CHAPTER 2 THE MAXIMUM PRINCIPLE: CONTINUOUS TIME. Chapter2 p. 1/67

Transcription

1 CHAPTER 2 THE MAXIMUM PRINCIPLE: CONTINUOUS TIME Chapter2 p. 1/67

2 THE MAXIMUM PRINCIPLE: CONTINUOUS TIME Main Purpose: Introduce the maximum principle as a necessary condition to be satisfied by any optimal control. Necessary conditions for optimization of dynamic systems. General derivation by Pontryagin et al. in A simple (but not completely rigorous) proof using dynamic programming. Examples. Statement of sufficiency conditions. Computational method. Chapter2 p. 2/67

3 2.1 STATEMENT OF THE PROBLEM Optimal control theory deals with the problem of optimization of dynamic systems. Chapter2 p. 3/67

4 2.1.1 THE MATHEMATICAL MODEL State equation ẋ(t) = f(x(t),u(t),t), x(0) = x 0, (1) where the vector of state variables, x(t) E n, the vector of control variables, u(t) E m, and the function f : E n E m E 1 E n. The function f is assumed to be continuously differentiable. Chapter2 p. 4/67

5 2.1.1 THE MATHEMATICAL MODEL CONT. The path x(t),t [0,T], is called a state trajectory and u(t), t [0, T], is called a control trajectory. Admissible control u(t), t [0,T], is piecewise continuous and satisfies, in addition, u(t) Ω(t) E m, t [0,T]. (2) Chapter2 p. 5/67

6 2.1.3 THE OBJECTIVE FUNCTION Objective function is defined as follows J = T 0 F(x(t),u(t),t)dt + S[x(T),T], (3) where the functions F : E n E m E 1 E 1 and S : E n E 1 E 1 are assumed to be continuously differentiable. Chapter2 p. 6/67

7 2.1.4 THE OPTIMAL CONTROL PROBLEM The problem is to find an admissible control u, which maximizes the objective function (3) subject to the state equation (1) and the control constraints (2). We now restate the optimal control problem as: max u(t) Ω(t) subject to { J = T ẋ = f(x,u,t), x(0) = x 0. 0 } F(x,u,t)dt + S[x(T),T] The control u is called an optimal control and x is called the corresponding optimal trajectory. The optimal value of the objective function will be denoted as J(u ) or J. (4) Chapter2 p. 7/67

8 THREE CASES Case 1. The optimal control problem (4) is said to be in Bolza form. Case 2. When S 0, it is said to be in Lagrange form. Case 3. When F 0, it is said to be in Mayer form. If F 0 and S is linear, it is in linear Mayer form, i.e., max {J = cx(t)} u(t) Ω(t) subject to ẋ = f(x,u,t), x(0) = x 0, (5) where c = (c 1,c 2,,c n ) E n. Chapter2 p. 8/67

9 The Bolza form can be reduced to the linear Mayer form: FORMS Introduce a new state vector y = (y 1,y 2,...,y n+1 ), having n + 1 components defined as follows: y i = x i for i = 1,...,n, and ẏ n+1 = F(x,u,t) + S(x,t) x f(x,u,t) + S(x,t), t y n+1 (0) = S(x 0, 0), (6) so that the objective function is J = ĉy(t) = y n+1 (T), where ĉ = (0, 0,...,0, 1). If we now integrate (6) from 0 to T, we have }{{} n J = ĉy(t) = y n+1 (T) = T 0 F(x,u,t)dt + S[x(T),T], (7) which is the same as the objective function J in (4). Chapter2 p. 9/67

10 PRINCIPLE OF OPTIMALITY Statement of Bellman s Optimality Principle "An optimal policy has the property that whatever the initial state and initial decision are, the remaining decisions must constitute an optimal policy with regard to the state resulting from the first decision." Chapter2 p. 10/67

11 PRINCIPLE OF OPTIMALITY CONT. A B J J BE AB JBCE E ASSERTION: If ABE (shown in blue) is an optimal path from A to E, then BE (in blue) is an optimal path from B to E PROOF. Suppose it is not. Then there is another path (existence is assumed here) BCE (in red), which is optimal from B to E, i.e., J BCE > J BE. But then J ABE = J AB + J BE < J AB + J BCE = J ABCE. This contradicts the hypothesis that ABE is an optimal path from A to E. C Chapter2 p. 11/67

12 A DYNAMIC PROGRAMMING EXAMPLE Stagecoach Problem Stages Costs (c i,j ): Chapter2 p. 12/67

13 A DYNAMIC PROGRAMMING EXAMPLE CONT. Stagecoach Problem Stages Note that is a greedy (or myopic) path that minimizes cost at each stage. However, this may not be the minimal cost solution. For example, yields a cheaper cost than Chapter2 p. 13/67

14 SOLUTION TO THE DP EXAMPLE Let 1 u 1 u 2 u 3 10 be the optimal path. Let f n (s,u n ) be the minimal cost at stage n given that the current state is s and the decision taken is u n. Let f n(s) be the minimal cost at stage n if the current state is s. Then f n(s) = min u n f(s,u n ) = min u n { } c s,un + fn+1(u n ). This is the Recursion Equation of DP. It can be solved by a backward procedure, which starts at the terminal stage and stops at the initial stage. Chapter2 p. 14/67

15 SOLUTION TO THE DP EXAMPLE CONT. Stage 4. s f 4 (s) u Stage 3. s f 3 (s, u 3 ) = C s,u3 + f4 (u 3) f3 (s) u = = = = = = Chapter2 p. 15/67

16 SOLUTION TO THE DP EXAMPLE CONT. Stage 2. Stage 1. s f 2 (s, u 2 ) = C s,u2 + f3 (u 2) f2 (s) u = = = , = = = = = = , s f 1 (s, u 1 ) = C s,u1 + f 2 (u 1) f 1 (s) u = = = ,4 The optimal paths: Total Cost = 11. Chapter2 p. 16/67

17 2.2 DYNAMIC PROGRAMMING AND THE MAXIMUM PRINCIPLE The Hamilton-Jacobi-Bellman Equation { T V (x,t) = max u(s) Ω(s) t where for s t, } F(x(s),u(s),s)ds + S(x(T),T), (8) dx ds = f(x(s),u(s),s), x(t) = x. Chapter2 p. 17/67

18 FIGURE 2.1 AN OPTIMAL PATH IN THE STATE-TIME SPACE Chapter2 p. 18/67

19 TWO CHANGES IN THE OBJECTIVE FUNCTION (i) The incremental change in J from t to t + δt is given by the integral of F(x,u,t) from t to t + δt. (ii) The value function V (x + δx,t + δt) at time t + δt. In equation form this is { t+δt V (x,t) = F[x(τ),u(τ),τ]dτ max u(τ) Ω(τ) τ [t,t+δt] t } +V [x(t + δt),t + δt], (9) Since F is a continuous function, the integral in (9) is approximately F(x,u,t)δt so that we can rewrite (9) as V (x,t) = max {F(x,u,t)δt + V [x(t + δt),t + δt]}+o(δt). u Ω(t) (10) Chapter2 p. 19/67

20 TWO CHANGES IN THE OBJECTIVE FUNCTION CONT. Assume that the value function V is continuously differentiable. Use the Taylor series expansion of V with respect to δt and obtain V [x(t + δt),t + δt] = V (x,t) + [V x (x,t)ẋ + V t (x,t)]δt + o(δt), (11) Substituting for ẋ from (1), we obtain { V (x,t) = max F(x,u,t)δt + V (x,t) u Ω(t) } +V x (x,t)f(x,u,t)δt + V t (x,t)δt + o(δt). (12) Cancelling V (x,t) on both sides and then dividing by δt we get { } 0 = max F(x,u,t) + V x (x,t)f(x,u,t) + V t (x,t) + o(δt). u Ω(t) δt (13) Chapter2 p. 20/67

21 TWO CHANGES IN THE OBJECTIVE FUNCTION CONT. Letting δt 0 in (10) gives the equation 0 = max {F(x,u,t) + V x (x,t)f(x,u,t) + V t (x,t)}, (14) u Ω(t) for which the boundary condition is V (x,t) = S(x,T). (15) V x (x,t) can be interpreted as the marginal contribution vector of the state variable x to the maximized objective function. Denote it by the adjoint (row) vector λ(t) E n, i.e., λ(t) = V x (x (t),t) := V x (x,t) x=x (t). (16) Chapter2 p. 21/67

22 We introduce the so-called Hamiltonian or, simply, HAMILTON-JACOBI-BELLMAN EQUATION H[x,u,V x,t] = F(x,u,t) + V x (x,t)f(x,u,t) (17) H(x,u,λ,t) = F(x,u,t) + λf(x,u,t). (18) We can rewrite equation (14) as the following equation, 0 = max [H(x,u,V x,t) + V t ], (19) u Ω(t) called the Hamilton-Jacobi-Bellman equation or, simply, the HJB equation. Chapter2 p. 22/67

23 HAMILTONIAN MAXIMIZING CONDITION From (19), we can get the Hamiltonian maximizing condition of the maximum principle, H[x (t),u (t),λ(t),t] + V t (x (t),t) H[x (t),u,λ(t),t] Cancelling the term V t on both sides, we obtain for all u Ω(t). +V t (x (t),t). (20) H[x (t),u (t),λ(t),t] H[x (t),u,λ(t),t] (21) Remark: H decouples the problem over time by means of λ(t), which is analogous to dual variables or shadow prices in Linear Programming. Chapter2 p. 23/67

24 2.2.2 DERIVATION OF THE ADJOINT EQUATION Let x(t) = x (t) + δx(t), (22) where δx(t) < ε for a small positive ε. Fix t and use the HJB equation in (19) as H[x (t),u (t),v x (x (t),t),t] + V t (x (t),t) H[x(t),u (t),v x (x(t),t),t] + V t (x(t),t). (23) LHS= 0 from (19) since u (t) maximizes H + V t. RHS will be zero if u (t) also maximizes H + V t with x(t) as the state. In general x(t) x (t), and thus RHS 0. But then RHS x(t)=x (t) = 0 RHS is maximized at x(t) = x (t). Chapter2 p. 24/67

25 2.2.2 DERIVATION OF THE ADJOINT EQUATION CONT. Since, x(t) is unconstrained, we have RHS x x(t)=x (t) = 0, or H x [x (t),u (t),v x (x (t),t),t] + V tx (x (t),t) = 0. (24) By definition of H, F x +V x f x +f T V xx +V tx = F x +V x f x +(V xx f) T +V tx = 0. (25) Note: (25) assumes V to be twice continuously differentiable. See (1.16) or Exercise 1.9 for details. Chapter2 p. 25/67

26 2.2.2 DERIVATION OF THE ADJOINT EQUATION By definition (1.16) of λ(t), the row vector λ(t) is dv x dt = ( ) dvx1 dt, dv x2 dt,, dv xn dt = (V x1 xẋ + V x1 t,v x2 xẋ + V x2 t,,v xn xẋ + V xn t) = ( n i=1 V x 1 x i x i, n = (V xx ẋ) T + V xt = (V xx f) T + V tx. Note V x1 x = (V x1 x 1, V x1 x 2,..., V x1 x n ), and CONT. i=1 V x 2 x i x i,, n i=1 V ) x n x i x i + (Vx ) t (26) ¼ V x1 x 1 V x1 x 2... V x1 x n ¼ ½ ẋ 1 ½ V xx ẋ = V x2 x 1 V x2 x 2... V x2 x n.. ẋ 2.,a column vector. V xn x 1 V xn x 2... V xn x n ẋ n Chapter2 p. 26/67

27 2.2.2 DERIVATION OF THE ADJOINT EQUATION CONT. Using (25) and (26) we have Using (16) we have dv x dt = F x V x f x. (27) λ = F x λf x. From (18) we have λ = H x. (28) Chapter2 p. 27/67

28 TERMINAL BOUNDARY OR TRANSVERSALITY CONDITION: λ(t) = S(x,T) x=x(t) = S x [x(t),t]. (29) x (28) and (29) can determine the adjoint variables. From (18), we can rewrite the state equation as ẋ = f = H λ. (30) From (28), (29), (30) and (1), we get ẋ = H λ, x(0) = x 0, λ = H x, λ(t) = S x [x(t),t], (31) called a canonical system of equations or canonical adjoints. Chapter2 p. 28/67

29 FREE AND FIXED END POINT PROBLEMS In our problem x(t) is free. In this case, S 0 (no salvage value) λ(t) = 0 However, if x(t) is not free, that is, x(t) fixed (given) λ(t) = a constant to be determined. In other words, λ(t) is free. Chapter2 p. 29/67

30 2.2.3 THE MAXIMUM PRINCIPLE The necessary conditions for u to be an optimal control are: ẋ = f(x,u,t),x (0) = x 0, λ = H x [x,u,λ,t], λ(t) = S x [x (T),T], H[x (t),u (t),λ(t),t] H[x (t),u,λ(t),t], (32) for all u Ω(t), t [0,T]. Chapter2 p. 30/67

31 2.2.4 ECONOMIC INTERPRETATION OF THE MAXIMUM PRINCIPLES J = T 0 F(x,u,t)dt + S[x(T),T], where F is the instantaneous profit rate per unit of time, and S[x,T] is the salvage value. Multiplying (18) formally by dt and using the state equation (1) gives Hdt = Fdt + λfdt = Fdt + λẋdt = Fdt + λdx. F(x, u, t)dt: direct contribution to J from t to t + dt λdx: indirect contribution to J due to added capital dx Hdt: total contribution to J from time t to t + dt, when x(t) = x and u(t) = u in the interval [t, t + dt]. Chapter2 p. 31/67

32 2.2.4 ECONOMIC INTERPRETATION OF THE MAXIMUM PRINCIPLES CONT. By (28) and (29), we have λ = H x = F x λ f x, λ(t) = S x[x(t),t]. Rewrite the first equation as dλ = H x dt = F x dt + λf x dt. dλ: marginal cost of holding capital x from t to t + dt H x dt: marginal revenue of investing the capital F x dt: direct marginal contribution λf x dt: indirect marginal contribution Thus, the adjoint equation implies MC = MR. Chapter2 p. 32/67

33 EXAMPLE 2.1 Consider the problem: max { J = 1 0 } xdt (33) subject to the state equation and the control constraint ẋ = u, x(0) = 1 (34) u Ω = [ 1, 1]. (35) Note that T = 1, F = x, S = 0, and f = u. Because F = x, we can interpret the problem as one of minimizing the (signed) area under the curve x(t) for 0 t 1. Chapter2 p. 33/67

34 SOLUTION OF EXAMPLE 2.1 We form the Hamiltonian H = x + λu. (36) Because the Hamiltonian is linear in u, the form of the optimal control, i.e., the one that would maximize the Hamiltonian, is 1 if λ(t) > 0, u (t) = undefined if λ(t) = 0, 1 if λ(t) < 0. It is called Bang-Bang Control. In the notation of Section 1.4, (37) u (t) = bang[ 1, 1;λ(t)]. (38) Chapter2 p. 34/67

35 SOLUTION OF EXAMPLE 2.1 CONT. To find λ, we write the adjoint equation λ = H x = 1, λ(1) = S x [x(t),t] = 0. (39) Because this equation does not involve x and u, we can easily solve it as λ(t) = t 1. (40) It follows that λ(t) = t 1 0 for all t [0, 1]. So u (t) = 1, t [0, 1). Since λ(1) = 0, we can also set u (1) = 1, which defines u at the single point t = 1. We thus have the optimal control u (t) = 1 for t [0, 1]. Chapter2 p. 35/67

36 SOLUTION OF EXAMPLE 2.1 CONT. Substituting this into the state equation (34) we have ẋ = 1, x(0) = 1, (41) whose solution is x(t) = 1 t for t [0, 1]. (42) The graphs of the optimal state and adjoint trajectories appear in Figure 2.2. Note that the optimal value of the objective function is J = 1/2. Chapter2 p. 36/67

37 FIGURE 2.2 OPTIMAL STATE AND ADJOINT TRAJECTORIES FOR EXAMPLE 2.1 Chapter2 p. 37/67

38 EXAMPLE 2.2 Let us solve the same problem as in Example 2.1 over the interval [0, 2] so that the objective is to { 2 maximize J = 0 } xdt. (43) As before, the dynamics and constraints are (33) and (34), respectively. Here we want to minimize the signed area between the horizontal axis and the trajectory of x(t) for 0 t 2. Chapter2 p. 38/67

39 SOLUTION OF EXAMPLE 2.2 As before the Hamiltonian is defined by (36) and the optimal control is as in (38). The adjoint equation λ = 1, λ(2) = 0 (44) is the same as (39), except that now T = 2 instead of T = 1. The solution of (44) is easily found to be λ(t) = t 2, t [0, 2]. (45) Hence the state equation (41) and its solution (42) are exactly the same. The graph of λ(t) is shown in Figure 2.3. The optimal value of the objective function is J = 0. Chapter2 p. 39/67

40 FIGURE 2.3 OPTIMAL STATE AND ADJOINT TRAJECTORIES FOR EXAMPLE 2.2 Chapter2 p. 40/67

41 EXAMPLE 2.3 The next example is: max { J = } 2 x2 dt (46) subject to the same constraints as in Example 2.1, namely, ẋ = u, x(0) = 1, u Ω = [ 1, 1]. (47) Here F = (1/2)x 2 so that the interpretation of the objective function (46) is that we are trying to find the trajectory x(t) in order that the area under the curve (1/2)x 2 is minimized. Chapter2 p. 41/67

42 SOLUTION OF EXAMPLE 2.3 The Hamiltonian is H = 1 2 x2 + λu, (48) which is linear in u so that the optimal policy is The adjoint equation is u (t) = bang [ 1, 1;λ]. (49) λ = H x = x, λ(1) = 0. (50) Here the adjoint equation involves x so that we cannot solve it directly. Because the state equation (47) involves u, which depends on λ, we also cannot integrate it independently without knowing λ. Chapter2 p. 42/67

43 SOLUTION OF EXAMPLE 2.3 CONT. A way out of this dilemma is to use some intuition. Since we want to minimize the area under (1/2)x 2 and since x(0) = 1, it is clear that we want x to decrease as quickly as possible. Let us therefore temporarily assume that λ is nonpositive in the interval [0, 1] so that from (49) we have u = 1 throughout the interval. (In Exercise 2.5, you will be asked to show that this assumption is correct.) With this assumption, we can solve (47) as x(t) = 1 t. (51) Chapter2 p. 43/67

44 SOLUTION OF EXAMPLE 2.3 CONT. Substituting this into (50) gives λ = 1 t. Integrating both sides of this equation from t to 1 gives or 1 t λ(τ)dτ = 1 t (1 τ)dτ, λ(1) λ(t) = (τ 1 2 τ2 ) 1 t, which, using λ(1) = 0, yields λ(t) = 1 2 t2 + t 1 2. (52) Chapter2 p. 44/67

45 SOLUTION OF EXAMPLE 2.3 CONT. The reader may now verify that λ(t) is nonpositive in the interval [0, 1], justifying our original assumption. Hence, (51) and (52) satisfy the necessary conditions. In Exercise 2.6, you will be asked to show that they satisfy sufficient conditions derived in Section 2.4 as well, so that they are indeed optimal. Figure 2.4 shows the graphs of the optimal trajectories. Chapter2 p. 45/67

46 FIGURE 2.4 OPTIMAL TRAJECTORIES FOR EXAMPLE 2.3 AND EXAMPLE 2.4 * t - Chapter2 p. 46/67

47 EXAMPLE 2.4 Let us rework Example 2.3 with T = 2, i.e., solve the problem: { 2 max J = 1 } 2 x2 dt subject to the constraints ẋ = u, x(0) = 1, u Ω = [ 1, 1]. 0 (53) It would be clearly optimal if we could keep x (t) = 0, t [1, 2]. This is possible by setting u (t) = 0, t [1, 2]. Note u (t) = 0, t [1, 2], is a singular control, since it gives λ(t) = 0, t [1, 2]. See Fig Chapter2 p. 47/67

48 EXAMPLE 2.5 The problem is: subject to max { J = 2 0 } (2x 3u u 2 )dt (54) ẋ = x + u, x(0) = 5 (55) and the control constraint u Ω = [0, 2]. (56) Chapter2 p. 48/67

49 SOLUTION OF EXAMPLE 2.5 The Hamiltonian is H = (2x 3u u 2 ) + λ(x + u) = (2 + λ)x (u 2 + 3u λu). (57) The optimal control is The adjoint equation is u(t) = λ(t) 3. (58) 2 λ = H x = 2 λ, λ(2) = 0. (59) Chapter2 p. 49/67

50 SOLUTION OF EXAMPLE 2.5 CONT. Using the integrating factor as shown in Appendix A.6, its solution is λ(t) = 2(e 2 t 1). The optimal control is 2 if e 2 t 2.5 > 2, u (t) = e 2 t 2.5 if 0 e 2 t 2.5 2, 0 if e 2 t 2.5 < 0. (60) It can be written as u (t) = sat [0, 2;e 2 t 2.5]. Chapter2 p. 50/67

51 FIGURE 2.5 OPTIMAL CONTROL FOR EXAMPLE 2.5 u e 2 t u (t) 0 t t t Chapter2 p. 51/67

52 2.4 SUFFICIENT CONDITIONS The Envelope Theorem gives Here is why? H 0 (x,λ,t) = max H(x,u,λ,t). (61) u Ω(t) H 0 (x,λ,t) = H(x,u 0,λ,t). (62) H 0 x(x,λ,t) = H x (x,u 0,λ,t). (63) H 0 x(x,λ,t) = H x (x,u 0,λ,t) + H u (x,u 0,λ,t) u0 x. (64) But H u (x,u 0,λ,t) u0 x = 0, (65) since either H ui = 0 or u 0 i/ x j = 0 or both. Chapter2 p. 52/67

53 THEOREM 2.1 (Sufficiency Conditions) Let u (t), and the corresponding x (t) and λ(t) satisfy the maximum principle necessary condition (32) for all t [0,T]. Then, u is an optimal control if H 0 (x,λ(t),t) is concave in x for each t and S(x,T) is concave in x. Chapter2 p. 53/67

54 PROOF OF THEOREM 2.1 By definition of H 0, H[x(t),u(t),λ(t),t] H 0 [x(t),λ(t),t]. (66) Since H 0 is differentiable and concave, H 0 [x(t),λ(t),t] H 0 [x (t),λ(t),t] (67) +Hx[x 0 (t),λ(t),t][x(t) x (t)]. Using the Envelope Theorem, we can write H[x(t),u(t),λ(t),t] H[x (t),u (t),λ(t),t] (68) +H x [x (t),u (t),λ(t),t][x(t) x (t)]. Chapter2 p. 54/67

55 PROOF OF THEOREM 2.1 CONT. By the definition of H and the adjoint equation, F[x(t),u(t),t] + λ(t)f[x(t),u(t),t] (69) F[x (t),u (t),t] + λ(t)f[x (t),u (t),t] λ(t)[x(t) x (t)]. Using the state equation, F[x (t),u (t),t] F[x(t),u(t),t] λ(t)[x(t) x (t)] + λ(t)[ẋ(t) ẋ (t)]. (70) Chapter2 p. 55/67

56 PROOF OF THEOREM 2.1 CONT. Since S is differentiable and concave S[x(T),T] S[x (T),T] + S x [x (T),T][x(T) x (T)] (71) or, S[x (T),T] S[x(T),T] + S x [x (T),T][x(T) x (T)] 0. Integrating (70) and adding (72), J(u ) J(u) + S x [x (T),T][x(T) x (T)] (72) λ(t)[x(t) x (T)] λ(0)[x(0) x (0)], (73) J(u ) J(u). (74) Chapter2 p. 56/67

57 EXAMPLE 2.6 Let us show that the problems in Examples 2.1 and 2.2 satisfy the sufficient conditions. We have from (36) and (61), H 0 = x + λu 0, where u 0 is given by (37). Since u 0 is a function of λ only, H 0 (x,λ,t) is certainly concave in x for any t and λ. Since S(x, T) = 0, the sufficient conditions hold. Chapter2 p. 57/67

58 FIXED-END-POINT PROBLEM The terminal values of the state variables are completely specified, i.e., x(t) = k E, where k is a vector of constants. Recall that the proof of the sufficient conditions requires (73) to hold, i.e., J(u ) J(u) + S x [x (T),T][x(T) x (T)] λ(t)[x(t) x (T)] λ(0)[x(0) x (0)]. Since x(t) x (T) = 0, the RHS vanishes regardless of the value of λ(t) in this case. This means that the sufficiency result would go through for any value of λ(t). Not surprisingly, therefore, the transversality condition is λ(t) = β. (75) Chapter2 p. 58/67

59 SOLVING A TPBVP BY USING SPREADSHEET SOFTWARE Example 2.7. Consider the problem: { 1 max J = 1 } 2 (x2 + u 2 )dt subject to 0 ẋ = x 3 + u, x(0) = 5. (76) Chapter2 p. 59/67

60 SOLVING A TPBVP BY USING SPREADSHEET SOFTWARE CONT. Solution: λ = x + 3x 2 λ, λ(1) = 0. (77) ẋ = x 3 + λ, x(0) = 5. (78) x(t + t) = x(t) + [ x(t) 3 + λ(t)] t, x(0) = 5, (79) λ(t + t) = λ(t) +[x(t) +3x(t) 2 λ(t)] t, λ(1) = 0. (80) Chapter2 p. 60/67

61 SOLVING A TPBVP BY USING SPREADSHEET Set t = 0.01, λ(0) = 0.2 and x(0) = 5. SOFTWARE CONT. Chapter2 p. 61/67

62 SOLVING A TPBVP BY USING SPREADSHEET SOFTWARE CONT. Chapter2 p. 62/67

63 SOLVING A TPBVP BY USING SPREADSHEET SOFTWARE CONT. Chapter2 p. 63/67

64 SOLVING A TPBVP BY USING SPREADSHEET SOFTWARE CONT. Chapter2 p. 64/67

65 SOLVING A TPBVP BY USING SPREADSHEET SOFTWARE CONT. Chapter2 p. 65/67

66 SOLVING A TPBVP BY USING SPREADSHEET SOFTWARE CONT. Chapter2 p. 66/67

67 SOLVING A TPBVP BY USING SPREADSHEET SOFTWARE CONT. Chapter2 p. 67/67