THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem for first-order differentil equtions, lso known s the Picrd-Lindelöf or Cuchy-Lipschitz theorem It ws written with specil ttention to both rigor nd clrity The proof is primrily bsed on the one given in the textbook I used in my Differentil Equtions clss (Differentil Equtions nd Their Applictions by Mrtin Brun, fourth edition)
2 RADON ROSBOROUGH 1 Gol Let f be function of two vribles nd let nd y 0 be rel numbers This defines n initil-vlue problem: (1) y (t) = f(t, y(t)) y( ) = y 0 Our gol is to prove tht under certin conditions, there is exctly one solution y to this differentil eqution Tht is, we would like to prove both the existence nd uniqueness of solutions to the eqution 2 Initil Steps We will strt our proof by trnsforming the differentil eqution (1) into more convenient form This is done by integrting both sides from to t: y (τ) dτ = f(τ, y(τ)) dτ Here, to void mbiguity, we re using the vrible τ s our vrible of integrtion insted of t The bove eqution reduces to y(t) y 0 = since y( ) = y 0, nd solving for y(t) gives the eqution (2) y(t) = y 0 + f(τ, y(τ)) dτ, f(τ, y(τ)) dτ By this resoning, ny function stisfying (1) must lso stisfy eqution (2) However, the converse sttement requires little more work Suppose tht function y stisfies eqution (2) Then it follows immeditely tht y( ) = y 0, becuse y( ) = y 0 + 0 f(τ, y(τ)) dτ = y 0 Now, the fundmentl theorem of clculus tells us tht if function f is continuous, then b f(x) dx is differentible with respect to b, nd d db b f(x) dx = f(b) Hence, if we ssume tht if both f nd y re continuous, then f(τ, y(τ)) is continuous function of τ nd we hve y (t) = d dt f(τ, y(τ)) dτ = f(t, y(t)) Therefore, if y is continuous function stisfying (2) nd f is continuous, then y is solution of the originl differentil eqution (1)
THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS 3 3 Outline Our proof will consist of the following mjor steps: () Construct sequence of functions {y 0 (t), y 1 (t), }, clled Picrd itertes, which pproximte solution to the eqution (2) (b) Show tht the sequence of functions converges nd define y(t) = lim y n (t) (c) Show tht the function y(t) stisfies eqution (2) (d) Show tht the function y(t) is continuous (e) Show tht there cn only be one solution to the eqution (2) Hving completed these tsks, our resoning bove will imply tht the function y(t) is the unique solution to (1) 4 Construction of the Picrd itertes As our first pproximtion to solution to the differentil eqution (1), we will choose the simplest possible function tht stisfies the condition y( ) = y 0 ; tht is, y 0 (t) = y 0 Our procedure for generting better pproximtions is motivted by the reltion (2) which is stisfied by ny solution y to (1), reprinted here: In prticulr, we will define y(t) = y 0 + (3) y n (t) = y 0 + f(τ, y(τ)) dτ f(τ, y n 1 (τ)) dτ for every n 1 Observe tht we hve y n ( ) = y 0 for every n 0, so every Picrd iterte obeys the initil condition As concrete exmple of this itertion process, consider the differentil eqution y (t) = y(t) y(0) = 1, whose unique solution is y(t) = e t For this eqution, we hve f(t, y) = y, = 0, nd y 0 = 1 This trnsforms the recurrence reltion (3) to therefore, y 0 (t) = 1 y n (t) = 1 + y 1 (t) = 1 + y 2 (t) = 1 + 0 0 0 y n 1 (τ) dτ; dτ = 1 + t 1 + τ dτ = 1 + t + t2 2 y n (t) = 1 + t + t2 2 + + tn n!
4 RADON ROSBOROUGH The stute reder will recognize y n (t) s the nth prtil sum of the Mclurin series for e t It follows esily, then, tht the sequence {y 0 (t), y 1 (t), } converges nd tht the limiting function y(t) = lim y n (t) = e t is continuous nd stisfies (2) We now show tht this conclusion is true in generl under suitble ssumptions 5 Bounding the Picrd itertes In generl, even if the differentil eqution (1) hs unique solution, the solution my only be vlid on specified intervl typiclly becuse f(t, y(t)) is not defined for one or more vlues of t Therefore, we will hve to restrict our resoning to limited intervl contining However, it is difficult to reson bout the lrgest possible intervl tht is, the lrgest intervl over which the differentil eqution hs solution Insted, we will pick smller intervl in such wy tht the behvior of the Picrd itertes is esy to nlyze over the intervl To construct this intervl, we will strt by picking two rbitrry positive rel numbers nd b These numbers define rectngle in the t-y plne tht hs vertices t (, y 0 b), (, y 0 + b), ( +, y 0 b), nd ( +, y 0 + b) This rectngle is illustrted in Figure 1 y y 0 + b (, y 0 + b) ( +, y b + b) y 0 (, y 0 ) y = y(t) y 0 b (, y 0 b) + ( +, y 0 b) t Figure 1 Let R denote the rectngle nd its interior, ie the set of ll points (t, y) such tht t + nd y 0 b y y 0 + b Since we re ssuming tht f is continuous, it follows tht f is continuous nd hs mximum vlue on R We let M denote this mximum vlue, ie M = mx f(t, y) (t,y) R Next, we consider the lines through the point (, y 0 ) tht hve slope M nd M, respectively These lines hve equtions y = y 0 ± M(t ), nd re shown in Figure 2 From the figure, it is esy to see tht depending on the vlue of M, the lines will leve the rectngle t either t = or t = b/m, whichever is smller We will denote this t-vlue by α, ie α = min (, We will now prove tht for t + α, every Picrd iterte lies between the two lines Tht is, until the lines leve the rectngle R, every y n (t) lies within the shded regions in Figure 2 We cn reformulte this hypothesis s follows: b M ) y 0 M(t ) y n (t) y 0 + M(t ) M(t ) y n (t) y 0 M(t )
THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS 5 (, y 0 + M) (y 0 + b/m, b) (, y 0 ) (, y 0 ) (, y 0 M) (y 0 b/m, b) Figure 2 y n (t) y 0 M(t ) Becuse M 0 nd t, we need no bsolute vlue brs on the right-hnd side To prove the hypothesis, we use induction on n The cse of n = 0 follows immeditely, s y 0 (t) y 0 = y 0 y 0 = 0 M(t ) For the inductive cse, we ssume for some n 0 tht y n (t) y 0 M(t ) for t α nd seek to prove tht y n+1 (t) y 0 M(t ) on the sme intervl We now use the definition (3) of the Picrd itertes, reprinted here: y n+1 (t) = y 0 + f(τ, y n (τ)) dτ In prticulr, we note tht y n+1 (t) y 0 = f(τ, y n (τ)) dτ Next, we use the following two elementry properties of definite integrls: b b f(x) dx f(x) dx b ( ) b f(x) g(x) dx mx f(x) g(x) dx x b Note tht we hve τ t + α Consequently, it follows from the inductive hypothesis tht y n (τ) lies between the lines nd hence within R on this intervl Thus, (τ, y n (τ)) lies within R for ll τ from to t, nd f(τ, y n (τ)) M over this intervl From these properties, we hve: f(τ, y n (τ)) dτ f(τ, y n (τ)) dτ M(t ) We hve therefore shown tht y n+1 (t) y 0 M(t ), which completes the proof tht y n (t) y 0 M(t ) for every n 0 6 Proof tht the Picrd itertes converge Now tht we hve obtined bound on the size of y n (t) on suitble intervl, we cn show tht the sequence {y 0 (t), y 1 (t), } converges on tht intervl We do this by rewriting y n (t) s telescoping series: y n (t) = y 0 (t) + y 1 (t) y 0 (t) + y 2 (t) y 1 (t) + + y n (t) y n 1 (t)
6 RADON ROSBOROUGH so tht = y 0 (t) + n y k (t) y k 1 (t), lim y n(t) = y 0 (t) + y k (t) y k 1 (t) If the infinite series y k(t) y k 1 (t) converges, then so does the sequence {y 0 (t), y 1 (t), } Now, if we replce every term of series with its bsolute vlue nd it still converges, then certinly the originl series must lso converge Thus, it suffices to show the convergence of y k(t) y k 1 (t) To do this, we will use series of pproximtions involving the quntity y k (t) y k 1 Firstly, we will use the definition (3) of the Picrd iterte, gin reprinted here: y n (t) = y 0 + f(τ, y n 1 (τ)) dτ In prticulr, we find tht ( ) ( ) y k (t) y k 1 (t) = y 0 + f(τ, y k 1 (τ)) dτ y 0 + f(τ, y k 2 (τ)) dτ = f(τ, y k 1 (τ)) f(τ, y k 2 (τ)) dτ f(τ, y k 1 (τ)) f(τ, y k 2 (τ)) dτ, provided tht k 2 Next we invoke the men vlue theorem, which sttes tht if function g is continuous on, b nd differentible on (, b) then there exists number ξ (, b) such tht Now, for ny given τ we cn define g (ξ) = g(b) g() b g(y) = f(τ, y) = y k 2 (τ) b = y k 1 (τ) If we ssume tht f is continuous nd the prtil derivtive f y = f/ y exists, ie tht g is differentible, then the men vlue theorem tells us tht there exists number ξ between y k 2 (τ) nd y k 1 (τ) such tht f y (τ, ξ) = f(τ, y k 1(τ)) f(τ, y k 2 (τ)) y k 1 (τ) y k 2 (τ) Rerrnging this eqution, we find the useful reltion f(τ, y k 1 (τ)) f(τ, y k 2 (τ)) = f y (τ, ξ) y k 1 (τ) y k 2 (τ) If we mke this rgument for every τ t, we my obtin different number ξ for ech τ Tht is, we must replce the number ξ with function ξ(τ) We then find tht f(τ, y k 1 (τ)) f(τ, y k 2 (τ)) dτ = f y (τ, ξ(τ)) y k 1 (τ) y k 2 (τ) dτ = f y (τ, ξ(τ)) y k 1 (τ) y k 2 (τ) dτ
THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS 7 Now, let us ssume tht f/ y not only exists over the rectngle R, but it is lso continuous 1 Then f/ y is lso continuous, nd therefore hs mximum vlue on R We will denote this mximum vlue by L, ie L = mx (t,y) R f y(t, y) Since ξ(τ) lies between the Picrd itertes y k 2 (τ) nd y k 1 (τ), our work in the previous section proves tht it lies between the lines y = y 0 ± M(t ) for τ α Hence, ll points (τ, ξ(τ)) lie within R for τ t, nd so f y (τ, ξ(τ)) L The sme elementry properties of definite integrls we used erlier pply gin, so tht f y (τ, ξ(τ)) y k 1 (τ) y k 2 (τ) dτ L y k 1 (τ) y k 2 (τ) dτ In summry, y k (t) y k 1 (t) L y k 1 (τ) y k 2 (τ) dτ for every k 2 We now switch to n inductive rgument on k For k = 1, recll we proved in the previous section tht y 1 (t) y 0 (t) M(t ) for t + α For k = 2, we hve y 2 (t) y 1 (t) L y 1 (τ) y 0 (τ) dτ For k = 3, we hve Inductively, we find tht L y 3 (t) y 2 (t) L M(t ) dτ = ML(t ) 2 2 L y 2 (τ) y 1 (τ) dτ ML(t ) 2 2 = ML2 (t ) 3 3! (4) y k (t) y k 1 (t) MLk 1 (t ) k for t + α But now we cn esily show tht the series y k(t) y k 1 converges, becuse ML k 1 (t ) k y k (t) y k 1 (t) = M L(t ) k L = M L(t t0 ) k 1 L 1 This is not strictly necessry All we need is tht f/ y is bounded k=0 dτ
8 RADON ROSBOROUGH = M L ( e L(t ) 1 ) As t + α, we hve t α nd e L(t t0) e Lα This shows tht y k (t) y k 1 (t) M ( e Lα 1 ), L which implies the series converges This completes our proof tht the sequence of Picrd itertes {y 0 (t), y 1 (t), } converges We thus cn define y(t) = lim y n (t) 7 Proof tht y(t) stisfies eqution (2) We will now show tht the function y(t) = lim y n (t) stisfies eqution (2), reprinted here: y(t) = y 0 + f(τ, y(τ)) dτ To do so, we strt with the definition (3) of the Picrd itertes, reprinted here: y n+1 (t) = y 0 + f(τ, y n (τ)) dτ Tking the limits of both sides s n gives us y(t) = y 0 + lim f(τ, y n (τ)) dτ; to show tht y(t) stisfies (2), we must demonstrte tht f(τ, y(τ)) dτ = lim f(τ, y n (τ)) dτ, t 0 or equivlently tht lim f(τ, y(τ)) dτ f(τ, y n (τ)) dτ = 0 We my now use roughly the sme procedure tht we used to show the convergence of the sequence {y 0 (t), y 1 (t), } In prticulr: t f(τ, y(τ)) dτ f(τ, y n (τ)) dτ = f(τ, y(τ)) f(τ, y n (τ)) dτ f(τ, y(τ)) dτ f(τ, y n (τ)) dτ Next, pplying the men vlue theorem shows tht for every τ between nd t, there is number ξ(τ) between y n (τ) nd y(τ) such tht or equivlently We then find tht f y (τ, ξ(τ)) = f(τ, y(τ)) f(τ, y n(τ)), y(τ) y n (τ) f(τ, y(τ)) f(τ, y n (τ)) = f y (τ, ξ(τ)) y(τ) y n (τ) f(τ, y(τ)) f(τ, y n (τ)) dτ = f y (τ, ξ(τ)) y(τ) y n (τ) dτ As y(t) is the limit of sequence of functions y n (t) which ll lie within the closed rectngle R for τ t, it follows tht y(t) lso lies within R on tht intervl Becuse ξ(τ) is between y n (τ) nd
THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS 9 y(τ), ll points (τ, ξ(τ)) lie within R for τ t, nd f y (τ, ξ(τ)) L We cn thus conclude tht f y (τ, ξ(τ)) y(τ) y n (τ) dτ L y(τ) y n (τ) dτ nd Now observe tht the reltions my be combined to obtin Also, reltion (4) tells us tht so y(τ) y n (τ) = y n (τ) = y 0 (τ) + y(τ) = y 0 (τ) + y(τ) y n (τ) = n y k (τ) y k 1 (τ) y k (τ) y k 1 (τ) y k (τ) y k 1 (τ) y k (τ) y k 1 (τ) MLk 1 (τ ) k, y k (τ) y k 1 (τ) nd using the fct tht τ t α gives (5) y(τ) y n (τ) y k (τ) y k 1 (τ) Substituting (5) yields: f y (τ, ξ(τ)) y(τ) y n (τ) dτ L ML k 1 α k ML k 1 α k dτ ML k 1 (τ ) k, Since every term of this series is nonnegtive, Tonelli s theorem gurntees tht we my swp the integrl nd summtion: ML k 1 α k ML k 1 α k L dτ = L dτ; this llows us to simplify s follows: ML k 1 α k L dτ M L k α k (t ) Mα (Lα) k Since the ltter summtion is the til end of series expnsion for e Lα, it pproches zero s n To prove this formlly, observe tht (Lα) k (Lα) k n (Lα) k n = = e Lα (Lα) k 1 1 1
10 RADON ROSBOROUGH nd Since nd it follows tht lim (Lα) k f(τ, y(τ)) dτ lim lim = e Lα lim n (Lα) k = 0 f(τ, y n (τ)) dτ Mα f(τ, y(τ)) dτ (Lα) k 1 = 0, f(τ, y n (τ)) dτ = 0, which completes the proof tht y(t) = lim y n (t) stisfies (2) (Lα) k 8 Proof tht y(t) is continuous To show tht y(t) is continuous, we must show tht for every ɛ > 0 there exists δ > 0 such tht h < δ implies y(t + h) y(t) < ɛ To do so, we observe tht nd consequently y(t + h) y(t) = y(t + h) y n (t + h) + y n (t + h) y n (t) + y n (t) y(t) y(t + h) y(t) y(t + h) y n (t + h) + y n (t + h) y n (t) + y n (t) y(t) for every n 0 By picking lrge enough n, we cn reduce the mgnitude of this sum to ɛ Since the summtion ML k 1 α k is the til end of convergent Mclurin series, we cn mke it s smll s we wish by selecting sufficiently lrge n In prticulr, we will choose n n such tht ML k 1 α k < ɛ 3 nd reltion (5) implies y(τ) y n (τ) < ɛ 3, for both τ = t nd τ = t + h This tkes cre of two out of the three terms For the third, y n (t + h) y n (t), note tht y n (t) is continuous for every n 0, nd so for every ɛ > 0 there exists δ > 0 such tht h < δ implies y n (t + h) y n (t) < ɛ We let ɛ = ɛ/3 nd define δ = δ Thus, ech of the three terms is strictly less thn ɛ/3, nd This completes the proof tht y(t) is continuous y(t + h) y(t) < ɛ 3 + ɛ 3 + ɛ 3 = ɛ
THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS 11 9 Proof tht the solution to eqution (2) is unique Hving shown tht solution to (1) exists, we now show tht it is unique Supposing tht two solutions of (1) re given by y(t) nd z(t), we define w(t) = y(t) z(t) ; it is sufficient, then, to show tht w(t) = 0 for ll t If y(t) nd z(t) re solutions of (1), then they re lso solutions of (2); tht is, Subtrcting, we find tht y(t) z(t) = w(t) y(t) = y 0 + z(t) = y 0 + f(τ, y(τ)) dτ f(τ, z(τ)) dτ f(τ, y(τ)) dτ f(τ, z(τ)) dτ f(τ, y(τ)) f(τ, z(τ)) dτ, nd by the sme resoning with the men vlue theorem tht we used twice before, We now define w(t) L y(τ) z(τ) dτ = L w(τ) dτ U(t) = w(τ) dτ, so tht U (t) = w(t) L w(τ) dτ = LU(t), or equivlently U (t) LU(t) 0 Multiplying both sides by the strictly positive integrting fctor e L(t t0) gives U (t)e L(t t0) LU(t)e L(t t0) 0 d U(t)e L(t ) 0 dt Now, U( ) = 0 w(τ) dτ = 0, so the function U(t)e L(t t0) is zero t t = Furthermore, w(t) is nonnegtive nd t, so U(t) is lso nonnegtive Since the function U(t)e L(t t0) is zero t t =, is never less thn zero, nd is never incresing, it must be zero for ll t > s well We must then conclude tht U(t) = 0 nd therefore tht w(t) = 0 This completes the proof