LINEARIZATION OF NONLINEAR EQUATIONS By Dominick Andrisani A. Linearization of Nonlinear Functions A.1 Scalar functions of one variable . We are given the nonlinear function g(x). We assume that g(x) can be represented using a Taylor series expansion about some point x R as follows 2 dg() x 1 dgx() 2 gx()=+ gx ()|==|(xx −+ ) |( =xx − ) xxRRdx xx R 2! dx 2 xx RR + higher order terms A linear approximation for g(x) involves taking only the first two terms dg() x gx()≈+ gx ()|= ()xx − xxR R = dx xxR − This approximation is most accurate if ()xxR is small so that the neglected higher order terms are negligible. Example: gx()= x2 = Expanding g(x) about x R 2 gives dg gx()≈+ gx ()|== | (xx − ) xxRRdx xx R 2 ≈+22xx|(= − 2 ) xxR =+44()xx − 2 =−+ 44 Notice that this is a linear function of x. The simplification resulted because we evaluated == all nonlinear terms at the number xxR 2. Because we evaluated the terms on the == right hand side of the equation above at xxR 2, the only term that depends on x is the x-2 term, and this is a linear term. This approximation to g(x) is valid near x=2. We can generate another approximation to g(x) by expanding g(x) about x R =1 gives =+2 −=+− gx()12 x |x=1 ( x 1121 ) ( x ) =−12x + This second approximation is valid near x=1. Clearly the linear approximation depends on the choice of reference point x R . A.2 Scalar function of 2 variables Given the nonlinear function gx(,1 x2 ). This function can be represented by a Taylor series expansion about xx, as follows 1R 2R ∂ ∂ =+g −+g − gx(,1 x2 ) gx (1 , x2 ) |xx==,, xx (xx11 ) | xx == xx (x 2 R R ∂ 11R 22R R ∂ 11R 22R x1 x 2 ∂ ∂ ∂ ∂ 1 g − 2 g − 2 |()|()xx==,, xxxx11 + xx == xx xx22 ∂ ∂ 11R 22R R ∂ ∂ 11R 22R R 2! x11x x 22x ∂ ∂ + g −−+ |()()...xx==, xx xx11 xx22 hot ∂ ∂ 11R 22R R R x1 x 2 A linear approximation of g can be obtained by retaining the first three terms above (underlined). The two variables in this problem can be associated together in a vector xas follows x1 gx() where x = x 2 Example: = 2 gx(1 , x2 ) x1 cos x2 can be approximated about xx==20, as follows 1R 2R =+2 − gx(1 , x2 ) ( x1 cos x2 ) |xx==20,, (22 x 1 cos x2 ) | xx ==201 ( x ) 1R 2R 1R 2R −−2 (xx1 sin2 ) |xx==20, ( x2 0 ) 1R 2R =+ − +=−+ 44()xx11 2 0 44 The same function can be approximated about x = 2 x =π/ 4 1R 2R π =+2 − gx(1 , x2 )2 cos( ) (22 x1 cos x2 ) |π ( x1 ) x1 ==2, x2 4 R R 4 π −−2 (xx1 sin2 ) |π ( x2 ) xx==2, 1R 2R 4 4 π =+−004()xx − =−π 4 224 Notice again how important the linearization point or reference point is to the linearized result. A.3 Vector function of a vector of variables . Let gx() be an nx1 vector of nonlinear functions. Let x be an nx1 vector of variables g1 x1 g x g = 2 , x = 2 MM gnn x A linear approximation about xR is ∂g ggx≈+() |= (xx− ) R ∂x xxR R where g (,xx ) 11R L nR = gx()R M =nx1 vector g (,xx ) n 1 L nR R ∂g ∂g ∂g 1 1 1 ∂x ∂x L∂x ∂g 1 2 n = ∂x ∂M L = Jacobian Matrix = nxn matrix gn ∂g ∂g n L n ∂x ∂ ∂ 1 x 2 x n A.4 Accuracy of linearized solution . When we approximate gx() by retaining only the linear terms, we must guarantee that the deleted terms, i.e., the h.o.t. are negligible. This is true only when − xxR is small, i.e. when the perturbations from the reference point are small. B. Linearization on Nonlinear Differential Equations in First Order Form B.1 First order form Nonlinear differential equations in first order form can be written as xgxu˙ = (, ), x(0 ) where x˙ 1 u1 x g 11 x˙ u x = ,, g = x˙ = 2 , u = 2 MM M M x nn g x˙ n um Note that u represents specified forcing functions and x(0 ) is a specified initial condition vector. Example B.1a x xu2 − 2 x = 1 , g = 2 −+2 x 2 x1 1 2 − 2 x˙ 1 xu = 2 ˙ −+2 x 2 x1 1 B.2 The reference or trim solution When we were linearizing nonlinear functions, we saw how important the choice of reference point was. In linearizing nonlinear differential equations, we are also concerned with the reference about which we linearize. However, we are now interested in obtaining a linearized solution valid for all time. This requires that we linearize around a reference solution, which is valid for all time. Let xtR () be a known solution to the nonlinear differential equation with specified forcing function utR () and specified initial condition xR ()0 . i.e., ˙ = xt() gx (RR (), t u ()) t xR ()0 xtR () is said to be the reference solution to the nonlinear differential equation. Example B.1b For the differential equations given in Example B.1a 1 0 xt()= , ut ()= 1 , xt˙ ()= RR1 R 0 is a constant solution to the nonlinear differential equation. Verify this fact for yourself by substituting this solution into the differential equation given in Example B.1a. Please keep straight in your mind the difference between a differential equation (e.g. xx˙ = ) and a solution to a differential equation (e.g. x==0 for x˙ x) . Example B.1c For the differential equations given in Example B.1a −1 0 xt()= , ut ()=−1 x˙ = [] RR−1 R 0 is another constant solution to the nonlinear differential equations. Example B.1d For the differential equations given in Example B.1a =± x1 1 0 x = u= const x˙ = R =± = R R x2 uR const 0 is a constant solution to the nonlinear differential equations for any constant. B.3 Linearization about a reference solution Let xtutRR(), () be a reference solution. We now want to find a linearized solution to the nonlinear differential equation about this reference solution. We again expand gx() in a Taylor series expansion about xR and uR i.e., ∂g ∂g xgxu˙ =+(,) |== (xx−+ ) | == (uu− ) RR ∂x xxRR,, uu R ∂u xx RR uu R +h.o.t. − − The linear approximation is obtained by assuring that xxR and uuR are small enough that the h.o.t. can be neglected. B.4 Definition of small distribution variables Define δ =− xxxR δ =− uuuR δ˙˙˙=− xxxR For the linearized solution to be valid, these perturbations must be “small.” B.5 Separation of the linearized differential equations into two parts Assuming that the perturbations are small, we can write the approximation to the differential equations as ∂g ∂g xgxu˙ =+(,) |(xx−+ ) |(uu− ) RR∂x R R∂u R R we can now substitute the small perturbation variables ∂g ∂g xxgxu˙˙+=δδδ(,) + |x + | u RRRRR∂x ∂u In the equation above we have simplified the notation with | to denote | ==. R xxRR, uu Notice that the underlined terms are numerically equal from the definition of reference solution. Since they are equal, they can be cancelled out leaving ∂g ∂g δδx˙ = ||x + δu ∂x RR∂u This is a set of linear small perturbation differential equations. In summary, the original nonlinear problem xgxu˙ = (, ), x(0 ) with solution xt() for specified input u(t) has been decomposed into two separate problems. • The reference problem ˙ = r xgxuRRR(,) with initial condition xR ()0 with solution xtR () to input utR () • The small perturbation problem ∂g ∂g δδδx˙ = ||x + u ∂x RR∂u with initial condition δ =− xxx()00 ()R () 0 with solution δxt() to input δut(). Finally the total approximate solution is given by the entire solution procedure is shown in Figure 1. =+δ xt() xR () t xt () B.6 On picking a reference solution Any solution to xgxu˙ = (, ) makes a good reference solution but these solutions can be hard to find. An easier set of solutions are constant solutions i.e., ˙ = = = solutions so that xtR () 0 and xtR () constant for utR () constant. For constant reference solutions, finding the reference solution to a nonlinear differential equation becomes a problem of finding the solution to a nonlinear algebraic equation = gx(,)RR u 0 B.7 Linearization Example 2 − 2 x˙ 1 xu xt˙()= = 2 ˙ −+2 x 2 x1 1 a) Choice of Reference Solution To simplify our choice, assume that the reference solution is constant, == 2 −=2 i.e., xx˙˙1 2 0 . This requires that xu2 0 and −+=2 x1 10. These equations can be satisfied whenever 2 = 2 2 = xu2 and x1 1 Values of x1 and x 2 which satisfy these equations are =± xu2 where u is any constant =± x1 1 Exact Nonlinear Solution (x(t)) (e.g., from numerical integration) Nonlinear Differential Equations xgxux˙ = (,),() 0 yhxu= (,) u() t= specified Linear Small Perturbation Nonlinear Reference Problem Problem ∂g ∂g δδδx˙ = ||δδx + δu = r RR xgxu˙ RRR(,) ∂x ∂u ∂h ∂h x(),0 u specified δy ==+||δx + δu RR ∂x RR∂u where typically xR and uR δxxx()00==− ()− () 0 are constants R δ =− ut() ut () uR Linear Nonlinear Reference Solution Solution (i.e., from Laplace Transforms) (e.g., constant solution) δδδxt(), δ yt () xu, r RR Total Approximate Solution ==++δ xt() xR xt () ==++δ yt() yR yt () ==++δ xxx()00R () ==++δ ut() uR ut () Figure 1 Solution Procedures for Nonlinear Differential Equations We will consider two different reference solutions Ref .# 1 Ref .# 2 +1 −1 xt()= , ut ()=+1 xt()= , ut ()=−1 RR+1 RR−1 1 −1 x ()0 = x ()0 = R 1 R −1 b) Small Perturbation Equations of Motion ∂g ∂g δδx˙ = ||x + δu ∂x RR∂u δ =− δ =− where xxxR uuuR ∂ ∂ g1 g1 ∂g 1 ∂g ∂x ∂x ∂g ∂ A = = 1 2 , B = = u ∂ ∂g ∂g ∂ ∂g x 2 2 u 2 ∂ ∂ ∂ x1 x 2 u − ∂g 02x 2 ∂g 2u = , = ∂ − ∂ x 20x1 u 0 1 Using Ref.