LONG TERM BEHAVIOR OF SOLUTIONS FOR RICCATI INITIAL-VALUE PROBLEMS

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1 Electronic Journal of Differential Equations, Vol. 8(8, No. 4, pp. 8. ISSN: URL: or ftp ejde.math.txstate.edu (login: ftp LONG TERM BEHAVIOR OF SOLUTIONS FOR RICCATI INITIAL-VALUE PROBLEMS SARAH Y. BAHK, NADEJDA E. DYAKEVICH, STEFAN C. JOHNSON Abstract. The Riccati equation has been known since the early 7s. Numerous papers have been written on the solvability of its special cases. However, to the best of our knowledge, there are no papers that investigate the exact (equation specific conditions for unbounded growth in finite time of solutions for Riccati initial-value problems. In this paper, we first derive conditions that are necessary and sufficient for the solutions of Riccati problems with constant coefficients to grow unbounded in finite time. Then we use a comparison method to extend these results to Riccati problems with variable coefficients.. Introduction Count Jacopolo Francesco Riccati (May 8, April 5, 754 is famous for introducing and researching the solvability of the equation that now bears his name: y (t = a(ty + b(ty + c(t. (. The matrix form of this equation is very important in modern times since it is used extensively in design problems in filtering and control [, 3]. Even though the Riccati equation (. is not solvable in general, numerous methods have been developed for finding solutions for many of its special cases [], [6]-[]. In Section, we consider real solutions of the initial-value problem y (t = ay + by + c, y( = d, (. where a, b, c, and d are real numbers and t represents time. We determine conditions on the constants a, b, c, and d that are necessary and sufficient for y(t to approach either + or as t approaches some finite value t b. We provide exact values for the time t b for the cases when is positive, negative, or zero. In particular, we are interested in the first occurrence of blow-up. We do not consider behavior of y(t for t > t b. Mathematics Subject Classification. 34C. Key words and phrases. Riccati equation; unbounded growth in finite time; comparison. c 8 Texas State University - San Marcos. Submitted May 3, 8. Published October 4, 8.

2 S. Y. BAHK, N. E. DYAKEVICH, S. C. JOHNSON EJDE-8/4 In Section 3, we use a comparison theorem to extend the results to the more general initial-value problem y (t = a(ty + b(ty + c(t, y( = d, where a(t, b(t, c(t are continuous and differentiable functions for t, and d is a real number.. Riccati Problems with Constant Coefficients Theorem.. The following is true for the solution y(t of (.: ( Let >. If a >, then y(t +, while if a <, then y(t. ( Let =. If a > and d > b, then y(t +. If a < and d < b, then y(t. Otherwise, y(t is bounded for any finite t >. If d = b, then y(t d. (3 Let <. If a > and d > b+ b 4ac, then y(t +. If a < and d < b+ b 4ac, then y(t. Otherwise, y(t is bounded for any finite t >. If d = b± b 4ac, then y(t d. (4 If a =, then y(t is bounded for all t >. Proof. ( Let >. The solution of the initial value problem (. can be found using separation of variables: [ t ( b + d ] y(t = tan + arctan b. (. We can find the blow-up time t b by solving for t in equation t ( b + d + arctan = π, π t b = arctan [ b + d ]. Also, π < arctan [ b + d ] π <. This implies that t b is always positive and that the solution y(t of (. is guaranteed to blow-up as t approaches t b. Also, from equation (. we notice that if a >, then y(t +, while if a <, then y(t. Changing the initial value d cannot prevent blow-up from occurring. However, d influences the blow-up time t b. For example, if a <, then decreasing d will accelerate the blow-up. If a >, then increasing d will accelerate the blow-up. ( Let =. Using separation of variables, we obtain dy a(y + b dt. Integration leads to the solution: y(t = = d + b a( dt bt b. (.

3 EJDE-8/4 LONG TERM BEHAVIOR OF SOLUTIONS 3 To find the blow-up time, we set the denominator of the first term in (. equal to and solve for t in dt bt =, t b = d + b. From the inequality t b >, and from (. we obtain the following: If a > and d > b b, then y(t +, while if a < and d <, then y(t. Initial value d is very important since certain values can prevent blow-up from occurring. Also, d influences the blow-up time t b. If blow-up occurs for some value d, then decreasing d (if a < or increasing d (if a > will accelerate the blow-up. If d = b, then y(t d satisfies the initial-value problem (.. Therefore, in this special case, y(t is bounded for all finite t >. (3 Let <. We notice that if d = b ± b 4ac, then y d is the solution of the initial-value problem (.. Therefore, in this case y(t is bounded for all t >. Now let us consider the case when d b ± b 4ac. Using separation of variables, we have dy ay + by + c = dt, b 4ac ln y + b b 4ac y + b + = t + C, (.3 b 4ac where C is a constant of integration. We can find C by substituting the initial condition y( = d into equation (.3. Thus, C = b 4ac ln d + b b 4ac d + b +. (.4 b 4ac We will consider the two possible cases: d + b b 4ac d + b + b 4ac > (.5 and d + b b 4ac d + b + <. b 4ac (.6 We now substitute (.4 into (.3 and solve for y(t. In case (.5, we can omit the absolute value symbol: ( y + b b 4ac ln b 4ac y + b + ( d + b = t + b 4ac b 4ac ln b 4ac d + b +. b 4ac In the case (.6, we have ( b 4ac ln = t + ( b 4ac ln y + b b 4ac y + b + b 4ac d + b b 4ac d + b + b 4ac.

4 4 S. Y. BAHK, N. E. DYAKEVICH, S. C. JOHNSON EJDE-8/4 In both cases (.5 and (.6, the solution of (. is given by the formula b 4ac y(t = bd + d b 4ac c + (bd + d b 4ac + ce t d + b + b 4ac (d + b b 4ace t b 4ac To find the blow-up time we set the denominator equal to,. (.7 d + b + b 4ac (d + b b 4ace t b 4ac = (.8 and solve for t to obtain ( d + b + t b = b 4ac ln b 4ac d + b. b 4ac Since the blow-up time t b must be positive, we have d + b + b 4ac d + b >. (.9 b 4ac Let us observe that if (.6 holds, then (.8 can never be satisfied, thus, there is no blowup. On the other hand, if (.5 holds, then there are two possibilities: either or d + b b 4ac > and d + b + b 4ac > (. d + b b 4ac < and d + b + b 4ac <. (. Solving (. and (.9 simultaneously, we obtain the conditions on d that lead to blow-up in finite time: d > b + b 4ac, if a >, d < b + b 4ac, if a <. Solving (. and (.9 simultaneously, we obtain a contradiction which implies that there is no blow-up in this case. We notice from (.7 that if a > and d > b+ b 4ac, then y(t +. If a < and d < b+ b 4ac, then y(t. Thus, initial value d is very important since certain values can prevent blow-up from occurring. Also, d influences the value of t b at which blow-up occurs. (4 If a =, then the equation is linear. Using separation of variables, we obtain dy by + c = dt, y(t = (bd + cebt c. b Here we find that y(t is bounded for any finite time t >. Example. Let us investigate the blow-up property of the solution for the initialvalue problem y (t = 4y + 5y, (. y( = d with three different values of d as indicated below. First, we note that a = 4 <, b = 5, c =, and = 9 <. Also, ( b + b 4ac/( =.5. According to Theorem., we expect that

5 EJDE-8/4 LONG TERM BEHAVIOR OF SOLUTIONS 5 the solution of problem (. blows up for any d <.5 and is bounded otherwise. The solution of problem (. is given by formula (.7: y(t = d + + (8d e3t 8d + 8 ( 8d + e 3t. Let us notice that if we differentiate function y with respect to d, then the corresponding derivative is as follows: 9e 3t ( 4d de 3t e 3t. Therefore, the function y and its derivative with respect to d are both discontinuous when t = 3 ln ( 4d 4, 4d which holds only for d <.5. If d =, we have + 4e3t y(t = 8 + 4e 3t. This function is bounded for any finite time t >. If d =.5, we have y(t = + e3t 4 + e 3t. This function is bounded for any finite time t >. If d =, we have y(t = e3t 4 e 3t. Here y(t when t b = ln(4/3. 3. Riccati Problems with Variable Coefficients Let a(t, b(t, and c(t be continuous and differentiable functions for t. We consider the initial-value problem y (t = a(ty + b(ty + c(t, y( = d. (3. We notice that ϕ(t = exp ( t b(ˆtdˆt is the unique solution of the initial-value problem ϕ (t = b(tϕ(t, ϕ( =. Also, ϕ(t is bounded for any finite t. Let ψ(t satisfy the initial-value problem ψ (t = a(tψ (tϕ(t + c(t ϕ(t, ψ( = d. (3. Then y(t = ϕ(tψ(t satisfies the Riccati initial-value problem (3.. We now investigate conditions on a(t, c(t, and d that lead to unbounded growth of ψ(t, and therefore, of y(t.

6 6 S. Y. BAHK, N. E. DYAKEVICH, S. C. JOHNSON EJDE-8/4 Theorem 3.. Let a(tc(t. Then the following is true for the solution y(t of (3.: ( If a(t exp ( t b(ˆtdˆt k > and c(t exp ( t b(ˆtdˆt k > for all t >, then y(t + for any initial condition d. ( If a(t exp ( t b(ˆtdˆt k 3 < and c(t exp ( t b(ˆtdˆt k 4 < for all t >, then y(t for any initial condition d. (3 Let a(tc(t have zeroes at t = t, t, t 3,... and let g (t and g (t be nontrivial, continuous, and differentiable functions such that for all t, g (t min { a(te R t b(ˆtdˆt, c(te R t b(ˆtdˆt }, g (t max { a(te R t b(ˆtdˆt, c(te R t b(ˆtdˆt }. Then one of the following three statements is true: (a y(t + if for some t b >, tb (b y(t if for some t b >, tb g (tdt = π arctan(d. (3.3 g (tdt = π arctan(d. (3.4 (c y(t is bounded for all t > if the following two inequalities hold simultaneously: t g (ˆtdˆt < π arctan(d, t g (ˆtdˆt > π arctan(d. Proof. ( By the comparison theorem [4, pp. -3], the solution ψ(t of the initial-value problem ψ (t = k ψ (t + k, ψ( = d, where d d, is a lower solution for (3.. By Theorem., ψ(t approaches + as t approaches t b. Therefore, ψ(t ψ(t also approaches + as t approaches some t b t b. ( By the comparison theorem, solution ψ(t of the initial-value problem ψ (t = k 3 ψ (t + k 4, ψ( = d, where d d, is an upper solution for (3.. By Theorem., ψ(t approaches as t approaches t b. Therefore, ψ(t ψ(t also approaches as t approaches t b t b. (3(a. By the comparison theorem, the solution ˆψ(t of the initial-value problem ˆψ (t = g (t ˆψ (t + g (t, ˆψ( = ˆd, where ˆd d, is a lower solution for (3.. Using the separation of variables method, we obtain that ˆψ(t approaches + as t approaches ˆt b provided that (3.3 holds.

7 EJDE-8/4 LONG TERM BEHAVIOR OF SOLUTIONS 7 Therefore, ψ(t ˆψ(t also approaches + as t approaches t b ˆt b. If condition (3.3 does not hold, then ˆψ(t and ψ(t are bounded from above for all t >. (3(b By the comparison theorem, the solution ψ(t of the initial-value problem ψ (t = g (t ψ (t + g (t, ψ( = d, where d d, is an upper solution for (3.. Using the separation of variables method, we obtain that ψ(t approaches as t approaches t b provided that (3.4 holds. Therefore, ψ(t ψ(t also approaches as t approaches t b t b. If condition (3.4 does not hold, then ψ(t and ψ(t are bounded from below for all t >. (3(c The proof of this part follows directly from the proofs described in parts (a and (b above. Theorem 3.. Let c(t =. One of the following three statements is true for the solution y(t of problem (3.: ( y(t + provided d > and tb for some t b >. ( y(t provided d < and a(te R t b(ˆtdˆt dt = d (3.5 for some t b >. (3 If tb d < t a(te R t b(ˆtdˆt dt = d a( te R t b(ˆtdˆt d t < d for all t, then y(t is bounded for all t. Proof. ( Applying the separation of variables method to the problem ψ (t = a(tψ (t, ψ( = d, (3.6 we obtain that ψ(t approaches + as t approaches t b provided that d >, and (3.5 holds. Otherwise, ψ(t is bounded from above for all t >. ( Similarly, we obtain that ψ(t approaches as t approaches t b provided that d <, and (3.6 holds. Otherwise, ψ(t is bounded from below for all t >. (3 The proof of this part follows directly from the proofs of parts and above. Theorem 3.3. For the case a(tc(t <, one of the following two statements is true for the solution y(t of (3.: ( If a(t exp ( t b(ˆtdˆt k 5 > and > c(t exp ( t b(ˆtdˆt k 6 for all t >, then y(t + for any d > k 6 /k 5. ( If a(t exp ( t b(ˆtdˆt k 7 < and < c(t exp ( t b(ˆtdˆt k 8 for all t >, then y(t for any d < k 8 /k 7.

8 8 S. Y. BAHK, N. E. DYAKEVICH, S. C. JOHNSON EJDE-8/4 Proof. ( By the comparison theorem [4, pp. -3], the solution ψ(t of the initial-value problem ψ (t = k 5 ψ (t + k 6, ψ( = d, where k 6 /k 5 < d d, is a lower solution for (3.. By Theorem., ψ(t approaches + as t approaches t b. Therefore, ψ(t ψ(t also approaches + as t approaches some t b t b. ( By the comparison theorem, the solution ψ(t of the initial-value problem ψ (t = k ψ 7 (t + k 8, ψ( = d, where k 8 /k 7 > d d is an upper solution for (3.. By Theorem., ψ(t approaches as t approaches t b. Therefore, ψ(t ψ(t also approaches as t approaches t b t b. Acknowledgements. The authors are grateful to the College of Natural Sciences of California State University San Bernardino for its support of this project. The authors would like to thank the anonymous referee for his or her helpful suggestions. References [] H. Abou-Kandil, G. Freiling, V. Ionescu, G. Jank; Matrix Riccati Equations in Control and Systems Theory, Birkhuser Verlag, Basel, 3. [] J. L. Allen, F. M. Stein; On solutions of certain Riccati differential equations, Amer. Math. Monthly, 7 (964, 3-5. [3] S. Bittanti, J. C. Willems, A. J. Laub; The Riccati Equation, Springer Verlag, New York, 99. [4] R. H. Cole; Theory of Ordinary Differential Equations, Appleton-Century-Crofts Corp., New York, 968. [5] E. L. Ince; Ordinary Differential Equations, Dover Publications, New York, 956. [6] A. D. Polyanin, V. F. Zaitsev; Handbook on Exact Solutions for Ordinary Differential Equations, nd ed., Chapman and Hall/CRC, 3. [7] M. Rahman; On the Integrability and Application of the Generalized Riccati Equation, SIAM J. Appl. Math., Vol., No., (97, [8] P. R. P. Rao, V. H. Ukidave; Some separable forms of the Riccati equation, Amer. Math. Monthly, 75 (968, [9] G. N. Watson; A Treatise on the Theory of Bessel Function, Cambridge University Press, Cambridge, 966. [] E. T. Whittaker, G. N. Watson; A Course of Modern Analysis, Cambridge University Press, Cambridge, 996. Department of Mathematics, California State University San Bernardino, 55 University Parkway, San Bernardino, CA , USA address, Sarah Y. Bahk: sbahk@csusb.edu address, Nadejda E. Dyakevich: dyakevic@csusb.edu address, Stefan C. Johnson: steve@sevej.us