +u x ( D x ) k 2 k=2 u f u kx f u kx M ( D x ) k 1 k=1 M + u x ( D x ) k 1 f k=1 u kx M + u 2x ( D x ) k 2 k=2 f u kx f u kx
f u x +... + u Mx = u f u + u x M D x u ( D x ) k 1 = k=1 +... + u (M 1)x M M i=0 = Mf D x = 0 u ix f u ix D x M 1 i=0 k=m M 1 u ix f u Mx f u kx + u x ( D x ) k M i=0 M k=i+1 u ix M k=i+1 M ( D x ) k 2 k=2 f u kx ( D x ) k (i+1) ( D x ) k (i+1) f u kx f u kx f u kx
Mf = D x M 1 i=0 u ix M k=i+1 ( D x ) k (i+1) f u kx Apply M 1 and use M 1 D x = D x M 1. M 1 M f = D x M 1 ( D x ) k (i+1) u ix i=0 k=i+1 f u kx Apply Dx 1 and use the formula for M 1. 1 M 1 M F = Dx 1 f = u ix ( D x ) k (i+1) = 1 0 = H u(x) f k=1 0 i=0 k=i+1 M k 1 u ix ( D x ) k (i+1) i=0 f u kx f u kx [λu] dλ λ [λu] dλ λ
Application 1: Zakharov-Kuznetsov Equation Computation of Conservation Laws 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. Hence, the weights of the variables are W (u) = 2, W (D t ) = 3, W (D x ) = 1, W (D y ) = 1.
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 x 3u 2 ( α 4 u2 +βu 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.
Compute the density of selected rank, say, 6. 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
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
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 α)u x u yy + 2(3c 1 β + c 3 α)u y u xy ) +2c 4 αu x u xy +c 4 αu y u xx +3(3c 1 β +c 2 α)u x u xx 0
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)
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. )
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 = ( 3α 4 u4 + βu 2 (3u xx + 2u yy ) 2βu(3u 2 x + u 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 + 3 2 u yyy) + β2 α (3u2 xx + 5 2 u2 xy + 3 4 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 ) )
Application 2: Shallow Water Wave Equations Computation of Conservation Laws Quick Recapitulation Conservation law in (2+1) dimensions D t ρ + J = D t ρ + D x J 1 + D y J 2 = 0 conserved density ρ and flux J = (J 1, J 2 ) Example: Shallow water wave (SWW) equations u t + uu x + vu y 2 Ωv + 1 2 hθ x + θh x = 0 v t + uv x + vv y + 2 Ωu + 1 2 hθ y + θh y = 0 θ t + uθ x + vθ y = 0 h t + hu x + uh x + hv y + vh y = 0
Typical density-flux pair: ρ (5) = θ (v x u y + 2Ω) J (5) = 1 2 θ 4Ωu 2uu y + 2uv x hθ y 4Ωv + 2vv x 2vu y + hθ x
Step 1: Construct the form of the density The SWW equations are invariant under the scaling symmetries (x, y, t, u, v, θ, h, Ω) (λ 1 x, λ 1 y, λ 2 t, λu, λv, λθ, λh, λ 2 Ω) and (x, y, t, u, v, θ, h, Ω) (λ 1 x, λ 1 y, λ 2 t, λu, λv, λ 2 θ, λ 0 h, λ 2 Ω) Construct a candidate density, for example, ρ = c 1 Ωθ + c 2 u y θ + c 3 v y θ + c 4 u x θ + c 5 v x θ which is scaling invariant under both symmetries.
Step 2: Determine the constants c i Compute E = D t ρ and remove time derivatives E = ( ρ u x u tx + ρ u y u ty + ρ v x v tx + ρ v y v ty + ρ θ θ t) = c 4 θ(uu x + vu y 2Ωv + 1 2 hθ x + θh x ) x + c 2 θ(uu x + vu y 2Ωv + 1 2 hθ x + θh x ) y + c 5 θ(uv x + vv y + 2Ωu + 1 2 hθ y + θh y ) x + c 3 θ(uv x + vv y + 2Ωu + 1 2 hθ y + θh y ) y + (c 1 Ω + c 2 u y + c 3 v y + c 4 u x + c 5 v x )(uθ x + vθ y ) Require that L u(x,y) E = L v(x,y) E = L θ(x,y) E = L h(x,y) E 0.
Solution: c 1 = 2, c 2 = 1, c 3 = c 4 = 0, c 5 = 1 gives ρ = ρ (5) = θ (2Ω u y + v x ) Step 3: Compute the flux J E = θ(u x v x + uv xx + v x v y + vv xy + 2Ωu x + 1 2 θ xh y u x u y uu xy u y v y u yy v +2Ωv y 1 2 θ yh x ) + 2Ωuθ x + 2Ωvθ y uu y θ x u y vθ y + uv x θ x + vv x θ y Apply the 2D homotopy operator: J = (J 1, J 2 ) = Div 1 E = (H (x) u(x,y) E, H(y) u(x,y) E)
Compute I u (x) E = u E + u x ( ) 1 E 2 u yi 1 2 ud y u xy Similarly, compute = uv x θ + 2Ωuθ + 1 2 u2 θ y uu y θ I (x) v E = vv y θ + 1 2 v2 θ y + uv x θ I (x) θ E = 1 2 θ2 h y + 2Ωuθ uu y θ + uv x θ I (x) h E = 1 2 θθ yh
Next, J 1 = H (x) = = u(x,y) E 1 ( I u (x) 0 1 0 ) E + I(x) v E + I (x) θ E + I (x) h E [λu] dλ λ ( ( 4λΩuθ + λ 2 3uv x θ + 1 2 u2 θ y 2uu y θ + vv y θ + 1 2 v2 θ y + 1 2 θ2 h y 1 2 θθ yh) ) dλ = 2Ωuθ 2 3 uu yθ+ uv x θ+ 1 3 vv yθ+ 1 6 u2 θ y + 1 6 v2 θ y 1 6 hθθ y + 1 6 h yθ 2
Analogously, J 2 = H (y) u(x,y) E Hence, = 2Ωvθ + 2 3 vv xθ vu y θ 1 3 uu xθ 1 6 u2 θ x 1 6 v2 θ x + 1 6 hθθ x 1 6 h xθ 2 J = J (5) = 1 12Ωuθ 4uu yθ+6uv x θ+2vv y θ+(u 2 +v 2 )θ y hθθ y +h y θ 2 6 12Ωvθ+4vv x θ 6vu y θ 2uu x θ (u 2 +v 2 )θ x +hθθ x h x θ 2
There are curl terms in J Indeed, subtract K where Div K = 0 Here K= 1 6 (2uu yθ+2vv y θ+u 2 θ y +v 2 θ y +2hθθ y +h y θ 2 ) 2vv x θ+2uu x θ+u 2 θ x +v 2 θ x +2hθθ x +h x θ 2 Note that K = (D y φ, D x φ) with φ = (hθ 2 + u 2 θ + v 2 θ) (i.e., curl in 2D).
After removing the curl term K, J = J (5) = 1 2 θ 4Ωu 2uu y + 2uv x hθ y 4Ωv + 2vv x 2vu y + hθ x
PART II: DISCRETE CASE Problem Statement Discrete case in 1D: Example: f n = u n u n+1 v n vn+u 2 n+1 u n+2 v n+1 +vn+1+u 2 n+3 v n+2 u n+1 v n Question: Can the expression be summed by parts? If yes, find F n = 1 f n (so, f n = F n = F n+1 F n ) Result (by hand): F n =vn 2 + u n u n+1 v n + u n+1 v n + u n+2 v n+1 How can this be done algorithmically? Can this be done as in the continuous case?
Tools from the Discrete Calculus of Variations Definitions: D is the up-shift (forward or right-shift) operator DF n = F n+1 = F n n n+1 D 1 the down-shift (backward or left-shift) operator D 1 F n = F n 1 = F n n n 1 = D I is the forward difference operator F n = (D I)F n = F n+1 F n Problem to be solved: Given f n. Find F n = 1 f n (so f n = F n = F n+1 F n )
Analogy Continuous & Discrete Cases Euler Operators Continuous Case Discrete Case L u(x) = M ( D x ) k k=0 u k x L un = M k=0 = u n D k M k=0 u n+k D k
Analogy Continuous & Discrete Cases Homotopy Operators & Integrands Continuous Case Discrete Case 1 H u(x) f= 0 N (I u (j)f)[λu] dλ λ H u n f n = j=1 1 0 N j=1 (I u (j) n f n )[λu n ] dλ λ I u (j)f = k 1 i=0 u (j) M (j) k=1 ix ( D x) k (i+1) f u (j) kx I (j) u f n = n k i=1 M (j) k=1 D i u (j) n+k f n u (j) n+k
Euler Operators Side by Side Continuous Case (for component u) L u = u D x u x + D 2 x u 2x D 3 x Discrete Case (for component u n ) L un = u n + D 1 u n+1 + D 2 u n+2 + D 3 = u n (I + D 1 + D 2 + D 3 + ) u 3x + u n+3 +
Homotopy Operators Side by Side Continuous Case (for components u and v) with I u f = H u(x) f = M (1) k=1 1 0 i=0 (I u f + I v f) [λu] dλ λ k 1 u ix ( D x ) k (i+1) f u kx and I v f = M (2) k=1 k 1 i=0 v ix ( D x ) k (i+1) f v kx
Discrete Case (for components u n and v n ) with H un f n = I un f n = 1 0 M (1) k=1 (I un f n + I vn f n ) [λu n ] dλ λ k i=1 D i u n+k f n u n+k and I vn f n = M (2) k=1 k i=1 D i v n+k f n v n+k
Analogy of Definitions & Theorems Continuous Case (PDE) Semi-discrete Case (DDE) u t =F(u, u x, u 2x, ) u n =F(, u n 1, u n, u n+1, ) D t ρ + D x J = 0 D t ρ n + J n = 0 Definition: f n is exact iff f n = F n = F n+1 F n Theorem (exactness test): f n = F n iff L un f n 0 Theorem (summation with homotopy operator): If f n is exact then F n = 1 f n = H un (f n )
Example: Testing Exactness Discrete Case f n = u n u n+1 v n v 2 n+u n+1 u n+2 v n+1 +v 2 n+1+u n+3 v n+2 u n+1 v n f n is exact After summation by parts (done by hand): F n = v 2 n + u n u n+1 v n + u n+1 v n + u n+2 v n+1 Easy to verify: f n = F n = F n+1 F n
Exactness test with Euler operator: For component u n (highest shift 3): L un f n = ( I + D 1 + D 2 + D 3) f n u n = u n+1 v n u n 1 v n 1 + u n+1 v n v n 1 Similarly, 0 + u n 1 v n 1 + v n 1 L vn f n = v n ( I + D 1) f n = u n u n+1 + 2v n u n u n+1 2v n 0
Application of Discrete Homotopy Operator Example: f n = u n u n+1 v n v 2 n+u n+1 u n+2 v n+1 +v 2 n+1+u n+3 v n+2 u n+1 v n Here, M (1) = 3 and M (2) = 2. Compute I un f n = (D 1 ) u n+1 f n u n+1 + (D 1 + D 2 ) u n+2 f n u n+2 + (D 1 + D 2 + D 3 ) u n+3 f n u n+3 = 2u n u n+1 v n + u n+1 v n + u n+2 v n+1
I vn f n = (D 1 ) v n+1 f n v n+1 Finally, + (D 1 + D 2 ) v n+2 f n v n+2 = u n u n+1 v n + 2v 2 n + u n+1 v n + u n+2 v n+1 F n = = 1 0 1 0 (I un f n + I vn f n ) [λu n ] dλ λ ) (2λvn 2 + 3λ 2 u n u n+1 v n + 2λu n+1 v n + 2λu n+2 v n+1 dλ = v 2 n + u n u n+1 v n + u n+1 v n + u n+2 v n+1
Application: Computation of Conservation Laws System of DDEs u n =F(, u n 1, u n, u n+1, ) Conservation law D t ρ n + J n = 0 conserved density ρ n and flux J n
Example: Toda lattice u n = v n 1 v n v n = v n (u n u n+1 ) Typical density-flux pair: ρ (3) n = 1 3 u3 n + u n (v n 1 + v n ) J (3) n = u n 1 u n v n 1 + v 2 n 1
Computation Conservation Laws for Toda Lattice Step 1: Construct the form of the density The Toda lattice is invariant under scaling symmetry (t, u n, v n ) (λ 1 t, λu n, λ 2 v n ) Construct a candidate density, for example, ρ n = c 1 u 3 n + c 2 u n v n 1 + c 3 u n v n which is scaling invariant under the symmetry
Step 2: Determine the constants c i Compute E n = D t ρ n and remove time derivatives E n = (3c 1 c 2 )u 2 nv n 1 + (c 3 3c 1 )u 2 nv n + (c 3 c 2 )v n 1 v n +c 2 u n 1 u n v n 1 + c 2 v 2 n 1 c 3 u n u n+1 v n c 3 v 2 n Compute Ẽn = DE n to remove negative shift n 1 Require that L un Ẽ n = L vn Ẽ n 0 Solution: c 1 = 1 3, c 2 = c 3 = 1 gives ρ n = ρ (3) n = 1 3 u3 n + u n (v n 1 + v n )
Step 3: Compute the flux J n Ẽ n = DE n = u n u n+1 v n + vn 2 u n+1 u n+2 v n+1 vn+1 2 Apply the homotopy operator J n = DJ n = 1 (Ẽn) = H un (Ẽn) Compute I un Ẽ n = (D 1 Ẽn ) u n+1 + (D 1 + D 2 ) u n+2 u n+1 = 2u n u n+1 v n Ẽn u n+2 Likewise, I vn Ẽ n = (D 1 ) v n+1 Ẽn v n+1 = (u n u n+1 v n + 2v 2 n)
Next, compute J n = = 1 0 1 0 ) (I un Ẽ n + I vn Ẽ n [λu n ] dλ λ (3λ 2 u n u n+1 v n + 2λv 2 n) dλ = u n u n+1 v n + vn 2 Finally, backward shift J n = D 1 ( J n ) given J n = J (3) n = u n 1 u n v n 1 + v 2 n 1
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 (moving terms amongst the components of the flux; prevent curl terms)
Homotopy operator approach pays off for computing irrational fluxes 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 2 ) 1 + 6u 2 x = 0 Continue the implementation in Mathematica Software: http://inside.mines.edu/ whereman
Thank You
Publications 1. D. Poole and W. Hereman, Symbolic computation of conservation laws for nonlinear partial differential equations in multiple space dimensions, J. Symb. Comp. 46(12), 1355-1377 (2011). 2. W. Hereman, P. J. Adams, H. L. Eklund, M. S. Hickman, and B. M. Herbst, Direct Methods and Symbolic Software for Conservation Laws of Nonlinear Equations, In: Advances of Nonlinear Waves and Symbolic Computation, Ed.: Z. Yan, Nova Science Publishers, New York (2009), Chapter 2, pp. 19-79.
3. W. Hereman, M. Colagrosso, R. Sayers, A. Ringler, B. Deconinck, M. Nivala, and M. S. Hickman, Continuous and Discrete Homotopy Operators and the Computation of Conservation Laws. In: Differential Equations with Symbolic Computation, Eds.: D. Wang and Z. Zheng, Birkhäuser Verlag, Basel (2005), Chapter 15, pp. 249-285. 4. W. Hereman, B. Deconinck, and L. D. Poole, Continuous and discrete homotopy operators: A theoretical approach made concrete, Math. Comput. Simul. 74(4-5), 352-360 (2007).
5. W. Hereman, Symbolic computation of conservation laws of nonlinear partial differential equations in multi-dimensions, Int. J. Quan. Chem. 106(1), 278-299 (2006).