By Leonard Lovering Barrett

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**Extra resources for An introduction to tensor analysis **

**Sample text**

51) leads to ∂F0 (t) ∂F1 (x, t) −2 = 0, ∂t ∂x ∂F0 (t) ∂F1 (x, t) +u + F2 (x, t) + uF3 (x, t) ∂t ∂t ∂F3 (x, t) ∂ 2 F1 (x, t) ∂F1 (x, t) +2 − = 0, −u ∂x ∂x ∂x2 ∂F2 (x, t) ∂F2 (x, t) ∂ 2 F2 (x, t) −u − ∂t ∂x ∂x2 ∂F3 (x, t) ∂F3 (x, t) ∂ 2 F3 (x, t) +u −u − = 0. 53) we obtain F1 (x, t) = ∂F3 (x, t) = 0, ∂x 3. SYMMETRIES OF THE BURGERS EQUATION ∂F3 (x, t) ∂ 2 F3 (x, t) ∂F2 (x, t) − − = 0, ∂t ∂x2 ∂x ∂F2 (x, t) ∂ 2 F2 (x, t) − = 0. 56) 1 ∂F0 (t) , 2 ∂t ∂F4 (t) x ∂ 2 F0 (t) F2 (x, t) = − − . 56) now fulfill automatically, while the third equation is a polynomial with respect to x; hence we have F3 (x, t) = − ∂ 2 F4 (t) x ∂ 3 F0 (t) + = 0, ∂t2 2 ∂t3 from which we finally arrive at F0 (t) = c1 + c2 t + c3 t2 , F4 (t) = c4 + c5 t.

Namely, we compute generating functions (ϕi )jµ = Yi ωAjµ , j = 1, . . , 3, µ = 1, . . 119) 50 1. CLASSICAL SYMMETRIES while Yi refers to the fields X1 , X2 , X3 , [X1 , X2 ], [X1 , X3 ], [X2 , X3 ]. Then we impose additional equations ϕi j (x , . . , x4 , . . , Ajµ , . . , Ajµν , . . 120) and solve them together with the initial system. 119) we arrive at a system of 6 × 12 = 72 equations. 121) where 1 r = (x21 + x22 + x23 + x24 ) 2 . 120). 124) C being a constant. 124) is just the Belavin–Polyakov– Schwartz–Tyupkin instanton solution!

Let E ⊂ J k (π) be a differential equation of order k. Define the set E 1 = {θk+1 ∈ J k+1 (π) | πk+1,k (θk+1 ) ∈ E, Lθk+1 ⊂ Tπk+1,k (θk+1 ) E} and call it the first prolongation of the equation E. 2. NONLINEAR PDE 29 If the first prolongation E 1 is a submanifold in J k+1 (π), we define the second prolongation of E as (E 1 )1 ⊂ J k+2 (π), etc. Thus the l-th prolongation is a subset E l ⊂ J k+l (π). Let us redefine the notion of l-th prolongation directly. Namely, take a point θk ∈ E and consider a section ϕ ∈ Γloc (π) such that the graph of jk (ϕ) is tangent to E with order l.

### An introduction to tensor analysis by Leonard Lovering Barrett

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