III Second Order Linear ODEs

Transcription

1 III Second Order Linear ODEs Boyce & DiPrima, Chapter 3 H.J.Eberl - MATH*2170 0

2 III Second Order Linear ODEs III.0 Problem Formulation and First Examples Boyce & DiPrima, Chapter 3.2 H.J.Eberl - MATH*2170 1

3 Problem formulation for the next while we will be concerned with linear second order differential equation, i.e. differential equations of type (1) y +p(t)y +q(t)y = g(t) where p(t), q(t), g(t) are continuous functions that are defined on an interval I IR an important special case is g(t) 0, i.e. the equation (2) y +p(t)y +q(t)y = 0 in case (1) the equation is called non-homogeneous in case (2) it is called homogeneous the treatment of the non-homogeneous case builds on the homogeneous case, so this is where we will start H.J.Eberl - MATH*2170 2

4 Problem formulation, cont d in some applications linear second order equations arise in the form (3) s(t)y +p(t)y +q(t)y = g(t) if s(t) 0 everywhere on I, then it can be reduced to form (1) (4) y + p(t) s(t) y + q(t) s(t) y = g(t) s(t) if there is one or more t I with s(t ) = 0, this is not possible; one says (3)degeneratesint, meaningitisnotsecondorderinthosepoints; todeal with degenerate equations one could divide the interval I into subintervals that do not contain such points, treat them there as problems of type (1) and piece the solutions together; we will not be concerned with this too much in this course H.J.Eberl - MATH*2170 3

5 Applications linear second order problems models describe mechanical systems with vibrations and oscillations electrical systems stationary diffusion and mass transfer processes stationary financial systems example: undamped free vibrations are described by my +ky = 0 damped free vibrations by my +γy +ky = 0 forced vibrations with damping by my +γy +ky = cos(t) H.J.Eberl - MATH*2170 4

6 Example: my +γy +ky = cos(t) and special cases damped, forced y +y +y=cos(t) damped, free y +y +y=0 undampded, free y +y=0 0.1 y t H.J.Eberl - MATH*2170 5

7 Example: my +γy +ky = cos(t) and special cases wewilllearnhowtofindtheexactsolutionsoftheseandmorecomplicated equations, but the plots I just showed were computed with Euler s Method to this end, the second order method is converted into a system of two first order ODEs dy dt = z dz = g(t) p(t)z q(t)y dt with initial data y(0) = 0.1,z(0) = y (0) = 0.1 Euler s method with time-step h for the system of two first order ODEs: y i+1 = y i +hz I used here h = 1/1000 (very fine) z i+1 = z i +h ( g(t i ) p(t i )z i q(t i ) ) y i H.J.Eberl - MATH*2170 6

8 Example: my +γy +ky = cos(t) and special cases autonomous 2nd order equations show more complex behaviour than autonomous 1st order equations autonomous 1st order systems show more complex behaviour than single 1st order equations sometimes it is instructive to plot the solution in the y-z-plane instead in the native form as a function of t; this is called the phase plane, the 2D version of the phase line H.J.Eberl - MATH*2170 7

9 Example: my +γy +ky = cos(t) and special cases: phase plane y +y +y=cos(t) y +y +y=0 y +y=0 0.1 dy/dt y H.J.Eberl - MATH*2170 8

10 III Second Order Linear ODEs III.1 Homogeneous Equations: Solution Theory Boyce & DiPrima, Chapter 3.2 H.J.Eberl - MATH*2170 9

11 The Superposition Principle for now we concern ourselves with the homogeneous linear second order equation y +p(t)y +q(t)y = 0 equations of this type are characterized by the superposition principle which allows us to super-impose two solutions to obtain a third one neither nonhomogeneous linear equations nor non-linear equations allow a similar construction of solutions the superposition principle is sometimes also called the linearity principle or the Fundamental Theorem for Homogeneous Linear Equations H.J.Eberl - MATH*

12 Theorem (Superposition principle). Let p(t), q(t) be continuous functions on an interval I IR. Furthermore, let y 1 (t) and y 2 (t) be two solutions of the homogeneous linear second order equation ( ) y +p(t)y +q(t)y = 0 on I. Then also the function z(t) := c 1 y 1 (t)+c 2 y 2 (t) for any c 1,2 IR is a solution of ( ) on I. Proof. We substitute z into ( ), expand bracketed terms and rearrange to find z +p(t)z +q(t)z = Thus z satisfies ( ). = (c 1 y 1 +c 2 y 2 ) +p(t)(c 1 y 1 +c 2 y 2 )+q(t)(c 1 y 1 +c 2 y 2 ) = = c 1 y 1 +c 2 y 2 +c 1 p(t)y 1 +c 2 p(t)y 2 +c 1 q(t)y 1 +c 2 q(t)y 2 = = c 1 (y 1 +p(t)y 1 +q(t)y 1 )+c 2 (y 2 +p(t)y 2 +q(t)y 2 ) = = c 1 0+c 2 0 = 0 q.e.d. H.J.Eberl - MATH*

13 The Superposition principle: examples the nonhomogeneous linear differential equation y +y = 1 has the solutions y 1 (t) = 1 + cost and y 2 (t) = 1 + sint but z(t) = 2 + cost+sint is not the nonlinear differential equation y y xy = 0 has the solution y 1 (x) = x 2 and y 2 (x) 1, but z(x) = 1 x 2 is not a solution Exercise: Show that the superposition principle holds for the homogeneous first order linear equation y +p(t)y = 0 H.J.Eberl - MATH*

14 Linear combinations, linear dependence of solutions Let y 1 (t) and y 2 (t) be two functions on an interval I and c 1,2 IR then c 1 y 1 (t)+c 2 y 2 (t) is called a linear combination of y 1 and y 2 the function y 1 (t) and y 2 (t) are called linearly independent on I if ( ) c 1 y 1 (t)+c 2 y 2 (t) = 0 implies that c 1 = c 2 = 0. If ( ) also holds for some constants c 1 and c 2 not both zero, then functions y 1 (t) and y 2 (t) are linearly dependent this means: y 1 and y 2 are linearlyindependentifthey are not proportional to each other, i.e if there is no constant k such that y 1 (t) = ky 2 (t) we will see in a short while: if y 1 and y 2 is a pair of linearly independent solutions of the homogeneous second order linear differential equation, then all solutions y can be expressed as a linear combination of y 1 and y 2, i.e. for every solution y(t) we can find c 1,2 IR such that y(t) = c 1 y 1 (t)+c 2 y 2 (t) H.J.Eberl - MATH*

15 Existence and Uniqueness Theorem instead of giving an existence and uniqueness theorem for the 2nd order linear equation, we state a more general result. recall from Part I: A higher order differential equation can always be converted into a system of first order differential equations in the case of a second order equation (1) y = F(x,y,y ) the new dependent variable z := y allows to transform (1) in to the system (2) { y = z z = F(x,y,z) we will formulate an Existence and Uniqueness Result for the initial value problem of such systems of first order ODEs, which is in analogy to the existence and uniqueness result for single equations H.J.Eberl - MATH*

16 Existence and Uniqueness Theorem for Systems of ODEs. Let I IR be an interval and let D IR n be a rectangle. Let f : I D D,(x,y) f(x, y) be a function that is continuous w.r.t. x and continously differentiable w.r.t. y. Let x 0 I, and y 0 D. Then the initial value problem of the system of differential equations y = f(x,y), y(x 0 ) = y 0 possesses a unique solution in some interval J I that contains x 0. a stronger result can be stated if one introduces a multi-dimensional Lipschitz condition, but this requires some Linear Algebra; as in the scalar case one can show that a continuously differentiable f always satisfies such a Lipschitz condition we won t give a proof of the above theorem here: the mathematical key ideas are the same as the one used in our discussion of the scalar case, but it is technically more involved H.J.Eberl - MATH*

17 Implications for our 2nd Order Equations the 2nd order linear equation y +p(x)y +q(x)y = g(x) with continuous coefficient functions p, q, g and initial conditions has a unique solution to see this, notice that { y = z is differentiable w.r.t to y and z y(0) = y 0, y (0) = z 0 z = g(x) p(x)z q(x)y an important observation is: to obtain a unique solution we need to prescribe two initial conditions, one for y and one for y H.J.Eberl - MATH*

18 Theorem. Suppose that y 1 and y 2 are two solutions of (1) y +p(x)y +q(x)y = 0. Then it is possible to find constants c 1 and c 2 such that the solution (2) y = c 1 y 1 +c 2 y 2 satisfies the initial condition (3) y(t 0 ) = y 0, y (t 0 ) = y 0 if and only if W := y 1 (t 0 )y 2(t 0 ) y 1(t 0 )y 2 (t 0 ) 0 note: if solutions y 1 and y 2 are linearly dependent, then y 1 (t) = ky 2 (t) and W = 0; H.J.Eberl - MATH*

19 Proof. To find constants c 1 and c 2 such that y(t) satisfies the initial conditions means that we have to be able to solve the system of two equations for c 1 and c 2 c 1 y 1 (t 0 )+c 2 y 2 (t 0 ) = y 0, c 1 y 1(t 0 )+c 2 y 2(t 0 ) = y 0. Note that y 1 (t 0 ),y 2 (t 0 ),y 0,y 1(t 0 ),y 2(t 0 ),y 0 are given and only c 1 and c 2 are unknown Recall Cramer s Rule that says that solutions c 1 = y 0 y 2(t 0 ) y 0y 2 (t 0 ) y 1 (t 0 )y 2 (t 0) y 1 (t 0)y 2 (t 0 ) c 2 = which is possible if and only if y 0 y 1(t 0 )+y 0y 1 (t 0 ) y 1 (t 0 )y 2 (t 0) y 1 (t 0)y 2 (t 0 ) W := y 1 (t 0 )y 2(t 0 ) y 1(t 0 )y 2 (t 0 ) 0 q.e.d. H.J.Eberl - MATH*

20 Remark. the quantity W(t) = y 1 (t)y 2(t) y1 (t)y 2 (t) is called the Wronskian or the Wronskian determinant of y 1 and y 2 ; it can be written as ( ) y1 (t) y W(t) = det 2 (t) y 1(t) y 2(t) in terms of Linear Algebra or Matrix Algebra we can say: the statement or the assertion of our Theorem is true if and only if the matrix is non-singular ( ) y1 (t 0 ) y 2 (t 0 ) y 1(t 0 ) y 2(t 0 ) H.J.Eberl - MATH*

21 Theorem. Let y 1 and y 2 be two solutions of (1) y +p(x)y +q(x)y = 0. Then the family of solutions (2) y = c 1 y 1 +c 2 y 2 with arbitrary coefficients c 1 and c 2 includes every solution of (1) if and only if there is a point t 0 where the Wronskian of y 1 and y 2 is not zero. Proof. See Boyce & di Prima, p Theorem/Definition. y 1 and y 2 are linearly independent if and only if there is a point t 0 where the Wronskian of y 1 and y 2 is not zero. One calls then y 1 (t),y 2 (t) a fundamental system of (1), and (2) the general solution. H.J.Eberl - MATH*

22 Examples. suppose y 1 (t) = e r 1t and y 2 (t) = e r 2t are two solutions of a homogeneous second order linear equations, then they form a fundamental system if r 1 r 2 : ( e r 1 t e W(t) = det r 2t re r 1t re r 2t ) = (r 2 r 1 )exp[(r 1 +r 2 )t] which vanishes only if r 1 = r 2 H.J.Eberl - MATH*

23 Example: 2t 2 y +3ty y = 0 two solutions of this equation are y 1 (t) = t, y 2 (t) = 1/t we can verify this by substitution for now; later we will solve this equations y 1 (t) and y 2 (t) are linearly independent: W(t) = det which for t > 0 does not vanish ( ) t 1/t 1/(2 y) 1/t 2 = 3 2 t 3/2 this is a good example that two different solutions of one equations might not look anything alike H.J.Eberl - MATH*

24 Example: 2t 2 y +3ty y = 0, t > y1(t)=1/t y2(t)= t^0.5 1/x+x^0.5 6 y x H.J.Eberl - MATH*

25 Reduction of Order for Homogeneous 2nd Order Linear ODEs recall form Part II that the nonlinear 2nd order ODEs F(t,y,y ) = 0 and F(y,y,y ) = 0 could be reduced to 1st order ODEs and attempted to be solved with elementary techniques we show now how to obtain a second, linearly independent, solution of a homogeneous 2nd order ODE from a first solution by reduction to a homogeneous 1st order ODE as always, whether this is a viable approach to determine y 2 in terms of elementary functions depends on the integrals that appear during the procedure H.J.Eberl - MATH*

26 Theorem. Let y 1 (t) be a solution of the second order homogeneous differential equation (1) y +p(t)y +q(t)y = 0 on an interval I IR such that y 1 (t) 0 for t J I. Let u(t) be a non-constant solution of the differential equation (2) u + on the interval J. ( ) 2 y 1(t) y 1 (t) +p(t) u = 0 Then the function y 2 (t) := y 1 (t)u(t) is a solution of (1) in J that is linearly independent of y 1. H.J.Eberl - MATH*

27 Remarks equation (2) is formally a 2nd order ODE, but it reduces to a first order ODE: Let v(t) := u (t), then (2) becomes ( ) (3) v + 2 y 1(t) y 1 (t) +p(t) v = 0 equation (3) is a linear 1st order ODE that can be formally solved with integrating factors; we have ( ( ) (4) v(t) = exp 2 y 1(t) )dt+c y 1 (t) +p(t) 1 the function u of (2) is then obtained by integration ( ( ) u(t) = v(t)dt = exp 2 y 1(τ) y 1 (τ) +p(τ) ) dτ dt sometimes it is easier to not use these formulae but to follow directly the steps that we will outline in the proof below... in many cases things cancel out along the way!!! H.J.Eberl - MATH*

28 Proof of the Theorem. From y 2 = uy 1 we obtain y 2 = y 1u+y 1 u and y 2 = y 1u+2y 1u +y 1 u. Using this we find y 2 +p(t)y 2 +q(t)y 2 = y 1u+2y 1u +y 1 u +p(t)y 2 +q(t)y 2 = y 1u+2y 1u +y 1 u +p(t)(y 1u+y 1 u )+q(t)y 1 u = u(y 1 +p(t)y 1 +qy 1 )+u (2y 1 +p(t)y 1 )+y 1u Since y 1 solves (1) we have y 1 +p(t)y 1 +qy 1 = 0 and, therefore, y 2 +p(t)y 2 +q(t)y 2 = y 1 u +(2y 1 +p(t)y 1 )u. Hence, y 2 solves (1), if and only if u solves y 2 +p(t)y 2 +q(t)y 2 = 0, u + ( ) 2 y 1(t) y 1 (t) +p(t) u = 0. If u is not constant then, because of y 2 (t) = u(t)y 1 (t) we find that y 1 and y 2 are not proportional, hence, linearly independent. q.e.d. H.J.Eberl - MATH*

29 Example: The Legendre Differential Equation (1) y 2x x 2y 1 x2y = 0, for x < 1 we can guess (and verify easily) that y 1 (x) = x solves (1) we have y 1 (x) 0 on (0,1) a second solution is y 2 (x) = xu(x) where ( 2 u + x 2x ) 1 x 2 u = 0. solving this differential equation for u we find ( ) 2 u (x) = exp x dx+ 2x 1 x 2dx = exp ( 2lnx ln(1 x 2 ) ) = = 1 x 2 (1 x 2 ) = 1 x = ( 1 1 x x ) H.J.Eberl - MATH*

30 Example: The Legendre Differential Equation (cont d) and by integration u(x) = u dx = 1 x ln 1+x 1 x and finally the new solution of (1) y 2 (x) = xu(x) = x 2 ln 1+x 1 x 1 we derivedy 2 for theinterval(0,1) butwe can verifythatit actuallysolves (1) on the entire interval ( 1,1) H.J.Eberl - MATH*

31 Example: The Legendre Differential Equation (cont d) y1(x)=x u(x) y2=y1(x)*u(x) y x y 2x x 2y 1 x2y = 0, for x < 1 H.J.Eberl - MATH*

32 Example: 2t 2 y +3ty y = 0, t > 0 (Boyce & diprima, p.171) a first solution is y 1 (t) = t 1 a second one is y 2 (t) = u(t)t 1... we follow now directly the steps we took in the proof we find y 2 = u t u y t2, 2 = u t 2u t 2 = 2u t 3 substituting these expression into to ODE and collecting terms we find 2t 2 ( u t 2u t 2 +2u t 3 ) ( u +3t t u ) t 2 u t =... = 2tu u = 0 we find now u as solution of the separable differential equation for v := u v = v 2t H.J.Eberl - MATH*

33 Example: 2t 2 y +3ty y = 0, t > 0 (Boyce & diprima, p.171) Thus v = u = c t = u = 2 3 ct3/2 +k and y 2 (t) = u(t)y 1 (t) = 2 3 c t+ k t note that the second term is a multiple of y 1 ; therefore y 2 (t) = t would be a (simpler) linearly independent second solution H.J.Eberl - MATH*

34 Example: 2t 2 y +3ty y = 0, t > 0 (Boyce & diprima, p.171) 10 8 y1(t)=1/t y2(t)= t^0.5 1/x+x^0.5 6 y x H.J.Eberl - MATH*

35 The Riccati Differential Equation we consider the nonlinear first order ODE y = q 0 (x)+q 1 (x)y +q 2 (x)y 2 named after Jacopo Francesco Riccati (28 May April 1754) Riccati equations arise in many areas of Pure Mathematics, but also in many application, such as control theory, urban hydrology, etc for many special cases, i.e. particular coefficient functions, closed form solutions of the Riccati equation can be found another alternative approach to study Riccati equations is to reduce them to a (1st order) Bernoulli equation actually, if q 0 (x) 0, then the Riccati equation is a Bernoulli equation H.J.Eberl - MATH*

36 Reduction to a 2nd order homogenous linear equation we consider the case q 2 (x) 0 in I and the Riccati equation (1) y = q 0 (x)+q 1 (x)y +q 2 (x)y 2 as well as the related homogeneous second order linear equation (2) u ( ) q 1 (x)+ q 2(x) u +q 2 (x)q 0 (x)u = 0 q 2 (x) a solution of the Riccati equation (1) can be found then as (3) y(x) = u (x) q 2 (x)u(x) where u is a solution of (2) H.J.Eberl - MATH*

37 Reduction to a 2nd order homogenous linear equation (cont d) to verify this differentiate (3) to find (4) y = u q 2 u = u ( ) q 2 q2 2u + u qu 2 substitute (3) and (4) into (1) to find ( ) u q 2 q 2 u +u q2 2u + u u = q qu 2 0 q 1 q 2 u + (u ) 2 q 2 u multiply by q 2 u and collect terms to find that u has to satisfy u +u q 2 q 2 = q 0 q 2 u q 1 u re-arrange to find (2) u ( ) q 1 (x)+ q 2(x) u +q 2 (x)q 0 (x)u = 0 q 2 (x) H.J.Eberl - MATH*

38 Example: the logistic equation y = µ(1 y/k) the logistic equation is a Riccati equation with q 0 = 0,q 1 = µ,q 2 = µ/k the second order ODE (2) is and has solution u µu = 0 u = C 1 e µt = u = C 1 µ eµt +C 2 we find then for the solution of the logistic equation y = u µ K u = µ K C 1 eµt ( C1 µ eµt +C 2 ) = to satisfy the initial conditions y(0) = y 0 we find K 1+µC 2 /C 1 = y 0 or C 1 Ke µt C 1 e µt +µc 2 = Keµt e µt +µ C 2 C 1 C 2 = K y 0 C 1 y 0 µ H.J.Eberl - MATH*

39 Outlook: Homogeneous and nonhomogeneous equations consider the nonhomogeneous equation (1) y +p(t)y +q(t)y = g(t) and its homogeneous part (2) y +p(t)y +q(t)y = 0 if y n (t) is a solution of (1) and y h (t) is a solution of (1), then z(t) = y n (t)+y h (t) is a solution of (1). Proof: substitute z into (1), expand, rearrange, and use that y h solves (2) H.J.Eberl - MATH*

40 III Second Order Linear ODEs III.2 Equations with Constant Coefficients Boyce & DiPrima, Chapter 3.1, 3.3, 3.4 H.J.Eberl - MATH*

41 Problem formulation we consider the 2nd order homogeneous linear equation with constant coefficients y +py +qy = 0 two simple special cases for which we can guess solutions are (i) y +qy = 0, (ii) y +py = 0 (i) y = exp( pt) is a solution if p > 0 (i) y = sin( pt) and y = cos( pt) are solutions if p < 0 (ii) y = exp( pt) = y = exp( pt)/p is a solution let s see whether the general problem (1) with p 0 q also permits exponential functions as solutions, and if so under which conditions on the parameters H.J.Eberl - MATH*

42 Ansatz: y = e rt with this ansatz we obtain y = e rt = y = re rt = y = r 2 e rt substituting into (1) we find y +py +qy = r 2 e rt +pre rt +qe rt = = ( r 2 +pr+q ) e r t thus, for y = e rt to be a solution of y +py +qy = 0 r needs to be a root of the characteristic polynomial r 2 +pr +q = 0 we have r 1,2 = p± p 2 4q 2 H.J.Eberl - MATH*

43 Case I: Two Distinct Real Roots we consider the 2nd order homogeneous differential equation with constant coefficients (1) y +py +qy = 0 and characteristic polynomial with roots r 2 +pr +q = 0 r 1,2 = p± p 2 4q 2 if p 2 4q > 0 these are real and distinct we verify that the functions y 1,2 (t) = e r 1,2t are solutions of (1) H.J.Eberl - MATH*

44 Case I: Two Distinct Real Roots we have y 1,2 = e r 1,2t = y 1,2 = r 1,2 e r 1,2t = y 1,2 = r 2 1,2e r 1,2t substituting into (1) we find y 1,2 +py 1,2 +qy 1,2 = e r 1,2t ( r 2 +pr 1,2 +q ) = 0 y 1 and y 2 are linearly independent and form a fundamental system the general solution is y(t) = c 1 e r 1t +c 2 e r 2t H.J.Eberl - MATH*

45 Example: transport and decay of a dissolved substrate time-independent convective and diffusive transport and first order decay of a dissolved substrate is described by the differential equation vc = Dc kc where the dependent variable is the substrate concentration c [g/m 3 ], the independent variable is the position along the river x [m]; the parameters are all positive: D [m 2 /d] is the diffusion coefficient, v [m/d] the velocity of the stream, and k [1/d] the decay rate we bring this into standard form the characteristic polynomial is c v D k D c = 0 r 2 v D r k D = 0 H.J.Eberl - MATH*

46 Example: transport and decay of a dissolved substrate with roots r 1,2 = v ( D ± v ) 2 D +4 k D 2 because all parameters are positive, these are real and distinct; in fact r 1 > 0, r 2 < 0 we have the linearly independent solutions c 1,2 (t) = exp(r 1,2 x) = exp v ( D ± v ) 2 D +4 k D 2 x and the general solution c(t) = α 1 c 1 (t)+α 2 c 2 (t) H.J.Eberl - MATH*

47 Example: transport and decay of a dissolved substrate suppose the following initial conditions are given: upstream concentration c(0) = c 0 > 0 upstream flux vc(0) Dc (0) = j 0 > 0 the flux condition can be replaced by a condition on the derivative only c (0) = (vc 0 j 0 )/D =: c 0 we have then because of y 1,2 (0) = 1 α 1 +α 2 = c 0 r 1 α 1 +r 2 α 2 = c 0 thus α 2 = c 0 r 1 c 0 r 2 r 1 α 1 = c 0 c 0 r 1 c 0 r 2 r 1 H.J.Eberl - MATH*

48 Case II: Complex Roots again we consider the 2nd order homogeneous differential equation with constant coefficients (1) y +py +qy = 0 and characteristic polynomial with roots r 2 +pr +q = 0 r 1,2 = p± p 2 4q 2 if p 2 4q < 0 these are complex conjugate, r 1,2 = λ±iµ can these complex roots help us finding real solutions of (1)? H.J.Eberl - MATH*

49 Case II: Complex Roots and Euler s Formula recall the Taylor series expansion of the exponential function e z = k=0 z k k! for imaginary z = iµ this is e iµ = k=0 z k k! = k=0 ( 1) k µ 2k (2k)! +i k=1 ( 1) k 1 µ 2k 1 (2k 1)! both series on R.H.S. are real, in fact the first sum is the Taylor series of cosµ, the second one of sinµ this suggests e λ+iµ = e λ (cosµ+isinµ) this is known as Euler s formula H.J.Eberl - MATH*

50 Case II: Complex Roots and Euler s Formula with this in mind, the complex conjugate roots r 1,2 = p 2 }{{} λ define two complex-valued functions ± i 1 2 4q p 2 }{{} µ z 1,2 (t) = e λt( cos(µt)±isin(µt) ) and from this we can build the linear combinations z 1 (t)+z 2 (t) = 2e λt cos(µt), z 1 (t) z 2 (t) = 2ie λt sin(µt) z 1 (t)+z 2 (t) is a real function and z 1 (t) z 2 (t) a purely imaginary one this suggests the following real functions as candidates for solutions of (1): y 1 (t) = z 1 (t)+z 2 (t) and y 2 (t) = i(z 1 (t) z 2 (t)) H.J.Eberl - MATH*

51 Case II: Complex Roots and Fundamental Solutions y 1 (t) = e λt cos(µt) y 1(t) = e λt (λcos(µt) µsin(µt)) y 1(t) = λe λt (λcos(µt) µsin(µt))+e λt( λµsin(µt) µ 2 cos(µt) ) substituting into (1) we find y 1 +py 1 +qy 1 = [( λ 2 µ 2 +pλ+q ) cos(µt)+sin(µt)(µ[ 2λ+p]) ] e λt using λ = p/2 and µ = 1 2 4q p2 yields indeed y 1 +py 1 +qy 1 = 0 hence, we found a first solution y 1 (t) = e λt cos(µt) = e p 2 t cos ( ) q p2 4 t H.J.Eberl - MATH*

52 Case II: Complex Roots and Fundamental Solutions y 2 (t) = e λt sin(µt) y 2(t) = e λt (λsin(µt)+µcos(µt)) y 2(t) = λe λt (λsin(µt)+µcos(µt))+e λt( λµcos(µt) µ 2 sin(µt) ) substituting into (1) we find y 2 +py 2 +qy 2 = [( λ 2 µ 2 +pλ+q ) sin(µt)+cos(µt)(µ[ 2λ+p]) ] e λt using λ = p/2 and µ = 1 2 4q p2 yields indeed y 2 +py 2 +qy 2 = 0 hence, we found a second solution y 2 (t) = e λt sin(µt) = e p 2 t sin ( ) q p2 4 t H.J.Eberl - MATH*

53 Case II: Complex Roots and the General Solution y 1 (t) and y 2 (t) are linearly independent the general solution of (1) is y(t) = e λt (c 1 cos(µt)+c 2 sin(µt)) this is an oscillating function; if p > 0 the oscillations are damped and lim t y(t) = 0, if p > 0 then y blows up H.J.Eberl - MATH*

54 Example: Damped and undamped free vibrations, see Boyce & diprima, p.191 the vertical movement of a mass on a spring is described by (1) my +γy +ky = 0 where the independent variable is time t and the dependent variable is the location y relative to the position at rest; (1) is obtained as a force balance accounting for gravity, the spring force and a damping force; all parameters are positive, m is the mass of the object, k describes the stiffness of the spring and γ is a damping coefficient the characteristic polynomial is mr 2 +γr+k = 0 and roots r 1,2 = γ ± γ 2 4km 2m H.J.Eberl - MATH*

55 Example: undamped free vibrations we consider first the case of undamped vibrations, γ = 0: obviously 4km < 0 and the roots are purely imaginary pair k r 1,2 = ±i m = λ = 0, µ = k m a pair of linearly independent basis functions is k k y 1 (t) = cos( m t), y 2(t) = sin( m t) and the general solution k k y(t) = c 1 cos( m t)+c 2sin( m t) these are periodic solutions with frequency k/m H.J.Eberl - MATH*

56 Example: undamped free vibrations y1 y2 y=y1+y2 0.5 y t/sqrt{k/m} my +ky = 0 H.J.Eberl - MATH*

57 Example: damped free vibrations accounting for some damping γ > 0 leads to complex conjugate roots set r 1,2 = γ ± γ 2 4km 2m λ := γ 2m, µ := to obtain the fundamental solutions and the general solution k ( γ ) 2 m 2m y 1 (t) = e λt cosµt, y 2 (t) = e λt sinµt y(t) = c 1 e λt cosµt+c 2 e λt sinµt finally, if 4km < γ 2 the roots r 1,2 are real and distinct with r 1 > 0 < r 2 and r 2 > r 1 and the general solution becomes y(t) = c 1 e r 1t +c 2 e r 2t H.J.Eberl - MATH*

58 Example: damped free vibrations y1 y2 y1+y2 0.5 y t/sqrt{k/m} mr 2 +γr +k = 0 H.J.Eberl - MATH*

59 Example: damped free vibrations, varying γ y t/sqrt{k/m} mr 2 +γr +k = 0 H.J.Eberl - MATH*

60 Case III: Repeated Roots again we consider the 2nd order homogeneous differential equation with constant coefficients y +py +qy = 0 and characteristic polynomial r 2 +pr +q = 0 with roots r 1,2 = p± p 2 4q 2 the remaining case is p 2 = 4q and r 1,2 = p 2 in this case solutions e r 1t and e r 2t are (real and) identical, i.e. not independent recall that given one solution, say y 1, a second linearly independent solution can be obtained by reduction of the 2nd order equation to a 1st order equation H.J.Eberl - MATH*

61 Case III: Repeated Roots (cont d) this second, linearly independent, solution is given by y 2 = uy 1 (x) where u solves ( ) u + 2 y 1(t) y 1 (t) +p u = 0 using that in our case y 1 (t) = e rt = y 1(t) = re rt this becomes u +(2r +p)u = 0 using that r = p/2 this simplifies to thus, u = const and u = const t u = 0 finally, we obtain the new solution y 2 (t) = te rt the general solution is then y(t) = c 1 e rt +c 2 te rt H.J.Eberl - MATH*

62 Example: Damped free vibrations revisited we return to my +γy +ky = 0 with and roots r 1,2 = γ ± γ 2 4km 2m and the remaining case is γ 2 = 4km with solutions y 1 (t) = e γt/2, y 2 = te γt/2 = y(t) = c 1 e γt/2 +c 2 te γt/2 H.J.Eberl - MATH*

63 Example: damped free vibrations y1 y2 y1+y2 0.6 y t/2gamma mr 2 +γr+k = 0,γ 2 = 4km H.J.Eberl - MATH*

64 Summary I the behaviour of constant coefficient second order linear equations y +py +qy = 0 is determined by the roots of the characteristic polynomial, namely r 2 +rp+q = 0, r 1,2 = p± p 2 4q 2 if r 1 and r 2 are real and distinct, then the general solution is y(t) = c 1 e r 1t +c 2 e r 2t if r 1 = r 2 =: r is a double root, then the general solution is y(t) = c 1 e rt +c 2 te rt if r 1 and r 2 are complex, then we can write them as r 1,2 = λ±iµ and the general solution is y(t) = c 1 e λt cosµt+c 2 e λt sinµt H.J.Eberl - MATH*

65 Summary II the behaviour of constant coefficient second order linear equations y +py +qy = 0 is determined by the roots of the characteristic polynomial, r 1,2 = p± p 2 4q 2 the quantity θ := p 2 4q is a bifurcation parameter: θ > 0: the fundamental solutions are monotonic exponential functions θ < 0: the fundamental solutions oscillate in the case θ < 0 the quantity λ = p/2 is another bifurcation parameter λ < 0: solutions dampen off exponentially λ > 0: solutions amplify exponentially H.J.Eberl - MATH*

66 III Second Order Linear ODEs III.3 Some nonhomogeneous equations and the method of undetermined coefficients Boyce & DiPrima, Chapter 3.5 H.J.Eberl - MATH*

67 Preliminaries we will now find out how to solve second order linear equations of type y +py +qy = g(t) where p, q are constant coefficients and g(t) is an exponential, trigonometric, polynomial function or certain combinations of those before we embark on this we want to recall some major results from before reading week: the superposition principle for homogeneous equations and what it implies for non-homogeneous equations in the homogenous case we found the general solution by guessing the form of the solution and then determining its coefficients such that the differential equation is satisfied H.J.Eberl - MATH*

68 Superposition for homogeneous equations and implications for non-homogeneous equations let y 1 (t), y 2 (t) be solutions of the homogeneous problem (1) y +py +qy = 0 and Y(t) be a solutions of the nonhomogeneous problem (2) y +py +qy = g(t) then z(t) = Y(t)+c 1 y 1 (t)+c 2 y 2 (t) is also a solution of (2). H.J.Eberl - MATH*

69 Superposition and implications for (2) to show this, we take z (t) = Y (t)+c 1 y 1(t)+c 2 y 2(t) and z (t) = Y (t)+ c 1 y 1(t)+c 2 y 2(t) and substitute into (2) z +pz +qz = = Y +c 1 y 1 +c 2 y = Y +py +qy }{{} =g(t) = g(t) 2 +p(y +c 1 y 1 +c 2 y 2 )+q(y +c 1 y 1 +c 2 y 2 ) +c 1 (y 1 +py 1 +qy 1 ) +c 2 (y 2 +py 2 +qy 2 ) = }{{}}{{} =0 =0 hence, to solve the nonhomogeneous equation (2) y +py +qy = g(t) we have to first solve the homogeneous equation (1) y +py +qy = 0 H.J.Eberl - MATH*

70 Solutions for the homogeneous equations with constant coefficients y +py +qy = 0 we found solutions y 1, y 2 by making the ansatz y(t) = e rt based on some basic considerations the correct solution was found by finding the coefficient r such that this function indeed satisfies the differential equation ( r 2 +pr+q ) e rt = 0 }{{} =0 we found that key to the solution of the problem were the roots of the characteristic polynomial r 2 +pr +q = 0 H.J.Eberl - MATH*

71 Solutions in the nonhomogeneous case for nonhomogeneous problems of type y +py +qy = g(t) with p,q constant and for certain forcing function g(t) we can take a similar approach underlying is the observation: we are looking for a function y such that g(t) can be represented as a linear combination of y, y, and y H.J.Eberl - MATH*

72 Take I: g(t) an exponential function g(t) = ae bt if y is an exponential function, then so are y and y if y,y,y are exponential functions, then so are their linear combinations lets try the ansatz y(t) = Ae bt y(t) = Ae bt = y (t) = Abe bt = y (t) = Ab 2 e bt substituting into y +py +qy = g(t) ( ) hence ( Ab 2 +pab+qa ) e bt = ae bt Ab 2 +pab+qa = a or A = a b 2 +pb+q H.J.Eberl - MATH*

73 Take I: g(t) an exponential function g(t) = ae bt, cont d thus we found the solution of ( ) Y(t) = ae bt b 2 +pb+q this solution is also called a particular solution of ( ) to distinguish it from the general solution c 1 y 1 (t)+c 2 (t)+y(t) wherey 1 andy 2 are asystemof fundamentalsolutionsofthehomogeneous y +py +qy = 0 which we can find with out techniques from two weeks ago H.J.Eberl - MATH*

74 Example: y 2y +y = e 3t we first find the general solution of the homogeneous equation y 2y +y = 0 via the roots of its characteristic polynomial r 2 2r +1 = 0 = r 1,2 = 1 =: r we have a double root r = 1, thus the fundamental system y 1 (t) = e t, y 2 (t) = te t for the nonhomogeneous problem we make the ansatz Y(t) = Ae 3t = Y (t) = 3Ae 3t = Y (t) = 9Ae 3t substituting into the differential equation Y +2Y +Y = 9Ae 3t 2 3Ae 2t +Ae 2t = 4Ae 2t = e 2t H.J.Eberl - MATH*

75 Example: y 2y +y = e 3t, cont d thus A = 1/4 gives the particular solution Y(t) = 1 4 e 3t throwing things together, the solution of the nonhomogeneous problem y(t) = c 1 e t +c 2 te t e 3t if we aim to satisfy initial conditions y(0) = 1,y (0) = 1 we find with y (t) = c 1 e t +c 2 e t (1+t) 3 4 e 3t that y(0) = c = 1 = c 1 = 3 4 y (0) = c 1 +c = 1 = c 2 = 1 thus the solution of the initial value problem y(t) = 3 4 et te t e 3t H.J.Eberl - MATH*

76 Example: y 2y +y = e 3t, cont d 6 5 y(t)=0.75*exp(t)-t*exp(t)+0.25*exp(-3t) Y(t)= 0.25*exp(-3t) 4 3 y t H.J.Eberl - MATH*

77 Take II: g(t) a polynomial g(t) = a m t m +a m 1 t m a 1 t+a 0 if y is a polynomial of degree m, then y is a polynomial of degree m 1, and y is a polynomial of degree m 2 any linear combination of y, y and y is then a polynomial of degree m for simplicity, take a look at g(t) = at and ( ) y +py +qy = at let s make the ansatz y(t) = A 1 t+a 0 = y (t) = A 1 = y (t) = 0 and substitute into ( ) to obtain y +py +qy = pa 1 +q(a 1 t+a 0 ) = at H.J.Eberl - MATH*

78 Take II: g(t) = at, cont d thus qa 1 = a, pa 1 +qa 0 = 0 or we found a particular solution A 1 = a q, A 0 = pa q 2 Y(t) = a q t pa q 2 the same trick works for polynomials or monomials of higher degree, but it involves then more algebraic manipulation than I want to do to demonstrate the idea H.J.Eberl - MATH*

79 Example: y y = t we make the ansatz Y(t) = A 1 t+a 0 = Y (t) = A 1 = Y (t) = 0 from which Y Y = A 1 t A 0 = t thus A 1 = 1,A 0 = 0 and a particular solution Y(t) = t the general solution is then y(t) = Y(t)+c 1 y 1 (t)+c 2 y 2 (t) where y 1,y 2 are a fundamental system of the homogeneous y y = 0 H.J.Eberl - MATH*

80 Example: y y = t, cont d recall that we find y 1,y 2 from the roots of the characteristic polynomial r 2 1 = 0 = r 1,2 = ±1 as y 1 (t) = e t, y 2 (t) = e t finally the general solution of our non-homogenous problem y(t) = c 1 e t +c 2 e t t let s impose the initial conditions y(0) = 1, y (0) = 0: y (t) = c 1 e t c 2 e t 1 and, therefore, y(0) = c 1 +c 2 = 1 = c 1 = 1 c 2 y (0) = c 1 c 2 1 = 2c 2 = 0 = c 2 = 0 = c 1 = 1 and y(t) = e t t H.J.Eberl - MATH*

81 Example: y y = t, cont d y y(t)=exp(t)-t -t exp(t) t H.J.Eberl - MATH*

82 Take III: g(t) a trigonometric function, say g(t) = asinbt if y is a sin function, then so is y but y will be a cos function on the other hand, if y is a cos function, then so is y and y is a sin function if y is a combination of sin and cos then so are y and y... let s try that we make the ansatz y(t) = A 1 cosbt+a 2 sinbt = y (t) = A 1 bsinbt+a 2 bcosbt = = y (t) = A 1 b 2 cosbt A 2 b 2 sinbt substituting into the differential equation and sorting by cos and sin terms ( A1 b 2 +A 2 bp+a 1 q ) cosbt+ ( A 2 b 2 pa 1 b+qa 2 ) sinbt = asinbt thus, we obtain two equations for the unknown A 1, A 2 A 1 (q b 2 )+A 2 bp = 0, A 1 bp+a 2 (q b 2 ) = a H.J.Eberl - MATH*

83 Example: damped forced vibration my +γy +ky = δcosωt we first rewrite our problem in standard form y + γ m y + k m y = δ m cosωt then make the ansatz for the particular solution hence then Y(t) = Acosωt+Bsinωt Y (t) = ωasinωt+ωbcosωt, Y (t) = ω 2 Acosωt ω 2 Bsinωt Y + γ m Y + k m Y = ω2 Acosωt ω 2 Bsinωt + γ m ( ωasinωt+ωbcosωt)+ k m (Acosωt+Bsinωt) = [ ( ) k = A m ω2 + ωγ ] [ ( ) k m B cosωt+ B m ω2 ωγ ] m A sinωt H.J.Eberl - MATH*

84 Example: damped forced vibration my +γy +ky = δcosωt cont d for this to be a solution of our damped force vibration model we need to choose A and B such that [ ( ) k A m ω2 + ωγ ] m B cosωt+ [ ( ) k B m ω2 this leads to a linear system of the form aa+bb = 1 ba+ab = 0 with a = 1 ( k mω 2 ), b = ωγ δ δ the particular solution Y is then found as with Y(t) = Acosωt+Bsinωt B = 1 a 2 b +b, A = a b B ωγ ] m A sinωt = δ m cosωt H.J.Eberl - MATH*

85 Example: damped forced vibration my +γy +ky = δcosωt cont d the general solution is found as y(t) = Y(t)+c 1 y 1 (t)+c 2 y 2 (t) where y 1,y 2 are a fundamental system of the homogeneous equation my +γy +ky = 0 we found two weeks back, for γ not too large, y 1 (t) = e γ 2 t sin k m γ2 4m 2t, y 2(t) = e γ k 2 t cos m γ2 4m 2t H.J.Eberl - MATH*

86 Example: damped forced vibration my +γy +ky = δcosωt cont d for illustration purposes, we take m = γ = k = ω = δ = 1 whence, and y(t) = c 1 e 1 2 t sin a = 0, b = 1 = A = 0, B = 1 ( ) 3 4 t +c 2 e 1 2 t cos ( ) 3 4 t +sint for simplicity lets choose c 1 = c 2 = 1 H.J.Eberl - MATH*

87 Example: damped forced vibration my +γy +ky = δcosωt cont d y y(t)=y(t)+y1(t)+y2(t) t H.J.Eberl - MATH*

88 Overview of basic functions g(t) for which the method works forcing term g(t) ae bt ansatz y(t) Ae bt at m A m t m +A m 1 t m A 1 t+a 0 acosbt asinbt A 1 cosbt+a 2 sinbt A 1 cosbt+a 2 sinbt (a m t m +...a 1 t+a 0 )e ax cosbt e ax [(A m t m +...A 1 t+a 0 )cosbt+ +(B m t m +...B 1 t+b 0 )sinbt] (a m t m +...a 1 t+a 0 )e ax sinbt e ax [(A m t m +...A 1 t+a 0 )cosbt+ +(B m t m +...B 1 t+b 0 )sinbt] the first four cases, as well as sin and cos alone can be considered special cases of the last two H.J.Eberl - MATH*

89 Observations and comments I: additivity let us consider two forcing terms g 1 (t) and g 2 (t) and let Y 1 and Y 2 be two particular solutions of y +py +qy = g 1 (t) y +py +qy = g 2 (t) then Y(t) = Y 1 (t) + Y 2 (t) is the particular solution that belongs to the forcing term g 1 (t)+g 2 (t), i.e Y(t) is a solution of y +py +qy = g 1 (t)+g 2 (t) to verify this, substitute Y(t) = Y 1 (t)+y 2 (t) into the L.H.S of the differential equation this observation may allow us to reduce the solution of the problem y +py +qy = g(t) with a more complicated g(t) = g 1 (t) g n (t) to solving n smaller problems with simpler R.H.S.s g i (t), i = 1,...,n H.J.Eberl - MATH*

90 Example: y y = t+cost we can write g(t) = t+cost as g(t) = g 1 (t)+g 2 (t), g 1 (t) = t, g 2 (t) and solve the two problems (1) (2) y y = t y y = cost from before: a fundamental system of the homogeneous equation is y 1 (t) = e t, y 2 (t) = e t this is the same for (1) and (2) from before: a particular solution of (1) is Y 1 (t) = t we still need to find a particular solution Y 2 of (2) H.J.Eberl - MATH*

91 Example: y y = t+cost, cont d to solve y y = cost we make the ansatz whence Y 2 (t) = Asint+Bcost Y 2(t) = Acost Bsint = Y 2 (t) = Asint Bcost substituting into the differential equation (2) hence, A = 0,B = 1 2 and overall we find the solution Y 2 (t) Y 2 (t) = 2Asint 2Bcost = cost Y 2 (t) = 1 2 cost y(t) = c 1 e t +c 2 e t t 1 2 cost H.J.Eberl - MATH*

92 Example: y y = t+cost, cont d to verify that this is indeed the correct solution, take the derivative: y (t) = c 1 e t c 2 e t sint and build y (t) = c 1 e t +c 2 e t cost y y = c 1 e t +c 2 e t cost ( c 1 e t +c 2 e t t 1 2 cost )+t+cost = t+cost H.J.Eberl - MATH*

93 Observations and comments II: overlap of homogeneous and nonhomogenous problem consider the equation y y = e t in this case our normal ansatz would be Y(t) = Ae t but this is a solution of the underlying homogeneous equation = the method breaks down. instead, let s try Y(t) = Ate t = Y (t) = Ae t (1 t) = Y (t) = Ae t (2 t) and find Y Y = 2Ae t = e t = A = 1 2 = Y(t) = t 2 e t H.J.Eberl - MATH*

94 Observations and comments II, cont d this gives a workaround for the more general problem y +py +qy = e rt where r is also a root of the characteristic polynomial, i.e. r 2 +pr +q = 0 the ansatz is now Y(t) = At s e rt where s = 1 if r is a single root, s = 2 if r is a double root that this indeed works in a general manner can be verified by substituting the Y(t) intothe differentialequationand usingthat r is a root tosimplify H.J.Eberl - MATH*

95 Observations and comments II, cont d in the case of a single root we have Y(t) = Ate rt = Y (t) = Ae rt (1+rt) = Y (t) = Ae rt( 2r +r 2 t ) substituting into y +py +qy and sorting by terms in t gives Y +py +q = Ae rt t(r 2 +pr+q) }{{} =0 +2r +p = e rt hence, if 2r p, the choice A = 1/(2r +p) yields a particular solution if 2r = p, we have a double root and make the ansatz Y(t)= At 2 e rt = Y (t)= Ae rt( 2t+rt 2) = Y (t)= Ae rt( r 2 t 2 +4rt+2 ) substituting into y +py +qy gives, after sorting by powers of t Y +py +qy = Ae rt and we find A = 1/2 t 2 (r 2 +pr+q) }{{} =0 +2t(2r +p) }{{} =0 +2 = e rt H.J.Eberl - MATH*

96 Observations and comments III: method is self-correcting if the ansatz chosen has too many degrees of freedom, it will lead to many coefficients being 0; the result will still be correct but unnecessary work was done if the ansatz chosen has not enough degrees of freedom (or is otherwise wrong), the method will get stuck; it will not yield a wrong result but it will become clear where additional degrees of freedom are required in any case, if the calculations are performed correctly, the method will not deliver a wrong result H.J.Eberl - MATH*

97 III Second Order Linear ODEs III.4 More nonhomogeneous equations and variation of constants Boyce & DiPrima, Chapter 3.6 H.J.Eberl - MATH*

98 Preliminaries on y +p(t)y +q(t)y = g(t) so far we know how to solve 2nd order linear equations with constant coefficients p, q and special types of forcing terms g(t) the Method of Undetermined Coefficients is based on knowing the form of the solution beforehand not a generally applicable method, but conceptually easy to use (although calculations might become cumbersome) we will now discuss Variation of the Constants more generally applicable, but more complicated; still requires us to know a fundamental system of the homogeneous equation both methods complement each other this method is due to Lagrange H.J.Eberl - MATH*

99 Joseph-Louis Lagrange ( ) born as Giuseppe-Lodovico Lagrangia in Turin, died in Paris inventor of the metric decimal system calculus of variations, optimization and control, analytical mechanics buried in the Pantheon If I had inherited a fortune I should probably not have cast my lot with mathematics H.J.Eberl - MATH*

100 Variation of the constants. Let y 1 and y 2 be linearly independent solutions of the homogenous second order equation y +p(t)y +q(t)y = 0 where p(t),q(t) are continuous functions on some interval I IR. Then a particular solution Y(t) of the nonhomogeneous equation y +p(t)y +q(t)y = g(t) with continuous forcing function g(t) on I is given by y2 (t)g(t) Y(t) = y 1 (t) W(t) where W(t) bis the Wronskian of y 1 and y 2, dt+y 2 y1 (t)g(t) W(t) W(t) = y 1 (t)y 2(t) y 1(t)y 2 (t). dt H.J.Eberl - MATH*

101 Formal proof of method. We take the derivatives of y2 (t)g(t) y1 (t)g(t) Y(t) = y 1 (t) dt+y 2 dt. W(t) W(t) Y y 2 (t)g(t) = y 1 y 1 y2 (t)g(t) dt+y y1 (t)g(t) y 1 (t)g(t) 2 dt+y 2 W(t) W(t) W(t) W(t) = y 1 y2 (t)g(t) dt+y 2 y1 (t)g(t) dt W(t) W(t) Y = y 1 y 2 (t)g(t) y 1 y2 (t)g(t) dt+y y1 (t)g(t) 2 dt+y y 1 (t)g(t) 2 W(t) W(t) W(t) W(t) }{{}}{{} =:u =:v and substitute this into Y +py +qy, re-arrange, and sort by u and v: Y +py +qy = v(y 2 +py 2 +qy 2 ) }{{} =0 = g(t) u(y 1 +py 1 +qy 1 ) + }{{} =0 W(t) {}}{ y 2y 1 y 1y 2 W(t) g(t) q.e.d H.J.Eberl - MATH*

102 Comments still requires us to know a fundamental system y 1,y 2, which for nonconstant p(t), q(t) can be a severe limitation for practical applicability recall that reduction of order sometimes can help in finding a solution of the homogeneous system y 1 and y 2 are linearly independent, so W(t) 0 and the integrands are well defined two integrals need to be evaluated; how easy this is depends on the functions involved; this will be a second severe limitation for practical applicability H.J.Eberl - MATH*

103 Motivation/derivation recall: if y 1,y 2 are lin. indep. sol s of the homogeneous equation, then y(t) = cy 1 (t)+c 2 y 2 (t) for any constants c 1 and c 2 is a solution of the homogeneous equation can we derive a solution of the inhomogeneous solution by varying the constant in every t, i.e. can we find functions u(t), v(t) such that ( ) Y(t) = u(t)y 1 (t)+v(t)y 2 (t) is a solution of the inhomogeneous equation? if we substitute ( ) into the differential equation then (i) the resulting expression will contain u, v,u, v (ii) we have one equation for two unknown functions (iii) u(t), v(t) have to depend on g(t) question: can we use the degree of freedom in (ii) to find an additional condition so that u and v can be determined? answer: yes, but it s not pretty H.J.Eberl - MATH*

104 Motivation/Derivation cont d revisiting the formal proof we see that a major simplification was that for our chosen functions u, v the expression of Y was simplified by (1) y 1 u +y 2 v = 0 which is such additional condition on u,v our problem reduces thus to finding u,v such that (1) and (2) y 1u +y 2v = g for each t, (1) and (2) are a system of two linear equations for two unknowns u and v which can be solved, e.g. with Cramer s rule u and v are then obtained by integration in many cases solving (1), (2) and integrating might be easier than substituting into the formula of the theorem H.J.Eberl - MATH*

105 Example: y +y = tant, π 2 < t < π 2 the homogeneous equation has constant coefficients; with the methods of III.2 we find the linearly independent solutions in this example we have y 1 (t) = cost, y 2 (t) = sint g(t) = tant = sint cost to which the method of undetermined coefficients of III.3 does not apply = try variation of the constants the ansatz for a particular solution is to find u,v we need to solve Y(t) = u(t)cost+v(t)sint u (t)cost+v (t)sint = 0 u (t)sint+v (t)cost = tant H.J.Eberl - MATH*

106 Example: y +y = tant, π 2 < t < π 2, cont d we find u (t) = tantsint, v (t) = tantcost = sint and by integration (likely using an integral table) sin 2 t u(t) = tantsintdt = dt = sint ln cost v(t) = sintdt = cos+c 2 for the solution Y(t) = u(t)y 1 (t)+v(t)y 2 (t) after substituting and some algebra Y(t) = cost ln 1+sint cost and, thus, the general solution y(t) = c 1 tantsint+c 2 sint cost ln 1+sint cost 1+sint cost +C 1 H.J.Eberl - MATH*

107 Example: y +y = tant, π 2 < t < π 2, cont d 10 8 Y(x) y1(x) y2(x) y1(x)+y2(x)+y(x) H.J.Eberl - MATH*

108 Example: Euler-Cauchy equation (1) x 2 y +axy +by = g(t) first we need to solve the homogeneous equation (2) x 2 y +axy +by = 0, which is known as Euler-Cauchy equation inspection of (2) suggests that we might want to test for polynomial solutions (recall: if we differentiate powers x m the order decreases by 1, if we take then the second derivative it decreases again by order 1) we make the ansatz y = x m = y = mx m 1 = y = m(m 1)x m 2 substitution inton (2) gives m(m 1)x m +amx m +b = 0 H.J.Eberl - MATH*

109 Example: Euler-Cauchy equation cont d for x 0 we find thus that m has to be chosen such that (3) m 2 +(a 1)m+b = 0 or m 1,2 = 1 a± (a 1) 2 4b 2 if (a 1) 2 > 4b then we find two distinct m, and thus linearly independent solutions y 1 = x m 1, y 2 = x m 2 if (a 1) 2 = 4b we find one solution y 1 = x m and can obtain a second one be reduction of order if (a 1) 2 < 4b then we have two complex conjugate roots, m 1,2 = µ±iν; we can write x m 1,2 = x µ x ±iν, write x ±iν = exp(±iνlnx) apply Euler s formula and find y 1 = x µ cos(νlnx), y 2 = x µ sin(νlnx) H.J.Eberl - MATH*

110 nonhomogeneous Example: Euler-Cauchy equation cont d specifically, we we look at x 2 y +4xy +2y = g(x), x > 0 with m 1,2 = 1 a± (a 1) 2 4b 2 = 3± = { 1 2 thus y 1 (x) = 1 x, y 2(x) = 1 x 2 = y 1(x) = 1 x 2, y 2(x) = 2 x 3 to find a particular solution by variation of the constant we first have to bring our equation into normal form y +p(x)y +q(x) = g(x): we divide by x 2 and obtain y +4 y x +2 y x 2 = g(x) x 2, x > 0 H.J.Eberl - MATH*

111 nonhomogeneous Example: Euler-Cauchy equation cont d the Variation of the Constant ansatz is then or or y 1 u +y 2 v = 0 y 1u +y 2v = g(x) x 2 u x + v x 2 = 0 u x 2 2v x 3 = g(x) x 2 xu +v = 0 xu 2v = xg(x) H.J.Eberl - MATH*

112 nonhomogeneous Example: Euler-Cauchy equation cont d thus v = xg(x) = v(x) = xg(x)dx u = v = g(x) = u(x) = g(x)dx x integral tables tell us for which g(x) these integrals can be solved; for example, take g(x) = lnx and we find v(x) = putting things together u(x) = xlnx x [ x 2 xlnxdx = 2 lnx x2 4 y(t) = c 1 y 1 +c 2 y 2 +u(x)y 1 (x)+v(x)y 2 (x) = = c 1 x + c 2 x 2 +lnx lnx+ 1 4 = c 1 x + c 2 x lnx 3 4 ] H.J.Eberl - MATH*

113 Example: nonhomogeneous Euler-Cauchy equation cont d to verify that this is indeed correct we build and find after substituting y(x) = c 1 x + c 2 x lnx 3 4 y (x) = c 1 x 2 2c 2 x x y (x) = 2 c 1 x 3 +6c 2 x 4 1 2x 2 x 2 y +4xy +2y = = 2 c 1 x +6c 2 x c 1 x 8c 2 x c 1 x +2c 2 x 2 +lnx 6 4 = = lnx H.J.Eberl - MATH*

114 Example: nonhomogeneous Euler-Cauchy equation cont d 10 8 y1(x) y2(x) Y(x) Y(x)+y1(x)+y2(x) x x 2 y +4xy +2y = lnx H.J.Eberl - MATH*

115 Summary the methods of undetermined coefficients and variations of the constant complement each other: the former is easier but only applies to specific problems (constant p, q, selected g(t)), the latter is generally applicable but much more involved both methods require a fundamental system of the homogeneous equation for homogeneous equations with constant coefficients, this is easy otherwise not: if we can find a first solution of the homogenous system, the reduction of order can be used to obtain a second one... but this requires integration, which in general is not easy variation of the constant requires integration, which in general is not easy H.J.Eberl - MATH*