Lecture Notes 1. First Order ODE s. 1. First Order Linear equations A first order homogeneous linear ordinary differential equation (ODE) has the form

Transcription

1 Lecture Notes 1 First Order ODE s 1. First Order Linear equations A first order homogeneous linear ordinary differential equation (ODE) has the form This equation we rewrite in the form or From the last equation we get y (t) + a(t)y(t) =. y (t) y(t) = a(t) (ln y(t) ) = a(t). y(t) = Ce a(t)dt. The general first order nonhomogeneous ODE has the form y (t) + a(t)y(t) = b(t). (1) 1gf Multiplying the equation ( 1) 1gf by a nonzero function µ(t), which is called an integrating factor, we obtain an equivalent equation Let us choose µ(t) such that µ(t)y (t) + µ(t)a(t)y(t) = µ(t)b(t) (2) 1gf2 µ (t) = µ(t)a(t), i.e. µ(t) = e a(t)dt + C. (3) 1gf3 Then the equation ( 2) 1gf2 takes the form (mu(t)y(t)) = µ(t)b(t). Integrating the last equation and using ( 3) 1gf3 we obtain y(t) = Ce a(t)dt + e a(t)dt b(t)e a(t)dt dt. Example 1.1. Show that the solution of the Cauchy problem { y (t) + a(t)y(t) = b(t) y() = y (4) 1Cpr has the form y(t) = y e a(τ)dτ + e a(τ)dτ 1 b(τ)e τ a(s)ds dτ. (5) 1Cpr1

2 ex1 2 Solution. Multiplying the equation ( 4) 1Cpr by e a(τ)dτ we obtain d (e ) t a(τ)dτ y(t) = b(t)e a(τ)dτ. dt Integrating last equality over the interval (, t) we obtain ( 5). 1Cpr1 Example 1.2. Solve the initial value problem for t >. ty + 4y = 6t 2, y(1) = 1 Solution. Rewrite the equation in the form y + 4 t y = 6t, exlop and multiply both sides by e 4 t dt = t 4 : Integrating the last equality we get By using the initial condition we obtain: Hence the solution is: (y(t)t 4 ) = 6t 5. y(t)t 4 = t 6 + C. C =. y(t) = t 2. Example 1.3. Let h(t) be a continuous function defined on [, ) and lim h(t) =. t Show that the solution of the problem y (t) + 5y(t) = h(t), y() = 2 Definition 2.1. An equation of the form 2. Separable Equations. is called a separable equation. If y = φ(x) is a solution of the equation ( sep1 6) then f(y)y = g(x) (6) sep1 f(φ(x))φ (x) = g(x). Hence f(φ(x))φ (x)dx = g(x)dx. (7) se1 Since dy = φ (x)dx we can rewrite ( 7) se1 as follows f(y)dy = g(x)dx

3 3 sepe1 sepe2 exhom1 or F (y) = G(x) + C, where F (y) is antiderivative of f(y), G(x) is antiderivative of g(x) and C is arbitrary constant. Example 2.2. Solve the Cauchy problem dy dx = (x 1)y2, y() = 2. dy = (x 1)dx, y2 1 y = x2 2 x + C. By using the initial condition y() = 2 we get C = 1 2. Hence y(x) = Example 2.3. Solve the Cauchy problem 2 2x + 1 x 2. (1 + e t )yy = e t, y() = 1. separating variables we get ( ) 1 2 y2 = et 1 + e. t Inegrating last equation over the interval (, t) and taking into account the initial condition we obtain 1 2 y2 (t) 1 2 = e τ 1 + e dτ = ln(1 + τ et ) ln 2. Thus y 2 = ln 1 + et. 2 Since y() = 1 > we have y(t) = ln 1 + et Homogeneous equations. An equation of the form dy ( y ) dx = f (8) heq1 x is called homogeneous equation. To solve the homogeneous equation ( 8) heq1 we make change of variables y = v and reduce the equation to a separable equation of the form x Example 3.1. Solve the equation xv + v = f(v). 2xyy = 4x 2 + 3y 2.

4 4 Dividing both sides of this equation by xy we obtain: dy dx = 2x y + 3 y 2 x. Hence this equation is a homogeneous equation. Therefore we make the change y = xv: v + xv = 2 1 v v, x dv dx = 4 + 3v2, 2v 2v dx v dv = x, d(v 2 + 4) v = dx x, ln(v 2 + 4) = ln x + ln C, v = C x y 2 x = C x, y 2 = Cx 3 4x 2, y = ± Cx 3 4x 2, x >. exhom1 Example 3.2. Solve the Cauchy problem We rewrite the equation in the form: x dy dx = y + x 2 y 2 ; y(1) =. dy dx = y x + ( y ) 2. 1 x So this equation is also a homogeneous equation. The change of variables v = y/x gives: v + xv = v + 1 v 2, dv 1 v 2 = dx x, arcsin v = ln x + C, v = sin(ln x + C), y(x) = x sin(ln x + C). Due to initial condition C = nπ, N = ±1, ±2,... Hence y(x) = x sin(ln x + nπ).

5 4. Exact Differential Equations and Integrating Factors Let us onsider the following equation This equation is neither linear nor separable. But it is not difficult to see that (4x 3 y + 3x 2 y 2 ) + (x 4 + 2x 3 y)y = (9) f1 4x 3 y + 3x 2 y 2 = ( x 4 y + x 3 y 2) x, x 4 + 2x 3 y = ( x 4 y + x 3 y 2) y. Thus the equation is an exact equation. Therefore we can write this equation in the form Ψ x (x, y) + Ψ y (x, y) dy dx =, or d Ψ(x, y) =. (1) fex1 dx Hence Ψ(x, y) = C. That is x 3 y + x 2 y 2 = C. The last equality defines the solution of the equation ( 9) f1 implicitly. Definition 4.1. An equation of the form M(x, y) + N(x, y)y = (11) exct1 is called an exact equation if there exists a function U(x, y) such that U x (x, y) = M(x, y), U y (x, y) = N(x, y). In other words an equation of the form ( 11) exct1 is an exact equation if the vector filed is a conservative vector filed. F(x, y) = M(x, y)i + N(x, y)j tex1 Theorem 4.2. Let the functions M, N, N x, M y be continuous functions in a region D. Then the equation M(x, y) + N(x, y)y = (12) fex2 is an exact equation if and only if M y (x, y) = N x (x, y) (13) fex3 at each point of the region D. Proof. Assume that ( fex2 12) is an exact equation. Then there exists a function U such that These equalities imply Thus Assume that U x = M, U y = N. U xy = M y, U yx = N x. M y = N x. M y (x, y) = N x (x, y) 5

6 6 and let us find U(x, y) such that U x = M, U y = N. It follows from U x (x, y) = M(x, y) that U(x, y) = M(x, y)dx + g(y). From this equality we get U y (x, y) = M y (x, y)dx + g (y), and the equation is an exact equation if N(x, y) = or N(x, y) M y (x, y)dx + g (y) M y (x, y)dx = g (y). Hence the function N(x, y) M y (x, y)dx must be a function of y. Due to the condition ( 12) fex2 [N(x, y) M y (x, y)dx] = N x (x, y) M y (x, y) =. x Therefore the function [ ] U(x, y) = N(x, y)dy + N(x, y) M y (x, y)dx dy. excta is the required function. Example 4.3. Find the value of b for which the equation (ye 2xy + 4x 3 )dx + bxe 2xy dy = is exact and then solve the equation. An equation of the form M(x, y)dx + N(x, y)dy = is exact if and only if M y (x, y) = N x (x, y). Thus our equation is exact if and only if M y = e 2xy + 2xye 2xy = b[e 2xy + 2xye 2xy ] = N x (x, y). Hence it is exact if and only if b = 1. Now we consider the equation (ye 2xy + 4x 3 )dx + xe 2xy dy =. Since this equation is an exact equation there exists a function F (x, y) such that F x (x, y) = ye 2xy + 4x 3, (14) ex1a F y (x, y) = xe 2xy. (15) ex2

7 Integrating ( 15) ex2 with respect to y we get F (x, y) = xe 2xy dy = 1 2 e2xy + h(x). 7 From the last equality we obtain: exact1 exact1a ( ex2 15) and ( ex3 16) imply: Thus the solution is Example 4.4. Solve the equation Since F x (x, y) = xe 2xy + h (x). (16) ex3 h (x) = 4x 3 h(x) = x 4 + C. 1 2 e2xy + x 4 + C = (e x sin y 2y sin x)dx + (e x cos y + 2 sin x)dy =. M y = e x cos y 2 sin x = N x this equation is an exact equation. Therefor U(x, y) = (e x cos y 2y sin x)dy = e x sin y y 2 sin x + g(y), U y (x, y) = e x sin y 2y sin x + g (y) = e x cos y + 2 sin x. Hence g (y) = and g = const. Therefore the solution has the form e x sin y y 2 sin x = C. Example 4.5. Solve the equation ( ) ( ) x x2 + y x + 1 y dx + y x2 + y y x dy =. y 2 ( ) x U x (x, y) = x2 + y x + 1 dx, y U y (x, y) = U(x, y) = x 2 + y 2 + ln x + x y + g(y), y x2 + y 2 x y 2 + g (y) = y x2 + y y x y 2. g (y) = 1, g(y) = ln y. y Hence the solution has the form x2 + y 2 + ln x + x y + ln y = C.

8 8 5. Integrating Factors Sometimes we can convert a differential equation of the form M(x, y)dx + N(x, y)dy = (17) intf1 which is not an exact equation to an exact equation by multiplying the equation by appropriate function which we call integrating factor. A function m(x, y) is called an integrating factor for the equation ( 17) intf1 if the equation is an exact equation. m(x, y)m(x, y)dx + m(x, y)n(x, y)dy = We consider just the case when m is a function of a single variable m(x) or m(y). Assume that the equation ( intf1 17) has an integrating factor m(x) depending only on x. Then or (m(x)m(x, y)) y = (N(x, y)m(x)) x, m(x)m y (x, y) = m(x)n x (x, y) + m (x)n(x, y), m (x) m(x) = M y(x, y) N x (x, y). N Hence ( 17) intf1 has an integrating factor depending on x if the expression My Nx N depending on y. is not Similarly we can show that ( intf1 17) has an integrating factor m(y) depending only on y if the expression Nx My M is not depending on x. exif1 Example 5.1. Solve the equations (3x 2 y + 2xy + y 3 )dx + (x 2 + y 2 )dy =. M y N x = 3. N Therefore this equation has an integrating factor depending on x. Thus we have m (x) = 3m(x), m(x) = e 3x. (x 2 + y 2 )e 3x = U y (x, y),

9 9 U(x, y) = yx 2 e 3x + y3 3 e3x + h(x), U x = 2xye 3x + 3yx 2 e 3x + y 3 e 3x + h (x) = 3x 2 ye 3x + 2xye 3x + y 3 e 3x, U(x, y) = yx 2 e 3x + y3 3 e3x exif1a Solution is an implicit function: Example 5.2. Solve the equation h (x) =, h = C yx 2 e 3x + y3 3 e3x = C. (1 + 3x 2 sin y)dx x cot ydy =. N x M y M = cot y 3x2 cos y = cot y[1 + 3x2 sin y] = cot y x 2 sin y 1 + 3x 2 sin y Hence the equation has an integrating factor depending only on y: m(y) = e cot ydy = e ln(sin y) = 1 sin y, m (y) + (cot y)m(y) =, exif2 m(y) = 1 sin y. So the equation ( ) 1 sin y + 3x2 dx x cos y sin 2 y dy = is an exact equation. Thus we have U(x, y) x (x, y) = 1 sin y + 3x2, U(x, y) = x 1 sin y + x3+g(y), U y (x, y) = cos y sin 2 y x + g (y). Hence g is a constant, and the solution has the form: x 1 sin y + x3 = C. Example 5.3. Show that the following equation has an integrating factor depending on xy and solve it xy 2 dx + (x 2 y x)dy =

10 1 (m(xy)xy 2 ) y = ((x 2 y x)m(xy)) x m (xy)x 2 y 2 = x 2 y 2 m (xy) xym (xy) + 2m(xy)xy m(xy) + 2xym(xy), m (xy)xy + m(xy) =, m (s) + 1 m(s) =, s m(s) = 1 s, ydx + (x 1 )dy =, y U x (x, y) = y, U(x, y) = xy + g(y), U y (x, y) = x + g (y), Hence x ln y = c is a general solution. g (y) = 1, g(y) = ln y. y 1ordex1 6. Existence and uniqueness. Theorem 6.1. If p(t) and h(t) are continuous on the interval a < t < b containing the point t = t, then there exists a unique function y = ϕ(t) that satisfies { y + p(t)y = h(t), t (a, b), y(t ) = y. (18) exun Proof. Suppose that z(t) is also a solution of the problem ( 18)i.e. exun { y + p(t)y = h(t), t (, T ), y(t ) = y. (19) 1exun The the function y(t) z(t) is a solution of the problem { w + p(t)w =, t (, T ), w() =. Multiplying the equation ( exun1a 2) by e d dt p(s)ds we obtain (w(t)e ) t p(s)ds =. Therefore w(t)e p(s)ds = e() =, t [, T ]. Hence w(t) = or y(t) = z(t), t [, T ]. (2) exun1a

11 Ex1 Theorem 6.2. (Existence and uniqueness) Suppose that the functions f(t, y) and f y (t, y) are continuous on the rectangle Then the initial value problem R := {(t, y) : t a, y y b} has a unique solution defined on the interval where y = f(t, y), y() = y (21) exun1 t h = min { } b a,, M M = max f(t, y), (t, y) R. Proof. Note that since f y (t, y) L, (t, y) R, the function f(, ) satisfies the Lipschitz condition in R : f(t, y 1 ) f(t, y 2 ) L y 1 y 2, (t, y 1 ), (t, y 2 ) R. (22) exlip The problem ( 21) exun1 is equivalent to the integral equation: y(t) = y + 11 f(s, y(s))ds. (23) exa Let us show that there exists a continuous function y(t) that satisfies( 23) exa (it follows from ( 23) exa that the function y(t) is differentiable, and satisfies ( 21)). exun1 We are going to show existence of such a function employing the method of successive approximations (Picard s iteration method), and start by choosing an initial function y (t) = y, t [ h, h] The subsequent iteration we choose as follows: y 1 (t) = y + y 2 (t) = y + y n (t) = y +... f(s, y )ds, f(s, y 1 (s))ds, f(s, y n 1 (s)ds, (24) exa4 Let us show that the graph of y n (t) is lying on R when t h (remember that f is defined only on R and f(t, y n (t)) has a meaning only when y n (t) y b, t [ h, h]). For n = the statement is clear For n = 1 and t [, h] we have y 1 (t) y = (t, y ) R, t [ h, h] f(s, y )ds f(s, y ) ds M t Mh = b

12 12 Suppose that the graph of y n (t) is in R, i.e. Then y n+1 (t) y y n (t) y b, t [ h, h] f(s, y n (s) ds M t Mh = b For t [ h, ] we have same estimate. It remains to show that there exists a continuous function y(t) defined on [ h, h], such that lim y n(t) = y(t), t [ h, h] n It is easy to see that the sequence {y n (t)} converges if and only if the series y + (y n (t) y n 1 (t)) (25) exa3 n=1 converges, and lim n y n (t) is equal to the sum of the series ( 25). exa3 To show that ( 25) exa3 converges to some continuous function (of course, defined on [ h, h]) we need the following. Proposition A Suppose that the functions f n (t), n = 1, 2,... are continuous on some interval [a, b], and f n (t) C n, n = 1, 2,... t [a, b], where the number series C n is convergent. Then the functional series f n (t) is n=1 absolutely convergent on [a, b], and the function f(t) = f n (t) is a continuous function. Morever b a f(t)dt = It is not difficult to see that y 1 (t) y = y 2 (t) y 1 (t) = n=1 n=1 b [f(s, y 1 (s))ds f(s, y )]ds f(s, y 1 (s)) f(s, y ds L y 1 (s) y ds ML t2 2, a f n (t)dt f(s, y )ds Mt, n=1

13 13 y 3 (t) y 2 (t) LML y n (t) y n 1 (t) L L y 2 (s) y 1 (s) ds s 2 t3 ds = ML2 2 3!,... y n 1 (s) y n 2 (s) ds ML n 1 s n 1 tn ds = MLn 1 (n 1)! n! ML(Lh)n. n! We know that the series (Lh) n converges. n! n=1 Therefore, due to the Proposition A, the series ( 24) exa4 converges to some continuous function y(t), which is the limit of the sequence {y n (t)} on [ h, h]. Hence passing to the limit in (3) as n we obtain y(t) = y + f(s, y(s))ds. Suppose that y(t) is a solution of the problem ( 18) exun and z(t)is a solution of the problem z (t) = f(t, z(t)), z() = z (26) exunz Assume that y(t)and z(t) exists on the interval [, T ]. Then subtracting from ( 23) exa the equation z(t) = z + which is equivalent to ( 26) exunz we get: y(t) z(t) y z + f(s, z(s))ds f(s, y(s)) f(s, z(s)) ds y z + L By using the Gronwall s lemma we obtain from ( 27) exun5 the following estimate: y(s) z(s) ds. (27) exun5 y(t) z(t) y z e LT, t [, T ] (28) exun6 It follows from the inequality ( 28) exun6 that if y is sufficiently close to z, then y(t) is sufficiently close to z(t) on [, T ], i.e. if z y is small enough, then max y(t) z(t) t [,T ] is small enough. In particular ( 28) exun6 implies that the problem ( 21) exun1 has a unique solution. Lemma (Gronwall)* If u(t), v(t) are non-negative continuous functions on [, T ], C is a non-negative number, and u(t) C + u(s)v(s)ds, t [, T ], (29) Gr7

14 14 1ordpr1 then Proof. Let us denote u(t) Ce v(s)ds, t [, T ]. (3) Gr8 Ψ(t) := C + u(s)v(s)ds Then multiplying both sides of ( 29) Gr7 by v(t) we get ( ) u(t)v(t) C + u(s)v(s)ds v(t) or Ψ (t) v(t)ψ(t), Ψ() = C Multiplying this inequality by e v(s)ds we obtain (e v(s)ds Ψ(t)) It follows from last inequality that the function is non-increasing on [, T ]. Therefore Hence e v(s)ds Ψ(t) e v(s)ds Ψ(t) Ψ() = C, t [, T ] Ψ(t) Ce v(s)ds Since Ψ(t) u(t), t [, T ] we see that ( 29) Gr7 holds true. You can find the proof of Gronwall s lemma and some of its generalizations in a famous book of E.F Beckennbach and R. Bellman. 1 Proposition 6.3. If f(y) is an increasing and continuous function, then has a unique solution. y + f(y) =, y(t ) = y (31) exu1 Suppose that the problem ( exu1 31) has two different solutions y(t) and ỹ(t). Then the function w(t) = y(t) ỹ(t) is a solution of the problem w (t) + f(y(t)) f(ỹ(t)) =, (32) exu2 w(t ) =. (33) exu3 Multiplying ( 32) exu2 by w(t) we obtain ( ) d 1 dt 2 w2 (t) = (f(y(t)) f(ỹ(t))(y(t) ỹ(t)). (34) exu4 1 E.F.Beckenbach and R.Belllman, Inequalities, Springer Verlag, 1961.

15 15 Since the function f(y) is an increasing function for each y and ỹ (f(y) f(ỹ))(y ỹ). Therefore ( 34) exu4 implies: ( ) d 1 dt 2 w2 (t). Integrating this inequality and using ( 33) exu3 we get w 2 (t) w 2 (t ) =, t t. Therefore w(t) =, t t. The following problem has infinitely many solutions. y = y; y() = For each positive L the following function is a solution of this problem {, for x L, and y L (t) = 1 (t 4 L)2, for x > L. The problem has a solution which tends to + as t 1. y = y 2 y() = 1 y(t) = 1 1 t 7. Some famous first order equations The equation y + p(t)y = q(t)y n (35) Ber is called the Beournulli equation. It was first considered by Jacob Bernoulli. The change v = y n 1 is reducing the Bernoulli equation to the linear first order equation. v = y 1 n, y n y + y n+1 p(t) = q(t), v = (1 n)y n y 1 1 n v + p(t)v = q(t). The following equation is called a Riccati equation Properties of the Riccati equation: y (t) = a(t)y 2 (t) + b(t)y(t) + c(t). (36) Ric i) If y 1 (t) is a particular solution of the Riccati equation, then setting y(t) = y 1 (t) + z(t)

16 16 we obtain: y 1 + z a(y y 1 z + z 2 ) b(y 1 + z) c = z (2ay 1 + b)z az 2 = z (2ay 1 + b)z az 2 = ii) If y, y 1, y 2, y 3 are solution of the Riccati equation, then y y 2 : y 3 y 2 = const. y y 1 y 3 y 1 The following equation y (t) = ry(t)(1 y(t)) is called the logistic equation. An equation of the form y = ty (t) + f(y (t) is called the Clairot equation.