Chapter 6 Partial Differential Equations

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48 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS or [ ( √ ) ][ ( √ ) ] b + b2 − ac b − b2 − ac a dy − dx dy − dx =0. a a Hence the characteristic curves can be found by solving ( √ ) dy b ± dx = b2 − ac , a just as we saw earlier. In an attempt to further illustrate the significance of characteristics consider the following example. Example 6.3.13. Consider the Cauchy problem that we looked at earlier Lu = ∂u ∂x =0, u(x, 0) = f(x), (x, y) ∈ R2 . From the PDE we must have u(x, y) =F (y) and the initial condition gives u(x, 0) = f(x) =F (0) so that f must be a constant or the problem has no solution. If f(x) =c is a constant then we can take any F (y) such that F (0) = c so we get infinitely many solutions. Note that the characteristics for this example are determined by dy dx =0, ⇒ y = constant. So we have prescribed our initial data along a characteristic curve. Note that y is constant on a characteristic. 6.3.2 The Cauchy Problem for Higher Order <strong>Equations</strong> Let S be a hypersurface of class C k .Ifu ∈ C k−1 defined near S, then the Cauchy data of u on S is the set {u, ∂ ν u, ··· ,∂ν k−1 u}. We observe that we can consider the so-called Cauchy problem in a neighborhood of a single point and then introduced a change of variables to transform the problem so that S contains the origin and near the origin S coincides with the hyperplane x n =0. Thusit is often the case when treating higher order initial value problems that we distinguish the variable x n and denote it by t and then define x =(x 1 , ··· ,x n−1 ) ∈ R n−1 . Also we then consider ∂ t and ∂ x and multi-indices α =(α 1 , ··· ,α n−1 ). With this the Cauchy Problem is:
6.3. CHARACTERISTICS AND HIGHER ORDER EQUATIONS 49 Given functions {φ j (x)} k−1 j=0 solve F ( x, t, (∂ α x ∂ j t u) |α|+j≤k ) =0 with ∂ j t u(x, 0) = φ j (x), 0 ≤ j ≤ (k − 1). This problem is far to general since the Cauchy data determine many derivatives on S. Indeed, up to order k the only unknown is ∂t k u. For the Cauchy problem to be well-posed it must be assumed that F = 0 can be solved for ∂t k . This is a characteristic condition on S. Numerous examples were given to illustrate this situation. We could embark on an in-depth treatment of the Quasi-linear case. In particular, we could consider the notion of noncharacteristic surface for the Cauchy problem: An initial value problem ∑ ( ) a α,j x, t, ((∂ β x ∂tu) i |β|+i≤k−1 ∂ α x ∂ j t u = b ( ) x, t, (∂x β ∂tu) i |β|+i≤k−1 |α|+j=k ∂ j t u(x, 0) = φ j (x), 0 ≤ j ≤ (k − 1) is called non-characteristic if a 0,k ( x, 0, (∂ α x ∂ j t u) |α|+j≤k ) ≠0 With this we could be in a position to prove the main fundamental existence and uniqueness result (such as it is) for PDE’s with analytic data. Rather that take such a deep excursion we will simple state the main result – the Cauchy-Kovalevski Theorem. Theorem 6.3.14. If G, {φ j } k−1 j=0 are analytic near the origin, the Cauchy problem ⎧ ( ) ⎨ ∂t k u = G x, t, (∂x α ∂ j t u) |α|+j≤k j
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 47: 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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