Chapter 6 Partial Differential Equations

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46 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS For the wave equation in R 2 we again have the equation and principal part are the same L = P = ∂ (2,0) 1 − ∂ (0,2) 2 = ∂2 − ∂2 . ∂x 2 1 ∂x 2 2 For the heat operator in R 2 we have the equation and principal part which in this case are not the same L = ∂ (2,0) 1 − ∂ (0,1) 2 = ∂2 − ∂ , ∂x 2 1 ∂x 2 and P = ∂ (2,0) 1 = ∂2 . ∂x 2 1 The characteristic form (or principal symbol) atx ∈ Ω is the homogeneous polynomial of degree k defined for ξ ∈ R n by χ L (x, ξ) = ∑ a α (x)ξ α . |α|=k A vector ξ is characteristic or is called a characteristic direction, for L at x if χ L (x, ξ) =0. The characteristic variety is the set of all characteristic vectors ξ, i.e., Char x (L) ={ξ ≠0:χ L (x, ξ) =0}. As we already have defined, a surface (or curve) S is said to be characteristic with respect to L at x 0 if the normal vector ν(x 0 )atx 0 defines a characteristic directions. Example 6.3.10. Consider L = a(x, y) ∂ ∂x + b(x, y) ∂ ∂y = a (1,0)∂ (1,0) + a (0,1) ∂ (0,1) . The characteristic form is χ L (x, ξ) =a(x, y)ξ 1 + b(x, y)ξ 2 . Let C be a characteristic curve given parametrically by (x(t),y(t)). The tangent to this curve is (x ′ (t),y ′ (t)) and the normal is (y ′ (t), −x ′ (t)). Since C is a characteristic curve we must have a(x(t),y(t))y ′ (t) − b(x(t),y(t))x ′ (t) =0. Thus the characteristic curve can be found from dx dt = a(x(t),y(t)), dy dt = b(x(t),y(t)) just as we defined them earlier.
6.3. CHARACTERISTICS AND HIGHER ORDER EQUATIONS 47 Example 6.3.11. Reconsider the wave equation in R 2 Lu = ∂ (2,0) 1 u − ∂ (0,2) 2 u = ∂2 u − ∂2 u =0. ∂x 2 1 ∂x 2 2 If ξ =(ξ 1 ,ξ 2 ) defines a characteristic direction, then we must have ξ 2 1 − ξ 2 2 =0. Now if C : (x(t),y(t)) is a characteristic curve, let ξ 1 = dy dt , ξ 2 = − dx dt denote a normal to C. Then we have ( ) 2 ( ) 2 ( dy dx dy − = dt dt dt − dx )( dy dt dt + dx ) =0 dt This equation is satisfied if we take or dy dx = ±1, y − x = c, y + x = c, and we see that the characteristic curves are straight lines (just as we saw earlier). Example 6.3.12. For the equation (6.3.4), the principal part is A characteristic curve (x(t),y(t)) has normal Pu = a ∂2 u ∂x 2 +2b ∂2 u ∂x∂y + c∂2 u ∂y 2 . that satisfies or or a ( ) 2 dy − 2b dt ξ 1 = dy dt , ξ 2 = − dx dt aξ 2 1 +2bξ 1 ξ 2 + cξ 2 2 =0, ( dy dt )( ) dx + c dt ady 2 − 2bdydx + cdx 2 =0, ( ) 2 dx =0, dt
Page 1 and 2: Chapter 6 Partial Differential Equa
Page 3 and 4: 6.1. INTRODUCTION 3 In this case we
Page 5 and 6: 6.1. INTRODUCTION 5 Solutions to La
Page 7 and 8: 6.1. INTRODUCTION 7 t u=0 u=1/2 u=1
Page 9 and 10: 6.2. LINEAR AND QUASILINEAR EQUATIO
Page 31 and 32: 6.3. CHARACTERISTICS AND HIGHER ORD
Page 45: 6.3. CHARACTERISTICS AND HIGHER ORD
Page 53 and 54: Bibliography [1] R. Dautray and J.L

Chapter 6 Partial Differential Equations

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