introduction to symmetry analysiscantwell/aa218_course_material/... · 2018-05-08 · introduction...
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Introduction to Symmetry Analysis
Brian Cantwell Department of Aeronautics and Astronautics
Stanford University
Chapter 9 - Partial Differential Equations
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!Finite transformation of partial derivatives
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The p-th extended finite group is
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!Variable count
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Infinitesimal transformation of first partial derivatives
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!Substitute
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The (p-1)th order extended infinitesimal transformation is
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!The number of terms in the infinitesimal versus the derivative order
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!Isolating the determining equations of the group - the Lie algorithm
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!The classical point group of the heat equation
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Invariance condition
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The fully expanded invariance condition is
Apply the constraint that the solution must satisfy theheat equation and gather terms
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The determining equations of the point group of the heat equation
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Series solution of the determining equations
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The classical six parameter group of the heat equation
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!Impulsive source solutions of the heat equation
Boundary conditions
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Is the integral conserved?
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!This problem is invariant under the three parameter group of dilationsin the dependent and independent variables and translation in time.
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!Similarity variables are the invariants of the infinitesimal transformation.
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!Now substitute the similarity form of the solution into the heat equation.The result is a second order ODE of Sturm-Liouville type.
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!Flow in porous media - A modified problem of an instantaneous heat source.
Darcy's law
Pressure diffuses in a porous medium.
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9.135 9.131
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9.135 9.131
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Example computation from Barenblatt Scaling, self-similarity and intermediate asymptotics
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ξ = − C2 coth C2 x +C1 C2⎡⎣ ⎤⎦
Note that
Also satisfies
ξxx − 2ξξx = 0
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!What solutions do these groups correspond to?
dx− C2 Tanh C2 x +C1 C2⎡⎣ ⎤⎦
= dt1= du0
ordx
− C2 Coth C2 x +C1 C2⎡⎣ ⎤⎦= dt1= du0
Characteristic equations
Similarity variables (invariants)
θ = eC2tSinh[ C2 x +C1 C2 ]U = uor
θ = eC2t Cosh[ C2 x +C1 C2 ]U = u
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What solution does this group correspond to?
Let u =U θ[ ]Substitute into the heat equation
ut =U ' θ[ ]θtux =U ' θ[ ]θ x
uxx =U '' θ[ ] θ x( )2 +U ' θ[ ]θ xx
ut − uxx =U ' θ[ ] θt −θ xx( )−U '' θ[ ] θ x( )2 = 0θt −θ xx = 0⇒U '' θ[ ] = 0U θ[ ] = Aθ + B
u = AeC2t Sinh[ C2 x +C1 C2 ]+ B
u = AeC2t Cosh[ C2 x +C1 C2 ]+ B
Solutions
θ = eC2tSinh[ C2 x +C1 C2 ]U = uor
θ = eC2t Cosh[ C2 x +C1 C2 ]U = u
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Let’s look for a solution of the heat equation using separation of variables
ut = uxx
Let u = g t[ ] f x[ ]ut = gt fux = gfxuxx = gfxx
Substitute gt f = gfxx
This equality implies gtg= fxx
f= λ
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The solution of the heat equation derived by separation of variables matches the solution generated from non-classical symmetries.
gtg= fxx
f= λ
g = Aeλt
f = Sinh λ x + a⎡⎣ ⎤⎦
f = Cosh λ x + a⎡⎣ ⎤⎦
u = Aeλt Sinh λ x + a⎡⎣ ⎤⎦ + B
u = Aeλt Cosh λ x + a⎡⎣ ⎤⎦ + B
Letλ = C2
a = C1 C2
u = AeC2t Sinh C2 x +C1 C2⎡⎣ ⎤⎦ + B
u = AeC2t Cosh C2 x +C1 C2⎡⎣ ⎤⎦ + B
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Include also the case c=constant. Identify the Lorentz transformation