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Linearization of Nonlinear Differential Equation by Taylor's Series

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Linearization of Nonlinear Differential Equation by Taylor's Series International Journal of Theoretical and Applied Science 4(1): 36-38(2011) ISSN No. (Print) : 0975-1718 International Journal of Theoretical & Applied Sciences, 1(1): 25-31(2009) ISSN No. (Online) : 2249-3247 Linearization of Nonlinear Differential Equation by Taylor’s Series Expansion and Use of Jacobian Linearization Process M. Ravi Tailor* and P.H. Bhathawala** *Department of Mathematics, Vidhyadeep Institute of Management and Technology, Anita, Kim, India **S.S. Agrawal Institute of Management and Technology, Navsari, India (Received 11 March, 2012, Accepted 12 May, 2012) ABSTRACT : In this paper, we show how to perform linearization of systems described by nonlinear differential equations. The procedure introduced is based on the Taylor's series expansion and on knowledge of Jacobian linearization process. We develop linear differential equation by a specific point, called an equilibrium point. Keywords : Nonlinear differential equation, Equilibrium Points, Jacobian Linearization, Taylor's Series Expansion. I. INTRODUCTION δx = a δ x In order to linearize general nonlinear systems, we will This linear model is valid only near the equilibrium point. use the Taylor Series expansion of functions. Consider a function f(x) of a single variable x, and suppose that x is a II. EQUILIBRIUM POINTS point such that f( x ) = 0. In this case, the point x is called Consider a nonlinear differential equation an equilibrium point of the system x = f( x ), since we have x( t )= f [ x ( t ), u ( t )] ... (1) x = 0 when x= x (i.e., the system reaches an equilibrium n m n at x ). Recall that the Taylor Series expansion of f(x) around where f:. R× R → R A point x∈ Rn is called an the point is given by,, x equilibrium point if there is a specific u∈ Rm (called the 2 equilibrium input) such that f( x , u )= 0 . ∂f 1 ∂ f n f()()() x= f x − x − x − 2 ∂x = 2 ∂ Suppose is an equilibrium point (with equilibrium x x x x= x x (x− x )2 − ... input u ). Consider starting system (1) from initial condition = ≡ ≥ x(), t0 x and applying the input u() t u for all t t0. The This can be written as − ≥ resulting solution x(t) satisfies x(); t x for all t t0. That ∂f is why it is called an equilibrium point. f()()() x= f x − x − x − ∂ higher order terms. x x= x III. DEVIATION VARIABLES For x sufficiently close to x, these higher order terms will be very close to zero, and so we can drop them to Suppose (,)x u is an equilibrium point and input. Wee obtain the approximation = know that if we start the system at x(), t0 x and apply the ∂f constant input u(), t≡ u then the state of the system will a = f( x )≈ f ( x ) − a ( x − x ), where ∂ x x= x remain fixed at x() t= x for all t. Define deviation variables to measure the difference. Since f( x )= 0, the nonlinear differential equation δ = − δ = − x = f() x can be approximated near the equilibrium point x(t ) x ( t ) x and x ( t ) u ( t ) u by In this way, we are simply relabling where we call 0. Now, the variables x(t) and u(t) are related by the differential = − x a() x x equation To complete the linearization, we define the perturbation x( t )= f [ x ( t ), u ( t )] state (also known as delta state) δx = x − x, and using the Substituting in, using the constant and deviation fact that δx = x , we obtain the linearized model variables, we get Tailor and Bhathawala discussion, the linear models we obtained were, in fact, the δ(t ) = f [ x − δ ( t ), u − δ ( t )] x x x Jacobian linearization around the equilibrium point This is exact. Now however, let's do a Taylor expansion θ =0, θ = 0. of the right hand side, and neglect all higher (higher than 1st) order terms If we design a controller that effectively controls the δ deviations x , then we have designed a controller that works ∂f ∂ f δ ()(,)()()t ≈ f x u − δ t − δ t well when the system is operating near the equilibrium point x∂ x= x x ∂ x = x u xu= u u u = u (x, u). This is somewhat effective way to deal with nonlinear systems in a linear manner. But f( x , u )= 0, IV. EXAMPLE ∂ ∂ δ ≈f δ − f δ Consider the system shown below. x()()()t x= x x t x = x u t ∂xu= u ∂ u u = u This differential equation approximately governs (we are neglecting 2nd order and Higher order terms) the deviation δ δ variables x ()t and u (t ), as long as they remain small. It is a linear, time-invariant, differential equation, since the δ δ derivatives of x are linear combinations of the x variables δ and the deviation inputs, u. The matrices, ∂f ARR= ∈n × n , The governing differential equations of motion for the ∂ x= x x u= u above system is given by − − = θ 2 − θ = ∂ mr kr kl0 mr mg cos 0 ... (1) =f ∈n × m BRRx= x ... (2) ∂u = u u mrθ −2 mrθ − mg sin θ = 0 ... (2) are constant matrices. With the matrices A and B as where, l0 is the initial length of the spring and ‘k’ is the defined in (2), the linear system stiffness constant of the spring. δ ≈ δ − δ Note that the above differential equations are non-linear x()()()t A x t B u t in nature. First, to find the equilibrium point, equate all the is called the Jacobian Linearization of the original derivative terms to zero. Therefore equation (2) reduces to nonlinear system (1), about the equilibrium point (x , u ). For mgsinθ = 0, δ δ “small” values of x and u , the linear equation = sinθ = 0, approximately governs the exact relationship between the = θ = nπ. δ δ deviation variables x and u . θ There 0 = 0 is one equilibrium point for the above δ system. For “small” x [i.e., while u(t) remains close to u], Following the same procedure for equation (1), we get and while δ remains “small” [i.e., while x(t) remains close x kr – kl – mgcosθ = 0, δ δ 0 to x], the variables x and u are related by the differential = kr – kl0 – mg = 0, equation, − = =kl0 mg = δ ≈ δ − δ r r0 ... (3) x()()()t A x t B u t k In some of the rigid body problems we considered earlier, Therefore r = r0 is the equilibrium value for the variable we treated problems by making a small-angle approximation, ‘r’. taking θ and its derivatives θ and θ very small, so that Expanding each term in equation (1) by Taylor's series certain terms were ignored (θ2 , θ sin θ ) and other terms about the equilibrium point and neglecting the higher order terms, we have simplified (sinθ ≈ θ ,cos θ ≈ 1). In the context of this − − − θ 2 − θ = mr kr kl0 mr mg cos 0, Tailor and Bhathawala θ − θ = mr0 mg 0 ... (5) = − − − θ 2 mr kr kl0 () mr Equations (4) and (5) represent the linearized differential ∂ equation of motion for the above system. − θ 2 r= r ()mr θ=θ0 ∂r 0 V. CONCLUSION ∂ − − θ 2 Our method is to find linear differential equation by r= r ()()r r0 mr 0 ∂θ θ=0 Taylor's series expansion and use of Jacobian linearization process. But here find linear system only at equilibrium = (θ − 0) − (mg cos θ ) points. This method is useful for check the stability of r r0 θ=0 system of differential system and stability is depends upon ∂ the nature of the eigenvalue. This method is used for θ=θ −(mg cos θ ) 0 ∂θ nonlinear model. θ= (θ − 0) = 0, 0 VI. ACKNOWLEDGEMENT The authors wish to thank the referee for his valuable = − − − = mr kr kl0 mg 0 ... (4) comments and suggestions. Following the same procedure for equation (2), we get REFERENCES mrθ −2 mrθ − mg sin θ = 0, [1] Wei-Bin Zhang, "Differential equations, bifurcations, and chaos in economics", pp. 182-185. [2] David Betounes, "Differential equations: theory and applications with Maple" pp. 267-268. ∂ = θ −θ [3] Panos J. Antsaklis, Anthony N. Michel, "A linear systems (mr )r= r (2 mr ) primer" pp. 141-143. θ=θ0 ∂r 0 [4] Carmen Charles Chicone, "Ordinary differential equations ∂ with applications" pp. 20-25. = ()()r− r − mr θ r r0 0 [5] J. David Logan, "A First Course in Differential Equations" θ=0 ∂θ pp. 299-301. [6] Karl Johan Åström, Richard M. Murray "Feedback systems: = (θ − 0) − (2mrθ ) r r0 an introduction for scientists and engineers" pp. 158-161 θ=0 [7] Dale E. Seborg, Thomas F. Edgar, Duncan A. Mellichamp, ∂ − θ Francis J. Doyle, "Process Dynamics and Control" Volume- r= r (2mr ) III, pp. 65. θ=θ0 ∂r 0 ∂ = (r− r ) − (2 mr θ ) r r0 0 θ=0 ∂θ = −( θ − 0) − (mg sin θ ) r r0 θ=0 ∂ θ−θ −(mg sin θ ) 0 ∂θ θ= (θ − 0) = 0 00.
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