# ODE: Existence and Uniqueness of a Solution

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1 Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() = α Just integrte both sides: It is not obvious how to solve du(t) dt u(t) = α + f(s)ds = f(x, u(t)) with initil condition u() = α becuse the unknown, u(t), is on both sides of the eqution In mny prticulr cses, by using specil devices one cn find formuls for the solutions but it is fr from obvious tht solution exists or is unique In fct, there re simple exmples showing tht unless one is creful, solution my not exist, nd even if one exists, it my not be unique Just becuse one my wnt something to hppen doesn t men tht it will hppen It is esy to hve presumptions tht turn out to be not quite true The results tht we present re clssicl They will use most of the ides we hve covered this semester We investigte one illuminting cse nd will prove the existence nd uniqueness of solution of the system of inhomogeneous liner eqution d U(t) dt Here we seek vector U(t) = nd vector F(t) = f 1 (t) = A(t) U(t) + F(t) with U() = α (1) u 1 (t) u n (t)) given the input n n mtrix A(t) = ( ij (t)), The elements of A(t) nd F(t) re ssumed to depend f n (t)) continuously on t for t b Also α R n is the initil condition Although we will not pursue it, there is firly strightforwrd extension of the method we use to the more generl nonliner cse d U dt = F(t, U(t)) with U() = α The ides re lredy cptured in our specil cse of eqution (1) 1

2 Exmple Here we hve system of two equtions u 1(t) = u 2 (t) u 2(t) = u 1 (t) with initil conditions u 1 () = 1 nd u 2 () = The (unique) solution of this hppens to be u 1 (t) = cos t, u 2 (t) = sint, but the point of these notes is to consider equtions where there my not be simple formuls for the solution The primry reson we re presenting the more generl mtrix cse n 1 is pply to the stndrd second order sclr initil vlue problem y (t) + p(t)y (t) + q(t)y(t) = f(t) with y() = nd y () = b, (2) where p(t), q(t), nd f(t) re continuous rel-vlued functions To reduce the problem (2) to problem (1), let u 1 = y nd u 2 = y then tht is, ( u1 u 2 ) = ( 1 q p u 1 = y = u 2 u 2 = y = pu 2 qu 1 + f ) ( u1 u 2 ) + ( ) f with ( ) u1 () = u 2 () ( ) (3) b This exctly hs the form of the system (1) If we cn solve the system (3), then u 1 (t)is the solution of eqution (2) Exmple Before plunging hed to prove tht eqution (1) hs exctly one solution, we present the ide with the simple exmple u = u with u() = 1 (4) whose solution we lredy know is u(t) = e t The first step is to integrte both sides of eqution (4) to obtin the equivlent problem of finding function u(t) tht stisfies u(t) = 1 + u(s)ds (5) We will solve this by successive pproximtions Let the initil pproximtion be u (t) = u() = 1 nd define the subsequent pproximtions by the rule u k+1 (t) = 1 + u k (s)ds, k =, 1, 2, (6) 2

3 Then u 1 (t) =1 + u 2 (t) =1 + u 3 (t) =1 + 1 ds = 1 + t (1 + s)ds = 1 + t t2 (1 + s s2 )ds = 1 + t t ! t3 u k (t) =1 + t t k! tk We clerly recognize the Tylor series for e t emerging This gives us hope tht our successive pproximtion pproch to eqution (6) is plusible technique With this s motivtion we integrte eqution (1) nd obtin U(t) = α + [ A(s) U(s) + F(s) ] ds (7) We immeditely observe tht if continuous function U(t) stisfies this, then by the Fundmentl Theorem of Clculus pplied to the left side, this U(t) is differentible nd is solution of our eqution (1) Therefore we need only find continuous U(t) tht stisfies eqution (7) As in our exmple we use successive pproximtions To mke the issues clerer, I will give the proof twice First for the specil cse n = 1 u (t) = (t)u(t) + f(t) with u() = α (8) so there re no vectors or mtrices In this prticulrly specil cse there hppens to be n explicit formul for the solution but we won t use it since it will not help for the generl mtrix cse We re ssuming tht (t) nd f(t) re continuous on the intervl [, b] Consequently, they re bounded there nd we use the uniform norm := sup (t), t [,b] f := sup f(t), (9) t [,b] Just s with eqution (1) we integrte (8) to find u(t) = α + [(s)u(s) + f(s)] ds (1) As we observed fter eqution (7), if we hve continuous function u(t) tht stisfies this, then by the Fundmentl Theorem of Clculus the u(t) on the left side is differentible nd is the desired solution of eqution (8) 3

4 For our initil pproximtion let u (t) = (this prticulr choice is not very importnt) nd recursively define u k+1 (t) = α + [(s)u k (s) + f(s)] ds, k =, 1, (11) We will show tht the u k converge uniformly to the desired solution of eqution (1) Notice since u k (t) is continuous in the intervl [, b], so is u k+1 (t) We will lso use the uniform norm u, s in eqution (9), for these Now subtrcting we obtin u k+1 (t) u k (t) = (s)[u k (s) u k 1 (s)] ds (12) Therefore u 2 (t) u 1 (t) u 1 u ds = u 1 u t (13) Using this estimte, we next find u 3 (t) u 2 (t) Similrly, u 4 (t) u 3 (t) Repeting this we obtin u 2 (t) u 1 (t) ds 2 u 1 u s ds = 2 u 1 u t2 2 (14) u 3 (t) u 2 (t) ds 3 s 2 u 1 u 2 ds = 3 u 1 u t3 3! (15) Becuse t [, b] then u k+1 (t) u k (t) k u 1 u tk k! u k+1 (t) u k (t) ( b)k u 1 u (16) k! Since the series ( b) k u 1 u converges, by the Weierstrss M-test the series k! uk+1 (t) u k (t) of continuous functions converges bsolutely nd uniformly in the intervl [, b] to some continuous function But N [u k+1 (t) u k (t)] =[u N+1 (t) u N (t)] + [u N (t) u N 1 (t)] + + [u 1 (t) u (t)] k= =u N+1 (t) u (t) = u N+1 (t) Consequently the sequence of continuous functions u N (t) converges uniformly in the intervl [, b] to some continuous function u(t) We use this to let k in eqution (11) nd 4

5 find tht u(t) is the desired solution of eqution (1) In this lst step we used the uniform convergence to interchnge limit nd integrl in eqution (11) This completes the existence proof for the specil cse of eqution (8) We now repet the proof for the generl system eqution (7), using, where needed, stndrd fcts bout vectors nd mtrices from the Appendix t the end of these notes It is remrkble tht the only chnges needed re chnges in nottion Just s in eqution (11) we let U (t) = nd recursively define U k+1 (t) = α + [ A(s) U k (s) + F(s) ] ds, k =, 1, Given the continuous U k (t) this defines the next pproximtion, Uk+1 (t) We will show tht the U k converge uniformly to the desired solution of eqution (7) Now U k+1 (t) U k (t) = A(s)[ Uk (s) U ] k 1 (s) ds To estimte the right hnd side we use the inequlities (23) nd (22) from the Appendix This results in U 2 (t) U 1 (t) A Using this we next find U 3 (t) U 2 (t) A U k+1 (t) U k (t) A Repeting this procedure we obtin U 1 (s) U (s) ds A U 1 U A U 2 (s) U 1 (s) ds A 2 U 1 U U k+1 (t) U k (t) A k U 1 U tk k! U k (s) U k 1 (s) ds (17) ds = A U 1 U t s ds = A 2 U 1 U t2 2 The reminder of the proof goes exctly s in the previous specil cse nd proves the existence of solution to eqution (7) nd hence our differentil eqution (1) Uniqueness There re severl wys to prove the uniqueness of the solution of the initil vlue problem (1) None of them re difficult Sy U(t) nd V (t) re both solutions Let W(t) := U(t) V (t) Then W = AW with W() = Let E(t) := 1 2 W 2 = 1 2 W, W Tke the derivtive nd use the Cuchy inequlity (19) nd inequlity (21): E (t) = W, W = A W, W A W W = 2 A E(t) 5

6 For simplicity write c = 2 A Then ( e ct E(t) ) = e ct (E ce) Therefore for t we see tht e t E(t) E() But E() = Hence E(t) for ll t Since E(t) = 1 2 W 2, the only possibility is E(t) This implies tht W(t) Remrk: There is useful conceptul wy to think of the proof If v(t) is continuous function, define the mp T : C([, b) C([, b]) by the rule T(u)(t) := α + [(s)u(s) + f(s)] ds Then eqution (1) sys the the solution we re seeking stisfies u = T(u) In other words u is fixed point of the mp T The solution to mny questions cn be usefully ttcked by viewing them s seeking fixed point of some mp Appendix on Norms of Vectors nd Mtrices This is review of few items concerning vectors nd mtrices Let u := (u 1, u 2,,u n ) nd v := (v 1,,v n ) be points (vectors) in R n Their inner product (lso clled their dot product is defined s In prticulr, the Eucliden length This gives the useful formul u, v = u, v = u 1 v 1 + u 2 v u n v n u 2 = u, u = u u u 2 n u v 2 = u v, u v = u, u 2 u, v + v, v = u 2 2 u, v + v 2 (18) For vectors in the plne, R 2, the inner product is interpreted geometriclly s u, v = u v cos θ, where θ is the ngle between u nd v Since cos θ 1, this implies the Cuchy inequlity u, v u v (19) The following is direct nlytic proof the Cuchy inequlity in R n without geometric considertions which gve us vluble insight We begin by noting tht using the inner product, from eqution (18) for ny rel number t u t v 2 = u 2 2t u, v + t 2 v 2 (2) 6

7 Pick t to tht the right side is s smll s possible (so tke the derivtive with respect to t) We find t = u, v / v 2 Substituting this in inequlity (2) gives Cuchy s inequlity (19) Next we investigte stndrd system of liner equtions A u = v where A = ( ij ) is n n n mtrix: 11 u u n u n =v 1 21 u u n u n =v 2 Then n1 u 1 + n2 u nn u n =v n A u 2 = v 2 =v v 2 n =( 11 u n u n ) ( n1 u nn u n ) 2 Applying the Cuchy inequlity to ech of the terms on the lst line bove we find tht ( n A u 2 j=1 ( n = i,j=1 2 1j = A 2 u 2 2 ij ) ( n u ) u 2 j=1 2 nj ) u 2, where we defined A 2 = n of A) Thus i,j=1 2 ij (this definition of A is often clled the Frobenius norm A u A u (21) If the elements of mtrix A = ij (t) nd the vector u = (v 1 (t),, v n (t)) re continuous functions of t for t in some intervl J R, we mesure the size of A nd u over the whole intervl J s follows: The inequlity (21) thus implies A J := sup A(t) nd u J := sup u(t) t J t J A u J A J u J (22) There is one more fct we will need bout integrting continuous vector-vlued function v(t) on n intervl [, b] It is the inequlity v(s) ds v(s) ds (23) 7

8 To prove this, note tht we define the integrl of vector-vlued function v(t) = (u 1 (t),,u n (t)) s integrting ech component seprtely Using this nd the Cuchy inequlity, we find tht for ny constnt vector V, V, v(s) ds = V, v(s) ds V v(s) ds In prticulr, if we let let V be the constnt vector V = v(s)ds, then V 2 = V, V = V, v(s) ds V v(s) ds After cnceling V from both sides this is exctly the desired inequlity (23) 8

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