# lecture6 - Differential Equations& Linear Algebra...

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Unformatted text preview: Differential Equations & Linear Algebra Lecture 6 Jianli XIE Email: [email protected] Department of Mathematics, SJTU Fall 2010 Contents Structure of solutions of linear homogeneous equations Existence and uniqueness of solutions Principle of superposition Fundamental set of solutions Contents Structure of solutions of linear homogeneous equations Existence and uniqueness of solutions Principle of superposition Fundamental set of solutions Linear independence and the Wronskian A differential operator We now study the structure of the solutions of all second order linear homogeneous equations. The nice properties of the structure will assist us in finding the solutions. It is helpful to introduce a differential operator notation. Let p and q be continuous functions on an interval I . For any twice differentiable function φ , we define the differential operator L by L [ φ ] = φ 00 + pφ + qφ. (1) Note that L [ φ ] is a function on I and the value of L [ φ ] at a point t is L [ φ ]( t ) = φ 00 ( t ) + p ( t ) φ ( t ) + q ( t ) φ ( t ) . Differential equation in operator form The operator L is often written as L = D 2 + pD + q , where D = d dt is the derivative operator. Using the operator L , the second order linear homogeneous equation can be written as L [ y ] = y 00 + py + qy = 0 . (2) With equation (2) we associate a set of initial conditions y ( t ) = y , y ( t ) = y , (3) where t is some point in I , and y and y are given real numbers. Existence and uniqueness Theorem 1 Consider the initial value problem y 00 + p ( t ) y + q ( t ) y = g ( t ) , y ( t ) = y , y ( t ) = y , where p, q , and g are continuous on an open interval I that contains the initial point t . Then there is exactly one solution y = φ ( t ) of this problem, and the solution exists throughout the interval I . The theorem says that the initial value problem has a unique solution and the solution is defined throughout the interval I where the coefficients are continuous. Example Example Find the longest interval in which the solution of the initial value problem ( t 2- 3 t ) y 00 + ty- ( t + 3) y = 0 , y (1) = 2 , y (1) = 1 is certain to exist. Solution Compared with the standard form in theorem 1, we have p ( t ) = 1 / ( t- 3) , q ( t ) =- ( t + 3) /t ( t- 3) , and g ( t ) = 0 . The only points of discontinuity of the coefficients are t = 0 and t = 3 . Therefore, the longest open interval, containing the initial point t = 1 , in which all the coefficients are continuous is < t < 3 . Thus, (0 , 3) is the longest interval in which the solution is certain to exist. Example Example Solve the initial value problem with homogeneous initial conditions y 00 + p ( t ) y + q ( t ) y = 0 , y ( t ) = 0 , y ( t ) = 0 , where p and q are continuous in an open interval I containing t ....
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lecture6 - Differential Equations& Linear Algebra...

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