Math 1152 Project #4 Solving Initial Value Problems using Power Series and the Laplace Transform Due: Tuesday, April 11 Name(s): - - - - - - - - - - - - - - - - - - - - Description - - - - - - - - - - - - - - - - - - - Power series solutions to initial value problems Using Power Series to approximate solutions to initial value problems is one of the most important applications in many areas of theoretical and applied mathematics as well as the applied sciences. However, many of the applications (like finite element analysis, fluid mechanics, and heat transfer to name a few) require prerequisite material that you will develop in later courses in your academic career. However, all of these applications have a common theme: Taylor Series can be used to approximate otherwise difficult phenomena to a very high degree of accuracy! The Laplace Transform The Laplace transform is also a very important tool in the theory of differential equations. In this course, the linear differential equations that we study are of the form: y (n) (t) + pn1 (t)y (n1) (t) + . . . + p1 (t)y 0 + p0 (t)y = F (t) and all of the functions p0 (t), p1 (t), . . . , pn1 (t), and F (t) have been infinitely differentiable. In many physical situations, including the analysis of linear time-invariant systems such as electrical circuits, harmonic oscillators, optical devices, and mechanical systems, the forcing function F (t) may not even be continuous! For instance, the force a punter exerts on a football is treated as a \"point force\"; gravitational force is the only force acting on the football before and after the moment of impact, but at that the instant when it is kicked, the force the punter exerts clearly changes the motion of the ball!. The Laplace transform is a fundamental tool in solving differential equations with discontinuous forcing functions, and the methodology needed to perform many of the calculations has been covered this semester! 1 - - - - - - - - - - - - - - Purpose of the Assignment - - - - - - - - - - - - - - To introduce theoretical considerations of convergence of Taylor series. To explore fundamental techniques necessary in applications of Taylor series. To incorporate Matlab or Desmos (or a program of your choice) into your math education. To intorduce the Laplace Transform. - - - - - - - - - - - - - - - - - - - - - Directions - - - - - - - - - - - - - - - - - - - - This assignment is worth 20 pts. You are STRONGLY encouraged to work in groups of up to 3 students with the following restraints: - The students in your group must have the same recitation instructor. - Each group will submit one copy of this assignment; group members should NOT submit individual assignments! - Each group member's name should appear on the top of this page. - Each member of the group will receive the same grade. You may of course work with students outside of your recitation instructor, but they must hand in their own version of the assignment to their instructor. If you need more space than what is provided, feel free to use extra sheets pf paper. For work on extra sheets to be considered, you must staple them to your assignment, clearly indicate to which problem any work belongs, and restaple the assignment! 2 Part I: Finding Taylor Series Solutions to Differential Equations Perhaps the most important modern application of Taylor series lies in finding Taylor series solutions to ordinary differential equations. It is impossible to write simple closed form solutions to many of the equations, like Bessel's equation and Airy's equation, that arise in science and engineering applications. However, it is possible to search for the Taylor series of the solution. For worked examples, please see the \"Projects\" folder! Since many of these equations require more actual scientific context than what many first or second year students may have seen, we will consider only simple differential equations here. The techniques used to solve these problems can be used in problems of physical interest you will see in future courses. Problem 1: (Approximating the solution to a differential equation with a Taylor polynomial) Consider the initial value problem: dy (x) y(x) = 0, dx y(0) = 1. A function y = f (x) is called a classical solution to this problem on (a, b) if: 1. y(x) is differentiable on (a, b). 2. The differential equation is satisfied for all values of x when the expression for y(x) and y 0 (x) are substituted into it. 3. The initial condition y(0) = 1 holds. It can be shown that there is a unique classical solution to this problem for all x. I. Verify by substitution that the function y = ex is a classical solution to this initial value problem. 3 II. A. Now, let's search for a Taylor Series solution to this problem. X Suppose that y(x) = ak xk is a solution to the initial value problem. Calculate k=0 the Taylor series for y 0 (x) and show that the differential equation becomes: X [(k + 1)ak+1 ak ] xk = 0. k=0 4 B. Conclude that the coefficients in the Taylor series satisfy the recursion relation: ak+1 = 1 ak . k+1 C. Use the initial condition y(0) = 1 to show that a0 = 1. Hint: Write out the first several terms in the series y = P k=0 ak xk . What is y(0)? D. Calculate a1 , a2 , and a3 using the recursion relation in B. Use these to write down the Taylor polynomials T1 (x), T2 (x), and T3 (x) for the solution y(x)a . T1 (x) = T2 (x) = T3 (x) = If you did this correctly, the Taylor polynomials you found in D. match the Taylor polynomials for the function y = ex ! a Recall that since we are centering this Taylor polynomial at x = 0, Tn (x) := n X k=0 5 ak xk . Problem 2: Consider the initial value problem: dy (x) + 2xy(x) = 0, dx y(0) = 2. 2 I. Verify by substitution that the function y = 2ex is a classical solution. II. Suppose y(x) = X ak xk is a solution to this initial value problem. Using the k=0 steps in the previous problem, show that the the first four nonzero terms in the Taylor series expansion for y(x)a are: 1 y(x) = 2 2x2 + x4 x6 + . . . 3 Use the space on the next page for additional work. a It can be shown that the function y(x) does equal the limit of its Taylor polynomials, so the \"=\" sign below is justifiable! 6 1 Denote f1 (x) = 2 2x2 and f2 (x) = 2 2x2 + x4 x6 . 3 Note that these are Taylor polynomials for the solution y(x) centered at x = 0. III. Write out the first four nonzero terms in the Taylor series for centered at x = 0 2 for 2ex . Is there a clear relationship between these and f2 (x)? IV. Using Matlab, Desmos, or a computer program of your choice, plot the graphs of 1 and f2 (x) = 2 2x2 + x4 x6 3 2 y = 2ex , f1 (x) = 2 2x2 , for 2 x 2 on the same set of axes. Attach a printout of your plots to the back of this sheet; do NOT just draw the graphs by hand! Note the higher order approximation f2 (x) is a better approximation than f1 (x)! 7 Problem 3: (Approximating a forcing function with a Taylor polynomial) Consider the initial value problem: dy (x) y(x) = ex , dx y(0) = 1. This is an example of a nonhomogeneous linear first order differential equationa . I. Verify by substitution that the function y = ex + xex is a classical solution to this initial value problem. II. We can approximate solutions to this initial value problem by approximating the forcing function ex with its Taylor polynomials. Suppose that y(x) = X ak xk is a solution to the initial value problem. In Problem k=0 1, Part II, you showed that the Taylor series for the lefthand side of the above differential equation is: 0 y (x) y(x) = X [(k + 1)ak+1 ak ] xk . k=0 We can write out the first several nonzero terms in the Taylor series for the forcing function ex and use them to find the coefficients ak . a The general form for a linear first order differential equation is: y 0 (x) + p(x)y(x) = F (x). If F (x) 6= 0, the equation is called nonhomogeneous. The term F (x) is frequently referred to as a forcing function. 8 A. Replace ex by its Taylor polynomial of order 1a . The initial value problem is now: dy (x) y(x) = 1 + x, dx y(0) = 1, After replacing the lefthand side with its Taylor series, we have: X [(k + 1)ak+1 ak ] xk = 1 + x. k=0 Call the solution to this f1 (x). i. Show that a0 = 1. ii. Show that a1 = 2 by the constant term on both sides. 1 1 1 1 3 . iii. Show that a2 = , a3 = , a4 = , a5 = , and a6 = 2 2 8 40 240 Thus, the first six nonzero terms in the approximate solution f1 (x) are given by: 3 1 1 1 1 6 f1 (x) = 1 + 2x + x2 + x3 + x4 + x5 + x + ... 2 2 8 40 240 a This is called the linearization of the differential equation since we have replaced the nonlinear function ex with its linear approximation! 9 B. Replace ex by its Taylor polynomial of order 3. The initial value problem is now: dy 1 1 (x) y(x) = 1 + x + x2 + x3 , dx 2 6 y(0) = 1. After replacing the lefthand side with its Taylor series, we have: X 1 1 [(k + 1)ak+1 ak ] xk = 1 + x + x2 + x3 . 2 6 k=0 Call the solution to this f2 (x). i. Use the initial condition to show that a0 = 1. ii. Repeat the steps in the previous problem to conclude that the first six nonzero terms in the Taylor series expansion for f2 (x) are given by: 2 5 1 1 6 3 x + ... f2 (x) = 1 + 2x + x2 + x3 + x4 + x5 + 2 3 24 24 144 10 C. Using Matlab, Desmos, or a computer program of your choice, plot the graphs of: y = ex + xex 3 f1 (x) = 1 + 2x + x2 + 2 3 f2 (x) = 1 + 2x + x2 + 2 1 3 x + 2 2 3 x + 3 1 4 1 1 6 x + x5 + x 8 40 240 5 4 1 1 6 x + x5 + x 24 24 144 for 4 x 4 on the same set of axes. Attach a printout of your plots to the back of this sheet; do NOT just draw the graphs by hand! D. Write out the coefficients of the Taylor series for y(x) = ex + xex up to the x6 term. Do the coefficients match those in the series for f1 (x) and f2 (x)? Notice that the coefficients of each power of f2 (x) are closer to the coefficients in the Taylor series for y(x) = ex + xex than the coefficients of each power of f1 (x)! Note that when we approximate a forcing term in a differential equation, we actually make two approximations to the solution: 1. The approximation to the forcing function affects the Taylor series of the solution to the approximate differential equation. 2. The Taylor polynomial to the approximate solution is an approximation of the approximate solution! Despite this, you can still see that the approximations to the approximate solutions are close to the actual solution! Proving that these approximate solutions actually converge to the real solution is much trickier, and is typically not studied until an advanced course in differential equations or analysis. The argument there does not parallel the computational techniques described here. 11 Part II: Applications in the Physical Sciences/Further Readings While the equations studied here were relatively simple, the arguments developed do arise in interesting physical problems. Here are references to a few such problems: 1. Airy's equation: http://www.sosmath.com/diffeq/series/series04/series04.html 2. The Simple Pendulum: i. The linearized equation: (You've likely seen this in an introductory physics course!) http://www.cems.uvm.edu/~tlakoba/AppliedUGMath/notes/lecture_6.pdf ii. Higher order corrections: http://www.cems.uvm.edu/~tlakoba/AppliedUGMath/notes/lecture_8.pdf 3. Spherical Harmonics: These arise in quantum mechanics/physical chemistry when studying the shape of electron orbitals. i. Theory/Calculations: http://scipp.ucsc.edu/~haber/ph116C/SphericalHarmonics_12.pdf There is math here that you will not see until Math 2173, 2177, or 2415 (depending on your major) involving separation of variables for partial differential equations. Instead, focus on equations (5) and (6) and note that a way to solve these equations is via the techniques discussed here. Indeed, Eqn (7) is a more efficient way to handle Eqn. (6), but the analysis can be done via the techniques discussed in this project as well! ii. Summary/Nice Pictures: http://users.physik.fu-berlin.de/~pascual/Vorlesung/SS06/Slides/ AMOL-L1d.pdf Click on any of the following links to be directed to the appropriate website! Points of Consideration While Studying These Examples These are highly nontrivial problems that require involved computations. If you cannot follow every step in these examples, don't be discouraged! There is some material that does require some outside knowledge or more advanced mathematical techniques. However, an essential technique necessary in all of these problems is using Taylor series to approximate solutions to differential equations! When you see these, or other, examples in future engineering and science courses, remember that you have learned the necessary mathematical techniques to analyze these problems! 12 Part III : (The Laplace Transform) The Laplace transform F (s) of a given function f (t) is given by: Z f (t)est dt F (s) := 0 when the integral exists. Note the Laplace transform is a function of a new (frequency) input s, which should be treated like a constant when computing the improper integral. Laplace transformations are particularly useful in solving many initial value problems that arise in mathematics, physics, electrical engineering, control engineering, optics, and signal processing; you will likely see these again! They are worth mentioning now as a special type of improper integral! 1 I. Show that the Laplace transformation of f (t) = 1 is given by F (s) = . s II. Show that the Laplace transformation of f (t) = t is given by F (s) = 1 . s2 III. Show that the Laplace transformation of f (t) = eat is given by F (s) = 13 1 . s+a IV. (Transforming a linear first order ODE) Consider the initial value problem: dy + y(t) = t, dt y(0) = 1. A. Verify that y(t) = 2et + t 1 is a solution. (Don't forget to verify this satisfies the differential equation and the initial condition!) Z y(t)e B. Let Y (s) = 0 st Z dt. Show that 0 dy st e dt = sY (s) 1. dt Hint: Think integration by parts. You may assume lim y(t)est = 0. The initial t condition y(0) = 1 should be important in your work somewhere! 14 C. Apply the Laplace transform to both sides of the differential equation, that is write: \u0015 Z Z \u0014 dy st test dt. + y(t) e dt = dt 0 0 R Letting Y (s) = 0 y(t)est dt, show that the above differential equation in y(t) becomes the algebraic equation: 1 1 + (s + 1)Y (s) = 2 s in Y (s). Hint: You have already done all of the computations necessary to show this! This should require you to compute NO integrals! D. Solve the above equation for Y (s) and show that: Y (s) = 1 1 2 + 2 . s+1 s s 15 We can now \"invert\" the Laplace transform1 as follows. f (t) = 1, then the Laplace transform of f (t) is F (s) = 1 (from Part I). s f (t) = t, then the Laplace transform of f (t) is F (s) = 1 (from Part I). s2 f (t) = eat , then the Laplace transform of f (t) is F (s) = 1 (from Part I). s+a so, we can invert the Laplace transform by assuming if: 1 F (s) = , then the inverse Laplace transform of F (s) is f (t) = s 1 F (s) = 1 , then the inverse Laplace transform of F (s) is f (t) = s2 F (s) = 1 , then the inverse Laplace transform of F (s) is f (t) = s+1 . . . Since the Laplace transform of y(t) is by definition Y (s), by definition, the inverse Laplace transform of Y (s) is y(s). Thus, by taking the inverse Laplace transform L1 of each term2 in Y (s) = 1 1 2 + 2 s+1 s s it follows that: 1 L [Y (s)] = L 1 \u0014 \u0015 \u0014 \u0015 \u0014 \u0015 2 1 1 1 1 +L L s+1 s2 s y(t) = This should agree precisely with the result in part A! 1 The theory behind the Laplace transform and its inverse requires a fair amount of care to establish and is typically discussed in great detail in a graduate level course in asymptotics. Specifically for this problem, it requires care to show that the Laplace transform has an inverse; i.e. if F (s) is identified with f (t), then there is some \"unique\" way of identifying f (t) with F (s). 2 It can be shown that the inverse Laplace transform is indeed linear, so this is justifiable. 16 The Laplace transform technique outlined above is incredibly useful when there are discontinuous forcing functions. For example, you drop a ball then kick it as it is falling; the kick is modeled mathematically as an instantaneous applied force! Or, you have a circuit with a charged capacitor and various resistors or inductors. The amount of charge on the capacitor can be modeled by a differential equation. If there is a surge in the circuit (e.g. a large voltage applied in a short time, which usually is the cause of a \"short\" circuit), the effect is once again treated as instantaneous. In both cases, a Laplace transform is very useful in solving the differential equation! The Laplace transform will appear again in these situations and more in more advanced science courses as well as in Math 4512 (and possibly Math 2415 depending on your instructor). There are several special techniques you will learn that are not covered within this project, but it is pivotal that you understand the computations and concepts contained here before learning more advanced techniques! For more applications of the Laplace transform, click the link below: http://sces.phys.utk.edu/~moreo/mm08/sarina.pdf 17