Module 3: Second-Order Partial Differential Equations

Transcription

1 Module 3: Second-Order Partial Differential Equations In Module 3, we shall discuss some general concepts associated with second-order linear PDEs. These types of PDEs arise in connection with various physical problems such as the motion of a vibration string, heat flow, electricity, magnetism and fluid dynamics. The second-order partial differential equations are classified into three different types. Further, the simplified canonical/normal forms are obtained for second order linear equations in two independent variables. The lectures are organized as follows. The first lecture is devoted to the classification of second-order linear PDEs in two or more independent variables. These equations are classified into hyperbolic, parabolic and elliptic types. The reduction to the canonical form (or normal form) of second order equations in two independent variables are discussed in the second lecture. Finally, the third lecture is devoted to wellposedness, superposition principle and method of factorization for these types of equations. 1

2 MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS 2 Lecture 1 Classification of Second-Order PDEs Classification of PDEs is an important concept because the general theory and methods of solution usually apply only to a given class of equations. Let us first discuss the classification of PDEs involving two independent variables. 1 Classification with two independent variables Consider the following general second order linear PDE in two independent variables: A 2 u x 2 + B 2 u x y + C 2 u y 2 + D u x + E u + F u + G = 0, (1) y where A, B, C, D, E, F and G are functions of the independent variables x and y. The equation (1) may be written in the form where Au xx + Bu xy + Cu yy + f(x, y, u x, u y, u) = 0, (2) u x = u x, u y = u y, u xx = 2 u x 2, u xy = 2 u x y, u yy = 2 u y 2. Assume that A, B and C are continuous functions of x and y possessing continuous partial derivatives of as high order as necessary. The classification of PDE is motivated by the classification of second order algebraic equations in two-variables ax 2 + bxy + cy 2 + dx + ey + f = 0. (3) We know that the nature of the curves will be decided by the principal part ax 2 +bxy+cy 2 i.e., the term containing highest degree. Depending on the sign of the discriminant b 2 4ac, we classify the curve as follows: If b 2 4ac > 0 then the curve traces hyperbola. If b 2 4ac = 0 then the curve traces parabola. If b 2 4ac < 0 then the curve traces ellipse. With suitable transformation, we can transform (3) into the following normal form x 2 a 2 y2 b 2 = 1 (hyperbola). x 2 = y (parabola). x 2 a 2 + y2 b 2 = 1 (ellipse).

3 MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS 3 Linear PDE with constant coefficients. Let us first consider the following general linear second order PDE in two independent variables x and y with constant coefficients: Au xx + Bu xy + Cu yy + Du x + Eu y + F u + G = 0, (4) where the coefficients A, B, C, D, E, F and G are constants. The nature of the equation (4) is determined by the principal part containing highest partial derivatives i.e., Lu Au xx + Bu xy + Cu yy. (5) For classification, we attach a symbol to (5) as P (x, y) = Ax 2 + Bxy + Cy 2 (as if we have replaced x by x and y by y ). Now depending on the sign of the discriminant (B2 4AC), the classification of (4) is done as follows: B 2 4AC > 0 = Eq. (4) is hyperbolic B 2 4AC = 0 = Eq. (4) is parabolic B 2 4AC < 0 = Eq. (4) is elliptic (6) (7) (8) Linear PDE with variable coefficients. The above classification of (4) is still valid if the coefficients A, B, C, D, E and F depend on x, y. In this case, the conditions (6), (7) and (8) should be satisfied at each point (x, y) in the region where we want to describe its nature e.g., for elliptic we need to verify B 2 (x, y) 4A(x, y)c(x, y) < 0 for each (x, y) in the region of interest. Thus, we classify linear PDE with variable coefficients as follows: B 2 (x, y) 4A(x, y)c(x, y) > 0 at (x, y) = Eq. (4) is hyperbolic at (x, y) B 2 (x, y) 4A(x, y)c(x, y) = 0 at (x, y) = Eq. (4) is parabolic at (x, y) B 2 (x, y) 4A(x, y)c(x, y) < 0 at (x, y) = Eq. (4) is elliptic at (x, y) (9) (10) (11) Note: Eq. (4) is hyperbolic, parabolic, or elliptic depends only on the coefficients of the second derivatives. It has nothing to do with the first-derivative terms, the term in u, or the nonhomogeneous term. EXAMPLE u xx + u yy = 0 (Laplace equation). Here, A = 1, B = 0, C = 1 and B 2 4AC = 4 < 0. Therefore, it is an elliptic type.

4 MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS 4 2. u t = u xx (Heat equation). Here, A = 1, B = 0, C = 0. B 2 4AC = 0. Thus, it is of parabolic type. 3. u tt u xx = 0 (Wave equation). In this case, A = 1, B = 0, C = 1 and B 2 4AC = 4 > 0. Hence, it is of hyperbolic type. 4. u xx + xu yy = 0, x 0 (Tricomi equation). B 2 4AC = 4x. Given PDE is hyperbolic for x < 0 and elliptic for x > 0. This example shows that equations with variable coefficients can change form in the different regions of the domain. 2 Classification with more than two variables Consider the second-order PDE in general form: a ij 2 u + i=1 b i u x i + cu + d = 0, (12) where the coefficients a ij, b i, c and d are functions of x = (x 1, x 2,, x n ) alone and u = u(x 1, x 2,, x n ). Its principal part is L a ij 2. (13) It is enough to assume that A = [a ij ] is symmetric if not, let ā ij = 1 2 (a ij + a ji ) and rewrite L ā ij 2. (14) Note that 2 u = 2 u x j x i. As in two-space dimension, let us attach a quadratic form P with (14) (i.e., replacing u x i by x i ). P (x 1, x 2,, x n ) = a ij x i x j. (15) Since A is a real valued symmetric (a ij = a ji ) matrix, it is diagonalizable with real eigenvalues λ 1, λ 2,..., λ n (counted with their multiplicities). In other words, there exists a corresponding set of orthonormal set of n eigenvectors, say σ 1, σ 2,, σ n with R =

5 MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS 5 [σ 1, σ 2,, σ n ] as column vectors such that λ 1 λ 2 R T AR = = D (16) We now classify (12) depending on sign of eigenvalues of A: λ n EXAMPLE 2. (a) If λ i > 0 i or λ i < 0 i then (12) is elliptic type. (b) If one or more of the λ i = 0 then (12) is parabolic type. (c) If one of the λ i < 0 or λ i > 0, and all the remaining have opposite sign then (12) is said to be of hyperbolic type u = u xx + u yy + u zz = 0. In this case, λ i = 1 > 0 for all i = 1, 2, 3. It is an elliptic PDE since all eigenvalues are of one sign. 2. It is an easy exercise to check that u t 2 u = 0 is of parabolic type. 3. The equation u tt 2 u = 0 is of hyperbolic type. EXAMPLE 3. Classify u x1 x 1 + 2(1 + cx 2 )u x2 x 3 = 0. To symmetrize, write it as u x1 x 1 + (1 + cx 2 )u x2 x 3 + (1 + cx 2 )u x3 x 2 = 0 i.e., x T A x c x2 = 0, where A = cx cx 2 0 x = Eigenvalues are λ 1 = 1, λ 2 = 1 + cx 2, λ 3 = (1 + cx 2 ) and normalized eigenvectors 1 0 σ 1 = 0 σ 2 = 1/ / σ 3 = 1/ 2 2 1/ 2 x1 x2 x3

6 MODULE 3: SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS 6 So R = / 2 1/ 2 0 1/ 2 1/ 2 Note that R = R T = R 1. R T AR = cx (1 + cx 2 ) = D Equation is parabolic if x 2 = 1 c (c 0), hyperbolic if x 2 > 1 c and x 2 < 1 c. For c = 0, λ 1 = λ 2 = 1 and λ 3 = 1, it is hyperbolic type. Practice Problems 1. Classify the following equations into hyperbolic, elliptic or parabolic type. (A) 5u xx 3u yy + (cos x)u x + e y u y + u = 0. (B) e x u xx + e y u yy u = 0. (C) xu xx + u yy = 0. (D) 8u xx + u yy u x + [log(2 + x 2 )]u = 0. (E) sin 2 xu xx + sin 2xu xy + cos 2 xu yy = x. 2. Classify the following equations into elliptic, parabolic, or hyperbolic type. (A) u xx + 2u yz + (cos x)u z e y2 u = cosh z. (B) u xx + 2u xy + u yy + 2u zz (1 + xy)u = 0. (C) e z u xy u xx = log[x 2 + y 2 + z 2 + 1]. 3. Determine the regions where u xx 2x 2 u xz + u yy + u zz = 0 is of hyperbolic, elliptic and parabolic.