M´ ethodes formelles pour les ´ equations aux d´ eriv´ ees partielles
Daniel Robertz
Centre for Mathematical Sciences Plymouth University
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M ethodes formelles pour les equations aux d eriv ees - - PowerPoint PPT Presentation
M ethodes formelles pour les equations aux d eriv ees partielles Daniel Robertz Centre for Mathematical Sciences Plymouth University JNCF Luminy 2018 Desiderata Given a system of differential equations, we would like to be able
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n
m
n
m
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x ∂j y ,
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x
√ 2 2 u2 , ∂x
√ 2 2 u2
√ 2 2 u2
x
√ 2 2 u2
√ 2 2 u2 one may also choose uy + √ 2 2 u2.) JNCF Luminy 2018
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2! + a1,1 xy 1!1! + a0,2 y2 2! + . . .
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2! + a1,1 xy 1!1! + a0,2 y2 2! + . . .
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∂ ∂x
∂x∂y − ∂u ∂y
∂y
∂x2 − ∂u ∂y
∂2u ∂y2 − ∂u ∂y = 0
2! + a1,1 xy 1!1! + a0,2 y2 2! + . . .
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∂ ∂x
∂x∂y − ∂u ∂y
∂y
∂x2 − ∂u ∂y
∂2u ∂y2 − ∂u ∂y = 0
2! + a1,1 xy 1!1! + a0,2 y2 2! + . . .
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∂ ∂x
∂x∂y − ∂u ∂y
∂y
∂x2 − ∂u ∂y
∂2u ∂y2 − ∂u ∂y = 0
2! + a1,1 xy 1!1! + a0,2 y2 2! + . . .
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A B
A B
1 2 (∂2 x − y∂2 z)A − 1 2 ∂2 yB
∂xA
1 2 (∂4 x − 2y∂2 x∂2 z + y2∂4 z)A − 1 2 (∂2 x∂2 y − y∂2 y∂2 z + 2∂y∂2 z)B 1 2 (∂3 x − y∂x∂2 z)A − 1 2 ∂x∂2 yB 1 2 (∂5 x − 2y∂3 x∂2 z + y2∂x∂4 z)A − 1 2 (∂3 x∂2 y + y∂x∂2 y∂2 z − 2∂x∂y∂2 z)B
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2,
1∂2,
1
2, ∂3 1∂2, ∂4 1 M
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1 · · · ∂an n .
1 · · · ∂d j · · · ∂bn n
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2, ∂3 1∂2, ∂4 1 M
2
1∂2 2
1∂2
1
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1 · · · ∂an n
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j=0 {yi−jg | g ∈ Gj},
i=0 Ti
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lc(d) lm(p) lm(d) d
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2
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x
√ 2 2 u2 , ∂x
√ 2 2 u2
√ 2 2 u2
x
√ 2 2 u2
√ 2 2 u2 one may also choose uy + √ 2 2 u2.) JNCF Luminy 2018
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r
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2, {∂2}), (∂2 1∂2 2, {∂2}), (∂3 1∂2, {∂2}), (∂4 1, {∂1, ∂2}) }
2
1∂2 2
1∂2
1
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r
r
j≥0
r
1 + ∂2 1∂2 + ∂3 1
6
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x,
x,
1 (1−∂y)(1−∂z) + ∂x 1−∂z + ∂2
x
1−∂z + ∂3
x
1−∂x .
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k αi,j,k pk,
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lm(d) d
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r
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t2+1, t2−1 t2+1
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∂v ∂t + v · ∇v − ν∆v + 1 ρ∇p
∂ρ ∂t + ∇ · (ρv)
r
r 2(t+c2), (θ+c1)r t+c2
2r2
t+c2
((θ+c1)2− 3
4 )r2
2(t+c2)2
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2! + a3 t3 3! + . . .
1 − 4 a0 = 0
2 − 6 a2 + 8 = 0
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1 · · · ∂in n u | i ∈ (Z≥0)n] = K[u, uz1, ..., uzn, uz1,z1, ...]
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x,x,y + . . .
x,x,y + . . .
x,x,y,y + . . .
x,x,y + . . .
∂q ∂ux,x,y ux,x,y,y + . . .
∂q ∂ux,x,y · p − u2 x,x,y,y · ∂y q
∂q ∂ux,x,y = 0
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n
n−(i−1) −
n−i : (ai, ai+1, . . . , an) −
i , . . . , a(e) i
i , ai+1, . . . , an) ∈ Vi,
i , . . . , a(f) i
i , ai+1, . . . , an) ∈ Vi,
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n
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∂p ∂ ˙ u = 2 ˙
∂ ˙ u, ˙
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y p2 − 6 u2 p2 − 2 u p3
y − u4 needs to be considered, which implies vanishing or
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y p2 − 6 u2 p2 − 2 u p3
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t.
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x+c1 −6t+c2
u(0)
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y,z
y,zux,x + 2uy,zux,zux,y − uy,yu2 x,z
y,z
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y,z = 0
y,z = 0
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y,z
y,z
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> with(DifferentialThomas): > ivar := [t]:
> ComputeRanking(ivar,
> L := [x1[t]-cx3*u1, x2[t]-sx3*u1, sx3[t]-cx3*u2,
> LL := Diff2JetList(Ind2Diff(L, ivar, dvar));
> TD := DifferentialThomasDecomposition(LL, [cx3]);
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> Print(TD[1]);
2 − y2 t 2 = 0,
2 u2 + y2 t 2u2 − y1 ty2 t,t + y2 ty1 t,t = 0,
2 + y2 t 2 = 0,
> collect(%[6], u2, factor);
2 + y2 t 2
> Print(TD[2]);
2 − y2 t 2 = 0,
2 + y2 t 2 = 0]
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> Print(TD[3]);
> Print(TD[4]);
> Print(TD[5]);
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> with(DifferentialThomas): > ivar := [t]:
> ComputeRanking(ivar, [[F1,F2],[sV,c],[c1,c2]]): > L := [2*sV[t]*sV-F1-F2+k*sV,
> LL := Diff2JetList(Ind2Diff(L, ivar, dvar));
2 − c2 0F2 0 + c0ksV 0 − c1 0F1 0 + 2 c0sV 1sV 0,
> TD := DifferentialThomasDecomposition(LL,
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> Print(TD[1]);
> collect(%[1], F1);
> collect(%%[2], F2);
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> Print(TD[2]);
> Print(TD[3]);
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> with(DifferentialThomas): > ivar := [t]:
> ComputeRanking(ivar, [[T,s,c,d,R],[x,z]]): > TD := DifferentialThomasDecomposition(
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> Print(TD[2]);
> collect(%[5], R, factor);
2 + zt,t 2 − 2 gzt,t + g2
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> Print(TD[1]);
> Print(TD[3]);
> Print(TD[4]);
> Print(TD[5]);
> Print(TD[6]);
> Print(TD[7]);
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> with(DifferentialThomas): > ivar := [x1,x2,x3]:
> ComputeRanking(ivar, [[xi1,xi2,xi3],[a]]): > L := [-a*xi1[x1]+xi3[x1]-a[x2]*xi2
> LL := Diff2JetList(Ind2Diff(L, ivar, dvar));
> TD := DifferentialThomasDecomposition(LL, []);
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> Print(TD[1]);
> Print(TD[2]);
> Print(TD[3]);
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