Proving the nonexistence of algebraic solutions of diﬀerential equations

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Proving the nonexistence of algebraic solutions of differential equations

S. C. Coutinho

Universidade Federal do Rio de Janeiro

RWTH-Aachen–2011

S. C. Coutinho Proving the nonexistence of algebraic solutions of differential equations

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Part I The problem

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Stating the problem

Solve the system of differential equations

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Stating the problem

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Stating the problem

˙

x=a(x,y)

˙

y=b(x,y),

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Stating the problem

˙

x=a(x,y)

˙

y=b(x,y), where aandb are polynomials inx and y.

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Stating the problem

˙

x=a(x,y)

˙

y=b(x,y),

where aandb are polynomials inx and y. More concisely,

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Stating the problem

˙

x=a(x,y)

˙

y=b(x,y),

where aandb are polynomials inx and y. More concisely, X˙ =F(X),

where X = (x,y) andF = (a,b) is a polynomial vector field.

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What does it mean to solve an equation?

The canonical definition

Find a curve C(t) such that

C˙ =

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What does it mean to solve an equation?

Find a curve C(t) such that

C˙ =

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What does it mean to solve an equation?

Find a parameterized curve C(t) such that

C˙ =

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What does it mean to solve an equation?

Find a parameterized curve C(t) such that C˙ =

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What does it mean to solve an equation?

Find a parameterized curve C(t) such that C˙ =F(C(t)).

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What does it mean to solve an equation?

Find a parameterizedcurve C(t) such that C˙ =F(C(t)).

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What does it mean to solve an equation?

What if the function were also known implicitly?

Suppose we know a function H =H(x,y) whose set of zeros is C. Question

How can we say that the curve is a solution of the system using H instead of the parameterization?

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What does it mean to solve an equation?

Suppose we know a function H =H(x,y) whose set of zeros is C. Question

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What does it mean to solve an equation?

Suppose we know a function H =H(x,y) whose set of zeros isC.

Question

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What does it mean to solve an equation?

Suppose we know a function H =H(x,y) whose set of zeros isC. Question

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What does it mean to solve an equation?

Suppose we know a function H =H(x,y) whose set of zeros isC.

Question

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What does it mean to solve an equation?

By definition

H(C(t)) = 0. Thus,

a∂H

∂x +b∂H

∂y

(C(t)) = 0; which is equivalent to

(F · ∇H)(C(t)) = 0.

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What does it mean to solve an equation?

By definition

H(C(t)) = 0. Thus,

a∂H

∂x +b∂H

∂y

(F · ∇H)(C(t)) = 0.

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What does it mean to solve an equation?

By definition

H(C(t)) = 0.

Thus,

a∂H

∂x +b∂H

∂y

(F · ∇H)(C(t)) = 0.

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What does it mean to solve an equation?

By definition

H(C(t)) = 0.

Thus,

a∂H

∂x +b∂H

∂y

(F · ∇H)(C(t)) = 0.

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What does it mean to solve an equation?

By definition

H(C(t)) = 0.

Thus,

d

dtH(C(t)) = 0.

a∂H

∂x +b∂H

∂y

(F · ∇H)(C(t)) = 0.

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What does it mean to solve an equation?

By definition

H(C(t)) = 0.

Thus,

dx dt

∂H

∂x(C(t)) + dy dt

∂H

∂y(C(t)) = 0.

a∂H

∂x +b∂H

∂y

(F · ∇H)(C(t)) = 0.

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What does it mean to solve an equation?

By definition

H(C(t)) = 0.

Thus,

a(C(t))∂H

∂x(C(t)) +b(C(t))∂H

∂y(C(t)) = 0.

a∂H

∂x +b∂H

∂y

(F · ∇H)(C(t)) = 0.

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What does it mean to solve an equation?

By definition

H(C(t)) = 0.

Thus,

a∂H

∂x +b∂H

∂y

(C(t)) = 0;

which is equivalent to

(F · ∇H)(C(t)) = 0.

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What does it mean to solve an equation?

By definition

H(C(t)) = 0.

Thus,

a∂H

∂x +b∂H

∂y

(C(t)) = 0;

(F · ∇H)(C(t)) = 0.

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What does it mean to solve an equation?

By definition

H(C(t)) = 0.

Thus,

a∂H

∂x +b∂H

∂y

(C(t)) = 0;

(F · ∇H)(C(t)) = 0.

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What does it mean to solve an equation?

First integral

A function H(x,y) is a first integral of the system X˙ =F(X) if F(x,y)· ∇H = 0,

as a function of x and y .

Key property

If H is a first integral of X˙ =F(X) then every integral curve of this system is contained in a level curve of H.

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What does it mean to solve an equation?

First integral

Key property

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What does it mean to solve an equation?

First integral

Key property

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What does it mean to solve an equation?

First integral

Key property

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What does it mean to solve an equation?

The system ˙X =F(X), defined by the vector field F(x,y) = (2y,3x²) has first integral H(x,y) =y²−x³. Two of its level curves are

H(x,y) = 0

and H(x,y) = 1.

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What does it mean to solve an equation?

The system ˙X =F(X), defined by the vector field

F(x,y) = (2y,3x²) has first integral H(x,y) =y²−x³. Two of its level curves are

H(x,y) = 0

and H(x,y) = 1.

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What does it mean to solve an equation?

The system ˙X =F(X), defined by the vector field F(x,y) = (2y,3x²)

has first integral H(x,y) =y²−x³. Two of its level curves are H(x,y) = 0

and H(x,y) = 1.

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What does it mean to solve an equation?

The system ˙X =F(X), defined by the vector field F(x,y) = (2y,3x²)

has first integral H(x,y) =y²−x³. Two of its level curves are H(x,y) = 0

and H(x,y) = 1.

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What does it mean to solve an equation?

The system ˙X =F(X), defined by the vector field F(x,y) = (2y,3x²) has first integral H(x,y) =y²−x³.

Two of its level curves are H(x,y) = 0

and H(x,y) = 1.

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What does it mean to solve an equation?

H(x,y) = 0

and H(x,y) = 1.

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What does it mean to solve an equation?

H(x,y) = 0

and H(x,y) = 1.

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What does it mean to solve an equation?

H(x,y) = 0 and

H(x,y) = 1.

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What does it mean to solve an equation?

H(x,y) = 0 and H(x,y) = 1.

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Stating the problem

The problem

Given a vector field F(X), compute a first integral of the differential equationX˙ =F(X).

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Stating the problem

The problem

Given a vector field F(X), compute a first integral of the differential equationX˙ =F(X).

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Stating the problem

The problem

Given a polynomial vector field F(X),

compute a first integral of the differential equation X˙ =F(X).

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Stating the problem

The problem

Given a polynomial vector field F(X), compute a first integral of the differential equation X˙ =F(X).

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Stating the problem

The problem

Given a polynomialvector field F(X), compute a first integral of the differential equation X˙ =F(X).

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Polynomial differential equations

the Lotka-Volterra system in population dynamics; the Lorenz system in meteorology;

the Euler equations of rigid body motion; Bianchi models in cosmology;

etc.

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Polynomial differential equations

the Lotka-Volterra system in population dynamics;

the Lorenz system in meteorology; the Euler equations of rigid body motion; Bianchi models in cosmology;

etc.

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Polynomial differential equations

the Lorenz system in meteorology;

the Euler equations of rigid body motion; Bianchi models in cosmology;

etc.

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Polynomial differential equations

the Euler equations of rigid body motion;

Bianchi models in cosmology; etc.

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Polynomial differential equations

Bianchi models in cosmology;

etc.

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Polynomial differential equations

Bianchi models in cosmology;

etc.

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Part II

The 19th century

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C. G. J. Jacobi, 1842

Solves a differential equation with linear coefficients, with a long calculation.

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C. G. J. Jacobi, 1842

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C. G. J. Jacobi, 1842

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C. G. J. Jacobi, 1842

Solves a differential equation with linear coefficients,

with a long calculation.

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C. G. J. Jacobi, 1842

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Alfred Clebsch, 1872

Geometric interpretation of differential equations using homogeneous coordinates.

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Alfred Clebsch, 1872

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Alfred Clebsch, 1872

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Alfred Clebsch, 1872

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G. Darboux, 1878

Introduces the method that defined the research line we will pursue in this talk.

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G. Darboux, 1878

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G. Darboux, 1878

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G. Darboux, 1878

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Darboux’s key idea

IfC is an integral curve of ˙X =F(X) and also the set of zeroes of a functionH(x,y), then,

H(C(t)) = 0; thus, as before,

(F(x,y)· ∇H)(C(t)) = 0. so that,

(F(x,y)· ∇H)(p) = 0 whenever H(p) = 0.

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Darboux’s key idea

IfC is an integral curve of ˙X =F(X)

and also the set of zeroes of a functionH(x,y), then,

(F(x,y)· ∇H)(C(t)) = 0. so that,

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Darboux’s key idea

IfC is an integral curve of ˙X =F(X) and also the set of zeroes of a functionH(x,y),

then,

(F(x,y)· ∇H)(C(t)) = 0. so that,

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Darboux’s key idea

H(C(t)) = 0;

thus, as before,

(F(x,y)· ∇H)(C(t)) = 0. so that,

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Darboux’s key idea

H(C(t)) = 0;

thus,

as before,

(F(x,y)· ∇H)(C(t)) = 0. so that,

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Darboux’s key idea

H(C(t)) = 0;

thus, as before,

(F(x,y)· ∇H)(C(t)) = 0. so that,

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Darboux’s key idea

H(C(t)) = 0;

thus, as before,

(F(x,y)· ∇H)(C(t)) = 0.

so that,

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Darboux’s key idea

H(C(t)) = 0;

thus, as before,

(F(x,y)· ∇H)(C(t)) = 0.

so that,

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Darboux’s key idea

H(C(t)) = 0;

thus, as before,

(F(x,y)· ∇H)(C(t)) = 0.

so that,

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Darboux’s key idea

IfH andF are polynomial, then so is F(x,y)· ∇H. Therefore, the conclusion above implies that,

F(x,y)· ∇H =GH,

for some polynomial G =G(x,y), called the co-factor ofH.

Assuming that H is reduced, this follows from Hilbert’s Nullstellensatz.

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Darboux’s key idea

IfH andF are polynomial, then so is F(x,y)· ∇H.

Therefore, the conclusion above implies that, F(x,y)· ∇H =GH,

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Darboux’s key idea

Therefore,

the conclusion above implies that, F(x,y)· ∇H =GH,

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Darboux’s key idea

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Darboux’s key idea

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Darboux’s key idea

Invariant curve

An algebraic curve H(x,y) = 0 is invariant under the system X˙ =F(x,y) if

F(x,y)· ∇H =GH,

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Darboux’s key idea

Invariant curve

An algebraic curve H(x,y) = 0 is invariant under the system X˙ =F(x,y) if

F(x,y)· ∇H =GH,

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Darboux’s key idea

Invariant curve

An algebraic curve H(x,y) = 0 is invariant under the systemX˙ =F(x,y) if

F(x,y)· ∇H =GH,

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Darboux’s key Theorem

Existence of first integral

IfX˙ =F(X) has invariant curves, then it admits a first integral.

Degree of a vector field

If F = (a,b), for polynomials a and b, then deg(F) =

max{deg(a),deg(b)}

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Darboux’s key Theorem

IfX˙ =F(X) has invariant curves, then it admits a first integral.

max{deg(a),deg(b)}

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Darboux’s key Theorem

IfX˙ =F(X) has enough invariant curves, then it admits a first integral.

max{deg(a),deg(b)}

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Darboux’s key Theorem

IfX˙ =F(X) has enoughinvariant curves, then it admits a first integral.

max{deg(a),deg(b)}

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Darboux’s key Theorem

If F = (a,b), for polynomials a and b, then

deg(F) = max{deg(a),deg(b)}

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Darboux’s key Theorem

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Darboux’s key Theorem

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Darboux’s key Theorem

IfX˙ =F(X) has more thandeg(F)(deg(F)−1)/2invariant curves, then it admits a first integral.

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Where does this bound come from?

IfH is invariant under ˙X =F(X) then

F(x,y)· ∇H =GH,

Hence,

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Where does this bound come from?

F(x,y)· ∇H =GH,

Hence,

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Where does this bound come from?

F(x,y)· ∇H =GH,

Hence,

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Where does this bound come from?

F(x,y)· ∇H =GH,

Hence,

a∂H

∂x +b∂H

∂y =GH.

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Where does this bound come from?

F(x,y)· ∇H =GH,

Hence,

deg

a∂H

∂x +b∂H

∂y

= deg(GH).

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Where does this bound come from?

F(x,y)· ∇H =GH,

Hence,

max

deg

a∂H

∂x

,

b∂H

∂y

≥deg(GH).

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Where does this bound come from?

F(x,y)· ∇H =GH,

Hence, max

deg(a) + deg ∂H

∂x

,deg(b) + ∂H

∂y

≥deg(GH).

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Where does this bound come from?

F(x,y)· ∇H =GH,

Hence,

max{deg(a) + deg(H)−1,deg(b) + deg(H)−1} ≥deg(GH).

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Where does this bound come from?

F(x,y)· ∇H =GH,

Hence,

max{deg(a),deg(b)}+ deg(H)−1≥deg(GH).

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Where does this bound come from?

F(x,y)· ∇H =GH,

Hence,

deg(F) + deg(H)−1≥deg(GH).

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Where does this bound come from?

F(x,y)· ∇H =GH,

Hence,

deg(F) + deg(H)−1≥deg(G) + deg(H).

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Where does this bound come from?

F(x,y)· ∇H =GH,

Hence,

deg(F) +deg(H)−1≥deg(G) +deg(H).

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Where does this bound come from?

F(x,y)· ∇H =GH,

Hence,

deg(F)−1≥deg(G).

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Where does this bound come from?

F(x,y)· ∇H =GH,

Hence,

In particular, G is an element of the subspace of polynomials of degree

≤deg(F)−1,

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Where does this bound come from?

F(x,y)· ∇H =GH,

Hence,

In particular, G is an element of the subspace of polynomials of degree

≤deg(F)−1, which has dimension

(deg(F)−1) deg(F)

2 .

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Proof of Darboux’s key Theorem

d = (deg(F)−1) deg(F) 2

= dimension of the space of polynomials of degree≤deg(F)−1.

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Proof of Darboux’s key Theorem

d = (deg(F)−1) deg(F) 2

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Proof of Darboux’s key Theorem

d = (deg(F)−1) deg(F) 2

p₁, . . . ,p_k be curves invariant under X˙ =F(X);

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Proof of Darboux’s key Theorem

d = (deg(F)−1) deg(F) 2

∇p_j ·F =g_jp_j, where 1≤j ≤k and deg(g_j)≤deg(F)−1.

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Proof of Darboux’s key Theorem

d = (deg(F)−1) deg(F) 2

Ifk >d then g1, . . . ,gk are linearly dependent,

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Proof of Darboux’s key Theorem

d = (deg(F)−1) deg(F) 2

Ifk >d then g1, . . . ,gk are linearly dependent,

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Proof of Darboux’s key Theorem

d = (deg(F)−1) deg(F) 2

Ifk >d then g1, . . . ,gk are linearly dependent, so

c₁g₁+· · ·+c_kg_k = 0 for scalars c₁, . . . ,c_k.

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Proof of Darboux’s key Theorem

Hypotheses:

F · ∇p_j =g_jp_j, where 1≤j ≤k;

c₁g₁+· · ·+c_kg_k = 0 for scalarsc₁, . . . ,c_k.

Define

h =p₁^c¹· · ·p^c_k^k then

F · ∇h=p₁^c¹· · ·p^c_k^k(c₁g₁+· · ·+c_kg_k)

= 0.

Henceh is a first integral of F.

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Proof of Darboux’s key Theorem

Hypotheses:

c₁g₁+· · ·+c_kg_k = 0 for scalarsc₁, . . . ,c_k. Define

F · ∇h=p₁^c¹· · ·p^c_k^k(c₁g₁+· · ·+c_kg_k)

= 0.

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Proof of Darboux’s key Theorem

Hypotheses:

F · ∇h

F · ∇h=p₁^c¹· · ·p^c_k^k(c1g1+· · ·+c_kg_k)

= 0.

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Proof of Darboux’s key Theorem

Hypotheses:

F · ∇h =F · ∇(p^c₁¹· · ·ckp_k^c^k)

F · ∇h=p₁^c¹· · ·p^c_k^k(c1g1+· · ·+c_kg_k)

= 0.

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Proof of Darboux’s key Theorem

Hypotheses:

F · ∇h=F ·(c1p₁^c¹⁻¹· · ·p^c_k^k∇p₁+· · ·+p^c₁¹· · ·c_kp_k^c^k⁻¹∇p_k)

F · ∇h=p₁^c¹· · ·p^c_k^k(c₁g₁+· · ·+c_kg_k)

= 0.

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Proof of Darboux’s key Theorem

Hypotheses:

F · ∇h=c1p₁^c¹⁻¹· · ·p^c_k^kF · ∇p₁+· · ·+p₁^c¹· · ·c_kp^c_k^k⁻¹F · ∇p_k

F · ∇h=p₁^c¹· · ·p^c_k^k(c₁g₁+· · ·+c_kg_k)

= 0.

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Proof of Darboux’s key Theorem

Hypotheses:

F · ∇h=c1p₁^c¹⁻¹· · ·p_k^c^kg1p1+· · ·+p^c₁¹· · ·c_kp_k^c^k⁻¹g_kp_k

F · ∇h=p₁^c¹· · ·p^c_k^k(c₁g₁+· · ·+c_kg_k)

= 0.

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Proof of Darboux’s key Theorem

Hypotheses:

F · ∇h=c1p1c1−1· · ·p^c_k^kg1p1+· · ·+p₁^c¹· · ·c_kp_k^c^k⁻¹g_kp_k

F · ∇h=p₁^c¹· · ·p^c_k^k(c₁g₁+· · ·+c_kg_k)

= 0.

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Proof of Darboux’s key Theorem

Hypotheses:

F · ∇h=c1p^c₁¹· · ·p_k^c^kg1+· · ·+p₁^c¹· · ·ckp^c_k^kgk

F · ∇h=p₁^c¹· · ·p^c_k^k(c1g1+· · ·+c_kg_k)

= 0.

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Proof of Darboux’s key Theorem

Hypotheses:

F · ∇h=p₁^c¹· · ·p^c_k^k(c1g1+· · ·+ckgk)

= 0. Henceh is a first integral of F.

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Proof of Darboux’s key Theorem

Hypotheses:

F · ∇h=p₁^c¹· · ·p^c_k^k(c1g1+· · ·+ckgk) = 0.

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Proof of Darboux’s key Theorem

Hypotheses:

F · ∇h=p₁^c¹· · ·p^c_k^k(c1g1+· · ·+ckgk) = 0.

Henceh is a first integral ofF.

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An example: the Jacobi equation

Its invariant lines are y+ 2 with co-factor 1; 2x+ 2y+ 3 with co-factor 2. Since 2·1 + (−1)·2 = 0,

h= (y+ 2)²(2x+ 2y+ 3)⁻¹, is a first integral of ˙X =F(X).

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An example: the Jacobi equation

Suppose that F = (2x+y+ 1,y+ 2).

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An example: the Jacobi equation

Find `linear andc a scalar, such that

Its invariant lines are y+ 2 with co-factor 1;

2x+ 2y+ 3 with co-factor 2. Since 2·1 + (−1)·2 = 0,

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An example: the Jacobi equation

Find `linear andc a scalar, such that F · ∇`=c`.

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An example: the Jacobi equation

Find `linear andc a scalar, such that (2x+y+ 1)∂`

∂x + (y+ 2)∂`

∂y =c`.

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An example: the Jacobi equation

(2x+y+ 1) ∂

∂x + (y+ 2) ∂

∂y

(`) =c`.

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An example: the Jacobi equation

(2x+y+ 1) ∂

∂x + (y+ 2) ∂

∂y

(`) =c`.

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An example: the Jacobi equation

(2x+y+ 1) ∂

∂x + (y+ 2) ∂

∂y

(`) =c`.

Solve an eigenvalue problem for the linear operator whose matrix in the basis {x,y,1} is

Its invariant lines are y+ 2 with co-factor 1;

2x+ 2y+ 3 with co-factor 2. Since 2·1 + (−1)·2 = 0,

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An example: the Jacobi equation

(2x+y+ 1) ∂

∂x + (y+ 2) ∂

∂y

(`) =c`.





2 0 0 1 1 0 1 2 0





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An example: the Jacobi equation





2 0 0 1 1 0 1 2 0





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An example: the Jacobi equation





2 0 0 1 1 0 1 2 0





Thus,

the matrix the differential equation

eigenvector eigenvalue ` c

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An example: the Jacobi equation





2 0 0 1 1 0 1 2 0





Thus,

(0,1,2)

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An example: the Jacobi equation





2 0 0 1 1 0 1 2 0





Thus,

(0,1,2) 1

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An example: the Jacobi equation





2 0 0 1 1 0 1 2 0





Thus,

(0,1,2) 1 y+ 2