Lecture notes 8 - Math 212a(Fall 2013 Yum-Tong Siu 1 Greens...

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Math 212a (Fall 2013) Yum-Tong Siu 1 Green’s Kernel for Sturm-Liouville Equation by Method of Variation of Parameters In this brief set of notes we consider the following Sturm-Liouville equa- tion Lg = f on the interval [ a, b ] of R , where the Sturm-Liouville operator L is of the form ( Lg ) ( x ) = d dx ( p ( x ) d dx g ( x ) ) q ( x ) g ( x ) . Here the two functions p ( x ) and q ( x ) are both infinitely differentiable on [ a, b ] with p ( x ) > 0 on [ a, b ] and q ( x ) 0 on [ a, b ]. We will show that a solution g of Lg = f is given by g ( x ) = b y = a K ( x, y ) f ( y ) dy, where the Green kernel K ( x, y ) is defined by K ( x, y ) = φ - ( x ) φ + ( y ) W if a x y b, φ - ( y ) φ + ( x ) W if a y x b. Here φ ( x ) and φ + ( x ) are two C -linearly independent solutions of the ho- mogeneous differential equation L ( φ ) = 0 satisfying the initial conditions φ ( a ) = 0 , φ ( a ) ̸ = 0 , φ + ( b ) = 0 , φ + ( b ) ̸ = 0 , and W = p ( x ) φ ( x ) φ + ( x ) φ ( x ) φ + ( x ) is the “Wronskian” of two C -linearly independent solutions φ ( x ) and φ + ( x ) and is independent of x . For the special case of p ( x ) 1, we will derive the Green kernel by using the “method of variation of parameters”. For the general case we will simply verify that the Green kernel actually gives a solution. We will assign, as a homework problem, the derivation of the Green kernel for the general case from the “method of variation of parameters”.
Math 212a (Fall 2013) Yum-Tong Siu 2 Definition of Wronskian. For functions f 1 , · · · , f n on ( a, b ) the Wronskian W ( f 1 , · · · , f n ) is defined by W ( f 1 , · · · , f n ) = det ( f ( j ) k ) 0 j n 1 , 1 k n , where f ( j ) k is the derivative of f k of order j . Vanishing of Wronskian and Linear Dependency. If f 1 , · · · , f n are R - linearly dependent, then W ( f 1 , · · · , f n ) 0. On the other hand, if W ( f 1 , · · · , f n 1 ) is nowhere zero and W ( f 1 , · · · , f n ) 0, then f 1 , · · · , f n are R -linearly de- pendent. Proof. solve for c 1 , · · · , c n 1 in f ( j ) n = n 1 k =1 c j f ( j ) k for 0 j n 1 and differentiate both sides for 0 j n 2 to show that c k 0 for 1 k n 1. Remark. The condition W ( f 1 , · · · , f n 1 ) being nowhere zero cannot be dropped, as shown in the following counter-example. f 1 ( x ) = 1 for x 0 and f 1 ( x ) = 1 + x 3 for x 0. f 2 ( x ) = 1 for x 0 and f 2 ( x ) = 1 + x 3 for x 0. f 3 ( x ) = 3 + x 3 .