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Unformatted text preview: Lecture 4 Some theory of linear second order differential equations Let us now return to the general form of a second order linear ODE, expressed in standard form, y + P ( x ) y + Q ( x ) y = R ( x ) . (1) For convenience, we shall also write the homogeneous ODE associated with (1), y + P ( x ) y + Q ( x ) y = 0 . (2) (Of course, Eq. (2) is a special case of Eq. (1), with R ( x ) = 0.) It would be nice to establish some conditions that guarantee the existence and uniqueness of solutions to these ODEs. In many applications, P ( x ), Q ( x ) and R ( x ) are continuous on an interval of interest, except perhaps at isolated points. (Well have to deal with such situations later.) Existence and uniqueness of solutions to initial value problems Theorem: Suppose that P , Q and R are continuous for all x [ a, b ]. Let x [ a, b ]. Then for any pair of real numbers A, B , there exists a unique solution to Eq. (1) which satisfies the initial conditions y ( x ) = A, y ( x ) = B. A proof of this theorem is postponed until later. We shall rely on the theory of first order linear systems of DEs to prove this result. The conditions for existenceuniqueness assumed in the above theorem were satisfied by most, if not all, the DEs that you encountered in AMATH 250 or equivalent. For there, you studied linear second order DEs with constant coefficients, i.e., y + py + qy = r ( x ) , (3) where r ( x ) is continuous. Obviously the functions P ( x ) = p and Q ( x ) = q are continuous everywhere. Existence and uniqueness of solutions is thus guaranteed. 1 Problems can occur, however, if the conditions of the above theorem are not satisfied. For example, consider Bessels equation with order p = 0: x 2 y + xy + x 2 y = 0 . (4) We must first rewrite this DE in standard form by dividing by x 2 : y + 1 x y + y = 0 . (5) Here, P ( x ) = 1 /x and Q ( x ) = 1. For any interval I = [ a, b ] that does not contain the origin, P and Q are continuous on I so that the conditions of the above theorem are satisfied. Hence existenceuniqueness of solutions is guaranteed at any x I . But P ( x ) is not continuous at x = 0. Let us investigate the nature of solutions there. First, multiply the above equation by x to give xy + y + xy = 0 . (6) At x = 0, we shall have y (0) + y (0) + 0 y (0) = 0 . (7) This implies that y (0) = 0. In other words, all solutions y ( x ) that exist at x = 0 must have zero derivative there. Therefore there exist no solutions that satisfy the initial condition y (0) = B for B negationslash = 0. In conclusion, the existence part of existenceuniqueness fails to hold because of the discontinuity of P ( x ) at x = 0....
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 Spring '08
 SivabalSivaloganathan

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