eufm10 1

Hodge decomposition and conservation laws

Jan Sanders Jing Ping Wang

Vrije Universiteit, Amsterdam

March 7, 1996

Contents

1

Chapter 1

Introduction

This paper generalizes results in [?] for conservation laws of evolution equations to conservation laws for systems defined by differential forms (like physical laws, which usually derive from some conserved quantity). To this end, we give an algorithm for the Hodge decomposition of horizontal differential forms depending on arbitrary functions and their derivatives. This allows us to reformulate the problem of finding conservation laws as a problem for finding integrating factors. This makes it amenable for computer algebra.

In the calculus of variations, and especially in the inverse problem, i.e. the determination of the Lagrangian from the differential equations, the so called variational bicomplex plays a prominent role. We show how one can move in a controlled manner in this complex using the unique decompositions defined in this paper.

The basic algorithm

2.1 Theory

First we formulate a slight generalization of Algorithm 1 and 2 in [?]. Let U ⊂ R^{n} be an open domain. Instead of polynomials in the dependent variables and their derivatives with constant coefficients, we now allow the coefficients to be C^{∞}-functions on ker F. Clearly this does not affect the nilpotency of F, which is the main ingredient in the algorithm. The operator μ^{E} is defined by μ^{E} = ∑_{σ} μ^{σ}π_{σ}, where the σ’s are eigenvalues of E and the π_{σ} projections on the eigenspaces. In our context this means that the dependent variables are scaled with μ. Let T (R) be a tensor space of the space of variables u_{i}, i = 0, ⋯, ∞ and coefficients in C^{∞}(U, R). The algorithm applies to elements like

c(x)exp(uw^{2})w_{x}u_{xx}^{3} du ∧ du_{xxx} ∧ dw_{xx}

with c ∈ C^{∞}(U, R), exp(uw^{2}) ∈ C^{∞}(Ker F) and w_{x}u_{xx}^{3} du ∧ du_{xxx} ∧ dw_{xx} ∈ T (R).

Algorithm 1 Let v ∈ C^{∞}(Ker F) ⊗ T (R).

1. Put v^{(0)} = v, and let m = 0.

2. while v^{(m)} ≠ 0 do

(a) Put v^{(m+1)} = F v^{(m)}.

(b) Let m = m + 1.

od

3. Put w^{(m)} = 0.

4. while m ≠ 0 do (a) Put u^{(m−1)} = ∫ ∫ w^{(m)} μ^{−1}dμdλ.

(b) Put w^{(m−1)} = Du^{(m−1)} + μ^{E}(v^{(m−1)} − Du^{(m−1)} |_{λ=1,μ=1}).

(c) Let m = m − 1.

od

5. Put ∫ v = w^{(0)} |_{λ=0,μ=1} and ∂v = u^{(0)} |_{λ=1,μ=1}.

Then v = ∫ v + D∂v.

2.2 Example

Let v = Q(x; u)(u_{xx} + u_{x}^{2}). Then v^{(0)} = Q(x; u)(u_{xx} + u_{x}^{2}), v^{(1)} = 2Q(x; u)(u_{x} + uu_{x}), v^{(2)} = 2Q(x; u)u(1 + u) ∈ ker F. We now put w^{(3)} = 0. Then u^{(2)} = 0 and w^{(2)} = 2Q(x; μu)(μu + μ^{2}u^{2}) We decrease m from 3 to 2. Then u^{(1)} = ∫ ∫ w^{(2)} μ^{−1}dμdλ = 2λ ∫ Q(x; μu)(u + μu^{2})dμ. Now

Du^{(1)} = 2λu ∫ Q_{x}(x; μu)(1 + μu)dμ + 2λμQ(x; μu)(1 + μu)u_{x}

It follows that

w^{(1)} =
= 2λu ∫ Q_{x}(x; μu)(1 + μu)dμ + 2λμQ(x; μu)(1 + μu)u_{x}
−2u ∫ Q_{x}(x; μu)(1 + μu)dμ

We decrease m from 2 to 1. Then

u^{(0)} = ∫ ∫ 2λμQ(x; μu)(1 + μu)u_{x} μ^{−1}dμdλ
+ ∫ ∫ 2λu ∫ Q_{x}(x; νu)(1 + νu)dν μ^{−1}dμdλ
− ∫ ∫ 2u ∫ Q_{x}(x; νu)(1 + νu)dν μ^{−1}dμdλ
= λ^{2}u_{x} ∫ Q(x; μu)(1 + μu) dμ
+(2λ − λ^{2})u ∫ log(μ)Q_{x}(x; μu)(1 + μu) dμ

We find that

Du^{(0)} = λ^{2}u_{xx} ∫ Q(x; μu)(1 + μu) dμ

+\frac{u_{x}^{2}}{u} λ^{2}Q(x; μu)(1 + μu)

−\frac{u_{x}^{2}}{u} λ^{2} ∫ Q(x; μu)(1 + μu) dμ

+(2λ − λ^{2})log(μ)Q_{xx}(x; μu)(1 + μu)u

−2(λ − λ^{2})u_{x} ∫ Q_{x}(x; μu)(1 + μu) dμ

Therefore

w^{(0)} = u_{xx} ∫ λ^{2}Q(x; μu)(1 + μu) dμ

+\frac{u_{x}^{2}}{u} λ^{2}Q(x; μu)(1 + μu)

−\frac{u_{x}^{2}}{u} λ^{2} ∫ Q(x; μu)(1 + μu) dμ

+(2λ − λ^{2})log(μ)Q_{xx}(x; μu)(1 + μu)u

−2(λ − λ^{2})u_{x} ∫ Q_{x}(x; μu)(1 + μu) dμ

+μ^{E}(\frac{Q(x; u)}{u}(uu_{xx} − u_{x}^{2})

−\frac{1}{u}(uu_{xx} − u_{x}^{2}) ∫ Q(x; μu)(1 + μu) dμ

−u ∫ log(μ)Q_{xx}(x; μu)(1 + μu) dμ)

So

∫ v = \frac{1}{u}(uu_{xx} − u_{x}^{2})(Q(x; u) − ∫ Q(x; μu)(1 + μu) dμ)
−u ∫ log(μ)Q_{xx}(x; μu)(1 + μu) dμ

and

∂v = u_{x} ∫ Q(x; μu)(1 + μu) dμ
+u ∫ log(μ)Q_{x}(x; μu)(1 + μu) dμ.

The variational complex

3.1 Theory

Let U ⊂ R^{n}. We consider differential forms of the type

ω = ∑_{I} f_{I}dx_{I}

where f_{I} ∈ T (R^{n}), and T (R^{n}) is a tensor space of the space of variables u_{J}^{i}, with J = {j_{1}, ⋯, j_{n}} and coefficients in C^{∞}(U, R). The total differential operators D_{k} are defined as shifts on the variables: D_{k}u_{J}^{i} = u_{J+1_{k}}^{i}. Here J ± 1_{k} is short for {j_{1}, ⋯, j_{k} ± 1, ⋯, j_{n}}. In [?] it is shown how expressions in f ∈ T (R^{n}) can be uniquely decomposed in f = f_{k}^{0} + D_{k}f_{k}^{1} with f_{k}^{0} ∈ Ker F_{k}, where F_{k}u_{J}^{i} = j_{k}u_{J−1_{k}}^{i} We denote in the sequel f_{k}^{0} = ∫_{k} f and f_{k}^{1} = ∂_{k}f. We see that [∂_{k}, D_{k}] = ∫_{k}.

We define (as usual) an operator d_{h} : ⋀^{p} ⟶ ⋀^{p+1} by

d_{h}ω = ∑_{k∈I^{∁}} D_{k}f_{I}dx_{k}dx_{I}.

We now dualize by defining δ_{h} : ⋀^{p} ⟶ ⋀^{p−1} by

δ_{h}ω = ∑_{k∈I}(−1)^{#k+1}∂_{k}f_{I}dx_{I−1_{k}}.

Here #k denotes the position of k in I, to take account of the antisymmetry of the differential forms. For each f_{I} denote by Δ_{f} the set of all indices i ∈ I such that ∂_{i}f_{I} ≠ 0, and let #(f) denote #Δ_{f}.

We now compute Δ_{h} = δ_{h}d_{h} + d_{h}δ_{h} : ⋀^{p} ⟶ ⋀^{p}.

(δ_{h}d_{h} + d_{h}δ_{h})f_{I}dx_{I} =

= δ_{h} ∑_{k∈I^{∁}} D_{k}f_{I}dx_{k}dx_{I}

+d_{h} ∑_{m∈I}(−1)^{#m+1}∂_{m}f_{I}dx_{I−1_{m}}

= ∑_{m∈I}(−1)^{#m}∂_{m} ∑_{k∈I^{∁}} D_{k}f_{I}dx_{k}dx_{I−1_{m}}

+ ∑_{k∈I^{∁}} ∂_{k}D_{k}f_{I}dx_{I}

+ ∑_{k∈I^{∁}} D_{k} ∑_{m∈I}(−1)^{#m+1}∂_{m}f_{I}dx_{k}dx_{I−1_{m}}

+ ∑_{k∈I} D_{k}∂_{k}f_{I}dx_{I}

= ∑_{k∈I^{∁}} ∂_{k}D_{k}f_{I}dx_{I} + ∑_{k∈I} D_{k}∂_{k}f_{I}dx_{I}

= ∑_{k∈I^{∁}} f_{I}dx_{I} + ∑_{k∈I} f_{I}dx_{I} − ∑_{k∈I} ∫_{k} f_{I}dx_{I}

= (n − p + #(f_{I}))f_{I}dx_{I} − ∑_{k∈Δ(f_{I})} ∫_{k} f_{I}dx_{I}

Since #(∫_{k} f_{I}) < #(f_{I}), this allows us to express ω in terms of its components in im δ_{h} and im d_{h} plus terms with smaller #-value. (Algorithmically one might have to split ω first into |I|-homogeneous parts). Applying δ_{h}d_{h} + d_{h}δ_{h} to these terms will eventually lead to a decomposition of ω over im δ_{h} and im d_{h}, except in the case |I| = n (i.e. n − p = 0), where there are also harmonic terms, i.e. terms in Ker δ_{h} ⋂ Ker d_{h}. We remark that Ker Δ_{h} = Ker δ_{h} ⋂ Ker d_{h}, since both (n − p + #(f_{I}))f_{I}dx_{I} and ∑_{k∈Δ(f_{I})} ∫_{k} f_{I}dx_{I} have to vanish (since they have different #-values). Thus one has n − p = 0, implying d_{h}ω = 0 and #(f_{I}) = 0, implying δ_{h} = 0.

Theorem 1 Let ω ∈ ⋀^{p}. Then for p < n, ω can be written as ω = d_{h}α + δ_{h}β, while for p = n we have ω = d_{h}α + γ, where α ∈ ⋀^{p−1}, β ∈ ⋀^{p+1} and γ ∈ Ker d_{h} ⋂ Ker δ_{h}.

We now have to prove that this decomposition is unique. The same inductive reasoning leads us to the conclusion that the space of differential forms can be split into eigenspaces of Δ_{h} with positive integer eigenvalues. In particular we have a decomposition ker Δ_{h} ⊕ im ΔH_{h}. Since im d_{h} ⋂ imδ_{h} ⊂ ker Δ_{h}, we have im Δ_{h} = im d_{h} ⊕ im δ_{h}.

Theorem 2 ⋀^{p} = ker Δ_{h} ⊕ im ΔH_{h} and im ΔH_{h} = im d_{h} ⊕ im δ_{h}.

The proof of the Hodge decomposition is algorithmic. One can give a slight reformulation of it by using the fact that multiples of the form are to be substracted in the process.

Algorithm 2 Let ω ∈ ⋀^{p} and define ω^{(i)} = Δ_{h}^{i}ω for i = 0, ⋯, p. Let f(x) = ∏_{i=n−p+1}^{2n−p}(x − i) = ∑_{i=0}^{p} f_{i}x^{i}. Then

ω = ω^{(p+1)} − Δ_{h} ∑_{i=1}^{p} \frac{f_{i}}{f_{0}} ω^{(i)},

with ω^{(p+1)} ∈ Ker Δ_{h}, gives the desired decomposition over ker Δ_{h} ⊕ im ΔH_{h}.

We can now try to compute conservation laws as follows. Suppose we are given a vanishing form ω, e.g. 0 = ω = (u_{t} − u_{xx})dt ∧ dx for the heat equation. If we can find a Q such that Qω = d_{h}α, then we have a conservation law, since ∫_{∂U} α = ∫_{U} d_{h}α = ∫_{U} Qω = 0, since ω vanishes on U, which makes α a conserved quantity. Finding the integrating factor Q is highly nontrivial. We come back to this in section ??.

3.2 A horizontal example

In this section we consider the following example in two variables:

ω = u_{x}u_{yy} dx ∧ dy ∈ ⋀_{0}^{2}(T (R^{2})).

Clearly, dω = 0. Let f_{11} = u_{x}u_{yy}. We have F_{x}f_{11} = uu_{yy} ∈ Ker F_{x} and therefore ∂_{1}f_{11} = \frac{1}{2}uu_{yy}. Now the other variable: F_{y}f_{11} = 2u_{x}u_{y} and F_{y}^{2}f_{11} = 2uu_{x} ∈ Ker F_{y}. Denoting by the power in λ the distance from Ker F_{y}, we have F_{y}f_{11} = u_{x}u_{y} − uu_{xy} + λ(u_{x}u_{y} + uu_{xy}). It follows that ∂_{2}f_{11} = \frac{3}{4}u_{x}u_{y} − \frac{1}{4}uu_{xy}. We combine these results to obtain

δ_{h}ω = \frac{1}{2}uu_{yy}dy − \frac{1}{4}(3u_{x}u_{y} − uu_{xy})dx.

It follows that

d_{h}δ_{h}ω = \frac{1}{4}(5u_{x}u_{yy} + uu_{xyy} + 2u_{y}u_{xy}) dx ∧ dy.

We now let ω^{(1)} = 2ω − Δ_{h}ω = \frac{1}{4}(3u_{x}u_{yy} − uu_{xyy} − 2u_{y}u_{xy}) dx ∧ dy. Define f_{11}^{(1)} = \frac{1}{4}(3u_{x}u_{yy} −uu_{xyy} −2u_{y}u_{xy}). We now repeat the algorithm. We find that F_{x}f_{11}^{(1)} = \frac{1}{2}(uu_{yy} − u_{y}^{2}) ∈ Ker F_{x}. Thus ∂_{1}f_{11}^{(1)} = \frac{1}{4}(uu_{yy} − u_{y}^{2}). On the other hand F_{y}f_{11}^{(1)} = u_{x}u_{y} − uu_{xy} ∈ Ker F_{y} and ∂_{2}f_{11}^{(1)} = \frac{1}{2}(u_{x}u_{y} − uu_{xy}). We have

δ_{h}ω^{(1)} = \frac{1}{4}(uu_{yy} − u_{y}^{2})dy − \frac{1}{2}(u_{x}u_{y} − uu_{xy})dx.

It follows that

Δ_{h}ω^{(1)} = d_{h}δ_{h}ω^{(1)} = \frac{1}{4}(3u_{x}u_{yy} − uu_{xyy} − 2u_{y}u_{xy}) dx ∧ dy.

We define ω^{(2)} = ω^{(1)} − Δ_{h}ω^{(1)} and see that it is equal to zero. It follows that γ = 0 in Theorem 1. It follows that ω = \frac{1}{2}Δ_{h}(ω + ω^{(1)}). We find that \frac{1}{2}(ω+ω^{(1)}) = \frac{1}{8}(7u_{x}u_{yy}−uu_{xyy}−2u_{y}u_{xy}) dx∧dy. Using our earlier computations we see that

α = δ_{h} \frac{1}{2}(ω + ω^{(1)}) = \frac{1}{8}((3uu_{yy} − u_{y}^{2}) dy − (5u_{x}u_{y} − 3uu_{xy}) dx).