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 g x nn
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 ∂ L∂ ∂ 1 x 2 x n g = M L = Jacobian Matrix = nxn matrix ∂x ∂g ∂ ∂ n gn gn ∂ L x1 ∂x ∂x 2 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 xgxuú = (, ), x(0 )
where
xú 1 u1 x g 11xú u = = = 2 = 2 x MM,, g xú , u M M x g nnxú u n m 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 xú 1 xu = 2 ú −+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., ú = 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 , xtú ()= 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. xxú = ) and a solution to a differential equation (e.g. x==0 for xú x) .
Example B.1c
For the differential equations given in Example B.1a −1 0 xt()= , ut ()=−1 xú = [] 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 xú = 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 xgxuú =+(,) |== (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 δúúú=− 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 xgxuú =+(,) |(xx−+ ) |(uu− ) RR∂x R R∂u R R
we can now substitute the small perturbation variables ∂g ∂g xxgxuúú+=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 δδxú = ||x + δu ∂x RR∂u
This is a set of linear small perturbation differential equations. In summary, the original
nonlinear problem xgxuú = (, ), x(0 )
with solution xt() for specified input u(t) has been decomposed into two separate
problems.
• The reference problem xgxuú = (,)r RRR
with initial condition xR ()0 with solution xtR () to input utR ()
• The small perturbation problem ∂g ∂g δδδxú = ||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 xgxuú = (, ) makes a good reference solution but these
solutions can be hard to find. An easier set of solutions are constant solutions i.e., ú = = = 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 xú 1 xu xtú()= = 2 ú −+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., xxúú1 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 xgxuxú = (,),() 0 yhxu= (,) u() t= specified
Linear Small Perturbation Nonlinear Reference Problem Problem ∂g ∂g δδδxú = ||δδx + δu = r RR xgxuú 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 δδxú = ||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. #1 xu= , = 1 RR1
δxú 02+ δx −2 1 = 1 + δu δ − δ xú 2 20 x 2 0
−1 Using Ref. #2 xu= , = 1 RR−1
δxú 02− δx +2 1 = 1 + δu δ + δ xú 2 20 x 2 0
c. The Linear Solution for Reference #1 δδδú =− (1) xxu1 222 δδ=− (2) xxú 2 2 1
take Laplace transforms δδ−= δ − δ (1) sx11() s x ()02 x2 () s 2 us () δδ−=− δ (2) sx22() s x ()02 x1 () s
multiply (2) by s 2δδ−=− δ sxs22() sx ()02 sxs1 ()
multiply (1) by –2 −=−+−δδδδ 24420sxsxsusx1 ()2 () ()1 ()
set these equal 2δδ−=−+− δ δδ sxssx22() ()04 xsusx 2 () 4 () 201 () δδδδ2 + =−+ xss2 ()[]40204 sx2 () x1 () us () sxδδ()02− x () 0 4 δxs()= 2 1 + ()()δus 2 ss22+ 4 + 4
The first term on the right gives initial condition response. The second term on the right δxs() 4 contains the transfer function 2 = . δus() s2 + 4 δ To find xt1 () take the inverse Laplace transform. From (2) δδ−=− δ sx22() s x ()02 x1 () s 1 δδδxs()=−[] sxs () − x ()0 1 2 22 δ δ δ To find the solutions xt1 () and xt2 () you must be given the input ut() and the initial ()δδ conditions xx1 (),002 (). Then the solutions can be found using inverse Laplace transforms.
d) Total Solution for Reference #1 δ + δ xt1 () xt1 () xt1 () 1 xt1 () = R + = xt() δ 1 + δxt() xt2 () 2R xt2 () =+=+δδ ut() uR () t ut ()1 ut () 10+ δx () x ()0 = 1 1 + δ 10x 2 () e) Comment:
For this procedure to be valid the perturbations must be small, i.e.,
all must be small.
Suppose we have the nonlinear problem with = x1 ()0101 . = x 2 ()099 . ut( )=+10 . . 01 sinω t
then we can use Ref.#1 and then we can have δ = x1 ()001 . δ =− x 2 ()001 . δωut( )= .01 sin t
On the other hand if for the nonlinear problem we have =− x1 ()0101 . =− x 2 ()099 . ut( )=−101 − . sinω t
We would use Ref. #2 with δ =− x1 ()001 . δ = x 2 ()001 . δωut( )=− .01 sin t C. Output Equations
Often nonlinear differential equations are associated with nonlinear output equations. This may come about in modeling the sensors aboard an aircraft. The sensors are often nonlinear functions of the state vector and control vector. Output equations can be expressed as follows.
yhxu= (, ) where is y a px1 vector
This can also be linearized about reference solution xR and uR as follows. ∂h ∂h yy=+≈δ yhxu(,)| + | (xx − ) + | (uu − ) RRRRRR∂x ∂u
The underlined terms are equal by definition of y on the reference and can be cancelled out on both sides of the equation. That leaves the linear small perturbation output equations in terms of small perturbation variables. ∂h ∂h δy =+=⋅+⋅||δx δδδuCxDu ∂x RR∂u The total output equation in linear form is then given by the following. ≈+δ yt() yR yt ()
D. Stability One of many possible definitions of dynamic stability for nonlinear systems is given in terms of the ∂g ∂g 1 1 ∂g ∂x ∂x = = 1 2 eigenvalues of the Jacobian matrix, A |R ∂ ∂ . ∂x g2 g2 ∂x ∂x 1 2 R
If all the eigenvalues of A have negative real parts we say that the reference solution, (,)xuRR, is stable.
If at least one of the eigenvalue of A has a positive real part we say that the reference solution,
(,)xuRR, is unstable.
If at least one eigenvalues of A has a zero real part, and if all the other eigenvalues have negative
real parts, we can draw no conclusion about the stability of the reference solution, (,)xuRR.
E. Concluding Comments
We have seen how the solution to nonlinear differential equations can be found by decomposing
the problem into two simpler parts. The reference part is simpler because it is often a nonlinear algebraic
problem. The second small perturbation part is simpler because it often involves solving linear differential
equations with constant coefficients. The total approximate solution to the original nonlinear differential
equation was shown to be the sum of the two simpler parts.