Chapter 6 Partial Differential Equations

More documents

Recommendations

Info

$Math 1351 Sample Test 1$

$Math 3350 Fall 2004 , Sample Test # 1, Name Please note that I am ...$

$Math 3350 Sec. 6, Sample Test # 1, Name 1. Solve the separable ...$

$Matlab Lesson 6 for Math 4330 and 5344 - Texas Tech University$

$Math 4330 Sec. 1, Matlab Assignment # 1, March 24, 2006 Name 1 ...$

36 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS Moreover, and so ϕ(s, t) =s ϕ s =1=ϕ x x s + ϕ y y s . If ϕ y = 0, then it follows from (6.3.13) that ϕ x = 0, a contradiction. Now, as we have learned in our earlier work, the condition (6.3.16) guarantees that we can solve for (s, t) in terms of (x, y) to obtain ϕ(x, y). Similarly, we can obtain ψ(x, y) by solving dx dt =1, ( √ ) dy b + dt = b2 − ac , a For this problem we take the initial conditions dψ dt =0. x 0 (s) =x 0 ,y 0 (s) =y 0 + s, ψ 0 (s) =s. Again, ψ(s, t) =s and as above ψ y ≠0. ThusJ ( (ϕ, ψ)/(x, y) ) ≠0. In summary we have found a change of coordinates α = ϕ(x, y), β = ψ(x, y) that reduces the equation to 2˜bu αβ + l.o.t. = 0. It is easy to show that ˜b =2 ( ac − b 2 a ) ϕ y ψ y (6.3.17) and so ˜b ≠ 0 and we can divide by it to put (6.3.17) into the normal form Remark 6.3.4. u αβ = F (α, β, u, u α ,u β ). (a) The reduction to normal form is a local result. (b) The special initial curve can be replaced by any curve (x 0 (s),y 0 (s),ϕ 0 (s)) with ϕ 0 (s) =s and (x 0 (s),y 0 (s)) that specifies a curve in the xy-plane where the PDE is hyperbolic. (c) The curves ϕ(x, y) = constant, ψ(x, y) = constant are called characteristics of (6.3.4). Further, equation (6.3.11) is called the characteristic equation for the PDE.
6.3. CHARACTERISTICS AND HIGHER ORDER EQUATIONS 37 (d) Note that the characteristics for the first order PDE’s for ϕ and ψ are determined by ( √ ) dx dt =1, dy b ± dt = b2 − ac a or ( √ ) dy b ± dx = b2 − ac . a But if ϕ(x, y) = constant and ϕ solves (6.3.13), then and so dy dx = −ϕ x ϕ y = ϕ x + ϕ y dy dx =0 ( b − √ b2 − ac a ) . (6.3.18) Hence the characteristics for a 2nd order PDE (6.3.4) coincide with the characteristics of the associated 1st order PDE (6.3.13) (Similarly for ψ). (e) In order to obtain the other hyperbolic form we set ξ = α + β, η = α − β so that and Thus we have α = ξ + η 2 , β = ξ − η 2 , u α = u η η α + u ξ ξ α = u η + u ξ , u β = u η η β + u ξ ξ β = −u η + u ξ . u αβ =(−u ηη + u ηξ )+(−u ξη + u ξξ )=u ξξ − u ηη . Example 6.3.5. Consider the equation y 2 u xx − x 2 u yy = 0, x > 0, y > 0. Here b 2 − ac = x 2 y 2 > 0. The characteristics are given by ( see (6.3.18)) ( √ ) dy b + dx = b2 − ac = x a y , and ( √ ) dy b − dx = b2 − ac = − x a y .
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 35: 6.3. CHARACTERISTICS AND HIGHER ORD
Page 53 and 54: Bibliography [1] R. Dautray and J.L

Chapter 6 Partial Differential Equations

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?