Symmetry Classification of Third-Order Nonlinear Evolution Equations. Part I: Semi-simple Algebras

P. Basarab-Horwath; F. Güngör; V. Lahno

doi:10.1007/s10440-012-9773-4

Acta Applicandae Mathematicae

April 2013, Volume 124, Issue 1, pp 123–170 | Cite as

Symmetry Classification of Third-Order Nonlinear Evolution Equations. Part I: Semi-simple Algebras

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P. Basarab-Horwath
F. Güngör
V. Lahno

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First Online: 31 July 2012

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Abstract

We give a complete point-symmetry classification of all third-order evolution equations of the form u _t=F(t,x,u,u _x,u _xx)u _xxx+G(t,x,u,u _x,u _xx) which admit semi-simple symmetry algebras and extensions of these semi-simple Lie algebras by solvable Lie algebras. The methods we employ are extensions and refinements of previous techniques which have been used in such classifications.

Keywords

Symmetries Lie algebras Equivalence group Semi-simple Lie algebras Solvable Lie algebras

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Notes

Acknowledgements

V. Lahno was partially supported by the Swedish Research Council (grant number 624-2004-1073). The research of F. Güngör was supported by the Research Council of Turkey (TÜBİTAK).

Faruk Güngör and Peter Basarab-Horwath wish to acknowledge the great debt they owe to Viktor Lahno, for his friendship, mathematical insights and patient collaboration over the years. Viktor died on June 5th 2011, while we were preparing this article for submission. We shall miss him greatly.

Appendix A: Representations of so(3,ℝ) and sl(2,ℝ)

$\operatorname {sl}(2, \mathbb{K})$ with $\mathbb {K}=\mathbb{R}$ or $\mathbb{K}=\mathbb{C}$: The irreducible finite-dimensional representations of $\operatorname {sl}(2, \mathbb{K})$ are well-known (see for instance [22]). An irreducible representation space is defined by a half-integer $J=0,\frac{1}{2}, 1, \frac{3}{2}, 2,\ldots$ and a vector space k _J=〈Q ₁,…,Q _2J+1〉 and dimk _J=2J+1. Denoting by e _k⋅Q _l the representation of e _k on Q _l, we have the relations

$$Q_k=\frac{e_3^{k-1}}{(k-1)!}\cdot Q_1,\quad e_3 \cdot Q_{2J+1}=0, $$

as well as

$$Q_{2J+1-k}=\frac{e_2^k}{k!}\cdot Q_{2J+1},\quad e_2\cdot Q_1=0, $$

and

$$e_1\cdot Q_k=2(J-k+1)Q_k. $$

These relations hold for both real and complex vector spaces k _J.

$\operatorname {so}(3, \mathbb{R})$: $\operatorname {so}(3, \mathbb {R})=\langle e_{1}, e_{2}, e_{3}\rangle$ is defined by the commutation relations $[e_{1}, e_{2}]=e_{3},\; [e_{2}, e_{3}]=e_{1},\; [e_{3}, e_{1}]=e_{2}$. In this case, we use the trick of complexifying. If V is the real irreducible representation space of $\operatorname {so}(3, \mathbb{R})$, then define the complex vector space W=V+iV as the space spanned by vectors u+iv with u,v∈V. Next define $h=2ie_{1},\; x=e_{2}+ie_{3},\; y=-e_{2}+ie_{3}$. Then 〈h,x,y〉 is just $\operatorname {sl}(2, \mathbb {C})$ and it acts irreducibly on W. So we have W=〈Q ₁,…,Q _2J+1〉 and [h,Q _2J−k+1]=−2(J−k)Q _2J−k+1 for k=0,1,…,2J. Further, [y,Q _2J+1]=0 and we also have [h,Q _2J+1]=−2JQ _2J+1. With Q _2J+1=X+iY for some X,Y∈V, we then find from [h,Q _2J+1]=−2JQ _2J+1 that $[e_{1}, X]=-JY,\; [e_{1}, Y]=JX$. Note also, that if J is an integer, then there exists a vector Z∈V with [e ₁,Z]=0 (this follows from [h,Q _2J−k+1]=−2(J−k)Q _2J−k+1). Thus, in any irreducible representation of $\operatorname {so}(3, \mathbb{R})$ on a real Lie algebra A (whether abelian or not) we have dimA=2J+1 for some half-integer J, non-zero vectors X,Y with $[e_{1}, X]=\alpha Y, \; [e_{1}, Y]=-\alpha X$ for α>0 for any J>0 and non-zero vectors Z with [e ₁,Z]=0 if J is an integer.

Finally, we note that $\operatorname {so}(3, \mathbb{R})$ has no two-dimensional irreducible representations: in fact if this were the case then there would be a representation by three real 2×2 trace-free matrices of 〈e ₁,e ₂,e ₃〉. These would then form a basis for $\operatorname {sl}(2, \mathbb{R})$, contradicting the commutation relations for $\operatorname {so}(3, \mathbb{R})$.

Appendix B: Proof of Theorem 2.1

In this appendix we give a proof of Theorem 2.1. To this end we begin with some informal preliminaries on contact structures.

We work on k-th order jet spaces J ^k(ℝⁿ,ℝ^m), with local coordinates $(x,u,\underset{(1)}{u}, \underset {(2)}{u},\ldots, \underset{(k)}{u})$ where x=(x ¹,…,x ⁿ), u=(u ¹,…,u ^m) and where $\underset{(j)}{u}$ stands for the collection of all the j-th order partial derivatives of (u ¹,…,u ^m) with respect to the (x ¹,…,x ⁿ). On these spaces we introduce the contact one-forms

Open image in new window

for l=1,…,m and μ _i=1,…,n, and we sum over repeated indices. The symbols $u^{l}_{\mu_{i}},\ldots, u^{l}_{\mu_{1}\cdots\mu_{k}}$ are then the partial derivatives of the functions u ^l. These one-forms vanish on solutions of differential equations. They are known as the Cartan distribution (see [23]). Note that a k-th order contact form

$$\omega^l_{\mu_1\cdots\mu_k}=du^l_{\mu_1\cdots\mu_k}-u^l_{\mu_1\cdots\mu _k\mu_{k+1}}dx^{\mu_{k+1}} $$

is not a properly defined form on J ^k(ℝⁿ,ℝ^m), because the $u^{l}_{\mu_{1}\cdots\mu_{k}\mu_{k+1}}$ are coordinates on J ^k+1(ℝⁿ,ℝ^m). To remedy this, we may work on the infinite jet bundle J ^∞(ℝⁿ,ℝ^m) [23, 24, 25], which may be thought of as a space of infinite sequences $(x, u, \underset {(1)}{u},\ldots,\underset{(k)}{u},\ldots)$, and we have projections $\pi^{\infty}_{k}: J^{\infty}(\mathbb{R}^{n}, \mathbb{R}^{m})\to J^{k}(\mathbb {R}^{n}, \mathbb{R}^{m})$ as well as projections $\pi^{k}_{l}: J^{k}(\mathbb {R}^{n}, \mathbb{R}^{m})\to J^{l}(\mathbb{R}^{n}, \mathbb{R}^{m})$ for l≤k. In terms of the sequences $(x, u, \underset{(1)}{u},\ldots,\underset {(k)}{u},\ldots)$, we have $\pi^{\infty}_{k}(x, u, \underset{(1)}{u},\ldots ,\underset{(k)}{u},\ldots)=(x, u, \underset{(1)}{u},\ldots,\underset {(k)}{u})$ and $\pi^{k}_{l}(x, u, \underset{(1)}{u},\ldots,\underset {(k)}{u})=(x, u, \underset{(1)}{u}, \ldots,\underset{(l)}{u})$. The infinite jet space J ^∞(ℝⁿ,ℝ^m) is well-defined, as the inverse limit of the k-th order jet bundles J ^k(ℝⁿ,ℝ^m) (see [23, 24, 25]). A function F:J ^∞(ℝⁿ,ℝ^m)→ℝ is then defined to be smooth if F is a smooth function of only $(x, u, \underset {(1)}{u},\ldots,\underset{(k)}{u})$ for some natural number k∈ℕ. We work within this infinite jet bundle unless otherwise stated.

A tangent transformation is then a transformation

$$\bigl(x, u, \underset{(1)}{u},\ldots, \underset{(k)}{u} \bigr)\to \bigl(x', u', \underset {(1)}{u'}, \ldots, \underset{(k)}{u'} \bigr) $$

for any k≥1 such that the transformed contact forms

Open image in new window

vanish when $\omega^{l},\, \omega^{l}_{\mu_{i}},\ldots, \omega^{l}_{\mu_{1}\cdots\mu _{k}}$ vanish (see [13]). Then the following result holds [13]:

Theorem B.1

A tangent transformation is given by a prolongation of a point transformation

$$(x,u)\to \bigl(x', u' \bigr) $$

if u=(u ¹,…,u ^m) for m≥2. It is given by the prolongation of a contact transformation

$$\bigl(x,u, \underset{(1)}{u} \bigr)\to \bigl(x',u', \underset{(1)}{u'} \bigr) $$

if m=1, and then there is a (locally smooth) function $W(t,x,u, \underset{(1)}{u})$ such that

$$x'^l=-W_{q^l},\qquad u'=W-pW_p-q^lW_{q^l}, \qquad u'_{x'^l}=W_{x^l} + q^lW_u $$

with $q^{l}=u_{x^{l}},\; l= 1,\ldots, k$.

Now let us consider smooth, real-valued functions $F(x,u,\underset {(1)}{u},\ldots, \underset{(k)}{u})$. Then one can show [23, 25] that the exterior derivative dF may be written as

$$dF = DF + \sum_{j=1}^{k} \varLambda_l^{\mu_1\cdots\mu_j}\omega^l_{\mu_1\cdots \mu_j} $$

for some functions $\varLambda_{l}^{\mu_{1}\cdots\mu_{j}}$, when we work on the infinite jet bundle, where we sum over repeated indices. Here we have

$$DF=D_{x^1}Fdx^1 + \cdots+ D_{x^n}Fdx^n $$

and $D_{x^{i}}$ is the operator of total differentiation with respect to x ⁱ:

$$D_{x^{\mu_i}}F=F_{x^i} + u^l_{\mu_i}F_{u^l} + u^l_{\mu_i\mu}F_{u^l_{\mu }}+\cdots+ u^{l}_{\mu_i\mu_1\cdots\mu_k}F_{u^{l}_{\mu_1\cdots\mu_k}}. $$

From this, it follows easily that if $D_{x^{i}}F=0$ then $F_{u^{l}}=F_{u^{l}_{\mu_{1}\cdots\mu_{j}}}=0$ for l=1,…,m and j=1,…,k, and then DF=0 implies $F=\operatorname {constant}$, and so DF=0 if and only if dF=0. A further result in this direction is the following:

Lemma B.1

Suppose that

$$\bigl(x,u,\underset{(1)}{u},\ldots, \underset{(k)}{u} \bigr)\to \bigl(x',u',\underset {(1)}{u'},\ldots, \underset{(k)}{u'} \bigr) $$

defines an invertible tangent transformation, with x′¹=X ¹,…,x′ⁿ=X ⁿ, then we have

$$DX^1\wedge\cdots\wedge DX^n\neq0. $$

Proof

There are two cases: if u=(u ¹,…,u ^m) with m≥2, then the transformation is the prolongation of an invertible point transformation (x,u)→(X(x,u),U(x,u)). Thus we have dX ¹∧⋯∧dX ⁿ∧dU ¹∧⋯∧dU ^m≠ by invertibility.

Now suppose that DX ¹∧⋯∧DX ⁿ=0. We know from the previous remarks that DX ⁱ≠0 for i=1,…,n. Hence DX ¹∈I(DX ²,…,DX ⁿ) where I(DX ²,…,DX ⁿ) is the module generated by the forms DX ²,…,DX ⁿ, that is, I(DX ²,…,DX ⁿ) consists of linear combinations of the form λ ₂ DX ²+⋯+λ _n DX ⁿ for some functions λ ₂,…,λ _n. Now each X ⁱ is a function of (x ¹,…,x ⁿ,u ¹,…,u ^m) and then we have

$$dX^i=DX^i+\varLambda_l\omega^l. $$

From this it now follows that dX ¹∈I(dX ²,…,dX ⁿ,ω ¹,…,ω ^m). Further, since the transformation is a tangent transformation, we have $dU^{l}-U^{l}_{i}dX^{i}=\lambda^{l}_{j}\omega^{j}$ for l=1,…,m, so that dU ^l∈I(dX ¹,dX ²,…,dX ⁿ,ω ¹,…,ω ^m) and consequently dU ^l∈I(dX ²,…,dX ⁿ,ω ¹,…,ω ^m) since we know that dX ¹∈I(dX ²,…,dX ⁿ,ω ¹,…,ω ^m). Hence each of the m+n one-forms dX ¹,…,dX ⁿ,dU ¹,…,dU ^m is a sum of m+n−1 one-forms, so that dX ¹∧⋯∧dX ⁿ∧dU ¹∧⋯∧dU ^m=0, which contradicts the invertibility of the transformation.

The case of just one function is treated similarly: in this case the tangent transformation is a prolongation of an invertible contact transformation

$$\bigl(x^1,\ldots, x^n, u, u_1,\ldots u_n \bigr)\to \bigl(x'^1,\ldots, x'^n, u', u'_1, \ldots u'_n \bigr), $$

where $u_{i}=u_{x^{i}}$, which is a particular case of a point transformation with n+1 functions u,u ₁,…,u _n. Hence we must have DX ¹∧⋯∧DX ⁿ≠0.

We now come to the proof of Theorem 2.1: □

Theorem B.2

Any invertible contact transformation

$$(t,x,u,p,q)\to \bigl(t', x', u', p', q' \bigr) $$

with $p=u_{t},\;\; q=u_{1}$ and $p'=u'_{t'},\;\, q'=u'_{x'}$, preserving the form of an evolution equation of order n

$$u_0=F(t,x,u,u_1,\ldots, u_n), $$

with n≥2, is such that t′=T(t) with $\dot{T}(t)\neq0$. Then the contact transformation has the form

Open image in new window

for some smooth function W.

Proof

Our contact transformation is

Open image in new window

and there is a function W(t,x,u,p,q) so that

Open image in new window

(see [13]). An evolution equation of order n+1 becomes in our notation

$$p=F(t, x, u, q, q_1,\ldots, q_n) $$

where $q_{1}=D_{x}q, q_{2}=D^{2}_{x}q,\ldots, q_{n}=D^{n}_{x}q$. We shall show that D _t T≠0, D _x T=0, from which it follows that T=T(t).

We define the sequence of functions Q ^k, k≥0, by Q ⁰=Q and for k≥1 they are given by

$$DT\wedge DQ^k=Q^{k+1}DT\wedge DX. $$

The Q ^k are well-defined since we have DT∧DX≠0 by Lemma B.1. These functions are just the transformations of the functions q _k=∂ ^k q/∂x ^k defined by the contact conditions:

$$dQ^k-Q^{k+1}dX - Q^k_0dT=0\quad \text{mod}\ I $$

where $Q^{k}_{0}=D_{T}Q^{k}$, Q ^k+1=D _X Q ^k and where I is the module of contact one-forms. From the fact that dF=DF mod I we obtain $DQ^{k}-Q^{k+1}DX-Q^{k}_{0}DT=0$ on imposing the contact conditions. Then DT∧DQ ^k=Q ^k+1 DT∧DX follows easily.

Then we make the substitution $q'_{l}=Q^{l}$ in the evolution equation

$$p'=F' \bigl(t',x',u',q',q'_1 \ldots, q'_n \bigr). $$

The result is to be another evolution equation

$$p=F(t,x,u,q,q_1,\ldots, q_n) $$

so that the right-hand side contains the highest spatial derivative $q_{n}=D^{n}_{x}q$ and none of the derivatives $p_{n}=D^{n}_{x} p,\; p_{0}^{n}=D^{n}_{t}p,\; q_{0}^{n}=D^{n}_{t}q$ nor the mixed derivatives $q_{k,l}=D^{k}_{t}D^{l}_{x} q$ for k≥1, and $p_{k, l}=D^{k}_{t}D^{l}_{x} p$. Note that since $p=D_{t}u,\; q=D_{x} u$ we have p _k,l=q _k+1,l−1 and in particular q _1,0=q ₀=p _0,1=p ₁. Thus, we require that $Q^{n}_{p_{0}^{n}}=0$ and $Q^{n}_{p_{n}}=0$ as well as $Q^{n}_{q_{n}}\neq0$.

We now show that D _t T≠0. For if D _t T=0 then T=T(x). Then the relation Q ^k+1 DT∧DX=DT∧DQ ^k gives Q ^k+1 T′(x)D _t Xdt∧dx=D _t Q ^k T′(x)dt∧dx, so that $Q^{k+1}=\frac{D_{t}Q^{k}}{D_{t}X}$. We have for k=0 that $Q^{1}=\frac {D_{t}Q}{D_{t}X}$ which shows that Q ¹=Q ¹(t,x,u,p,q,p ₀,q ₀). Thus, by induction, we find that Q ⁿ is independent of the spatial derivatives of $p_{n},\; q_{n}$, and hence we contradict the requirement that $Q^{n}_{p_{n}}\neq0$. Consequently we must have D _t T≠0.

It follows from the fact that D _t T≠0 that $Q^{1}_{q_{1}}\neq0$: in fact, we note that if $Q^{k}_{q_{k}}\neq0$ then DQ ^k will depend linearly on q _k+1 and this appears only in the term D _x Q ^k as the coefficient of $Q^{k}_{q_{k}}$. So, differentiating Q ^k+1 DT∧DX=DT∧DQ ^k with respect to q _k+1 we find that $Q^{k+1}_{q_{k+1}}DT\wedge DX=Q^{k}_{q_{k}}D_{t}Tdt\wedge dx$. Since we have $Q^{n}_{q_{n}}\neq0$, D _t T≠0, DT∧DX≠0, it follows, by induction, that $Q^{1}_{q_{1}}\neq0$.

Our next step is to show that D _x T=0 and to this end we assume that D _x T≠0. Then $Q^{1}_{p_{0}}=0$. In fact, if $Q^{k}_{p^{k}_{0}}\neq0$ then DQ ^k will be linear in $p^{k+1}_{0}$, and this will appear in the coefficient of $Q^{k}_{p^{k}_{0}}$ in D _t Q ^k. So, differentiating $Q^{k+1}_{q_{k+1}}DT\wedge DX=Q^{k}_{q_{k}}D_{t}Tdt\wedge dx$ with respect to $p^{k+1}_{0}$ we find that $Q^{k+1}_{p^{k+1}_{0}}DT\wedge DX=-Q^{k}_{p^{k}_{0}}D_{x}Tdt\wedge dx$. Thus, if D _x T≠0 then $Q^{1}_{p_{0}}\neq0$ implies, by induction, that $Q^{n}_{p^{n}_{0}}\neq0$, and this contradicts our requirement $Q^{n}_{p^{n}_{0}}=0$. Hence $Q^{1}_{p_{0}}=0$.

Another consequence of D _x T≠0 is that $Q_{q_{0}}=Q_{p_{1}}=0$. For suppose $Q^{1}_{q_{0}}\neq0$. We know that $Q^{k}_{p_{0}^{k}}=0$ so that D _x Q ^k is independent of $q^{k+1}_{0}=D_{x}p^{k}_{0}$, and therefore, from Q ^k+1 DT∧DX=DT∧DQ ^k it follows that $Q^{k+1}_{q^{k+1}_{0}}DT \wedge DX=-Q^{k}_{q^{k}_{0}}D_{x}Tdt\wedge dx$ for k≥1. So, if $Q^{1}_{q_{0}}\neq0$, then $Q^{n}_{q^{n}_{0}}\neq0$, as follows from induction. Since we require $Q^{n}_{q^{n}_{0}}= 0$, we must have $Q^{1}_{q_{0}}=0$, because we also assume D _x T≠0. Note that q ₀=D _t q=D _t D _x u=D _x D _t u=D _x p=p ₁, and so we also have $Q^{1}_{p_{1}}=0$.

From all this we find that Q ¹=Q ¹(t,x,u,p,q,q ₁) and it then follows that Q ⁿ will contain the n-th order derivative $q_{n-1, 1}=D^{n-1}_{t}D_{x}q$ if D _x T≠0. In fact, putting $q_{k-1, 1}=D^{k-1}_{t}D_{x}q$ for k≥1, we have from Q ^k+1 DT∧DX=DT∧DQ ^k that

$$Q^{k+1}_{q_{k, 1}}DT\wedge DX=-Q^{k}_{q_{k-1, 1}}D_xTdt \wedge dx $$

for k≥1, since q _k,1 is a derivative of order k+1 which can only occur in D _t Q ^k. Now q _0,1=q ₁ and we know that $Q^{1}_{q_{1}}\neq0$, so it follows by induction that $Q^{n}_{q_{n-1, 1}}\neq0$ if D _x T≠0. However, this is a contradiction since we must have $Q^{n}_{q_{n-1, 1}}=0$ for our transformations to transform any given evolution equation into an evolution equation. Hence we conclude that D _x T=0 and it then follows easily that T=T(t).

Finally, noting that

Open image in new window

for some function W(t,x,u,p,q), we find, on integrating this system, that the contact transformation has the stated form. This proves the result.

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P. Basarab-Horwath
- 1
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F. Güngör
- 2
V. Lahno
- 3

1.Department of MathematicsLinköping UniversityLinköpingSweden
2.Department of Mathematics, Faculty of Arts and SciencesDoğuş UniversityIstanbulTurkey
3.Department of MathematicsPedagogical UniversityPoltavaUkraine

Symmetry Classification of Third-Order Nonlinear Evolution Equations. Part I: Semi-simple Algebras

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