ORDINARY DIFFERENTIAL EQUATIONS

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1 Page 1 of 20 ORDINARY DIFFERENTIAL EQUATIONS Lecture 11 Fundamental Solution Sets for Linear Homogeneous Second-Order ODEs (Revised 22 March 09:16) Professor Stephen H Saperstone Department of Mathematical Sciences George Mason University Fairfax, VA sap@gmu.edu Copyright 2009 by Stephen H Saperstone All rights reserved 11.1 EXISTENCE & UNIQUENESS OF SOLUTIONS The discussion immediately following the heading Initial Conditions in Lecture 10 suggests how to establish existence and uniqueness for IVPs when the underlying ODE is of the form In particular, the calculations in demonstrate that the constants of integration and are uniquely determined by the initial conditions. The following theorem is an analog to the FTEU for first-order IVPs. Because of the simple nature of a linear homogeneous second-order ODE with constant coefficients, we are able to provide a strong and conclusive theorem below for the existence and uniqueness of solutions. The proof of the theorem is not hidden in a hyperlink. The ideas developed in the proof are needed later in the development of properties of solutions to linear homogeneous second-order ODEs. Theorem 11.1 (Existence & Uniqueness - EU for Homogeneous Linear Second- Order IVP) For any values of the constants and the IVP has a unique solution whose maximal interval of definition is the entire -axis.

2 Page 2 of 20 PROOF: According to Theorem 10.9, the general solution to the ODE of Eqn. (11.1) has the form where the functions and are determined by the characteristic roots and from the following table: Table Component Solutions to Eqn. (11.1) Then The initial conditions imply Since we may regard and as numbers (remember, and are known from Table 11.1 and the initial time is given), we can rewrite as a system of linear algebraic equations with unknowns, the variables and According to Cramer's Rule this system has a unique solution for and if and only if the determinant of the coefficients The value of this determinant depends on the functions and the initial time Denote its value by Then We verify in each of the three cases that CASE 1: :

3 Page 3 of 20 Since the last expression is is never zero for any value of CASE 2: : Again, the last expression is is never zero for any value of CASE 3: : Since the last expression is is never zero for any value of Q.E.D.

4 Page 4 of THE WRONSKIAN The expression defined in Eqn. (11.2) extends to ANY pair of functions (not necessarily solutions to a linear second-order homogeneous ODE) and any time Definition 11.2 (Wronskian) Given any two functions and defined for all, the Wronskian of and is given by the formula and is defined for all In view of special solution pairs listed in Table 11.1 of Theorem 11.1, the following table summarizes the calculations made in that proof. Table The Wronskian for Eqn. (11.2) Since all solutions of are defined for all we can replace in the definition of by any time. Thus we have the important definition regarding solutions to the ODE Although and refer to the particular solutions identified in Table 11.2, the expression for the determinant can be based on ANY two solutions - not necessarily the pair for each case. The following Theorem of Abel characterizes the Wronskian. Theorem 11.3 (Abel - The Wronskian Condition) Suppose and are any two solutions to the ODE Then the Wronskian

5 Page 5 of satisfies the first-order linear (Abel's) ODE and 2. is either never zero for any or zero for all PROOF ABEL's THEOREM We see from Table 11.2 that for all Not all pairs of solutions have this property as shown in the next example We saw in 10.6 that the ODE has the general solution By first setting and we get the solution Now set and to get the solution The Wronskian of the pair is Thus just because and are both solutions, it does not follow that their Wronskian is nonzero. The problem here is that is a multiple of End of 11.4 A solution pair to the ODE for which is never zero is special. We will see later that the Wronskian distinguishes amongst pairs of solutions that can serve as a general solution and those that cannot. For now we single out those solution pairs for which the Wronskian is never zero. Definition 11.5 (FSS) A pair of solutions and to the ODE is called a Fundamental Set of Solutions (FSS) if is never zero on Thus each of the pairs following FSS Theorem. listed in Table 11.2 is a FSS. We summarize the preceeding discussion in the Theorem 11.6 (FSS) Every linear homogeneous ODE of the form

6 Page 6 of 20 has a FSS. Moreover, every solution to Eqn. (11.2) on can be expressed in the form for some choice of constants and The constants are uniquely determined from any set of initial conditions LINEAR INDEPENDENCE OF SOLUTIONS Suppose is an FSS for Eqn. (11.2).The Wronskian condition for the FSS has an important geometric interpretation. Since for all the columns of the matrix must be linearly independent for all the zero vector. Thus the only way for This means that no nonzero linear combination of the columns is to be satisfied for all is for In particular, with regard to the first component of this vector equation, the only way that is for and This property may be better understood when we examine its negation; that is suppose there are some nonzero and so that for all Then we could express one of the functions and as a multiple of the other. For instance, if then from we can write Thus one of the solutions is a multiple of the other. This would imply that Since for all the ONLY WAY that for all is for and

7 Page 7 of 20 Definition 11.7 (Linear Independence of Functions on ) Two functions and defined on are called linearly independent (abbreviated LI) if the ONLY WAY the linear combination is for and If and are not linearly independent, they are called linearly dependent (abbreviated LD) In view of the Wronskian condition (Abel's Theorem), we can summarize the preceeding in the following theorem. Theorem 11.8 (Linear Independence of Solutions) Suppose and are solutions to the ODE Then and are linearly independent if and only if the Wronskian on Show that the functions and are LI by using the definition of LI. Solution: Suppose for all We must show that As for all, it follows that this equation must hold when Thus Thus Finally, if for all, then must equal zero. Thus and are LI. End of Show that the functions and are LI by using Theorem Solution: Calculate

8 Page 8 of 20 As is never zero on, then the functions and are LI. End of Show that the functions and are LI by using the definition of LI. Solution: Suppose for all We must show that As for all, it follows that this equation must hold when and when (We choose these values so that we get a pair of equations for and Thus and Together these equations imply End of and Note that the hypothesis of Theorem 11.8 assumes that the functions and are solutions to the ODE In the case of and we know that these constitute a FSS for the ODE. (The reader should verify this.) Consequently, we are allowed to use the Wronskian to determine LI of the pair as we did in In the case of and we suspect that this pair isn't a FSS for any ODE of the form Eqn. (11.2). This is because neither function nor appears in Table None of the possibilities for the characteristic roots can give rise to either of these functions. Nevertheless, it is instructive to calculate in this case. Calculate where Solution: To evaluate we must differentiate the piecewise representation of It will be easiest to do this if we first express

9 Page 9 of 20 piecewise. Then Although on, we cannot conclude that and are LD. This conclusion would be valid if both functions are solutions to an ODE of the form As this is not the case, then Theorem 11.8 is silent regarding the LI and/or LD of and End of The definitions of the Wronskian and linear independence of functions and do not require them to be solutions to the ODE But Abel's theorem (the Wronskian condition) says that if and ARE solutions, then on. This leaves open the possibility that and can be linear independent on yet on providing that and ARE NOT solutions. This fact is confirmed by the following simple example. Indeed, we conclude that the functions and from CANNOT be solutions to ANY ODE of the form The functions and are linearly independent on, yet is neither zero for all nor nonzero for all Solution: To establish linear independence we must show that implies and As for all, it follows that this equation must hold when and when (We choose these values so that we get a pair of equations for and Thus and

10 Page 10 of 20 Together these equations imply and Hence is LI. Now calculate which is zero at just one point, This one exception is enough to break the rule. Thus and cannot be solutions to an ODE of the form End of Alert The hypotheses of Theorem 11.8 and Abel's theorem require that and be solutions to Otherwise the conclusion of the theorems is false. For instance we just saw in that the Wronskian of the functions and is zero for all. In order that this result does not contradict the conclusion of Abel's theorem, it must be that and cannot be solutions of a linear homogeneous second-order ODE. End of Alert The definition of linear independence can be extended to any number of functions. In the case of just two functions and we can extend the colinearity property of LD vectors in the plane to obtain a necessary and sufficient condition for LD of two functions. Theorem (Simple Test for LI) Two functions and defined on are linearly independent if and only if their quotient is not a constant function on. Using this characterization of LI, we readily see that in each of the three cases the pairs of functions that constitute the general solution to are LI on Table FSS for Eqn. (11.2) The reader should verify that each of the pairs in Table 11.3 is LI on is LI because For example the pair

11 Page 11 of 20 is certainly nonconstant on Show that the functions and are LI by using Theorem Solution: Calculate which is nonconstant on End of Show that the functions and are LI by using Theorem Solution: Calculate which is nonconstant on End of FUNDAMENTAL SOLUTION SETS It turns out that the pairs listed in Table 11.3 aren't the only bases for solutions to Theorem (Fundamental Solution Set) A pair of solutions to the ODE is a FSS if and only if the pair is LI. Verify that 11.19

12 Page 12 of 20 is a FSS for Solution: We see from Table 11.2 that so is never zero. Hence, is a FSS for To show that is a FSS, we need to show that (1) and are solutions to (2) and are LI on To show (1) we need only observe that both and are linear combinations of the solutions and To show (2) we apply the definition of LI to the pair Let Then Since and is are LI on then Simple algebra implies that and Thus and are LI. End of suggests that given a FSS for the ODE we can create another FSS by taking appropriate linear combinations of and The following theorem shows how to generate all possible fundamental solution sets for the ODE. Theorem (All Possible FSS) Suppose and are Fundamental solution sets for the ODE Then there is a invertible matrix C so that What are the implications of this theorem? 1. ALL fundamental solution sets are related by a invertible matrix that is easy to calculate. 2. There are infinitely many fundamental solution sets. We include the proof of this theorem because the steps involved are useful in problem solving. Proof of All Possible FSS: Suppose and are Fundamental solution sets for the ODE

13 Page 13 of 20 In particular, since is a solution, then must be a linear combination of and, say Likewise, since is a solution, then must also be a linear combination of and, say Since and are LI then Calculate: Then for all implies As are LI then But, are the only values of and that solves these two equations. In other words, the ONLY solution to the vector equation is According to Items 5 and 4 of the Invertible Matrix Theorem It follows that also has nonzero determinant so that C is invertible. Q.E.D Determine a FSS for the ODE Solution: The characteristic equation is Factor as to get roots According to Table 11.3 a FSS is

14 Page 14 of 20 End of Consider the ODE Determine a FSS that satisfies and Solution: Start with the FSS from Set First find values of and that satisfy Substitute to get Then and so that Next find values of and that satisfy Substitute to get Then and so that Thus the FSS we want is End of Solve the IVP Solution: The characteristic equation is Factorization is too difficult so we use the quadratic formula:

15 Page 15 of 20 From Table 11.1 we have and and a FSS is The general solution is given by To compute the values of and from the initial conditions, we first need a formula for The values for and imply Then and so that the solution to the IVP is End of TRANSLATION INVARIANCE Because the independent variable is not present in the ODE time-translations of solutions are solutions. That is, if is a solution, then is a solution for any This property is easily verified as follows. Since is a solution, then is true for all Hence for any must be true for all Hence is a solution. Thus if is a solution to the IVP there is no loss in generality by translating the solution so that the value of solution is at and the value of the derivative is at. This is the significance of the following theorem. All Clocks Start at Theorem Suppose (Translate solutions to start at ) is a solution to the IVP Then is a solution to the IVP PROOF: We saw immediately before the statement of Theorem that It is also easy to verify that the initial conditions must solve the ODE are also satisfied.

16 Page 16 of 20 Indeed since solves the IVP of Eqn. (11.1), then we must have Next differentiate to get Then Thus solves the IVP of Eqn. (11.3) Q.E.D. Visual evidence of Theorem is presented in Lecture 12 as well as in the following animation TIME TRANSLATION (& UNIQUENESS) ANIMATION 11.6 THE IDEAL FSS Frequently we need to calculate a special FSS with the following property Definition A FSS (Ideal FSS) that satisfies is called ideal. Note that an ideal FSS for an ODE must be unique. This is because solutions to any IVP of the form must be unique. In particular, the solutions to each of the IVPs and are unique. Moreover, these two solutions constitute an ideal FSS. Check to see that constitute an ideal FSS Solution: Calculate

17 Page 17 of 20 Then End of Check to see that and does not constitute an ideal FSS. Then show how can be transformed to an ideal FSS Solution: Calculate Then We leave it to the reader to check that yields an ideal FSS. End of Determine the ideal FSS for Solution: We see from Table 11.2 that is a FSS for Set and Denote by the ideal FSS called for. According to Theorem there is a invertible matrix C so that The requirements for an ideal FSS imply that In order to use these requirements we must first differentiate Eqn. (11.4) to obtain

18 Page 18 of 20 Set in Eqn. (11.4) and use Eqn. (11.5a) to get Similarly, set in Eqn. (11.6) and use Eqn. (11.5b) to get These two vector equations can be combined to get the single matrix equation It follows that the coefficients and we want are the entries of the inverse of the matrix In fact set Then we have the matrix equation which implies Use row reduction (or any other applicable) method to get Finally from Eqn. (11.4) we have the ideal FSS or The reader might recognize these functions from calculus, namely By definition (from calculus) Note that

19 Page 19 of 20 and Alternatively, the calculation of the coefficients may be done without matrix methods. You just need to solve the system of 4 scalar equations obtained from Eqn. (11.6a), namely and from Eqn. (11.6b) The first and third of these equations yield End of and Remark Because the IVP has a unique solution, the solution formulas for differing FSS must be algebraically equivalent. The next example illustrates Remark Solve the IVP using two different FSS: and. Solution: Each of the bases give rise to apparently distinct general solutions, namely and (We use different symbols for the coefficients as their values will differ when calculating them to solve the IVP.) The solution to the IVP for the FSS was obtained in To calculate the solution to the IVP for the FSS we first set in Eqn. (11.8): or (where we have used the definitions of and given above). It follow that Now differentiate Eqn. (11.8) with respect to

20 Page 20 of 20 Let in Eqn. (11.9): or It follow that So the solution to the IVP is As the solution to the IVP must be unique, we need to check that the two solutions and are algebraically equivalent. We get End of Theorem (Why an ideal FSS is called "ideal") If is the ideal FSS for the ODE Then for any initial condition the solution to the IVP has the form Solve the IVP using its ideal FSS. Solution: The ideal FSS is from It follows immediately from Theorem that the solution of the IVP is End of 11.31