Classroom Tips and Techniques: Eigenvalue Problems for ODEs - Part 1

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1 Classroom Tips and Techniques: Eigenvalue Problems for DEs - Part 1 Initializations restart with LinearAlgebra : with Student Calculus1 : with RootFinding : Introduction Robert J. Lopez Emeritus Professor of Mathematics and Maple Fellow Maplesoft Some boundary value problems for partial differential equations are amenable to analytic techniques. For example, the constant-coefficient, second-order linear equations called the heat, wave, and potential equations are solved with some type of Fourier series representation obtained from the Sturm-Liouville eigenvalue problem that arises upon separating variables. In this, and the next two articles in this series, we examine the role of Maple in the solution of such boundary value problems. We will show efficient techniques for separating variables, then show how to guide Maple through the solution of the resulting Sturm-Liouville eigenvalue problems. In four previous columns we have demonstrated how to implement the Fourier series calculations in Maple. Together, these articles represent the essence of the typical undergraduate course in boundary value problems. In fact, the author's Advanced Engineering Mathematics electronic book includes a complete course in the subject area. In subsequent articles, we will explore separation of variables for Laplace's equation in cylindrical coordinates, a process that leads to Bessel's equation as the DE in the Sturm-Liouville eigenvalue problem, which will now be singular and not regular as in this article. Finally, we will separate variables for Laplace's equation in spherical coordinates so that the resulting DE in the Sturm- Liouville eigenvalue problem will be Legendre's equation, again a singular problem. The difference between these two problems is that for Bessel's equation, the resulting Bessel functions themselves become eigenfunctions, whereas for Legendre's equation, it is far more difficult to pass from Legendre functions in a general solution to the Legendre polynomials that are the eigenfunctions.

2 Separating Variables in a PDE It is not difficult to implement separation of variables in Maple. For example, the one-dimensional wave equation v 2 u x, t = c2 v2 2 vt vx 2 u x, t v 2 vt 2 u x, t = v 2 c2 vx 2 u x, t separates under the assumption that u x, t can be written as the product (3.1) U := X x T t Substitution leads to U := X x T t (3.2) (3.1) u x, t = U X x d 2 dt 2 T t = d 2 c2 whereupon division by u x, t in its separated form yields dx 2 X x T t (3.3) simplify (3.3) U c 2 d 2 dt 2 T t T t c 2 = d 2 dx 2 X x X x Note that we have also divided through by c 2, the wave speed. The Bernoulli separation constant λ is now introduced, resulting in the two ordinary differential equations (3.4) q 1 d lhs (3.4) =λ q 2 d rhs (3.4) =λ more commonly written as q 1 := q 2 := d 2 dt 2 T t T t c 2 =λ d 2 dx 2 X x X x =λ (3.5) DE 1 d numer normal lhs q 2 Krhs q 2 = 0 DE 2 d numer normal lhs q 1 Krhs q 1 = 0

3 DE 1 := d2 dx 2 X x Kλ X x = 0 DE 2 := d2 dt 2 T t Kλ T t c2 = 0 The first equation governing X x is a formally self-adjoint DE that will become part of the Sturm- Liouville eigenvalue problems considered in the following section. Solving the Sturm-Liouville Eigenvalue Problem Homogeneous Dirichlet Conditions A Sturm-Liouville eigenvalue problem arises if we couple the differential equation (3.6) DE 1 d 2 dx 2 X x Kλ X x = 0 (4.1.1) with the homogeneous Dirichlet boundary conditions X 0 = X π = 0. This is an example of "separated" conditions for which we are guaranteed a complete set of orthogonal eigenfunctions. ne approach to solving this eigenvalue problem might be to invoke Maple's dsolve command, with the result dsolve DE 1, X 0 = 0, X π = 0, X x X x = 0 Unfortunately, Maple returns the zero-solution, and does not even begin to solve the eigenvalue problem. That is why we are writing this series of articles. Maple can actually be of great help in solving such problems, but it must be guided by the user. The characteristic equation for (4.1.1) is obtained by making the exponential guess EG := e m x EG := e m x Substitution and the obvious algebra then give 1 EG collect DE 1 X x = EG, exp m 2 Kλ = 0 It should be obvious that the characteristic roots are m =G λ, and that λ = 0 is the "repeated root" case. Since the form of the general solution of (4.1.1) changes for this case, we consider it first as a special case. The differential equation is then DE 1a d DE 1 λ = 0

4 DE 1a := d2 dx 2 X x = 0 and a solution of this equation with just the left-hand boundary condition is dsolve DE 1a, X 0 = 0, X x X x = _C1 x Applying the right-hand boundary condition to this solution leads to the equation (4.1.2) rhs (4.1.2) = 0 x =π _C1 π = 0 from which it is clear that the multiplicative constant has to be zero. Since this leads to the trivial solution X x = 0, we conclude that λ = 0 is not an eigenvalue. We next consider the case λs 0, and again use the strategy of applying just the left-hand boundary condition. The resulting solution is X 1 d rhs dsolve DE 1, X 0 = 0, X x X 1 :=K_C2 e λ x C_C2 e K λ x Because Maple could return this solution with _C1 as the arbitrary constant, we use the following device to rewrite the solution in terms of the constant c. No matter what name Maple returns for the arbitrary constant, our construction will always be in terms of c. paramdremove has, indets X 1, λ, x 1 : X 2 d X 1 param = c X 2 :=Kc e λ x Cc e K λ x Applying the right-hand boundary condition leads to the algebraic equation X 2 x =π=0 Kc e λ π Cc e K λ π = 0 Without intervention, Maple will provide just the zero solution for λ, as we confirm with (4.1.3) solve (4.1.3),λ 0 The requisite tool is the environment variable _EnvAllSolutions := true _EnvAllSolutions := true which, if set to the value true, enables Maple to return the solution

5 solve (4.1.3),λ K_Z1 2 Maple uses the notation _Zk, k = 1, 2,, to indicate an integer. Unfortunately, each time this solve command is executed, the index k will increment by 1, so again, we cannot reference this symbol in a fail-safe way. Thus, we resort to the following device temp d indets (4.1.4) 1 : (4.1.4) temp = n Kn 2 to write the eigenvalues with a unique notation. The eigenvalues are the negative numbers Kn 2, n = 1, 2,. It is also possible to lead Maple through the steps resulting in this solution for the eigenvalues. We begin by writing (4.1.3) in the form (4.1.4) factor (4.1.3) Kc e λ π Ke K λ π = 0 Since c = 0 would result in the trivial solution X x = 0, we must have cs 0, so that map coeff, (4.1.5), c Ke λ π Ce K λ π = 0 This difference of exponentials suggests rewriting the left-hand side as (4.1.5) (4.1.6) convert (4.1.6), trig K2 sinh λ π = 0 (4.1.7) from which we realize that if λ is real, the only solution of this equation will be λ = 0, again leading to the trivial solution. This inspires the substitution simplify evalc (4.1.7) assuming µ0 λ =Kµ 2 K2 I sin µ π = 0 from which it easily follows that µ is an integer, given in Maple by (4.1.8) solve (4.1.8),µ _Z2~ Homogeneous Neumann Conditions The Sturm-Liouville eigenvalue problem that arises when the homogeneous (and separated) Neumann boundary conditions X# 0 = X# π = 0 are applied to the differential equation

6 DE 1 d 2 dx 2 X x Kλ X x = 0 will have a different set of eigenvalues and eigenfunctions than we obtained with the Dirichlet conditions. Since we already know that λ = 0 must be handled as a special case, we begin by obtaining the general solution of the DE DE 1a d 2 dx 2 X x = 0 We again take precautions to write the solution with unique arbitrary constants, obtaining temp := rhs dsolve DE 1a, X x : X 0 d eval temp, _C1 = b, _C2 = a X 0 := b xca To this general solution, we apply the two boundary conditions, resulting in the algebraic equations X 0 ' = 0 x = 0 b = 0 and X 0 ' = 0 x =π b = 0 Consequently, λ = 0 is an eigenvalue, and the corresponding eigenfunction is X 0 x = a Now, consider the case λs 0. We again find the general solution, and write it with uniquely named arbitrary constants. temp := rhs dsolve DE 1, X x : X 1 d eval temp, _C1 = c 1, _C2 = c 2 X 1 := c 1 e λ x Cc 2 e K λ x To this general solution we apply the boundary conditions, obtaining the homogeneous algebraic equations Eq 1 d X 1 ' = 0 x = 0 Eq 1 := c 1 λ Kc 2 λ = 0 and Eq 2 d X 1 ' = 0 x =π Eq 2 := c 1 λ e λ π Kc 2 λ e K λ π = 0 Maple's solve command cannot determine values of λ for which this system has nontrivial solutions. We guide Maple by writing the system matrix as

7 GenerateMatrix Eq 1, Eq 2, c 1, c 2 1 λ K λ λ e λ π K λ e K λ π (4.2.1) and setting its determinant to zero, the condition under which the system will have nontrivial solutions for c 1 and c 2. Determinant (4.2.1) = 0 Kλ e K λ π Cλ e λ π = 0 Clearly, λ = 0 is a solution, but that belongs to the first case. "Canceling" λ and solving, gives (4.2.2) solve (4.2.2) λ,λ where again, we replace this notation by K_Z3 2 (4.2.3) temp d indets (4.2.3) 1 : (4.2.3) temp = n Kn 2 so that the eigenvalues are more easily seen to be λ =Kn 2, n = 1, 2,. Having determined the eigenvalues, we return to the algebraic equations determining c 1 and c 2, and make the obvious substitution for λ to obtain the equations Eq 3 d simplify Eq 1 λ =Kn 2 Eq 4 d simplify Eq 2 λ =Kn 2 assuming n T posint; assuming n T posint Eq 3 := I c 1 nki c 2 n = 0 The solution of these equations is now Eq 4 := I n K1 n c 1 Kc 2 = 0 solve Eq 3, Eq 4, c 1, c 2 c 2 = c 1, c 1 = c 1 so that the remaining eigenfunctions are temp := subs c 1 = C, c 2 = C,λ =Kn 2, X 1 : simplify evalc temp assuming n T posint 2 C cos n x

8 f course, the constant C can be taken as 1 π, or as 2 2 to obtain orthonormal eigenfunctions. Example: An DE that is Not Self-Adjoint We conclude this article with an example of an eigenvalue problem in which the differential equation is not self-adjoint. Such an example is not covered by the Sturm-Liouville theory, but it does illustrate how Maple can be guided through some of the more difficult calculations that arise in eigenvalue problems. In particular, note the more general separated (but still homogeneous) boundary conditions in the following problem. deq := d2 dx 2 y x C2 d dx y x Cλ y x = 0 bc 1 d 3 D y 0 C5 y 0 = 0 bc 2 d 2 D y π Cy π = 0 deq := d2 dx 2 y x C2 d dx y x Cλ y x = 0 bc 1 := 3 D y 0 C5 y 0 = 0 bc 2 := 2 D y π Cy π = 0 To obtain solutions of the differential equation, we seek a fundamental set of linearly independent members. Making the exponential guess leads to the characteristic equation 1 EG collect deq, exp y x = EG whose solutions are m 2 C2 mcλ = 0 (5.1) solve (5.1), m K1C 1Kλ, K1K 1Kλ From these characteristic roots, we can see that λ = 1 is the repeated-root case for which the solution of the DE assumes the special form (5.2) temp := dsolve deq, y x : λ = 1 Y 1 d unapply eval rhs temp, _C1 = r 1, _C2 = s 1, x : 'Y 1 ' x = Y 1 x Y 1 x = r 1 e Kx Cs 1 e Kx x For λs 1, the general solution of the DE can be written as temp := dsolve deq, y x :

9 Y 2 d unapply eval rhs temp, _C1 = r 2, _C2 = s 2, x : 'Y 2 ' x = Y 2 x Y 2 x = r 2 e K1C KλC1 x Cs 2 e K1K KλC1 x Even though Maple defaults to an exponential form for the members of its fundamental set, we should not assume that λ is characterized by anything other than the condition λs 1. In fact, at this point we don't even know if the eigenvalues are real. Consequently, we avoid separating the calculation into the cases λ! 1 and λ 1 because that would imply that we knew λ was not a complex number. We let the boundary conditions determine λ by considering just the cases λ = 1 and λs 1. In the first case, we get the algebraic equations for k to 2 do q k d bc k y = Y1 end do q 1 := 2 r 1 C3 s 1 = 0 q 2 :=Kr 1 e Kπ Ks 1 e Kπ πc2 s 1 e Kπ = 0 and in the second, for k to 2 do q 2Ck := bc k y = Y 2 end do q 3 := 3 r 2 K1C KλC1 C3 s 2 K1K KλC1 C5 r 2 C5 s 2 = 0 q 4 := 2 r 2 K1C KλC1 e K1C KλC1 π C2 s 2 K1K KλC1 e Cr 2 e K1C KλC1 π Cs 2 e K1K KλC1 π = 0 The determinants of the system matrix in each case are then K1K KλC1 π Q 1 := Determinant GenerateMatrix q 1, q 2, r 1, s 1 1 Q 1 :=K2 e Kπ πc7 e Kπ and (5.3) Q 2 := collect Determinant GenerateMatrix q 3, q 4, r 2, s 2 1, exp Q 2 := 8K7 KλC1 K6 λ e K1C KλC1 π C K8K7 KλC1 (5.4) K1C KλC1 π C6 λ e respectively. Since the determinant in (5.3) is nonzero, we must have r 1 = s 1 = 0, so the corresponding

10 solution is X x h 0, and λ = 1 is therefore not an eigenvalue. There are zeros of the determinant in (5.4), some of which are Analytic Q 2, re =K10..10, im =K , , , , The Analytic command in the RootFinding package finds all real and complex roots in a region of the complex plane. We already know that λ = 1 is not an eigenvalue, so we have the real eigenvalues remove w/is w K1! 0.001, (5.5) , , , While the Analytic command provides a guarantee that all the solutions in a fixed region have been found, it is slow. If we accept (5.5) as evidence that the eigenvalues are positive and real, then we can find them slightly faster by changing the form of the determinant in (5.4) to (5.5) (5.6) simplify evalc Q 2 assuming λ 1 K2 I e Kπ 7 λk1 cos λk1 π K8 sin λk1 π C6 sin λk1 π λ and using the Roots command to find (5.7) (5.7) L := Roots,λ = , numeric = true Kπ I e L := , , , , , For any such eigenvalue, the equations q 3 and q 4 are redundant, so we can solve one of them for s 2, say, in terms of r 2, obtaining (5.8) S := solve q 3, s 2 S := r 2 2C3 K2C3 KλC1 KλC1 For any eigenvalue λ, the corresponding eigenfunction is then given (up to a multiplicative constant) by Yd 1 collect Y r 2 x, 2 s 2 = S r 2, exp Y := e K1C KλC1 x 2C3 C KλC1 K1K KλC1 x e K2C3 KλC1 With λ 1, the exponential terms will have trigonometric equivalents, so we write simplify evalc Y assuming λ 1 K 1 K5C9 λ 2 ekx 9 cos λk1 x K9 cos λk1 x λc6 λk1 sin λk1 x (5.9)

11 and then K4 I sin λk1 x C6 I λk1 cos λk1 x collect (5.9), sin, cos K 2 ekx 6 λk1 K4 I sin λk1 x K5C9 λ (5.10) K 2 ekx 9K9 λc6 I λk1 cos λk1 x K5C9 λ The real and imaginary parts are respectively Ya := evalc R (5.10) assuming λ 1; Yb := evalc I (5.10) assuming λ 1 Ya :=K 12 ekx λk1 sin λk1 x K5C9 λ K 2 ekx 9K9 λ cos λk1 x K5C9 λ Yb := 8 ekx sin λk1 x K5C9 λ K 12 ekx λk1 cos λk1 x K5C9 λ In fact, each of the real and imaginary parts is separately a solution. But they cannot be independent solutions - they must be multiples of each other. This common multiple is f := simplify convert taylor Ya, x, 3, polynom Yb f :=K 3 2 λk1 and we can verify that indeed, the real and imaginary parts of Y x are linearly dependent by showing the linear combination f Y b KY a vanishes. simplify Yb fkya The common multiple can be found with 0 g := Ya x = 0 and a simplified form of the real part is then g :=K 2 9K9 λ K5C9 λ Yc := collect Ya g, exp, cos, sin Yc := cos λk1 x C Using Y c, we obtain the first six eigenfunctions as 6 λk1 sin λk1 x 9K9 λ e Kx

12 unassign 'φ' φ 1 := simplify expand convert Yc for k from 2 to nops L φ k d Yc do λ = L 1, exp assuming x 0; end do λ = L k φ 1 := e K x C e K x φ 2 := cos x K sin x e Kx φ 3 := cos x K sin x e Kx φ 4 := cos x K sin x e Kx φ 5 := cos x K sin x e Kx φ 6 := cos x K sin x e Kx A graph of the first six eigenfunctions is given in Figure 1, where our normalization has all eigenfunctions passing through 0, 1. The eigenfunctions φ k x, k = 1,, 6, are given in dotted black, then solid black, blue, red, green, and magenta. plot seq φ k, k = 1..6, x = 0..π, color = black, black, blue, red, green, magenta, linestyle = 2, 1$5

13 x K0.5 Figure 1 Eigenfunctions φ k, k = 2., 6 are in solid black, blue, red, green, and magenta, respectively, whereas φ 0 is in dotted black Since the differential equation is not self-adjoint, we are not assured of orthogonal eigenfunctions. The inner products φ i,φ j, 1 % i, j% 6, are computed and displayed in Table 1. If these eigenfunctions were orthogonal, the off-diagonal entries would be zero. That they are not is our evidence the eigenfunctions are not orthogonal. Matrix 6, 6, i, j / φ i φ j dx 0 π

14 Table 1 The inner products φ i,φ j, 1 % i, j% 6 We expended some energy separating real and imaginary parts in the solution Y x, more in showing that these parts were multiples of each other. A more direct way to find the simple trigonometric form we used to generate the eigenfunctions is to obtain the general solution of the differential equation under the assumption that λ 1. This leads to temp := dsolve deq, y x assuming λ 1 : Y 3 d unapply eval rhs temp, _C1 = r 3, _C2 = s 3, x : 'Y 3 ' x = Y 3 x Y 3 x = r 3 e Kx sin λk1 x Cs 3 e Kx cos λk1 x and, after applying the boundary conditions, to for k to 2 do q 4Ck := bc k y = Y 3 end do q 5 := 3 r 3 λk1 C2 s 3 = 0 q 6 :=Kr 3 e Kπ sin λk1 π C2 r 3 e Kπ cos λk1 π λk1 Ks 3 e Kπ cos λk1 π K2 s 3 e Kπ sin λk1 π λk1 = 0 If we now express r 3, say, in terms of s 3, as in isolate q 5, r 3 we can write this new general solution as r 3 =K 2 3 s 3 λk1 (5.11) Ydd 1 s 3 collect eval Y 3 x, (5.11), exp, s 3 Yd := K 2 3 sin λk1 x λk1 Ccos λk1 x e Kx

15 To show this is equivalent to our Y c x, we show the difference Y d KY c vanishes. simplify Yd KYc 0 Legal Notice: The copyright for this application is owned by the author(s). Neither Maplesoft nor the author are responsible for any errors contained within and are not liable for any damages resulting from the use of this material. This application is intended for non-commercial, non-profit use only. Contact the author for permission if you wish to use this application in for-profit activities.