Linear Homogeneous ODEs of the Second Order with Constant Coefficients. Reduction of Order

Transcription

1 Linear Homogeneous ODEs of the Second Order with Constant Coefficients. Reduction of Order October 2 6, 2017

2 Second Order ODEs (cont.) Consider where a, b, and c are real numbers ay +by +cy = 0, (1) Let us also seek exponential solutions of (1), i.e. in the form y(t) = e rt, where r is a parameter ( to be determined ) (TBD). Then y = re rt, y = r 2 e rt Plug back to (1): ar 2 +br +c e rt = 0, and since e rt 0 we have The characteristic equation for the differential equation (1) ar 2 +br +c = 0 (2) In other words, if r is a root of the polynomial equation (2), then y(t) = e rt is a solution of the ODE (1) We distinguish 3 cases based on the nature of the roots of (2), namely when 1 Roots of (2) are real and different: r 1 r 2 2 Roots of (2) are complex conjugate: r 1 = λ+iµ, r 2 = λ iµ 3 Roots of (2) are real and equal: r 1 = r 2 =: r

3 I. Real and Different Roots of (2) Consider the case of r 1 r 2, then two functions y 1(t) = e r1t and y 2(t) = e r 2t solve (1). Let us check their Wronskian W [ e r1t,e ] r 2t = er 1t e r 2t r 1e r 1t r 2e r 2t = e(r 1+r 2 )t [r 1 r 2] 0, because e (r 1+r 2 )t 0 and r 1 r 2 0. Hence, y 1(t) = e r 1t and y 2(t) = e r 2t form a fundamental set of solutions of (1) and its general solution is Example: y(t) = C 1e r 1t +C 2e r 2t Solve { y +2y 3y = 0 y(0) = 0, y (0) = 1 The corresponding characteristic equation is r 2 +2r 3 = 0 whose roots are r 1 = 3 and r 2 = 1. Then the general solution of the given ODE by (3) is y(t) = C 1e 3t +C 2e t. Applying given ICs we have C 1 +C 2 = 0 and 3C 1 C 2 = 1 (from y (t) = 3C 1e 3t C 2e t ) C 1 = 1 4 and C 2 = 1 4. Hence, the solution of the IVP is y(t) = 1 4 e3t 1 4 e t (3)

4 II. Complex Conjugate Roots of (2) Consider the case of r 1,2 = λ±iµ then two functions y 1(t) = e (λ+iµ)t and y 2(t) = e (λ iµ)t solve (1). We now recall Euler s Formula: apply it for y 1(t) and y 2(t): e α±iβ = e α (cosβ ±i sinβ) e (λ±iµ)t = e λt (cosµt ±i sinµt) Unfortunately, these solutions y 1 and y 2 are complex-valued functions, whereas in general we would prefer to have real-valued solutions, if possible, because the ODE (1) itself had real coefficients. Such solutions can be found as a consequence of Principle of Superposition of last lecture, namely, Claim: Functions also solve (1). u(t) = e λt cosµt and v(t) = e λt sinµt Check the validity of this claim yourself!

5 II. Complex Conjugate Roots (cont.) Let us check the Wronskian of u(t) and v(t): W [u(t),v(t)] = e λt cosµt e λt sinµt λe λt cosµt µe λt sinµt λe λt sinµt +µe λt cosµt = µe2λt 0, because e 2λt 0 and µ 0 (otherwise, the roots r 1,2 would be real). Hence, u(t) = e λt cosµt and v(t) = e λt sinµt form a fundamental set of solutions of (1) and its general solution is y(t) = C 1e λt cosµt +C 2e λt sinµt (4) where λ and µ are the real and imaginary parts of roots of (2), resp. Example: Solve { y +2y +2y = 0 y(0) = 0, y (0) = 1 The corresponding characteristic equation is r 2 +2r +2 = 0 whose roots are r 1,2 = 1±i. Then the general solution of the given ODE is y(t) = C 1e t cost +C 2e t sint. Applying given ICs we have C 1 = 0 and C 2 = 1 (from y (t) = C 1e t cost C 1e t sint C 2e t sint +C 2e t cost). Hence, the solution of the IVP is y(t) = e t sint

6 III. Real and Equal Roots of (2) Finally, consider the case of r 1 = r 2 =: r. The corresponding solution of (1) is y 1(t) = e rt but in order to find a general solution we need need a second solution y 2(t) that is not a multiple of y 1(t). This second solution can be found in several ways. Here, we use the method of reduction of order as follows: Assume that the second solution can be sought in the form y 2(t) = y 1(t)v(t) = e rt v(t), (5) where the function v(t) is TBD. Then computing y 2(t) = re rt v(t)+e rt v (t) and y 2 (t) = r 2 e rt v(t)+2re rt v (t)+e rt v (t) we substitue these back to (1): a [ r 2 e rt v +2re rt v +e rt v ] +b [ re rt v +e rt v ] +ce rt v = e [ rt v +(2ar +b)v +(ar 2 +br +c) ] = 0 First, we note that if r is a double root of (2) then r = b and ar 2 +br +c = 0 in (6) 2a therefore, since e rt 0, we have v = 0 or v(t) = C 1t +C 2, where C 1, C 2 are arbitrary constants Finally, substituting for v(t) in (5) we obtain the general solution of (1) as y(t) = C 1e rt +C 2te rt where C 1, C 2 are arbitrary constants (7) (6)

7 III. Real and Equal Roots (cont.) In other words, the general solution to the ODE (1) in the case of two equal roots of the characteristic equation (2) is a linear combination of e rt and te rt. As before, let us check their Wronskian: W [ e rt,te rt] = ert te rt re rt e rt +rte rt = e2rt 0, hence, y 1(t) = e rt and y 2(t) = te rt, indeed, form a fundamental set of solutions of (1) and its general solution is (7) Example: Solve { 9y +6y +1y = 0 y(0) = 1, y (0) = 1 The corresponding characteristic equation is 9r 2 +6r +1 = 0 whose roots are r 1,2 = 1. Then the general solution of the given ODE is 3 y(t) = C 1e t/3 +C 2te t/3. Applying given ICs we have C 1 = 1 and C 2 = 1 (from y (t) = 1 3 C1e t/3 +C 2e t/3 1 3 C2e t/3 ) C 2 = 4 3. Hence, the solution of the IVP is y(t) = e t/ te t/3

8 Reduction of Order Yuliya Gorb The procedure that we used above for the ODE (1) is more generally applicable. Suppose we know one solution y 1(t), that is not trivially zero, of the ODE y +p(t)y +q(t)y = 0 (8) To find a second solution y 2(t), assume y 2(t) = y 1(t)v(t) where v(t) is TBD. y 2 = v y 1 +vy 1 and y 2 = v y 1 +2v y 1 +vy 1. Substituting these in (10) and collecting terms, we find that y 1v +(2y 1 +py 1)v + ( y 1 +p(t)y 1 +q(t)y 1 ) v = 0 Since y 1 is a solution of (10), the coefficient of v then is zero, hence, obtain y 1v +(2y 1 +py 1)v = 0 (9) Despite its appearance, (9) is actually a 1st order ODE for v =: u and can be solved either as a 1st order linear equation (integrating factor method) or as a separable equation. Once v = u is found, then v is obtained by an integration v = u +C. Finally, y is determined from the assumption y 2(t) = y 1(t)v(t). This procedure is called the method of reduction of order because the crucial step is the solution of a 1st order ODE for v rather than the original 2nd order ODE for y.

9 Reduction of Order (cont.) Example: Find the general solution of the given ODE provided the first solution is known: (x 1)y xy +y = 0, x > 1, y 1(x) = e x As suggested above, assume y 2(x) = y 1(x)v(x) = e x v(x), where v(t) is TBD. Compute y 2 = e x v +e x v and y 2 = e x v +2e x v +e x v ans plug those back to the ODE: (x 1) [ e x v +2e x v +e x v ] x [ e x v +e x v ] +e x v = 0. Factoring our e x and collecting terms v, v and v we obtain (x 1)v +(x 2)v = 0 Substitute u := v, then obtain a 1st order linear ODE for u: (x 1)u +(x 2)u = 0, whose integrating factor is µ(x) = ex x 1 u(x) = Ce x (x 1) v = u +C 0 = Ce x x +C 0. Then the second solution is y 2(x) = e x v(x) = Cx +C 0e x Hence, the general solution of the given ODE is y(t) = C 1x +C 2e x, where C 1, C 2 are arbitrary constants

10 Abel s Theorem about Wronskian We start with Theorem (Abel) Suppose that y 1 (t) and y 2 (t) are solutions of y +p(t)y +q(t)y = 0, (10) with p(t) and q(t) that are continuous on some (a,b). Then the Wronkian of y 1 and y 2 is given by W[y 1,y 2 ](t) = Ce p(t)dt, (11) where the constant C depends on solutions y 1 and y 2. Note that since the Wronskian is a multiple of exponential function, then it is either trivially zero (when C = 0), or some non-zero function (when C 0)

11 Abel s Theorem about Wronskian (cont.) Example: Given solutions y 1(t) = t 1/2 and y 2(t) = t 1 of 2t 2 y +3ty y = 0, t > 0 Verify that their Wronskian is given by (11)? 1 By def: W [ t 1/2,t 1] = t1/2 t t 1/2 t 2 = 3 2 t 3/2 2 By (11) with p(t) = 3 2t : W[y1,y2](t) = Ce 3 2t dt = Ct 3/2 hence, (11) holds with C = 3 2

12 Reduction of Order Yuliya Gorb We apply (11) to reduction of order problem. Suppose one is given a solution y 1(t) of ODE (10). The goal is to find a general solution of (10). For that we need to seek for a second solution y 2(t) that together with y 1(t) forms a fundamental set of solutions to (10). For that we seek for such y 2 that provides the Wronskian (11) with const C = 1 together with the given y 1. 1 Then W[y 1,y 2](t) = y1(t) y2(t) y 1(t) y 2(t) = y1(t)y 2(t) y 1(t)y 2(t) = e p(t)dt, that is we obtain the 1st order linear ODE for the unknown y 2 that one can solve using the integrating factor method (dropping constants). This obtained solution y 2 does form a fundamental set of solutions to (10) since we chose C = 1 2 Alternative approach: With the same as above we seek for y 2 that gives C = 1 in (11) but assume that y 2(t) = y 1(t)v(t), where v(t) is TBD. Then W[y 1,y 2](t) = y1(t) y1(t)v(t) y 1(t) y 1(t)v (t)+y 1(t)v(t) = y2 1(t)v (t) = e or v (t) = e p(t)dt /y 2 1(t), which is solved by direct integration p(t)dt,

13 Reduction of Order (cont.) Example: Given known solution y 1(t) = t 1 find a general solution of 2t 2 y +3ty y = 0, t > 0 We already know that Abel s Thm applied for this ODE yields: W[y 1,y 2](t) = Ct 3/2. Assume C = 1, then 1 W[y 1,y 2](t) = t 3/2 = t 1 y 2 t 2 y 2 = t 1 y 2 t 2 y 2 = t 3/2. Solve this linear ODE y t 1 y 2 = t 1/2 for y 2 integrating factor is µ(t) = e dt t = t 1 d ] [t 1 y 2 = t 3/2 (drop a const) dt y 2(t) = t 1/2. Then a general solution is y(t) = C 1t 1 +C 2t 1/2 2 Here we assume that y 2(t) = t 1 v(t) and compute the corresponding Wronskian: W[y 1,y 2](t) = t 3/2 = t 1 t 1 v(t) t 2 t 1 v (t) t 2 v(t) = = t 2 v (t). Obtain ODE for v(t): t 2 v (t) = t 3/2 from which we find v(t) = t 3/2 (by dropping one constant and keeping const 1 in front of t 3/2 ). Then y 2(t) = t 1 t 3/2 = t 1/2 and a general solution is as above y(t) = C 1t 1 +C 2t 1/2