Introduction to Symmetry Analysis
Brian Cantwell Department of Aeronautics and Astronautics
Stanford University
Chapter 8 - Ordinary Differential Equations
The Extended Transformation is a Group Two transformations of the extended group
Compose the two transformations
The last relation is rearranged to read
Differentiating F and G gives
Comparing the expressions in parentheses we have
The composed transformation is in exactly the same form as the original transformation!
Infinitesimal transformation of the first derivative
Recall the infinitesimal transformation of coordinates
where
Substitute
Expand and retain only the lowest order terms
Once-extended infinitesimal transformation in the plane
where the infinitesimal function fully written out is
where
Expand and retain only the lowest order terms
Infinitesimal transformation of the second derivative
Expand and retain only the lowest order terms. The p times extended infinitesimal transformation is
The infinitesimal transformation of higher order derivatives
where
Construction of the general first order ODE that admits a given group - the Ricatti Equation
Let the first integral of the group be In principle we can solve for either x or y. Assume we solve for y. The equation takes the following form.
where
The general solution of the Ricatti equation can always be determined if a particular solution of the equation can be found. A particular solution in this case is
To demonstrate take the differential of f.
Dividing by yields the Ricatti equation in terms of f. ξdx
Now let.
and work out the equation that governs h[x].
The general second-order ordinary differential equation
is invariant under the twice-extended group if and only if
Consider the simplest second-order ODE
The invariance condition is
Fully written out the invariance condition is
For invariance this equation must be satisfied subject to the condition that y is a solution of
The determining equations of the group are
These equations can be used to work out the unknown infinitesimals.
Assume that the infinitesimals can be written as a multivariate power series
Insert into the determining equations
Solve yxx +1xyx − e
y = 0
Groups Xa = − x
2∂∂x
+ ∂∂y
Xb = x ln x[ ] ∂∂x
− 2 1+ ln x[ ]( ) ∂∂y
Lie algebra XiX j − X jXi =Xa Xb
Xa 0 −Xa
Xb Xa 0Xa is the ideal of the Lie algebra
Characteristic equations of Xa dxξ
= dyη
= dyxη 1{ }
⇒ dx−x / 2
= dy1
= dyxyx / 2
First two invariants dx−x / 2
= dy1
= dyxyx / 2
ln φ[ ] = y / 2 + ln x[ ]φ = xey/2
ln G[ ] = ln xyx[ ]G = xyx
Invariant groups are the same regardless of the sign
Example 8.2 - Solution
Solve yxx +1xyx − e
y = 0
Use the method of differential invariants
φ = xey/2
G = xyxFirst reduction
DG = ∂G∂x
dx + ∂G∂y
dy + ∂G∂yx
dyx
Dφ = ∂F∂x
dx + ∂F∂y
dy
DGDφ
=Gx +Gyyx +Gyx
yxxFx + Fyyx
= yx + xyxxey/2 + xyx
2ey/2
dGdφ
= yx + xyxxey/2 + xyx
2ey/2
=yx + x − 1
xyx + e
y⎛⎝⎜
⎞⎠⎟
ey/2 + xyx2ey/2
dGdφ
= xey/2
1+ xyx2
= φ
1+ G2
Second reduction integrate
1+ G2
⎛⎝⎜
⎞⎠⎟ dG = φdφ
G + G2
4− φ
2
2= C1 −1
G2 + 4G − 2φ 2 − 4 C1 −1( ) = 0
G φ[ ] = −2 ± 2 C1 +φ 2
2⎛⎝⎜
⎞⎠⎟
1/2
Example 8.2 – Solution continued; reduce the order twice
So far we have G φ[ ] = −2 ± 2 C1 +φ 2
2⎛⎝⎜
⎞⎠⎟
1/2
We still need to carry out one more integration.
φ = xey/2
G = xyx
dφ = ey/2dx + x2ey/2yxdx =
dxx
xey/2 + x2ey/2xyx
⎛⎝⎜
⎞⎠⎟ =
dxx
φ + φ2G⎛
⎝⎜⎞⎠⎟
dxx= dφ
φ 1+ G2
⎛⎝⎜
⎞⎠⎟
Nowxyx = G φ[ ]dy = G φ[ ]dx
x= G φ[ ] dφ
φ 1+ G φ[ ]2
⎛⎝⎜
⎞⎠⎟
Example 8.2 – Solution continued
G φ[ ] = −2 ± 2 C1 +φ 2
2⎛⎝⎜
⎞⎠⎟
1/2
dy =−2 ± 2 C1 +
φ 2
2⎛⎝⎜
⎞⎠⎟
1/2
1+−2 ± 2 C1 +
φ 2
2⎛⎝⎜
⎞⎠⎟
1/2⎛
⎝⎜
⎞
⎠⎟
2
⎛
⎝
⎜⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟⎟
dφφ
dy =−2 ± 2 C1 +
φ 2
2⎛⎝⎜
⎞⎠⎟
1/2
± C1 +φ 2
2⎛⎝⎜
⎞⎠⎟
1/2
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
dφφ
= 2 ∓ 2
C1 +φ 2
2⎛⎝⎜
⎞⎠⎟
1/2
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
dφφ
y = 2 ∓ 2
C1 +φ 2
2⎛⎝⎜
⎞⎠⎟
1/2
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
dφφ
+C2φ=xey∫
C2 = y −2C1
C1 ∓1( )Ln φ[ ] ± Ln 2C1 + 2C1 2C1 +φ2( )1/2⎡
⎣⎤⎦( )
φ=xey/2
Example 8.2 – Solution continued
Problem 8.9
yxx −yxy2
+ 1xy
= 0
Groups Xa = x2 ∂∂x
+ xy ∂∂y
Lie algebra XiX j − X jXi =Xa Xb
Xa 0 −Xa
Xb Xa 0Xa is the ideal of the Lie algebra
Characteristic equations of Xa dxξ
= dyη
= dyxη 1{ }
⇒ dxx2 = dy
xy= dyxy − xyx
First two invariants dxx2
= dyxy
= dyxy − xyx
φ = y / xG = y − xyx
Xb = x ∂∂x
+ y2
∂∂y
Solve
φ = y / xG = y − xyxFirst reductionDG = −xyxdx + dy − xdyx
Dφ = − yx2 dx +
1xdy
DGDφ
=Gx +Gyyx +Gyx
yxxFx + Fyyx
= −yx + yx − xyxx− yx2 +
1xyx
= −xyxx− yx2 +
1xyx
dGdφ
= −x3yxx−y + xyx
=x3 yx
y2 −1xy
⎛⎝⎜
⎞⎠⎟
y − xyx=
x2
y2 xyx − y( )y − xyx
dGdφ
= − 1φ 2
Second reduction integrate
G = 1φ+C1
Problem 8.9 – reduce the order twice
φ = y / xG = y − xyx
G φ[ ] = 1φ+C1
y − xyx =xy+C1
Considerd x / y( )dx
= 1y− xyxy2
xyx = y − y2 d x / y( )
dx= y − x
y−C1
y2 d x / y( )dx
= xy+C1
Let f = x / yx2
f 2
d f( )dx
= f +C1
x2 d f( )dx
= f 3 +C1 f2
dff 3 +C1 f
2 = dxx2 = − 1
C1 f+ 1C1
2 LnC1 + ff
⎡⎣⎢
⎤⎦⎥= − 1
x+C2
C2 =1x+ 1
C12 Ln 1+C1φ[ ]− φ
C1
⎛⎝⎜
⎞⎠⎟φ=y/x
Problem 8.9 – Solution
Carry out one more integration
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