# Functional Approach for Solving Reduced Order of Index-Four Hessenberg Differential-Algebraic Control System

Authors: Ghazwa F. Abd
Journal: Journal of Mathematics
 (2022)

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## Abstract

This research investigates differential-algebraic equations with higher index (index four). Specifically, a functional analytic approach is proposed to find the solution of (index four) Hessenberg differential-algebraic equations (DAEs). The approximate solution of the proposed functional approach for a suitable separable Hilbert space is obtained with the help of the direct optimisation technique and Ritz basis functions method. Illustrations have been ranked from an 8 × 8 test system of index-four Hessenberg linear DAEs to a 4 × 4 DAE of rotating masses as well as an 8 × 8 differential-algebraic generator model, where it reformulated index-four linear Hessenberg DAEs. Their approximate solutions were obtained using the present approach with comparisons. The numerical results demonstrate the simplicity of the proposed approach and express suitable accuracy and efficiency.


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## Body

## 1. Introduction

Many life applications can be modelled as differential-algebraic systems, including optimal control problems, constrained robotic systems, and constrained electrical networks [1–3]. In the literature, the index property, of which several definitions exist, is usually utilised to characterise the differential-algebraic system [4, 5]. To determine an equivalent ordinary differential equation, the minimum number of times that the algebraic constraint of DAEs must be differentiated to time is represented with the frequently used differentiation index [5]. The higher index DAEs present a greater challenge in being solved than those in the lower index system because it makes applications more cumbersome [6]. In general DAEs, with a special structure, Hessenberg DAEs are an exceedingly important class implemented in various engineering and scientific applications [7, 8].The authors in [2, 3, 9] presented numerical methods to solve index two and index three. These methods utilize the index reduction method to reduce the higher DAE index to be lower one [10–12].To solve higher index (index four) Hessenberg linear DAE systems, this article develops an analytical approximation method to be efficiently and easily implemented. It is based on the functional analytical theory, which will help to determine a functional with critical points that serve as solutions for the fourth index Hessenberg DAEs and vice versa. To present the solution as a linear combination of the basic elements of the given setting space (i.e., separable Banach space), this work introduces a parameterisation approach inserted into the obtained variational formulation. Subsequently, to determine the solution of the variational formulation for unknown parameters, the method uses a direct method of calculus of variation. Finally, solving the linear algebraic system obtains the unknown parameters.


## 2. Fourth-Order Hessenberg DAEs Problem Formulation

Index-four linear Hessenberg DAEs with control input are considered as follows:(1a)x˙1+A11x1+A12x2+A13x3+A14x4=B1u+f1,(1b)x˙2+A21x1+A22x2+A23x3=B2u+f2,(1c)x˙3+A32x2+A33x3=B3u+f3,(1d)A43x3=B4u+f4,where(1)
A43A32A21A14≜c is a square invertible matrix, and x1,x2,x3,x4T denotes independent state vectors(2)
Aij;i,j=1,..,4, are constant matrices of appropriate dimension, and Aij is not required to be invertible or a square matrix(3)
Bj,j=1,..,4 are suitable constant vectors, and u is the given single control input with u∈C3T;R≜�u as the class of admissible control. T=t0,tf,t0<tf; and fj;j=1,..,4 denotes differentiable functions, fj∈C3T;RUsing the implicit function theorem, which can be observed in [13, 14], (1d) can be differentiated with respect to t:(2)A43x˙3=B4u˙+f˙4.From equation (1c), the following is obtained:(3)−B4u˙−f˙4+A43B3u+f3−A32x2−A33x3=0.Now, in equation (3), a new constraint is conjugated with equations (1a)–(1d) (i.e., differential-algebraic system). The differentiability of equation (3) continues until the state x4 is estimated to have a class of differential equations with states x1,x2,x3 defined on a manifold of the resulting algebraic constraints.Furthermore,x4 (state variable) does not exist in equation (3); accordingly, equation (1c) can be differentiated (with respect to t) again.(4)−B4u¨−f¨4+A43B3u˙+f˙3−A32x˙2−A33x˙3=0,(5)−B4u¨+A43B3u˙−A43A32B2+A43A33B3u−f¨4+A43f˙3−A43A32f2−A43A33f3+A43A32A21x1+A43A32A22+A43A33A32x2+A43A32A23+A43A33A32x3=0.Hence, another algebraic constraint (inserted into the differential-algebraic system (1a)–(1d)) is represented by equation (5). To determine the state variable x4 and until the index-four condition (A43A32A21A14) is satisfied, the derivation is continued:(6)−B4u⃛+A43B3u¨−A43A32B2+A43A33B3u˙+A43A32A21B1+A43A32A22B2+A43A33A32B2+A43A32A23B3+A43A33A32B3u−f⃛4+A43f¨3−A43A32f˙2+A43A32A21f1A43A32A22+A43A33A32f2+A43A32A23+A43A33A32f3−A43A32A21A11+A43A32A22A21+A43A33A32A21x1−A43A32A21A12+A43A32A22A22+A43A33A32A22+A43A32A23A32+A43A32A23A32x2−A43A32A21A13+A43A32A22A23+A43A33A32A23+A43A32A23A33+A43A33A32A33x3−A43A32A21A14x4=0.Then, sinceA43A32A21A14 is invertible,(7)x4=c−1u˜+f˜+A˜1x1+A˜2x2+A˜3x3=Lx1+Lx2+Lx3+Lu˜,f˜,where c=A43A32A21A14,Lx1=c−1A˜1x1,Lx2=c−1A˜2x2,Lx3=c−1A˜3x3,Lu˜,f˜=c−1u˜+f˜,(8)u˜=−B4u⃛+A43B3u¨−A43A32B2+A43A33B3u˙+A43A32A21B1+A43A32A22B2+A43A33A32B2+A43A32A23B3+A43A33A32B3uf˜=−f⃛4+A43f¨3−A43A32f˙2+A43A32A21f1+A43A32A22+A43A33A32f2+A43A32A23+A43A33A32f3,A˜1=−A43A32A21A11+A43A32A22A21+A43A33A32A21,A˜2=−A43A32A21A12+A43A32A22A22+A43A33A32A22+A43A32A23A32+A43A32A23A32,A˜3=−A43A32A21A13+A43A32A22A23+A43A33A32A23+A43A32A23A33+A43A33A32A33.Since system (1) is the index-four Hessenberg DAEs system, the initial condition must be selected in such a way as to satisfy all algebraic constraints in equations (1d), (3), and (5). From the additional two hidden constraint equations (1c)–(1d) and from equation (7), over the class of C3T;R, the following form is the reduced type index-four Hessenberg DAEs (1) with control u,u˙,…,u⃛.(9)x˙1+A11+A14c−1A˜1x1+A12+A14c−1A˜2x2+A13+A14c−1A˜3x3=B1u+f1−A14Lu˜,f˜,(10)x˙2+A21x1+⋯+A23x3=B2u+f2,(11)x˙3+A32x2+A33x3=B3u+f3,(12)with algebraic constraint one:A43x3=B4u+f4,(13)algebraic constraint two:A43A32x2+A43A33x3=−B4u˙+A43B3u−f˙4+A43f3,(14)algebraic constraint three:A43A32A21x1+A˜2x2+A˜3x3=B4u¨−A43B3u˙+A43A32B2+A43A33B3u+f¨4−A43f˙3+A43A32f2+A43A33f3.SetJ={x1.,x2.,x3.∈C3T;R3|where (12)–(14) are satisfied. Ultimately, the following represents the class of consistent initial condition:(15)J0=x1t0,x2t0,x3t0∈R3|Jis satisfied att=t0.


## 3. Fourth Index Hessenberg DAE Solution via Variational Formulation Technique

The present approach is step by step represented as follows:Step 1: define the operatorL:DL⊂H⟶RL⊂H, with DL¯=H, where H is an appropriate Hilbert space (e.g., H≜L2T;T=t0,tf, t0<tf) by Lx=Ft;u˜,f˜. The operator L refers to a linear differential and algebraic operator can be written as follows:(16)Lx=I000000I000000I000dx1dtdx2dtdx3dt+A1A21000A6A2A22A210A4A21A3A23A21A21A5A21x1x2x3.whereA1=A11+A14c−1A˜1, A2=A12+A14c−1A˜2,A3=A13+A14c−1A˜3, A4=A43A32,A5=A43A33, A6=A43A32A21,and Ft;u˜,f˜≜Ft;u,u˙,u¨,u⃛,fj,f˙j,f¨j,,fj3, for all u∈C3T;R≜ΔuBy using the variational formulation of obtained functional corresponding to (13) on D(L)={x.=x1.,x2.,x3.T∈L2T;R3|xt∈J,xt0∈J0andxt satisfy (13)}.In almost everywhere inH, the DAEs are satisfied by the linearly invariant (reduced type) index Hessenberg DAEs system. Furthermore, almost everywhere over the setting class C1T;R3, the solution should be a continuous function that satisfies equations (12)–(14))Step 2: in equation (16), the operator L is not symmetric to the usual bilinear form x,x˜=∫​xTt.x˜tdt, x=x1t,x2t,x3t,x˜=x˜1t,x˜2t,x˜3t because of the existence of the operator d./dt. To satisfy the nondegenerate condition on the range and the domain of operator L, the bilinear form needs to be chosen (i.e., for every x∈DL,x,x˜=0, then x˜=0, and for every x˜∈DL,x,x˜=0 (then x=0. Therefore, a usual variational formulation exists only if the usual bilinear (inner product) or the linear operator is redefined form using x,x˜≜x,Lx˜,x,x˜∈DL, where DL⊂HL⊂H with xL=x,Lx˜, and HL,dx,x˜ is the complete Hilbert space (where L refers to the asymmetric linear operator and separable Hilbert space is denoted by HL). The reader is recommended to [10, 11] for more details. Jx is defined if L (on DL¯) is positive definite; otherwise, it is needed to determine JJ0x (the critical points). Over the class of the consistent initial conditions and u∈C3T;R, the augmented functional of JJ0x can be defined as follows:(17)JJ0x=12L1Txt,L1xt−F1T,L1xt+⋯+12L6Txt,L6xt−F6T,L6xt+12L4Txt0,L4xt0−F4Tt0,L4xt0+12L5Txt0,L5xt0−F5Tt0,L5xt0+12L6Txt0,L6xt0−F6Tt0,L6xt0.whereL1xt=x˙1+A1x1+A2x2+A3x3,L2xt=x˙2+A21x1+⋯+A23x3,L3xt=x˙3+A32x2+A33x3,L4xt=A43x3,L5xt=A4x2+A5x3,L6xt=A6x1+A2x2+A3x3, and F1=B1u+f1−A14Lu˜,f˜, F2=B2u+f2, F3=B3u+f3,F4=B4u+f4, F5=−B4u˙+A43B3u−f˙4+A43f3, and F6=B4u¨−A43B3u˙+A4B2+A5B3u+f¨4−A43f˙3+A4f2+A5f3.Thus, the critical points ofJJ0x is guaranteed by the functional as the solution of equations (12)–(14) and vice versa when x,x˜ is chosen in the form of nondegenerate bilinear.Step 3: the critical points of equations (12)–(14) are the solutions of equation (12). Furthermore, the critical points are a solution of equations (12)–(14) by the nondegenerate property on DL,RL. A countable linearly independent (bases) system φ1 , φ2, ... etc.) (complete set of functions; base) can be used to approximate the solution, where xjt=∑ζ=1naζφζ,nis a positive number,j=1,2,3 and aζ is determined so that Jx1t,x2t,x3t is extremized.Step 4: with respect toaζ,∀ζ=1,…,n, J can be derived. To determine these critical points as a function of aζ parameters, n is given a suitable positive number. To obtain the system of a solvable linear algebraic, one can set dJ/daζ=0,∀ζ=1,..,n. Furthermore, ∑ζ=1nΜζ,jaζ=Νj,∀j=1,…,n,M=Μζ,jζ=1,j=1n,3 is a nonsingular matrix, where the coefficients can be determined from dJ/daζ=0, if xt0∈J0. Accordingly, aζ=Μ−1ζ,jΝj, and the unique solution is guaranteed. Else, if xt0 is chosen as arbitrary, there may be infinitely solutions or no solution.


## 4. Illustration Examples

### 4.1. Test Example

The following test example is designed with a suitable exact solution in mind to confirm the validity of the aforementioned approach, where(1)
The system is8×8 DAEs with control(2)
The system is represented as fourth order Hessenberg linear DAE with control(18)x˙11+2x11+3x12+x31+x41=f11,(19)x˙12−2x11−2x21+x22+x31+x41=u+f12,(20)x˙21+4x11+x12−2x21+x31=3u+f21,(21)x˙22+x12−2x21+x32=f22,(22)x˙31+3x21−2x22+x31=f31,(23)x˙32+3x21+x32=−2u+f32,(24)x31=−u+f42,(25)x32=f42,with(26)f11=t4+t+3,f12=t5+t3−t2−sint−4,f21=5t+t4−2t2−3sint−2,f22=t5+t3−2t2−t+1,f31=t4+4t3+3t2−4t+4,f32=t5+5t4+t3+6t2+2sint+3,f41=t4−sint−1,f42=t5+t3,A11=23−20,A12=00−21,A13=1001,A14=1001,A21=4101,A22=−20−20,A23=1001,A32=3−230,A33=1001,A43=1001,B1=01,B2=30,B3=0−2,B4=−10,x1=x11x12,x2=x21x22,x3=x31x32,x4=x41x42.Note thatA43A32A21A14 is an invertible matrix.Significantly, in the DAE system, variablex4 does not appear. Therefore, the algebraic constraint equations (18)–(23) can be differentiated with respect to t. Then, to obtain the following, substitute the differential constraints to equations (24) and (25).(27)A43x˙3=B4u˙+f˙4,A43B3u+f3−A32x2−A33x3=B4u˙+f˙4.Thus, equation (27) is considered a new constraint inserted to equations (18)–(25). In the last constraint, variable x4 is not exist. Therefore, to obtain the following, equation (27) can be differentiated one more time with respect to t.(28)−B4u¨+A43B3u˙−A4B2+A5B3u−f¨4+A43f˙3−A4f2−A5f3+A4A21x1+A4A22+A5A32x2+A4A23+A5A33x3=0,where A4=A43A32,A4=A43A33.Next, one can differentiate equation (28) again for the t set: c≜A43A32A21A14−1=0.1250−0.0417−0.50.5.Step 1:x4 is estimated as x4=Lx1,x2,x3+Lu,u˙,u¨,u⃛,f1,f2,f3,f˙2,f˙3,f¨3, where the derivative of the class of the consistent initial conditions can be (29)J0=x1t0,x2t0,x3t0|A43x3t0=B4ut0+f4t0−A43A32x2t0−A43A33x3t0=B4u˙t0+f˙4t0−A43B3ut0−A43f3t0A4A21x1t0+A5A22x2t0+A4A23x3t0+A5A32x2t0+A5A33x3t0=B4u¨t0+f¨4t0−A43B3u˙t0−A43f˙3t0+A4B2ut0+A4f2t0+A5B3ut0+A5f3t0.Step 2: as can be seen in the following, the definition of the variational function with the class of the consistent initial conditions can be written as follows:(30)JJ0x=∫t0tf12x˙1+A1x1+A2x2+A3x3Tx˙1+A1x1+A2x2+A3x3+x˙2+A21x1+A22x2+A23x3Tx˙2+A21x1+A22x2+A23x3+x˙3+A32x2+A33x3Tx˙3+A32x2+A33x3+A43x3TA43x3+A4x2+A5x3TA4x2+A5x3+A6x1+A2x2+A3x3TA6x1+A2x2+A3x3+A43x3t0TA43x3t0+A4x2t0+A5x3t0TA4x2t0+A5x3t0+A6x1t0+A2x2t0+A3x3t0TA6x1t0+A2x2t0+A3x3t0−F1Tx˙1+A1x1+A2x2+A3x3+F2Tx˙2+A21x1+A22x2+A23x3+F3Tx˙3+A32x2+A33x3+F4TA43x3+F5TA4x2+A5x3+F6TA6x1+A2x2+A3x3+F4t0TA43x3t0+F5t0TA4x2t0+A5x3t0+F6t0TA6x1t0+A2x2t0+A3x3t0dt,whereF1=B1u+f1−cA43B3u¨−cA43f¨3+cB4u⃛+cf⃛4+cA4B2u˙+cA4f˙2−cA4A21B1u−cA4A21f1−cA4A22B2u−cA4A22f2−cA4A23B3u−cA4A23f3+cA5B3u˙+cA5f˙3−cA5A32B2u−cA5A32f2,−cA5A33B3u−cA5A33f3,F2=B2u+f2, F3=B3u+f3, F4=B4u+f4, F5=B4u˙+f˙4−A43B3u−A43f3, F6=B4u¨+f¨4−A43B3u˙−A43f˙3+A4B2u+A4f2+A5B3u+A5f3.Step 3: one can setx11t=∑i=05aiφit,x12t=∑i=05biφit,x21t=∑i=05ciφit,x22t=∑i=05diφit,x31t=∑i=05viφit,andx32t=∑i=05wiφit, with φit=ti to obtainx4t=Lx1,x2,x3,t+Lu,u˙,u¨,u⃛,f1,f2,f3,f˙2,f˙3,f¨3Step 4: as mentioned above, the functionalJ is a function of a⟶,b⟶,c⟶,d⟶,v⟶,andw⟶. As Ja⟶,b⟶,c⟶,d⟶,v⟶,w⟶ is extremized, one can set ∂J/∂a⟶=0, ∂J/∂b⟶=0, ∂J/∂c⟶=0,∂J/∂d⟶=0, ∂J/∂v⟶=0, and ∂J/∂w⟶. It leads to the linear algebraic equation Μa⟶b⟶c⟶d⟶v⟶w⟶=Ν, which is solvable directly for a⟶b⟶c⟶d⟶v⟶w⟶=Μ−1Ν. By deriving J with respect to a⟶,b⟶,c⟶,d⟶,v⟶, and w⟶, M (the matrix coefficient) can be determined. Hence, the approximate solution x1t,x2t,x3t is obtained. The numerical values of the unknown coefficients a⟶,b⟶,c⟶,d⟶,v⟶,andw⟶ are listed in Table 1.Table 1
Unknown coefficients of the representation ofx1t,x2t,andx3t.
Coef. ofx11 statesa5a4a3a2a1a0Numerical values1×10−39–8×10−391×10−38–6×10−3910Coef. ofx12 statesb5b4b3b2b1b0Numerical values1×10−38–3×10−383×10−38–1×10−38–11Coef. ofx21 statesc5c4c3c2c1c0Numerical values2×10−38–7×10−386×10−3811×10−391Coef. ofx22 statesd5d4d3d2d1d0Numerical values1×10−40–1×10−391×10−39–1×10−392–1Coef. ofx31 statesv5v4v3v2v1v0Numerical values2×10−401–3×10−392×10−39–3×10−40−1Coef. ofx32 statesw5w4w3w2w1w0Numerical values11×10−3819×10−39–1×10−390Figure1 depicts the numerical solutions with equality states compared with the exact solution x1=t1−t,x2=t2+12t−1,x3t=t4−1t3+t5,andx4t=2tt2, for a given ut≜sint.Figure 1
The comparisons between the exact solution and approximate solution of the test example.


### 4.2. Application of the Generator Model

Consider the model of a generator described with the following parameters. On the right axis, the input is the angleθ connected to mass with inertia K which is rotated about the angle θ with angular velocity ν. The torque acting on the right and left sides are T1 and T2, respectively. The mass is then connected to the second axis with a generator. The electrical quantities and the variables described in the second axis are then assumed as in [15]. The rest of the electrical circuit includes one inductor and resistor. The model is illustrated in Figure 2.Figure 2
A model of generator.The system is now modelled using the equations representing the different parts and their connections.(31)1000000001000000001000000001000000001000000001000000000000000000T˙1T˙2V˙∅˙I˙u˙1u˙2u˙3+12001010−101101012110100001100100001−11000001001000000100000000100T1T2V∅Iu1u2u3=01100−110θ,with θt≜sint∈C2I,R,(32)A11=12−10,A12=0011,A13=1001,A14=1001,A21=2101,A22=1010,A23=1001,A32=1−110,A33=1001,A43=1001,B1=01,B2=10,B3=0−1,B4=10,x1=T1T2,x2=V∅,x3=Iu1,x4=u2u3.and A43A32A21A14 is an invertible matrix.In the DAE system, variablex4 does not appear. Therefore, the algebraic constraint equation (31) can be differentiated with respect to t. Then, three times until x4 is yielded in the following:(33)A43x˙3=B4θ˙,(34)A43B3θ+f3−A32x2−A33x3=B4θ˙,(35)−B4θ¨+A43B3θ˙−A4B2+A5B3θ+A4A21x1+A4A22+A5A32x2+A4A23+A5A33x3=0,where A4=A43A32,A5=A43A33.Next, one can differentiate equation (27) again for the t and set c≜A43A32A21A14−1=011/20.:Step 1: it is mentioned above thatx4 is obtained as x4=Lx1,x2,x3+Lθ,θ˙,θ¨,The derivative of the class of the consistent initial conditions is as follows:(36)J0=x1t0,x2t0,x3t0|A43x3t0=B4θt0−A43A32x2t0−A43A33x3t0=B4u˙t0−A43B3θt0A4A21x1t0+A4A22x2t0+A4A23x3t0+A5A32x2t0+A5A33x3t0=B4θ¨t0−A43B3θ˙t0+A4B2θt0+A5B3θt0,Step 2: the following is the definition of the variational function with the class of the consistent initial condition.(37)JJ0x=∫t0tf12x˙1+A1x1+A2x2+A3x3Tx˙1+A1x1+A2x2+A3x3+x˙2+A21x1+A22x2+A23x3Tx˙2+A21x1+A22x2+A23x3+x˙3+A32x2+A33x3Tx˙3+A32x2+A33x3+A43x3TA43x3+A4x2+A5x3TA4x2+A5x3+A6x1+A2x2+A3x3TA6x1+A2x2+A3x3+A43x3t0TA43x3t0+A4x2t0+A5x3t0TA4x2t0+A5x3t0+A6x1t0+A2x2t0+A3x3t0TA6x1t0+A2x2t0+A3x3t0−F1Tx˙1+A1x1+A2x2+A3x3+F2Tx˙2+A21x1+A22x2+A23x3+F3Tx˙3+A32x2+A33x3+F4TA43x3+F5TA4x2+A5x3+F6TA6x1+A2x2+A3x3+F4t0TA43x3t0+F5t0TA4x2t0+A5x3t0+F6t0TA6x1t0+A2x2t0+A3x3t0dt,whereF1=B1θ−cA43B3θ¨+cB4θ⃛+cs1B2θ˙−cs1A21B1θ−cs1A22B2θ−cs1A23B3θ+cs2B3θ˙+cs2A32B2θ−cs2A33B3θ,F2=B2θ, F3=B3θ, F4=B4θ, F5=B4θ˙−A43B3θ, F6=B4θ¨−A43B3θ˙+s1B2θ+s2B3θ.Step 3: one can setx11t=∑i=05aiφit,x12t=∑i=05biφit,x21t=∑i=05ciφit,x22t=∑i=05diφit,x31t=∑i=05viφit,andx32t=∑i=05wiφit, with φit=ti to obtain(38)x4t=L∑i=05aiti,∑i=05biti,∑i=05citi,∑i=05diti,∑i=05viti,∑i=05witi,t+Lθ,θ˙,θ¨,θ�.Step 4:J is a function of a⟶,b⟶,c⟶,d⟶,v⟶,andw⟶, and as Ja⟶,b⟶,c⟶,d⟶,v⟶,w⟶ is extremized, one can set ∂J/∂a⟶=0, ∂J/∂b⟶=0, ∂J/∂c⟶=0,∂J/∂d⟶=0, ∂J/∂v⟶=0, and ∂J/∂w⟶=0. It is led to the linear algebraic equation Μa⟶b⟶c⟶d⟶v⟶w⟶=Ν, which is solvable directly for a⟶b⟶c⟶d⟶v⟶w⟶=Μ−1Ν. Hence, the approximate solution x1t,x2t,x3t,x4t is determined. Lastly, the derivative of J with respect to a⟶,b⟶,c⟶,d⟶,v⟶, and w⟶ is used to obtain the matrix coefficient M.Step 5: to evaluate the solution’s accuracy without knowing the exact solution, the model utilisesL2-norm by substituting the values of x11t=∑i=05aiti,x12t=∑i=05biti, x21t=∑i=05citi,x22t=∑i=05diti, x31t=∑i=05viti, and x32t=∑i=05witi in system (31). The error presented in the test example at each point along time t∈0,1 and for different kinds of θ are displayed (Figures 3–6).Figure 3
The error of the generator system solution forθ=e−t.Figure 4
The error of the generator system solution forθ=sint.Figure 5
The error of the generator system solution forθ=sint+cost.Figure 6
The error of the generator system solution forθ=t4+t2.


### 4.3. Application of the Rotating Masses System

Consider a system of two rotating masses that are described by the torquesT1t,T2t,T3t,andT4t and the angular velocities α1t and α2t. The masses have the moments of inertia γ1 and γ1. Figure 7 illustrates this scenario.Figure 7
Connected rotating masses.The system to matrically describes this model is provided as(39)γ10000γ20000100000α˙1α˙2T˙2T˙3+010000−1011−111000α1α1T2T3=10000001T1T4,with γ1=γ2=1 and T1=t5−4t,T4=2t2+3.Step 1: sinceA43A32A21A14 exists as an invertible matrix, the variable T3 does not appear and T˙2=4t through differentiation of the algebraic constraint of system (39), we can continue to differentiate until T3 obtained as T3=α1+t5−4t with consistent initial condition:(40)J0=α1t0,α2t0,T2t0|T2=2t02+3,α2+T2=4t0,−α1+α2=4.Step 2: using the variational formulation associated with this consistent initial condition,(41)JJ0x=∫t0tf12α˙1+T2+T3Tα˙1+T2+T3+α˙2+α1+T2Tα˙2+α1+T2+T˙2−α2−T2TT˙2−α2−T2+T2TT2+α2+T2Tα2+T2+−α1+α2T−α1+α2−,t5−4tTα˙1+T2+T3+2t2+3TT2+4tα2+T2+4−α1+α2dt.One can setα1t=∑i=05piti,α2t=∑i=05ηiti, T2t=∑i=05μiti to obtain(42)T3=L∑i=05piti,∑i=05ηiti,∑i=05μiti,t+LT1,T4,T˙4,T¨4,T⃛4.Step 3: as mentioned above, the functionalJ is a function of p⟶,η⟶, and μ⟶. As Jp⟶,η⟶,μ⟶ is extremized, one can set ∂J/∂p⟶=0, ∂J/∂η⟶=0, and ∂J/∂μ⟶=0. This yields the linear algebraic equation Μp⟶η⟶μ⟶=Ν, which is solvable directly for p⟶η⟶μ⟶=Μ−1Ν. Hence, the approximate solution α1t,α2t,T2t is determined. The derivative of J with respect to p⟶,η⟶,andμ⟶ is used to obtain the matrix coefficient M.Step 4: to evaluate the solution’s accuracy without knowing the exact solution, we useL2-norm where the values of α1t=∑i=05piti,α2t=∑i=05ηiti, and T2t=∑i=05μiti are substituted in system (39).For differential equations,(43)α˙1+T2+T3−t5+4t2=∫01α˙1+T2+T3−t5+4t2dt1/2=0.0035,α˙2+α1+T22=∫01α˙2+α1+T22dt1/2=2×10−5,T˙2−α2−T22=∫01T˙2−α2−T22dt1/2=7×10−3.For equality algebraic constraint,∫01T2−2t2−32dt1/2=2×10−3.


## 4.1. Test Example

The following test example is designed with a suitable exact solution in mind to confirm the validity of the aforementioned approach, where(1)
The system is8×8 DAEs with control(2)
The system is represented as fourth order Hessenberg linear DAE with control(18)x˙11+2x11+3x12+x31+x41=f11,(19)x˙12−2x11−2x21+x22+x31+x41=u+f12,(20)x˙21+4x11+x12−2x21+x31=3u+f21,(21)x˙22+x12−2x21+x32=f22,(22)x˙31+3x21−2x22+x31=f31,(23)x˙32+3x21+x32=−2u+f32,(24)x31=−u+f42,(25)x32=f42,with(26)f11=t4+t+3,f12=t5+t3−t2−sint−4,f21=5t+t4−2t2−3sint−2,f22=t5+t3−2t2−t+1,f31=t4+4t3+3t2−4t+4,f32=t5+5t4+t3+6t2+2sint+3,f41=t4−sint−1,f42=t5+t3,A11=23−20,A12=00−21,A13=1001,A14=1001,A21=4101,A22=−20−20,A23=1001,A32=3−230,A33=1001,A43=1001,B1=01,B2=30,B3=0−2,B4=−10,x1=x11x12,x2=x21x22,x3=x31x32,x4=x41x42.Note thatA43A32A21A14 is an invertible matrix.Significantly, in the DAE system, variablex4 does not appear. Therefore, the algebraic constraint equations (18)–(23) can be differentiated with respect to t. Then, to obtain the following, substitute the differential constraints to equations (24) and (25).(27)A43x˙3=B4u˙+f˙4,A43B3u+f3−A32x2−A33x3=B4u˙+f˙4.Thus, equation (27) is considered a new constraint inserted to equations (18)–(25). In the last constraint, variable x4 is not exist. Therefore, to obtain the following, equation (27) can be differentiated one more time with respect to t.(28)−B4u¨+A43B3u˙−A4B2+A5B3u−f¨4+A43f˙3−A4f2−A5f3+A4A21x1+A4A22+A5A32x2+A4A23+A5A33x3=0,where A4=A43A32,A4=A43A33.Next, one can differentiate equation (28) again for the t set: c≜A43A32A21A14−1=0.1250−0.0417−0.50.5.Step 1:x4 is estimated as x4=Lx1,x2,x3+Lu,u˙,u¨,u⃛,f1,f2,f3,f˙2,f˙3,f¨3, where the derivative of the class of the consistent initial conditions can be (29)J0=x1t0,x2t0,x3t0|A43x3t0=B4ut0+f4t0−A43A32x2t0−A43A33x3t0=B4u˙t0+f˙4t0−A43B3ut0−A43f3t0A4A21x1t0+A5A22x2t0+A4A23x3t0+A5A32x2t0+A5A33x3t0=B4u¨t0+f¨4t0−A43B3u˙t0−A43f˙3t0+A4B2ut0+A4f2t0+A5B3ut0+A5f3t0.Step 2: as can be seen in the following, the definition of the variational function with the class of the consistent initial conditions can be written as follows:(30)JJ0x=∫t0tf12x˙1+A1x1+A2x2+A3x3Tx˙1+A1x1+A2x2+A3x3+x˙2+A21x1+A22x2+A23x3Tx˙2+A21x1+A22x2+A23x3+x˙3+A32x2+A33x3Tx˙3+A32x2+A33x3+A43x3TA43x3+A4x2+A5x3TA4x2+A5x3+A6x1+A2x2+A3x3TA6x1+A2x2+A3x3+A43x3t0TA43x3t0+A4x2t0+A5x3t0TA4x2t0+A5x3t0+A6x1t0+A2x2t0+A3x3t0TA6x1t0+A2x2t0+A3x3t0−F1Tx˙1+A1x1+A2x2+A3x3+F2Tx˙2+A21x1+A22x2+A23x3+F3Tx˙3+A32x2+A33x3+F4TA43x3+F5TA4x2+A5x3+F6TA6x1+A2x2+A3x3+F4t0TA43x3t0+F5t0TA4x2t0+A5x3t0+F6t0TA6x1t0+A2x2t0+A3x3t0dt,whereF1=B1u+f1−cA43B3u¨−cA43f¨3+cB4u⃛+cf⃛4+cA4B2u˙+cA4f˙2−cA4A21B1u−cA4A21f1−cA4A22B2u−cA4A22f2−cA4A23B3u−cA4A23f3+cA5B3u˙+cA5f˙3−cA5A32B2u−cA5A32f2,−cA5A33B3u−cA5A33f3,F2=B2u+f2, F3=B3u+f3, F4=B4u+f4, F5=B4u˙+f˙4−A43B3u−A43f3, F6=B4u¨+f¨4−A43B3u˙−A43f˙3+A4B2u+A4f2+A5B3u+A5f3.Step 3: one can setx11t=∑i=05aiφit,x12t=∑i=05biφit,x21t=∑i=05ciφit,x22t=∑i=05diφit,x31t=∑i=05viφit,andx32t=∑i=05wiφit, with φit=ti to obtainx4t=Lx1,x2,x3,t+Lu,u˙,u¨,u⃛,f1,f2,f3,f˙2,f˙3,f¨3Step 4: as mentioned above, the functionalJ is a function of a⟶,b⟶,c⟶,d⟶,v⟶,andw⟶. As Ja⟶,b⟶,c⟶,d⟶,v⟶,w⟶ is extremized, one can set ∂J/∂a⟶=0, ∂J/∂b⟶=0, ∂J/∂c⟶=0,∂J/∂d⟶=0, ∂J/∂v⟶=0, and ∂J/∂w⟶. It leads to the linear algebraic equation Μa⟶b⟶c⟶d⟶v⟶w⟶=Ν, which is solvable directly for a⟶b⟶c⟶d⟶v⟶w⟶=Μ−1Ν. By deriving J with respect to a⟶,b⟶,c⟶,d⟶,v⟶, and w⟶, M (the matrix coefficient) can be determined. Hence, the approximate solution x1t,x2t,x3t is obtained. The numerical values of the unknown coefficients a⟶,b⟶,c⟶,d⟶,v⟶,andw⟶ are listed in Table 1.Table 1
Unknown coefficients of the representation ofx1t,x2t,andx3t.
Coef. ofx11 statesa5a4a3a2a1a0Numerical values1×10−39–8×10−391×10−38–6×10−3910Coef. ofx12 statesb5b4b3b2b1b0Numerical values1×10−38–3×10−383×10−38–1×10−38–11Coef. ofx21 statesc5c4c3c2c1c0Numerical values2×10−38–7×10−386×10−3811×10−391Coef. ofx22 statesd5d4d3d2d1d0Numerical values1×10−40–1×10−391×10−39–1×10−392–1Coef. ofx31 statesv5v4v3v2v1v0Numerical values2×10−401–3×10−392×10−39–3×10−40−1Coef. ofx32 statesw5w4w3w2w1w0Numerical values11×10−3819×10−39–1×10−390Figure1 depicts the numerical solutions with equality states compared with the exact solution x1=t1−t,x2=t2+12t−1,x3t=t4−1t3+t5,andx4t=2tt2, for a given ut≜sint.Figure 1
The comparisons between the exact solution and approximate solution of the test example.


## 4.2. Application of the Generator Model

Consider the model of a generator described with the following parameters. On the right axis, the input is the angleθ connected to mass with inertia K which is rotated about the angle θ with angular velocity ν. The torque acting on the right and left sides are T1 and T2, respectively. The mass is then connected to the second axis with a generator. The electrical quantities and the variables described in the second axis are then assumed as in [15]. The rest of the electrical circuit includes one inductor and resistor. The model is illustrated in Figure 2.Figure 2
A model of generator.The system is now modelled using the equations representing the different parts and their connections.(31)1000000001000000001000000001000000001000000001000000000000000000T˙1T˙2V˙∅˙I˙u˙1u˙2u˙3+12001010−101101012110100001100100001−11000001001000000100000000100T1T2V∅Iu1u2u3=01100−110θ,with θt≜sint∈C2I,R,(32)A11=12−10,A12=0011,A13=1001,A14=1001,A21=2101,A22=1010,A23=1001,A32=1−110,A33=1001,A43=1001,B1=01,B2=10,B3=0−1,B4=10,x1=T1T2,x2=V∅,x3=Iu1,x4=u2u3.and A43A32A21A14 is an invertible matrix.In the DAE system, variablex4 does not appear. Therefore, the algebraic constraint equation (31) can be differentiated with respect to t. Then, three times until x4 is yielded in the following:(33)A43x˙3=B4θ˙,(34)A43B3θ+f3−A32x2−A33x3=B4θ˙,(35)−B4θ¨+A43B3θ˙−A4B2+A5B3θ+A4A21x1+A4A22+A5A32x2+A4A23+A5A33x3=0,where A4=A43A32,A5=A43A33.Next, one can differentiate equation (27) again for the t and set c≜A43A32A21A14−1=011/20.:Step 1: it is mentioned above thatx4 is obtained as x4=Lx1,x2,x3+Lθ,θ˙,θ¨,The derivative of the class of the consistent initial conditions is as follows:(36)J0=x1t0,x2t0,x3t0|A43x3t0=B4θt0−A43A32x2t0−A43A33x3t0=B4u˙t0−A43B3θt0A4A21x1t0+A4A22x2t0+A4A23x3t0+A5A32x2t0+A5A33x3t0=B4θ¨t0−A43B3θ˙t0+A4B2θt0+A5B3θt0,Step 2: the following is the definition of the variational function with the class of the consistent initial condition.(37)JJ0x=∫t0tf12x˙1+A1x1+A2x2+A3x3Tx˙1+A1x1+A2x2+A3x3+x˙2+A21x1+A22x2+A23x3Tx˙2+A21x1+A22x2+A23x3+x˙3+A32x2+A33x3Tx˙3+A32x2+A33x3+A43x3TA43x3+A4x2+A5x3TA4x2+A5x3+A6x1+A2x2+A3x3TA6x1+A2x2+A3x3+A43x3t0TA43x3t0+A4x2t0+A5x3t0TA4x2t0+A5x3t0+A6x1t0+A2x2t0+A3x3t0TA6x1t0+A2x2t0+A3x3t0−F1Tx˙1+A1x1+A2x2+A3x3+F2Tx˙2+A21x1+A22x2+A23x3+F3Tx˙3+A32x2+A33x3+F4TA43x3+F5TA4x2+A5x3+F6TA6x1+A2x2+A3x3+F4t0TA43x3t0+F5t0TA4x2t0+A5x3t0+F6t0TA6x1t0+A2x2t0+A3x3t0dt,whereF1=B1θ−cA43B3θ¨+cB4θ⃛+cs1B2θ˙−cs1A21B1θ−cs1A22B2θ−cs1A23B3θ+cs2B3θ˙+cs2A32B2θ−cs2A33B3θ,F2=B2θ, F3=B3θ, F4=B4θ, F5=B4θ˙−A43B3θ, F6=B4θ¨−A43B3θ˙+s1B2θ+s2B3θ.Step 3: one can setx11t=∑i=05aiφit,x12t=∑i=05biφit,x21t=∑i=05ciφit,x22t=∑i=05diφit,x31t=∑i=05viφit,andx32t=∑i=05wiφit, with φit=ti to obtain(38)x4t=L∑i=05aiti,∑i=05biti,∑i=05citi,∑i=05diti,∑i=05viti,∑i=05witi,t+Lθ,θ˙,θ¨,θ�.Step 4:J is a function of a⟶,b⟶,c⟶,d⟶,v⟶,andw⟶, and as Ja⟶,b⟶,c⟶,d⟶,v⟶,w⟶ is extremized, one can set ∂J/∂a⟶=0, ∂J/∂b⟶=0, ∂J/∂c⟶=0,∂J/∂d⟶=0, ∂J/∂v⟶=0, and ∂J/∂w⟶=0. It is led to the linear algebraic equation Μa⟶b⟶c⟶d⟶v⟶w⟶=Ν, which is solvable directly for a⟶b⟶c⟶d⟶v⟶w⟶=Μ−1Ν. Hence, the approximate solution x1t,x2t,x3t,x4t is determined. Lastly, the derivative of J with respect to a⟶,b⟶,c⟶,d⟶,v⟶, and w⟶ is used to obtain the matrix coefficient M.Step 5: to evaluate the solution’s accuracy without knowing the exact solution, the model utilisesL2-norm by substituting the values of x11t=∑i=05aiti,x12t=∑i=05biti, x21t=∑i=05citi,x22t=∑i=05diti, x31t=∑i=05viti, and x32t=∑i=05witi in system (31). The error presented in the test example at each point along time t∈0,1 and for different kinds of θ are displayed (Figures 3–6).Figure 3
The error of the generator system solution forθ=e−t.Figure 4
The error of the generator system solution forθ=sint.Figure 5
The error of the generator system solution forθ=sint+cost.Figure 6
The error of the generator system solution forθ=t4+t2.


## 4.3. Application of the Rotating Masses System

Consider a system of two rotating masses that are described by the torquesT1t,T2t,T3t,andT4t and the angular velocities α1t and α2t. The masses have the moments of inertia γ1 and γ1. Figure 7 illustrates this scenario.Figure 7
Connected rotating masses.The system to matrically describes this model is provided as(39)γ10000γ20000100000α˙1α˙2T˙2T˙3+010000−1011−111000α1α1T2T3=10000001T1T4,with γ1=γ2=1 and T1=t5−4t,T4=2t2+3.Step 1: sinceA43A32A21A14 exists as an invertible matrix, the variable T3 does not appear and T˙2=4t through differentiation of the algebraic constraint of system (39), we can continue to differentiate until T3 obtained as T3=α1+t5−4t with consistent initial condition:(40)J0=α1t0,α2t0,T2t0|T2=2t02+3,α2+T2=4t0,−α1+α2=4.Step 2: using the variational formulation associated with this consistent initial condition,(41)JJ0x=∫t0tf12α˙1+T2+T3Tα˙1+T2+T3+α˙2+α1+T2Tα˙2+α1+T2+T˙2−α2−T2TT˙2−α2−T2+T2TT2+α2+T2Tα2+T2+−α1+α2T−α1+α2−,t5−4tTα˙1+T2+T3+2t2+3TT2+4tα2+T2+4−α1+α2dt.One can setα1t=∑i=05piti,α2t=∑i=05ηiti, T2t=∑i=05μiti to obtain(42)T3=L∑i=05piti,∑i=05ηiti,∑i=05μiti,t+LT1,T4,T˙4,T¨4,T⃛4.Step 3: as mentioned above, the functionalJ is a function of p⟶,η⟶, and μ⟶. As Jp⟶,η⟶,μ⟶ is extremized, one can set ∂J/∂p⟶=0, ∂J/∂η⟶=0, and ∂J/∂μ⟶=0. This yields the linear algebraic equation Μp⟶η⟶μ⟶=Ν, which is solvable directly for p⟶η⟶μ⟶=Μ−1Ν. Hence, the approximate solution α1t,α2t,T2t is determined. The derivative of J with respect to p⟶,η⟶,andμ⟶ is used to obtain the matrix coefficient M.Step 4: to evaluate the solution’s accuracy without knowing the exact solution, we useL2-norm where the values of α1t=∑i=05piti,α2t=∑i=05ηiti, and T2t=∑i=05μiti are substituted in system (39).For differential equations,(43)α˙1+T2+T3−t5+4t2=∫01α˙1+T2+T3−t5+4t2dt1/2=0.0035,α˙2+α1+T22=∫01α˙2+α1+T22dt1/2=2×10−5,T˙2−α2−T22=∫01T˙2−α2−T22dt1/2=7×10−3.For equality algebraic constraint,∫01T2−2t2−32dt1/2=2×10−3.


## 5. Results and Discussion

Using the test example to discuss and validate the presented method for the system of index-four Hessenberg differential algebraic equations, the results indicate how the exact and the approximate solutions match exceedingly well in a way that demonstrates the accuracy and efficiency of the model. Applications 4.1 and 4.2 show to employ differential algebraic equations as a modelling environment in system applications.In examples where a state space cannot effectively describe the whole system, the model system will be reduced to a standard state-space system on the manifold of the resulting algebraic constraints which is provide a simplest system to deal with and this reduction reduce accuracy of the results. These examples are taken as a test for the proposed method. A step-by-step implementation, the approximate solution is parameterised via polynomial base functions, which are dense inCI,R3. Despite reasonably small numbers of these polynomials with an unknown coefficient, the obtained solutions prove to be rather accurate and efficient.


## 6. Conclusions

This research presented an efficient and reliable to solve higher index (index four) linear Hessenberg control DAEs. To ensure a controllable and linearly independent set of functions in its base, the setting space of the solvability is assumed to be the separable Banach space. The differentiation index is adopted and used repeatedly in this study until we achieve the required conditions around the implicit function theorem. This outcome depends on the index of the system. Reduced Hessenberg DAEs are obtained with the help of the implicit function theorem. An operator form has been used to write the latest system. For this case, we developed the variational formulation. The representation of the solution has been as a linear combination of elements in the setting space. The cost for utilising these new measures consists of more smoothness conditions on the control signal.Generally, the treating of the system has been as a perturbation in the nonhomogenous functions. The example, index four, is efficiently illustrated. Despite a rather simple basis function type, the evaluation of the approach revealed excellent accuracy Due to the decision-maker, these bases can be modified to include spline and orthonormal basis.



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