LINEAR SECOND-ORDER EQUATIONS

Transcription

1 LINEAR SECON-ORER EQUATIONS Classification In two independent variables x and y, the general form is Au xx + 2Bu xy + Cu yy + u x + Eu y + Fu + G = 0. The coefficients are continuous functions of (x, y) in Ω. We will use the 1-1 transformation ξ = ξ(x, y), η = η(x, y) to transform the equation to one of three canonical forms determined by the sign of the discriminant B 2 AC. With w(ξ, η) = u(x(ξ, η), y(ξ, η)), the transformed equation takes the form aw ξξ + 2bw ξη + cw ηη + dw ξ + ew η + fw + g = 0, in which a = Aξ 2 x + 2Bξ x ξ y + Cξ 2 y, c = Aη 2 x + 2Bη x η y + Cη 2 y, and b = Aξ x η x + B(ξ x η y + ξ y η x ) + Cξ y η y. An important connection between the coefficients is b 2 ac = (B 2 AC)J 2 where J = ξ x η y ξ y η x ( = 0 in Ω) is the Jacobian of the 1-1 transformation. The sign of the discriminant is thus an invariant of the transformation. 1

2 The pde is classified according to the sign of the discriminant as following: (1) hyperbolic if B 2 AC > 0, canonical form u ξη +... = 0, (2) parabolic if B 2 AC = 0, canonical form u ξξ +... = 0, (3) elliptic if B 2 AC < 0, canonical form u ξξ + u ηη +... = 0, where... stands for terms containing at most first-order partials. The hyperbolic type To turn the equation into the canonical hyperbolic form, it is necessary to choose ξ and η that make the coefficients a and c both zero. A and C are not simultaneously zero or the equation is already in canonical form. Suppose A is not zero, then a is zero if one of the following equations is satisfied: ξ x + m ± ξ y = 0 roots are -m ± where m ± = (B ± B 2 AC)/A are two different real-value functions. The same equations, applied to η, can also make c = 0. One can pick either one of these equations for ξ and the another one for η. Therefore, there are two characteristic equations dy dx = m ± 2

3 that describe two families of characteristics. In terms of ξ and η, the canonical hyperbolic equation u ξη +... = 0 can be obtained. An alternate canonical form With ξ = ξ + η, η = ξ η, the canonical form of the hyperbolic type can be transformed to an alternate form: u ξ ξ u η η +... = 0. The simplest hyperbolic equation The equation u ξη = 0 has the general solution u = f(ξ) + g(η) where f, g are arbitrary differentiable functions. Solve u xx + 2u xy 3u yy = 0. u = f(y 3x) + g(y + x) Equation containing an inhomogeneous term Consider u ξη = h(ξ, η). The general solution is u = ( hdη ) dξ + f(ξ) + g(η) where f, g are arbitrary differentiable functions. 3

4 The parabolic type Suppose the discriminant B 2 AC = 0 in Ω, there is only one characteristic equation η x + B A η y = 0 (or dy/dx = B/A) which makes c = 0. As B 2 AC = 0 B/A = C/B Bη x + Cη y Aη x + Bη y = 0, b = (Aη x + Bη y )ξ x + (Bη x + Cη y )ξ y = 0. Thus, the equation can be written as u ξξ +... = 0. Note that as long as the Jacobian of the transformation is non-zero, ξ can be quite arbitrarily chosen. We normally adopt the simple choice ξ = x. Solve u xx + 4u xy + 4u yy = 0 η = y 2x, ξ = x, u = xf(y 2x) + g(y 2x) The elliptic type As B 2 AC < 0, there are no characteristics. While a general approach to reduce the equation is in principle possible, it can be cumbersome. We ll proceed by considering only cases with constant coefficients. 4

5 Equation with constant coefficients In such a case, conversion to canonical form can follow a simple procedure. Suppose that A 0 so that the equation can be divided throughout by A. (1) hyperbolic case Completing square gives ( x + B A y) 2 u ( A y) 2 u +... = 0 where = B 2 AC is the discriminant. By setting ξ = x + B A y and η = A y, the pde can be converted to the canonical form: Since u ξξ u ηη +... = 0. ξ = x ξ x + y ξ y and η = x η x + y η y, comparison of coefficients yields x y = 1, ξ ξ = B A, x y = 0, and η η = A. The transformation equations can be chosen as x = ξ and y = B A ξ + A η. 5

6 (2) parabolic case Rewrite the equation in the form: u ξξ ( A )2 u ηη +... = 0. As = 0, there is no constraint on x η and y η. Setting them to 0 and 1 respectively produce the transformation x = ξ, y = B ξ + η. The resultant equation is A u ξξ +... = 0. (3) elliptic case Use the transformation: x = ξ, y = B A ξ + η to get A u ξξ + u ηη +... = 0. Find a transformation to transform the equation to its canonical form. u xx + 2u xy + 3u yy + 4u = 0 ξ = x, η = (y x)/ 2 The second-order Cauchy problem Let Γ be a curve on which the values of u and its derivative u n in the direction normal to Γ are given. It is possible to find a unique solution if Γ is non-characteristic. The 6

7 Cauchy-Kowaleski theorem states that a unique solution exists in a neighborhood of a point of Γ if the coefficients of the pde, the functions specifying the solution and its normal derivative along Γ are analytic (possess power series expansions). Consider the following pde with the Cauchy data defined on the x-axis. u xx + u yy = 0 u(x, 0) = φ(x), u y (x, 0) = ψ(x) Use the pde and the Cauchy data to calculate the first few terms of the Taylor expansion of u about (x, 0): u(x, y) = u(x, 0) + u y (x, 0)y + 1 2! u yy(x, 0)y ! u yyy(x, 0)y ! u yyyy(x, 0)y u(x,y) = φ + ψy 1 2! φ y 2 1 3! ψ y ! φ(4) y The Cauchy-Kowalewski theorem is a local existence theorem. It only asserts that a solution exists in a neighborhood of the point, not in all space. A peculiar feature of the theorem is that the type of the differential equation (wether it is hyperbolic, parabolic, or elliptic) is irrelevant. Problem Set 2 7