The Method of Undetermined Coefficients.

Transcription

1 The Method of Undetermined Coefficients. James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University May 24, 2017

2 Outline 1 Annihilators 2 Finding The Annihilator L(D) 3 Linear Models with Forcing Functions

3 Abstract This lecture is about the method of undetermined coefficients for linear ODE.

4 Annihilators The standard exponential growth/ decay model x = r x; x(0) = A is the same as x r x = 0; x(0) = A (D r)(x) = 0; x(0) = A. Thus, x = 3 x; x(0) = A (D 3)(x) = 0; x(0) = A. x = 2 x; x(0) = A (D + 2)(x) = 0; x(0) = A.

5 Annihilators Definition Let L(D) be a differential operator for a linear ordinary differential equation. Then L(D) = a 0 D n + a n 1 D n a n 1 i.e. L(D) is a n th degree polynomial in D and so the characteristic equation has n roots. The kernel of L(D) is the set of all functions f such that L(D)(f ) = 0. We denote the kernel of L(D) by ker(l(d). L(D) = D 2. Then ker(d 2) = {f : f (t) = Ae 2t } where A is an arbitrary real number. Since every function in ker(d 2) is a multiple of e 2t we say e 2t is a basis for the kernel of D 2. We denote this by writing ker(d 2) =< e 2t >.

6 Annihilators L(D) = D + 2. Then ker(d + 2) = {f : f (t) = Ae 2t } where A is an arbitrary real number. Since every function in ker(d + 2) is a multiple of e 2t we say e 2t is a basis for the kernel of D + 2. We denote this by writing ker(d + 2) =< e 2t >. L(D) = D. Then ker(d) = {f : f (t) = Ae 0t } = {f : f (t) = A 1} where A is an arbitrary real number. Since every function in ker(d) is a multiple of the constant function 1 we say 1 is a basis for the kernel of D. We denote this by writing ker(d) =< 1 >.

7 Annihilators L(D) = D 2. Then since the root 0 occurs twice ker(d 2 ) = {f : f (t) = Ae 0t + Bte 0t } = {f : A 1 + B t} where A and B are arbitrary real numbers. Since every function in ker(d 2 ) is a combination of 1 and t, we say {1, t} is a basis for the kernel of D 2. We denote this by writing ker(d 2 ) =< 1, t >. L(D) = D 3. Then since the root 0 occurs three times, an argument similar to what we have already done for roots repeated two times shows ker(d 3 ) = {f : f (t) = Ae 0t + Bte 0t + Ct 2 e 0t } = {f : A 1 + B t + C t 2 } where A, B and C are arbitrary real numbers. Since every function in ker(d 3 ) is a combination of 1, t and t 2 we say {1, t, t 2 } is a basis for the kernel of D 3. We denote this by writing ker(d 3 ) =< 1, t, t 2 >.

8 Annihilators L(D) = (D 3) 2. Then since the root 3 occurs twice ker((d 3) 2 ) = {f : f (t) = Ae 3t + Bte 3t } where A and B are arbitrary real numbers. Since every function in ker((d 3) 2 ) is a combination of e 3t and te 3t, we say {e 3t, te 3t } is a basis for the kernel of (D 3) 2. We denote this by writing ker((d 3) 2 ) =< e 3t, te 3t >. L(D) = D The root now is pure imaginary, ±2i and so ker(d 2 + 4) = {f : f (t) = A cos(2t) + B sin(2t)} where A and B are arbitrary real numbers. Since every function in ker(d 2 + 4) is a combination of cos(2t) and sin(2t) we say {cos(2t), sin(2t)} is a basis for the kernel of D We denote this by writing ker(d 2 + 4) =< cos(2t), sin(2t) >.

9 Annihilators L(D) = (D 2 + 4) 2. The root now is pure imaginary, ±2i and is repeated. Hence, the complex solutions are combinations of e 2i t, te 2i t and e 2i t, te 2i t. Our argument from before allows us to find real solutions which are combinations of cos(2t), t cos(2t), sin(2t) and t sin(2t). Thus ker((d 2 + 4) 2 ) = {f : f (t) = A cos(2t) + B sin(2t) +Ct cos(2t) + Dt sin(2t)} where A, B, C and D are arbitrary real numbers. Since every function in ker((d 2 + 4) 2 ) is a combination of cos(2t), sin(2t), t cos(2t) and t sin(2t), we say {cos(2t), t cos(2t), sin(2t), t sin(2t)} is a basis for the kernel of (D 2 + 4) 2. We denote this by writing ker((d 2 + 4) 2 ) =< cos(2t), sin(2t), t cos(2t), t sin(2t) >.

10 Annihilators L(D) = D 2 2D + 5. The roots are 1 ± 2i and so ker(d 2 2D + 5) = {f : f (t) = A cos(2t)e t + B sin(2t)e t } where A and B are arbitrary real numbers. We say {cos(2t)e t, sin(2t)e t } is a basis for the kernel of D 2 2D + 5. Hence, ker(d 2 2D + 5) =< cos(2t)e t, sin(2t)e t >. L(D) = (D 2 2D + 5) 2. The roots 1 ± 2i that occurs twice and so ker((d 2 2D + 5) 2 ) = {f : f (t) = A cos(2t)e t + B sin(2t)e t +Ct cos(2t)e t + Dt sin(2t)} where A, B, C and D are arbitrary real numbers. We say {cos(2t)e t, sin(2t)e t, t cos(2t)e t, t sin(2t)e t } is a basis for the kernel of (D 2 2D + 5) 2. Thus, ker((d 2 + 4) 2 ) =< cos(2t)e t, sin(2t)e t, t cos(2t)e t, t sin(2t)e t >.

11 Annihilators Definition Let L(D) be a differential operator for a linear ordinary differential equation. Then L(D) = a 0 D n + a n 1 D n a n 1 i.e. L(D) is a n th degree polynomial in D and so the characteristic equation has n roots. The functions in the basis of L(D) are said to be annihilated by L(D). Equivalently, we say L(D) annihilates each function in the basis of ker(l(d)). We also say L(D) is an annihilator of each function in the basis of ker(l(d)). L(D) = D 2. Then ker(d 2) =< e 2t >. Hence, D 2 is an annihilator of e 2t.

12 Annihilators L(D) = D + 2. Then ker(d + 2) =< e 2t >. So D + 2 is an annihilator of e 2t. L(D) = D. Then ker(d) =< 1 >. So D is an annihilator of 1. L(D) = D 2. Then ker(d 2 ) =< 1, t >. So D 2 annihilates any polynomial A + Bt, ie D 2 annihilates 1 and t and any linear combination of them. L(D) = D 3. Then ker(d 3 ) =< 1, t, t 2 >. So D 3 annihilates any polynomial A + Bt + Ct 2, ie D 3 annihilates 1, t and t 2 and any linear combination of them.

13 Annihilators L(D) = (D 3) 2. Then ker((d 3) 2 ) =< e 3t, te 3t >. So (D 3) 2 annihilates e 3t and te 3t and any linear combination of them. L(D) = D Then ker(d 2 + 4) =< cos(2t), sin(2t) >. So D annihilates cos(2t) and sin(2t) and any linear combination of them. L(D) = (D 2 + 4) 2. Then ker((d 2 + 4) 2 ) =< cos(2t), sin(2t), t cos(2t), t sin(2t) >. So (D 2 + 4) 2 annihilates cos(2t), sin(2t), t sin(2t) and t cos(2t) and any linear combination of them.

14 Annihilators L(D) = D 2 2D + 5. Then ker(d 2 2D + 5) =< cos(2t)e t, sin(2t)e t >. So D 2 2D + 5 annihilates cos(2t)e t and sin(2t)e t and any linear combination of them. L(D) = (D 2 2D + 5) 2. Then ker((d 2 2D + 5) 2 ) =< cos(2t)e t, sin(2t)e t, t cos(2t)e t, t sin(2t)e t > So (D 2 2D + 5) 2 annihilates cos(2t)e t, sin(2t)e t, t cos(2t)e t and t sin(2t)e t and any linear combination of them. L(D) = D 2 9 = (D + 3)(D 3). Then ker(d 3) =< e 3t > and ker(d + 3) =< e 3t >. Thus, ker(d 2 9) =< e 3t, e 3t >. So D 2 9 annihilates e 3t and e 3t and any linear combination of them.

15 Annihilators L(D) = D 2 + 5D + 6 = (D + 2)(D + 3). Then ker(d + 2) =< e 2t > and ker(d + 3) =< e 3t >. Thus, ker(d 2 + 5D + 6) =< e 2t, e 3t >. So D 2 9 annihilates e 2t and e 3t and any linear combination of them. L(D) = (D 2 + 5D + 6)(D 2 + 4) = (D + 2)(D + 3)(D 2 + 4). Then ker(d + 2) =< e 2t >, ker(d + 3) =< e 3t > and ker(d 2 + 4) =< cos(2t), sin(2t) >. Thus, ker((d 2 + 5D + 6)(D 2 + 4)) =< e 2t, e 3t, cos(2t), sin(2t) >. So (D 2 + 5D + 6)(D 2 + 4) annihilates e 2t, e 3t, cos(2t) and sin(2t). and any linear combination of them. We can clearly find the kernel of differential operators of this type by factoring and finding the kernels of each factor. This tells us the functions the operator annihilates.

16 Annihilators Homework For L(D) = (D 2 + 9) 2, find its kernel and the basis of the kernel. and find the functions L(D) annihliates For L(D) = D + 4, find its kernel and the basis of the kernel. and find the functions L(D) annihliates For L(D) = (D 5) 2, find its kernel and the basis of the kernel. and find the functions L(D) annihliates For L(D) = (D + 3) , find its kernel and the basis of the kernel. and find the functions L(D) annihliates For L(D) = (D )(D 2 ), find its kernel and the basis of the kernel. and find the functions L(D) annihliates.

17 Finding The Annihilator L(D) What is the annihilator of 2 + 4t + 5t 2? Solution D 3 annihilates cubic polynomials. What is the annihilator of 2 + 4t + t cos(3t)? Solution We need D 2 to annihilate 2 + 4t and we need (D 2 + 9) 2 to annihilate t cos(3t). So the annihilator is D 2 (D 2 + 9) 2.

18 Finding The Annihilator L(D) Homework Find the annihilator of 3 + 2t 3 + e 3t cos(5t) Find the annihilator of sin(4t) + e 3t cos(5t) Find the annihilator of t 2 sin(5t) Find the annihilator of 10 + t Find the annihilator of 4 cos(t) + 5 sin(2t) 7t cos(5t).

19 Linear Models with Forcing Functions Definition For any linear differential operator L(D) which is a polynomial in D, the homogeneous model is L(D)(x) = 0, plus initial conditions The nonhomogeneous model has the form L(D)(x) = f, plus initial conditions where f is called the forcing function. Of course, if f = 0 the homogeneous and nonhomogeneous models are the same. Comment The forcing function is also the input function or just input.

20 Linear Models with Forcing Functions We already know how to solve the homogeneous IVP. If the degree of the polynomial L(D) is n, we need n IC s: x(0) = x 0, x (1) (0) = x 1 x (2) (0) = x 2, x (3) (0) = x 3. =. x (n 2) (0) = x n 2, x (n 1) (0) = x n 1 where x (i) is the i th derivative of x. Factor L(D) by finding the roots of the characteristic polynomial. The ker(l(d) is the general solution and involves n unknown constants. Use the ICS to find the values of the constants.

21 Linear Models with Forcing Functions Solve the problem x + 5x + 6 = 0, x(0) = 2 and x (0) = 3. Solution We know ker(d 2 + 5D + 6) = ker((d + 2)(D + 3)) and so any function in the kernel has the form Ae 2t + Be 3t which is exactly the form we have used for the general solution. Note the roots of the characteristic equation r 2 + 5r + 6 = 0 are just another way to find the kernel of this linear differential operator. We can then use the ICs to find A and B like usual. Solving a model with a third or higher derivative term in it is no different in principle. Once we have the roots to the characteristic polynomial we are essentially done.

22 Linear Models with Forcing Functions Definition For any linear differential operator L(D) which is a polynomial in D, the general solution to the homogeneous model L(D)(x) = 0 is denoted x h and any solution to the nonhomogeneous model, L(D)(x) = f, is called x p where the p means it is a particular solution to the nonhomogeneous model. At this point the ICs have not yet been used. The general solution to the nonhomogeneous model is then x = x h + x p. The ICS are then used with the general solution to determine the arbitrary constants. Solve the problem x + 5x + 6 = 2 + 3t, x(0) = 2 and x (0) = 3.

23 Linear Models with Forcing Functions Solution We know ker(d 2 + 5D + 6) = ker((d + 2)(D + 3)) and so any function in the kernel has the form Ae 2t + Be 3t which is exactly the form we have used for the general solution. Hence, x h (t) = Ae 2t + Be 3t To find x p, note that D 2 annihilates the forcing function f (t) = 2 + 3t. Then ker(d 2 ) = {1, t} and any function in the kernel has the form C + Dt. Apply D 2 to the inhomogeneous model to get D 2 (D 2 + 5D + 6) = D 2 (2 + 3t) = 0 The kernel of D 2 (D 2 + 5D + 6) =< 1, t, e 2t, e 3t >

24 Linear Models with Forcing Functions Solution The kernel of the annihilator of the forcing function and the kernel of the model do not overlap. So guess x p (t) = C + Dt which is the general form of the kernel of the annihilator of the forcing function. x p must solve the nonhomogeneous model, so calculate (D 2 + 5D + 6)(C + Dt) = 0 + 5D + 6(C + Dt) and set this equal to the forcing function 2 + 3t. We get. (5D + 6C) + 6Dt = 2 + 3t

25 Linear Models with Forcing Functions Solution The only way this is true is if the coefficients of each power of t match. So we get 5D + 6C = 2, 6D = 3 and so D = 1/2 and C = 1/12. So x p (t) = (1/2) + (1/12)t So x = x h + x p = x(t) = Ae 2t + B 3t + (1/2) + (1/12)t Now apply ICs. x(0) = 2 = A + B + (1/2) x (0) = 3 = 2A 3B + (1/12) or A + B = 3/2 2A 3B = 3 (1/12) = 37/12

26 Linear Models with Forcing Functions Solution [ ] [ ] 1 1 A 2 3 B = [ ] 3/2 37/12 By Cramer s rule ([ ]) ([ ]) 3/ A = det / det( 37/ = ( 9/2 + 37/12)/( 1) = 17/12 ([ ]) ([ ]) 1 3/2 1 1 B = det / det( 2 37/ = ( 37/12 + 6/2)/( 1) = 1/12 So x(t) = (17/12)e 2t + (1/12)e 3t + (1/2) + (1/12)t

27 Linear Models with Forcing Functions Solve the problem x + 5x + 6 = 2 + 3t + 10e 3t, x(0) = 2 and x (0) = 3. Solution We know ker(d 2 + 5D + 6) = ker((d + 2)(D + 3)) and so any function in the kernel has the form Ae 2t + Be 3t which is exactly the form we have used for the general solution. Hence, x h (t) = Ae 2t + Be 3t To find x p, note that D 2 (D + 3) annihilates the forcing function f (t) = 2 + 3t + 10e 3t.

28 Linear Models with Forcing Functions Solution Apply D 2 (D + 3) to the inhomogeneous model to get D 2 (D + 3)(D 2 + 5D + 6) = D 2 (D + 3)(2 + 3t + 10e 3t ) = 0 Factoring, we see D 2 (D + 3)(D 2 + 5D + 6) = D 2 (D + 3) 2 (D + 2) The kernel of D 2 (D + 3) 2 (D + 2) =< 1, t, e 2t, e 3t, te 3t >. The forcing function s annihilator involves D 2 and D + 3. So the part of D 2 (D + 3) 2 (D + 2) we need to keep is everything related to D and D + 3. So we keep D 2 (D + 3) 2. The kernel of D 2 (D + 3) 2 is < 1, t, e 3t, te 3t > and the general form of a function in this kernel is what we choose for x p. Thus, x p (t) = A + Bt + Ce 3t + Ete 3t

29 Linear Models with Forcing Functions Solution Apply D 2 + 5d + 6 to x p. We will skip some of the steps here, so do them yourself. D 2 + 5D + 6(A + Bt + (C + Et)e 3t ) = D 2 (A + Bt) + D 2 (C + Et)e 3t ) +5D(A + Bt) + 5D(C + Et)e 3t ) +6A + 6Bt + 6(C + Et)e 3t = (9C 6E + 9Et)e 3t + 5B + 5(E 3C 3Et)e 3t = +6A + 6Bt + 6(C + Et)e 3t Equating coefficients of 1, t, e 3t and te 3t, we have E = 10, A = 1/12 and B = 1/2. Hence x p (t) = (1/12) + (1/2)t 10te 3t

30 Linear Models with Forcing Functions Solution Now use ICs, Recall x h (t) = Ae 2t + Be 3t. So 2 = x(0) = x h (0) + x p (0) = A + B (1/12) 3 = x (0) + x (0) = 2A 3B 10 We then find A and B like usual

31 Linear Models with Forcing Functions Homework Solve x + 4x + 4x = 1 + 3t, x(0) = 2, x (0) = Solve x 4x 8x = 2 sin(2t), x(0) = 1, x (0) = Solve x + 4x + 13x = 40, x(0) = 2, x (0) = 4.