2 nd order Linear Homogeneous DEs with Non-Constant Coefficients

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1 Math 231, Wed 4-May Wed 4-May-2011 Wednesday, May 4th 2011 Topic:: DEs with Non-Constant Coeffs 2 nd order Linear Homogeneous DEs with Non-Constant Coefficients We consider linear 2 nd order homogeneous DEs of the form y 2 ppxqy 1 qpxqy 0 (1) [For some reason, Boyce & DiPrima begin, in Chapter 5, using x for the independent variable instead of t] The basic theory we learned in Chapter 3 holds for Equation (1) ie, it is a 2 nd order DE, and we seek a fundamental set of solutions ty 1 pxq, y 2 pxqu from which we may build a general solution ypxq c 1 y 1 pxq c 2 y 2 pxq, a form which all solutions take If the DE comes with ICs ypx 0 q y 0, y 1 px 0 q y 1 0, (2) then c 1, c 2 must be chosen so as to satisfy these conditions What makes Problem (1) new is that, without constant coefficients, we do not have a method for finding the solutions of a fundamental set [What we were able to do in Chapter 3 was this: given a first solution y 1 pxq for a fundamental set, use reduction of order and methods from Chapter 2 for 1 st order DEs to generate a 2 nd linearly independent solution y 2 pxq] The theorem (Theorem 321 in the text) on Existence and Uniqueness of solutions to an IVP (1) (2) says that, if p, q are continuous in an open interval I containing x 0, then the IVP has a unique solution that exists on I It is generally in the neighborhood of this initial time that we seek the solution The solutions of such problems are, however, unlikely to be expressible in terms of elementary functions (ie, polynomials, exponentials, logarithms, trig fns) Power Series Solutions As in other chapters, we will proceed as if our DE (1) has a solution of some specified form Since we seek a solution in the neighborhood of x 0, and since elementary functions do not prove

2 successful, we assume the form of a power series ypxq a n px x 0 q n (3) We often use summation notation for power series Summation notation is flexible in several ways which will prove useful when solving for the coefficients a n Included among these, are that it allows one to strip out as many terms as one likes: a n px x 0 q n a 0 a 1 px x 0 q a 2 px x 0 q 2 a n px x 0 q n, n3 it allows for a substitution rule involving the index of summation, much like the substitution rule for integration (changes to the dummy variable): a n px x 0 q n a 0 a 1 px x 0 q a 2 px x 0 q 2 a n px x 0 q n n3 a 0 a 1 px x 0 q a 2 px x 0 q 2 a j 3 px x 0 q j 3 (letting j n 3) j0 We employ ideas like this in the following examples Our first example is chosen so that we see how the method works when the problem is easily solved using other methods Example 1: Exercise 1, Section 52 Problem: Consider the DE y 2 y 0 Find a power series solution in an interval containing x 0 Answer: To guarantee the solution lives on an interval containing t 0, we seek a power series solution centered at 0: ypxq a n x n Taking derivatives and plugging them into the DE, get rpn 2qpn 1qa n 2 a n sx n 0 The only power series that gives zero is the one whose coefficients are all zero ie, 8 0 xn Thus, we can equate coefficients for various powers of x: pn 2qpn 1qa n 2 a n 0, or a n 2 a n, for n 0, 1, 2, pn 2qpn 1q 2

3 Show that a 2 a 0 p2qp1q, a 3 a 1 p3qp2q, a 4 a 2 p4qp3q a 0 4!, a 5 a 3 p5qp4q a 1 5!, a 2k a 0 p2kq!, a 2k 1 a 1 p2k 1q! This means our series really has just two free constants: a 0 and a 1, and looks like ypxq a 0 1 2! x2 4! x4 a 1 x 3! x3 5! x5 a 0 y 1 pxq where y 1 pxq x 2k p2kq! and y 2 pxq Note that, by Chapter 3 methods, we would get general solution x 2k 1 p2k 1q! a 1 y 2 pxq, ypxq c 1 e x c 2 e x Is there a discrepancy here? Using MacLaurin series expansions, we have c 1 e x c 2 e x c 1 x n n! c 1 1 x p xq n c 2 n! x 2 2! x 3 3! c 2 1 x pc 1 c 2 q1 pc 1 c 2 qx pc 1 c 2 q x2k p2kq! a 0 a 1 x 2k p2kq! where a 0 c 1 c 2 and a 1 c 1 c 2 x 2k 1 p2k 1q!, x 2 2! x3 3! x 2k 1 pc 1 c 2 q p2k 1q!, Example 2: Problem: How do we modify the solution strategy for the previous problem in the presence of ICs, say, yp0q 2, y 1 p0q 1? Answer: The fact that these ICs occur at the point where the power series is centered makes this very easy We have 2 yp0q 1 y 1 p0q a n x n x0 a 0 a 1 0 a 2 0 a 0, and na n x n 1 x0 a 1 2a 2 0 3a 3 0 a 1 3

4 Example 3: Airy s Equation: Example 2, p 255 (Section 52) Problem: Compute a series solution around x 0 for the DE y 2 xy 0 Answer: Start with 8 ypxq a nx n and use this and similar expressions for y, y 1 and y 2 to obtain 2a 2 rpn 2qpn 1qa n 2 a n 1 sx n 0, leading to the conclusions Show that a 2 0 and a n 2 a n 1 pn 2qpn 1q a 3 a 0 p3qp2q, a 4 a 1 p4qp3q,, for n 1, 2, 3, a 6 a 3 p6qp5q a 0 p6qp5qp3qp2q, a 7 a 4 p7qp6q a 1 p7qp8qp4qp3q, a 3k a 0 p2qp3qp5qp6q p3k, 1qp3kq a 3k 1 a 1 p3qp4qp7qp8q p3kqp3k while a 3k 2 0, for k 0, 1, 2, Thus, the general solution is ypxq a 0 u 0 pxq u 0 pxq 1 k1 x 3k p2qp3qp5qp6q p3k 1qp3kq and u 1 pxq x k1 1q a 1 u 1 pxq, where x 3k 1 p3qp4qp6qp7q p3kqp3k 1q One cannot really get a computer to compute the full infinite series that define u 0 pxq and u 1 pxq In practice, one takes a truncated approximation, keeping the first N terms for some reasonably-large N and leaving off the rest In the code below, I have called the 50-term truncated approximation of u 0 pxq f1(x), and the 50-term truncated approximation of u 1 pxq f2(x) var( x ) numtermstokeep = 50 f1(x) = 1 f2(x) = x for j in range(1,numtermstokeep): f1(x) = f1(x) + xˆ(3*j) * prod([3*k+1 for k in range(j)]) / factorial(3*j) f2(x) = f2(x) + xˆ(3*j+1) * prod([3*k+2 for k in range(j)]) / factorial(3*j+1) p1=plot(f1(x), -10, 3, color= red ) p2=plot(f2(x), -10, 3, color= blue ) show(p1+p2) 4

5 Example 4: Airy s Equation again, different center for power series: Example 3, p 257 Problem: For the same DE as the previous example, find the first five terms for a power series solution centered at x 1 Answer: Use the series expression 8 ypxq a npx 1q n and the fact that xy px 1qy y to obtain the equation pn 2qpn 1qa n 2 px 1q n a n 1 px 1q n a n px 1q n 0, or 2a 2 pn 2qpn 1qa n 2 px 1q n a n 1 px 1q n a 0 a n px 1q n 0 Proceed from there to obtain the relationships 2a 2 a 0 0, pn 2qpn 1qa n 2 a n 1 a n 0, or a a n 1 a n n 2, n 1, 2, pn 1qpn 2q After a little more work, get ypxq a 0 1 a 1 x 1 2 px 1 1q2 6 px 1 3q3 24 px 1 1q4 1 6 px 1 1q3 12 px 1 1q4 120 px 1q5 30 px 1q5 Knowledge to have for Exam 3: practically anything - discussed in class or class notes This includes, but is not limited to a working knowledge of concepts and how they are defined some sample concepts (to give the flavor of what ideas are important): linear DEs: homogeneous and nonhomogeneous convolution Laplace transform phase plain and portrait equilibrium/critical point a working knowledge of theorems: what they tell you, under what conditions 5

6 - that has come up in homework Skills to have for Exam 3: - linear 1st order homogeneous systems x = Ax (A is n-by-n matrix) eigenpairs of A find eigenvalues and their algebraic/geometric multiplicities find basis of corresponding eigenvectors for each find general soln in settings where GM = AM or GM = 1 for each eigenvalue this includes the complex e-value case this includes the 1 = GM < AM (for one or more eigenvalues) case when A is 2-by-2 with nonzero determinant sketch phase portraits classify the critical/equilibrium point at the origin - special functions: know what they model, be able to use their properties unit step function be able to express piecewise-defined functions using it unit impulse function - various algebraic procedures partial fraction decompositions completing the square determining if a quadratic is irreducible manipulate expressions into forms like those in the transform column, p Laplace transforms find the Laplace transform for a given input function from the definition itself given Entries 1-6, 12-14, and from p 319 find inverse Laplace transform for a given input function use the Laplace transform to solve DEs in the presence of initial conditions various nonhomogeneous terms - for a given linear DE, be able to identify its transfer function identify its impulse response express its solution (when zero ICs) as convolution of two functions - power series methods algebraic manipulation of summation expressions, including separating out some terms from the rest of the summation substitutions term-by-term differentiation solving DEs 6

7 expressing power series coefficients in terms of free variables finding a fundamental set of solutions, an writing a general solution using ICs to nail down a specific solution Wednesday, May 4th 2011 HW:: PS22 7