Lie Symmetry Methods for Pdes and P∆Es

Lecture 3: Lie symmetry methods for PDEs and P∆Es

Lecture 3: Lie symmetry methods for PDEs and P∆Es Symmetry Methods for Diﬀerential and Diﬀerence Equations

Peter Hydon

University of Kent Lecture 3: Lie symmetry methods for PDEs and P∆Es Outline

1 Lie symmetries of scalar PDEs

2 Lie symmetries of scalar P∆Es

3 Invariant solutions

4 Linearization using Lie point symmetries

5 Summary: the main results in Lecture 3 Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar PDEs

Lie symmetries of scalar PDEs Point symmetries of PDEs are defined in much the same way as those of ODEs. For simplicity, let us start by considering PDEs with one dependent variable, u, and two independent variables, x and t. A point transformation is a locally-defined diffeomorphism

Γ:(x, t, u) 7→ xˆ(x, t, u), ˆt(x, t, u), uˆ(x, t, u).

This transformation maps any surface u = f (x, t) to the following surface (which is parameterized by x and t):

xˆ =x ˆx, t, f (x, t), ˆt = ˆtx, t, f (x, t), uˆ =u ˆx, t, f (x, t). Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar PDEs

Lie point transformations can be written in the form: xˆ = x + εξ(x, t, u) + O(ε2), ˆt = t + ετ(x, t, u) + O(ε2), uˆ = u + εη(x, t, u) + O(ε2). For such transformations, the surface u = f (x, t) is mapped to uˆ = f (x, t) + εηx, t, f (x, t) + O(ε2), x =x ˆ − εξxˆ, ˆt, f (ˆx, ˆt) + O(ε2), t = ˆt − ετxˆ, ˆt, f (ˆx, ˆt) + O(ε2), which amounts to uˆ = f (ˆx, ˆt) + εQxˆ, ˆt, f (ˆx, ˆt) + O(ε2), where Q = η(x, t, u) − u,x ξ(x, t, u) − u,t τ(x, t, u) is the characteristic. Any surface on which Q = 0 is invariant. Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar PDEs

Now consider the effect of a point transformation on a given scalar PDE, A(x, t, [u]) = 0; (1) for differential equations, square brackets around an expression denote the expression and its derviatives. The transformation is prolonged to derivatives of u via the chain rule, by differentiatingx ˆ, ˆt andu ˆ with respect the parameters x and t. To do this, use the total derivative operators

Dx = ∂x + ux ∂u + uxx ∂ux + uxt ∂ut + ··· ,

Dt = ∂t + ut ∂u + uxt ∂ux + utt ∂ut + ··· .

(Total derivatives treat the dependent variable and its derivatives as functions of the independent variables.) Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar PDEs

The transformation is a symmetry if it maps the set of solutions to itself. This gives the symmetry condition,

A(ˆx, ˆt, [ˆu]) = 0 when (1) holds. (2)

For Lie point symmetries, the prolonged inﬁnitesimal generator is

x t xx xt X = ξ ∂x + τ ∂t + η ∂u + η ∂u,x + η ∂u,t + η ∂u,xx + η ∂u,xt + ··· ,

where ξ, τ and η stand for ξ(x, t, u), τ(x, t, u) and η(x, t, u) respectively, and where the prolongation terms are

x t η = Dx Q + ξuxx + τuxt , η = Dt Q + ξuxt + τutt , xx 2 xt η = Dx Q + ξuxxx + τuxxt , η = Dx Dt Q + ξuxxt + τuxtt ,

and so on. Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar PDEs

Explicitly,

x 2 η = ηx + (ηu − ξx )ux − τx ut − ξuux − τuux ut ,

t 2 η = ηt − ξt ux + (ηu − τt )ut − ξuux ut − τuut ,

xx 2 η = ηxx + (2ηxu − ξxx )ux − τxx ut + (ηuu − 2ξxu)ux 3 2 − 2τxuux ut − ξuuux − τuuux ut + (ηu − 2ξx )uxx − 2τx uxt − 3ξuux uxx − τuut uxx − 2τuux uxt ,

xt 2 η = ηxt + (ηtu − ξxt )ux + (ηxu − τxt )ut − ξtuux 2 2 2 + (ηuu − ξxu − τtu)ux ut − τxuut − ξuuux ut − τuuux ut − ξt uxx − ξuut uxx + (ηu − ξx − τt )uxt − 2ξuux uxt

− 2τuut uxt − τx utt − τuux utt . Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar PDEs

The linearized symmetry condition (LSC) for Lie point symmetries is X A = 0 when A = 0, where A is shorthand for A(x, t, [u]). Just as for ODEs, the LSC is solved by ﬁrst using the given PDE to reduce the LSC to an identity on (x, t, u) and some derviatives of u, then splitting this identity according to each term’s dependence on the remaining derivatives of u. This process works when the PDE can be put into the following (generalized) Kovalevskaya form: it can be written (possibly after a change of independent variables) as

p A = Dx u − ω(x, t, [u]),

where ω has no derivatives of u of order ≥ p with respect to x. Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar PDEs

Example The LSC for Burgers’ equation,

uxx − ut − uux = 0, (3) is xx t x η − η − uη − ηux = 0 when (3) holds.

Once uxx has been replaced by the left hand side of (3), the highest-order derivative terms in the LSC have a factor uxt . We start by writing down those terms alone:

−2(τx + τuux )uxt = 0 −→ τx = τu = 0. This early result removes many terms from the LSC, leaving only

0 = ηxx − ηt − uηx + (2ηxu + ξt − ξxx − uξx − η)ux 2 3 + (ηuu − 2ξxu − 2uξu)ux − ξuuux + τt − 2ξx − 2ξuux ut . Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar PDEs

In particular, the terms multiplied by ut split into

1 0 ξu = 0, ξx = 2 τ (t). Hence 1 0 ξ = 2 τ (t)x + α(t), for some locally smooth function α.The remaining terms in the LSC are split by equating powers of ux . A routine calculation then shows that the Lie algebra of point symmetry generators is spanned by

X1 = ∂x , X2 = ∂t , X3 = t∂x + ∂u,

2 X4 = x∂x + 2t∂t − u∂u, X5 = xt∂x + t ∂t + (x − ut)∂u. Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar PDEs

Generalized symmetries of PDEs For Lie point symmetries, the inﬁnitely prolonged inﬁnitesimal generator decomposes into X = ξDx + τDt + Xe, where the red terms generate trivial symmetries and the nontrivial component is 2 Xe = Q∂u+(Dx Q)∂u,x +(Dt Q)∂u,t +(Dx Q)∂u,xx +(Dx Dt Q)∂u,xt +··· . A useful notation, which generalizes immediately to an arbitrary number of independent variables, is

X ∂ j k Xe = (DJQ) , J = (j, k), DJ = Dx Dt . ∂u,J J

Generalized symmetries are generators Xe whose characteristic Q has any given dependence on x, t and [u]. The LSC in all cases is Xe A = 0 when [A = 0]. Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar PDEs

Example: The potential Burgers equation,

2 A = u,xx − u,t + u,x = 0,

has a generalized symmetry characteristic,

3 Q = u,xxx + 3u,x u,xx + u,x . (Olver)

Deﬁnition Two generalized symmetry characteristics for a given PDE, A = 0, are equivalent if their diﬀerence vanishes on all solutions of A = 0. The characteristic in the example above is not equivalent to a characteristic of Lie point symmetries; it is genuinely higher-order. Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar P∆Es

Lie symmetries of scalar P∆Es Notation (scalar P∆Es with two independent variables)

u n m

Independent variables: m, n; dependent variable: u. Shift operators:

Sm :(m, n) → (m + 1, n); Sn :(m, n) → (m, n + 1).

Shorthand: ui,j = u(m + i, n + j). Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar P∆Es

The method for ﬁnding Lie symmetries of a given P∆E is much the same as the method for O∆Es. For example, to ﬁnd the Lie point symmetries of a P∆E of the form

u1,1 = ω(m, n, u, u1,0, u0,1), (4)

substitute uˆ = u + εQ(m, n, u) + O(ε2) into (4) and expand to ﬁrst order in ε. This yields the LSC,

Q(m + 1, n + 1, ω) = ω,1Q(m, n, u) + ω,2Q(m + 1, n, u1,0)

+ ω,3Q(m, n + 1, u0,1).

Just as for O∆Es, the LSC is a functional equation, which can be solved by the method of diﬀerential elimination. Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar P∆Es

Example For the autonomous dpKdV equation, 1 u1,1 = u + , (5) u1,0 − u0,1

the LSC can be written as an identity in u, u1,0 and u0,1:

Q(m, n + 1, u0,1) − Q(m + 1, n, u1,0) Q(m+1, n+1, ω) = Q(m, n, u)+ 2 , (u1,0 − u0,1) where ω is the right-hand side of (5). Applying the operator ∂/∂u1,0 + ∂/∂u0,1 to the LSC, then simplifying, we get

0 0 Q (m, n + 1, u0,1) − Q (m + 1, n, u1,0) = 0. (6)

00 Thus Q (m + 1, n, u1,0) = 0, so Q is linear in its third argument: Q(m, n, u) = A(m, n) u + B(m, n)

for some functions A and B. Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar P∆Es

The intermediate equation (6) gives

A(m, n + 1) − A(m + 1, n) = 0 −→ A(m, n) = α(m + n).

Finally, the LSC gives the following (simpliﬁed) conditions:

α(m + n + 1) = −α(m + n),

B(m + 1, n + 1) = B(m, n), B(m + 1, n) = B(m, n + 1). Therefore the set of all characteristics of Lie point symmetries is spanned by

m+n m+n Q1 = 1, Q2 = (−1) , Q3 = (−1) u.

Note This is a relatively simple example: diﬀerential elimination usually removes only one term at a time. Even so, it illustrates the general method. Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar P∆Es

The method is easily extended to generalized symmetries, for which Q depends on m, n and [u]. (For diﬀerence equations, square brackets around an expression denote the expression and its shifts.) Exercise (Fill in the gaps!) The P∆E

u (u − 1) u = 0,1 1,0 + 1 (7) 1,1 u has no Lie point symmetries, so let us look for characteristics of the form Q = Q(m, n, u, u1,0). The P∆E (7) prolongs to the following, whose right-hand side is written in terms of a set of ‘initial variables’, I = {ui,j : ij = 0}:

{u0,1(u1,0 − 1) + u}(u2,0 − 1) u2,1 = + 1. (8) uu1,0 Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar P∆Es

Now write the LSC as an identity in terms of I, diﬀerentiate the result with respect to u2,0 and simplify to obtain

u1,1 u1,0 Q,2(m+1, n+1, u1,1, u2,1) = Q,2(m+1, n, u1,0, u2,0). u1,1 − 1 u1,0 − 1 (9) Here u1,1 and u2,1 are defined by (7) and (8). To eliminate the left-hand side, differentiate with respect to u1,0, keeping u1,1 and u2,1 fixed. This yields the PDE ∂ u2,0 − 1 ∂ u1,0 + Q,2(m + 1, n, u1,0, u2,0) = 0, ∂u1,0 u1,0 ∂u2,0 u1,0 − 1 whose solution, taking (9) into account, is u2,0 − 1 Q(m+1, n, u1,0, u2,0) = (u1,0−1) A m + 1, +B(m+1, n, u1,0). u1,0 Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar P∆Es

Having split Q, it is now fairly easy to continue the diﬀerential elimination and obtain the general solution of the LSC:

f (m + 1)u(1 − u) Q(m, n, u, u1,0) = f (m)u + . (10) u1,0 − 1 Here f is an arbitrary function, so there is an inﬁnite family of characteristics. This may seem surprising, given that the P∆E (7) has no Lie point symmetries. (The reason is that the P∆E is Darboux integrable.) Note An essential step in ﬁnding symmetry characteristics for P∆Es (resp. PDEs) is that we can eliminate some shifts (resp. derivatives) of u to write the LSC as an identity in some initial variables. For a PDE that can be put in Kovalevskaya form, the existence of such a set is guaranteed. There is an analogue for P∆Es. Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar P∆Es

An important technicality: Kovalevskaya form

stencil points are black j vertices are ringed i

The stencil is the set of points (i, j) such that ui,j appears in the P∆E. Each of the following P∆Es has the above stencil:

2 u2,0 = (u1,1 − 2u + u0,2) , (11)

2 u1,1 = (u2,0 − 2u + u0,2) . (12) A scalar P∆E can be put into Kovalevskaya form if u at a vertex is uniquely determined by the values of u elsewhere on the stencil. So (11) can be put into Kovalevskaya form, but (12) cannot. Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar P∆Es

shear rotate

2 Lemma A transformation preserves the product structure of Z if and only if it is affine. A stencil point can be made rightmost by an affine transformation if and only if it is a vertex. Consequently, a given P∆E that can be put into Kovalevskaya form may be written (possibly after an affine transformation of the independent variables) as

A = up,0 − ω(m, n, [u]),

where ω contains only those ui,j for which 0 ≤ i < p. Then the LSC can be written as an identity in I0 = {ui,j : 0 ≤ i < p}. Lecture 3: Lie symmetry methods for PDEs and P∆Es Lie symmetries of scalar P∆Es

Note Kovalevskaya form guarantees the existence of a set of initial values. However, for some classes of P∆E, it is convenient (for computational reasons) to choose a set of initial values that diﬀers from I0. If a given P∆E of the form

2 u1,1 = ω(m, n, u, u1,0, u0,1), (m, n) ∈ Z ,

can also be written as equations for each of u, u1,0 and u0,1, it is usually easiest to use the initial values I = {ui,j : ij = 0}. Lecture 3: Lie symmetry methods for PDEs and P∆Es Invariant solutions

Invariant solutions of a given PDE

Main idea: Given a PDE and its Lie symmetries, seek an invariant solution that satisﬁes Q = 0. For Lie point symmetries, the condition Q = 0 amounts to

η(x, t, u) − u,x ξ(x, t, u) − u,t τ(x, t, u) = 0,

which is solvable by the method of characteristics. The solution is an arbitrary (locally smooth) function of the ﬁrst integrals of the Pfaﬃan system dx dt dx = = . ξ(x, t, u) τ(x, t, u) η(x, t, u) Lecture 3: Lie symmetry methods for PDEs and P∆Es Invariant solutions

Example The Lie symmetries of the Thomas equation,

u,xt = u,x u,t − 1, (13)

include ξ(x, t, u) = x, τ(x, t, u) = −t, η(x, t, u) = −1/2. Therefore, the Pfaﬃan system governing invariant solutions is dx dt du √ = = −→ r = xt, v = u − 1 ln |t|. x −t −1/2 2

As r = c cannot be a solution, every invariant solution must be of 1 the form v = F (r). Substitute u = 2 ln |t| + F (r(x, t)) into (13):

00 0 2 2r −2r F (r) − F (r) + 4 = 0 −→ F (r) = − ln c1e + c2e . Lecture 3: Lie symmetry methods for PDEs and P∆Es Invariant solutions

Generalized symmetries can also be used to give invariant solutions of a given PDE, A = 0. In general, the method of characteristics no longer works, but one may be able to use diﬀerential algebra to reduce the overdetermined system

A = 0, Q = 0

to something simpler. Lecture 3: Lie symmetry methods for PDEs and P∆Es Invariant solutions

Example: The potential Burgers equation,

2 A = u,xx + u,x − u,t = 0, has a generalized symmetry characteristic,

3 Q = u,xxx + 3u,x u,xx + u,x . (Olver) Although Q = 0 is a parametrized ODE, it is not easy to solve explicitly! (This is even though it is a second-order ODE for v = ux with two independent Lie point symmetry generators.) However, we seek only those solutions that also satisfy A = 0. Use this PDE to eliminate u,xxx and u,xx from Q = 0, giving a PDE that linearizes to the wave equation (using u = ln |v|):

u,xt + u,x u,t = 0 −→ v,xt = 0 −→ u = ln |f (x) + g(t)|.

2 The constraint A = 0 yields u = ln |c1(x + 2t) + c2x + c3|. Lecture 3: Lie symmetry methods for PDEs and P∆Es Invariant solutions

Invariant solutions of a given P∆E Exactly the same idea works for P∆Es. The main distinction is that, for Lie point symmetries, Q = 0 is an algebraic equation. Example Earlier, we found all Lie point symmetries of the autonomous dpKdV equation, 1 u1,1 = u + . (14) u1,0 − u0,1 Every characteristic that involves u is a multiple of

m+n m+n Q = (−1) u + c1 + c2(−1) , so any invariant solutions must be of the form

m+n u = −c1(−1) − c2.

No such solution exists, as it would give u1,0 − u0,1 = 0. Lecture 3: Lie symmetry methods for PDEs and P∆Es Invariant solutions

However, the autonomous dpKdV equation (14) also has (among many others) the generalized symmetry characteristic

−1 Q = 1 − 2(u1,0 − u−1,0) , which is zero if and only if

u = m + f (n) + g(n)(−1)m+n,

for some functions f and g. Then (14) holds if and only if

2 2 m+n 0 = (u1,1−u)(u0,1−u1,0)+1 = (f1−f ) −(g1−g) −2(−1) (g1−g). Comparing terms with equal dependence on m splits this condition:

m+n g1 − g = 0, f1 − f = 0, −→ u = m + c1 + c2(−1) . This is among the simplest of a rich family of invariant solutions that stem from generalized symmetry characteristics. Lecture 3: Lie symmetry methods for PDEs and P∆Es Linearization using Lie point symmetries

Linearization using Lie point symmetries As well as yielding invariant solutions, Lie point symmetries may be used to detect whether a given nonlinear PDE or P∆E is linearizable by a point transformation. By the superposition principle, every linear homogeneous scalar PDE (with the variables x, t and u) is invariant under the one-parameter Lie group

(ˆx, ˆt, uˆ) = x, t, u + εU(x, t),

where u = U(x, t) is an arbitrary solution of the PDE. Hence, the Lie algebra includes the family of generators XU = U(x, t)∂u that involves all solutions of the linear PDE. Lecture 3: Lie symmetry methods for PDEs and P∆Es Linearization using Lie point symmetries

The set of point symmetries is preserved under any point transformation (which can be regarded as a change of coordinates). Consequently, if a nonlinear PDE A = 0 can be mapped to a linear homogeneous PDE by a point transformation, the Lie algebras of the two equations must be related by this change of variables. In particular, the Lie algebra for A will contain generators that depend on solutions of the linear PDE, enabling one to identify the linearizing transformation. The same idea works equally for PDEs with any number of variables and for P∆Es. Lecture 3: Lie symmetry methods for PDEs and P∆Es Linearization using Lie point symmetries

Example For the Thomas equation, u,xt = u,x u,t − 1, the Lie algebra L of point symmetry generators is spanned by

X1 = ∂x , X2 = ∂t , X3 = x∂x − t∂t , X4 = ∂u,

u XV = V (x, t)e ∂u, where V,xt = V .

Having found that there is an underlying linear equation, v,xt = v, calculate its Lie algebra, L0:

0 0 0 0 X1 = ∂x , X2 = ∂t , X3 = x∂x − t∂t , X4 = v∂v ,

X 0 = V˜ (x, t)∂ , where V˜ = V˜ . V˜ v ,xt Use the change of variables formula to determine a transformation that maps L to L0; for simplicity, choose v = exp(−u). This is a linearizing transformation. Lecture 3: Lie symmetry methods for PDEs and P∆Es Linearization using Lie point symmetries

A slight modiﬁcation is to use the characteristics, as follows. Example Every characteristic of Lie point symmetries for the P∆E uu1,0 − 2u1,0 + 1 u1,1 = (15) u − u1,0 is of the form

2 Q(m, n, u) = c1(u − 1) + V (m, n)(u − 1) , (16) where v = V (m, n) is an arbitrary solution of

v1,1 = v1,0 − v. (17) The Lie point symmetries of (17) have characteristics

Q˜ (m, n, v) =c ˜1v + V˜ (m, n), (18) where v = V˜ (m, n) is an arbitrary solution of (17). Lecture 3: Lie symmetry methods for PDEs and P∆Es Linearization using Lie point symmetries

If the point transformation v = ψ(m, n, u) linearizes (15) to (17), every pair (c1, V ) must correspond to a pair (˜c1, V˜ ). The change of variables formula

Q˜ (m, n, v) = X ψ(m, n, u) = Q(m, n, u) ψ0(m, n, u),

amounts to

2 0 c˜1ψ(m, n, u)+V˜ (m, n) = c1(u − 1) + V (m, n)(u − 1) ψ (m, n, u).

This holds for all pairs (c1, V ) if and only if it splits as follows:

0 2 0 c˜1ψ(m, n, u) = c1(u − 1) ψ , V˜ (m, n) = V (m, n)(u − 1) ψ . It is convenient to choose V˜ (m, n) = V (m, n), which yields

ψ(m, n, u) = 1/(1 − u), c˜1 = −c1. This point transformation maps (15) to (17). Lecture 3: Lie symmetry methods for PDEs and P∆Es Summary: the main results in Lecture 3

Summary: the main results in Lecture 3 Provided that the LSC can be written as an identity, one can ﬁnd its Lie symmetries by the same methods as for ODEs or O∆Es. Besides Lie point symmetries, PDEs and P∆Es may also have generalized symmetries. Invariant solutions, which satisfy the given equation and Q = 0, may be constructed from Lie point or generalized symmetries. Lie point symmetries can be used to construct linearizing point transformations for a given PDE or P∆E, provided such transformations exist.