STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0.

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1 STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA STEPHEN SCHECTER. The unit step function nd piecewise continuous functions The Heviside unit step function u(t) is given by if t <, u(t) = if t >. The function u(t) is not defined t t =. Often we will not worry bout the vlue of function t point where it is discontinuous, since often it doesn t mtter. u(t) u(t ) u(t b) u(t ) u(t b) b b Figure.. Heviside unit step function. The function u(t) turns on t t =. The function u(t ) is just u(t) shifted, or drgged, so tht it turns on t t =. The function u(t b) turns off t t = b. The function u(t ) u(t b), with < b, turns on t = nd turns off t t = b. We cn use the unit step function to crop nd shift functions. Multiplying f(t) by u(t ) crops f(t) so tht it turns on t t = : if t <, u(t )f(t) = f(t) if t >. Multiplying f(t) by u(t ) u(t b), with < b, crops f(t) so tht it turns on t t = nd turns off t t = b: if t <, (u(t ) u(t b))f(t) = f(t) if < t < b, if t > b. Dte: August 7, 26.

2 2 SCHECTER We cn lso crop f(t) t t = nd drg it to t = : if t <, u(t )f(t ) = f(t ) if t >. f(t) u(t )f(t) u(t)f(t) u(t )f(t ) Figure.2. Cropping nd drgging. A function f(t) is sid to hve jump discontinuity t t = if () there re numbers t < < t such tht f is defined nd continuous on [t, ) nd on (, t ]; (2) lim t f(t) nd lim t + f(t) both exist s finite numbers; nd (3) lim t f(t) lim t + f(t). A function f(t) is clled piecewise continuous on the rel line if () f is defined nd continuous except t certin points where it hs jump discontinuity, nd (2) the number of jump discontinuities in ny closed intervl is finite. Usully we leve piecewise continuous functions undefined t the points where they hve jump discontinuities, becuse usully their vlues there don t mtter. We will use the nottion f( ) = lim f(t), f(+) = lim f(t). t t + These numbers re the sme t point where f is continuous, but they re different t point where f hs jump discontinuity. f(+) f( ) f(t) Figure.3. A piecewise continuous function. The functions pictured so fr re piecewise continuous on the rel line.

3 STEP FUNCTIONS, DELTA FUNCTIONS 3 2. The Dirc delt function Piecewise continuous functions hve continuous ntiderivtives. For exmple, n ntiderivtive of u(t) is the unit rmp function if t <, u(t) dt = t if t. A function with jump discontinuity cnnot possibly hve derivtive t tht point. Nevertheless, let s sk the question: if the unit step function u(t) hd derivtive u (t), wht would its properties be? Actully, wherever t, u(t) hs derivtive, nd it s. So the first property is: () u (t) = for t. The second property comes from the Fundmentl Theorem of Clculus: if t < < t, we should hve u (t) dt = u(t)] t t = u(t ) u(t ) = =. t Thus our second property is: (2) If t < < t, then t u (t) dt =. No ordinry function hs properties () nd (2)! Nevertheless, we will go hed nd define generlized function tht does hve these properties. It s clled the Dirc delt function δ(t), nd its properties re: () δ(t) = for t. (2) If t < < t, then t δ(t) dt =. We sy δ(t) is concentrted t t =. There re certin opertions tht one cn legitimtely do with δ(t). One cn shift it: δ(t ) is concentrted t t =. One cn lso multiply δ(t ) by function f(t) tht is continuous on n open intervl tht contins t =. The resulting product hs the following properties: () δ(t )f(t) = for t. (2) If t < < t, then t δ(t )f(t) dt = f(). Notice tht the vlue of the integrl only depends on the vlue of f t t =. A consequence of property (2) is: If t < < t, then kδ(t ) dt = k. t There re other opertions tht I do not recommend trying with δ(t). For exmple, you should not try to squre it, nd you should not multiply it by function tht hs jump discontinuity t t =. 3. First-order liner differentil equtions Piecewise continuous functions nd delt functions cn be used s forcing terms in liner differentil equtions. We ll just consider liner differentil equtions with constnt coefficients. We ll often consider rest initil conditions: y(t) = for t <.

4 4 SCHECTER Let s first look t first-order liner differentil equtions (3.) my + ky = f(t). Rules bout solutions of (3.): () If f(t) is piecewise continuous, ny solution y(t) is continuous. (2) If f(t) is delt function, ny solution y(t) hs jump discontinuity where the delt function is concentrted. An importnt consequence of rule (2) is the following: Theorem 3.. Any solution y(t) of (3.2) my + ky = δ(t ) hs y(+) y( ) = m. Here is why this formul for the jump is correct. Let t < < t, let y(t) be solution of (3.2), nd integrte both sides of (3.2) from t = t to t = t. We get m(y(t ) y(t )) + k t y(t) dt =. Now let t nd t pproch. By rule (2), y(t) is n ordinry piecewise continuous function. When we integrte it over shorter nd shorter intervls, the integrl pproches. On the other hnd, y(t ) y(t ) pproches y(+) y( ). Pssing to the limit, we hve Therefore m(y(+) y( )) =. y(+) y( ) = m. Exmple. A new territory is discovered. One million people immeditely move there. Therefter the popultion grows nturlly with growth rte of 4% per yer. Wht is the popultion fter t yers? Solution. Let p denote the popultion in millions. Due to the immigrtion, the popultion t time is. Thus we hve dp dt =.4p, p() =. The solution of this initil vlue problem is p = e.4t. We should only use this formul for t. Solution 2. Let p denote the popultion in millions. Then dp dt =.4p + δ(t), p(t) = for t <. Since the forcing function is delt function concentrted t t =, by rule (2) p(t) hs jump discontinuity t t =. By Theorem 3. the jump is p(+) p( ) = =. Since p( ) =, we hve p(+) =. Therefore, for t > we solve the initil vlue problem dp dt =.4p, p() =

5 STEP FUNCTIONS, DELTA FUNCTIONS 5 exctly s before. The solution to our problem is if t <, p(t) = e.4t if t >. 4. Second-order liner differentil equtions Now let s consider second-order liner differentil equtions (4.) my + by + ky = f(t). Rules bout solutions of (4.): () If f(t) is piecewise continuous nd y(t) is solution, then y(t) is differentible nd y (t) is continuous. (2) If f(t) is delt function nd y(t) is solution, then y(t) is continuous, but y (t) hs jump discontinuity where the delt function is concentrted. An importnt consequence of rule (2) is the following: Theorem 4.. Any solution y(t) of (4.2) my + by + ky = δ(t ) hs y (+) y ( ) = m. To derive this formul, let t < < t nd integrte both sides of (4.2) from t = t to t = t. We get m(y (t ) y (t )) + b(y(t ) y(t )) + k y(t) dt =. t Now let t nd t pproch. By rule (2), y(t) is continuous function, so y(t ) y(t ) nd the integrl pproch. On the other hnd, y (t ) y (t ) pproches y (+) y ( ). Pssing to the limit, we hve Therefore m(y (+) y ( )) =. y (+) y ( ) = m. When force f(t) is pplied to mss from time t = t to time t = t, chnge in momentum results: mv(t ) mv(t ) = m dv t dt dt = f(t) dt. t The integrl t f(t) dt is clled the impulse due to the force. It is the chnge in momentum tht the force produces. Suppose mss m is t rest nd we strike it t time t = with hmmer. The mss tkes off with velocity v. Clerly the hmmer imprted force over very short time tht cused the chnge in momentum. How do we represent this force mthemticlly? Let s mke mthemticl ideliztion: v(t) = for t < nd v(t) = v for t >. Let f(t) be the force imprted by the hmmer. Then for ny t < < t, we hve (4.3) t f(t) dt = mv(t ) mv(t ) = mv = mv.

6 6 SCHECTER Therefore f(t) = mv δ(t). In mechnics problems in which force is imprted instntneously t time t =, the force is represented mthemticlly by constnt times δ(t ). The constnt represents the impulse due to the force. Such n instntneously pplied force is often itself clled n impulse. Exmple. A spring-mss system is t rest. We strike the mss with hmmer, hrd enough to cuse n instntneous positive chnge in momentum of unit. Wht is the response? Solution. After t =, the differentil eqution is my + by + ky =. The system strts t rest so y() =. Becuse of the hmmer blow, we instntly get momentum of, so we might s well tke my () =, i.e., y () =. Now we hve n initil m vlue problem to solve. Cll the solution h(t). Of course, we only use h(t) for t. Solution 2. The differentil eqution nd initil conditions re my + by + ky = δ(t), y(t) = for t <. Recll our rule for solving second-order liner differentil equtions with delt function forcing term: the solution is continuous, nd its first derivtive hs jump discontinuity where the delt function is concentrted. Since the solution is continuous, we hve y() =. By Theorem 4., the jump in the derivtive is y (+) y ( ) =. Since m y ( ) =, we hve y (+) =. Therefore, for t >, we solve the initil vlue problem m my + by + ky =, y() =, y () = m s before. The solution to our problem is if t <, y(t) = h(t) if t. 5. Vrition of prmeters formul for first-order differentil equtions Definition. The solution to the problem y + y = δ(t), y(t) = for t <. is clled the impulse response function for the differentil opertor y + y. It is given by if t <, y(t) = e t if t. Theorem 5.. The solution of (5.) y + y = f(t), y() = y is (5.2) y(t) = y e t + e (t s) f(s) ds.

7 STEP FUNCTIONS, DELTA FUNCTIONS 7 Formul (5.2) is clled the vrition of prmeters formul. The reson for this nme is not importnt. We will give two derivtions of (5.2). Derivtion. Solve (5.) using the method we lerned in Section 2.3. Multiply (5.) by e t, rewrite the left-hnd side s derivtive, nd integrte: e t y + e t y = e t f(t), d ( e t y ) = e t f(t), dt e t y = e s f(s) ds + c. To integrte the right-hnd side, we hd to choose n ntiderivtive of e t f(t); we chose t es f(s) ds. Multiply by e t : y = ce t + e t e s f(s) ds = ce t + e (t s) f(s) ds. Set t = nd y = y to determine c: c = y. Derivtion 2. This derivtion uses the impulse response function. It should give you good intuitive understnding of why the formul is true. To estimte y(t) t some t >, divide the intervl [, t] into n equl prts of length h: = t < t < < t n = t. The impulse given to the system between t = t i nd t = t i is i t i f(t) dt, which is pproximtely f(t i )h. Insted of pplying the continuous force f(t), let s pply the force in discrete impulses t the times t i. Then our force will be n f h (t) = δ(t t i )f(t i )h. i= If h is smll, the force f h (t) should produce pproximtely the sme result s f(t). The response to the impulse δ(t t i )f(t i )h is of course e (t ti) f(t i )h, t t i. (The stimulus response function is shifted to t = t i nd multiplied by the pproprite constnt.) To get the response to f h (t) t the prticulr time t mentioned erlier, the responses to the little blows t ll the times t i must be dded. Therefore the response is n y h (t) = e (t ti) f(t i )h. i= This is Riemnn sum for the integrl e (t s) f(s) ds. Therefore, s h, y h (t) pproches e (t s) f(s) ds. This is therefore the response y(t) of the system to the continuous forcing function f(t). Actully, this is the response when the initil condition is y() =. It is therefore prticulr solution of (5.). To get the generl solution, dd the generl solution of the homogeneous problem, ce t, nd determine c from the initil condition. 6. Vrition of prmeters formul for second-order differentil equtions Definition. The solution to the problem my + by + ky = δ(t), y(t) = for t <

8 8 SCHECTER is clled the impulse response function for the differentil opertor my +by +ky. It is given by if t <, y(t) = h(t) if t, where h(t) is the solution of the initil vlue problem my + by + ky =, y() =, y () = m. Sometimes we will cll h(t), defined for t, the impulse response function. I hope this won t be confusing. Theorem 6.. The solution of (6.) my + by + ky = f(t), y() = y, y () = y is (6.2) y(t) = y c (t) + h(t s)f(s) ds. In this formul, y c (t) is the solution of the homogeneous initil vlue problem nd h(t) is the impulse response function. my + by + ky =, y() = y, y () = y, Formul (6.2) is nother version of the vrition of prmeters formul. We will only give one derivtion of (6.2). It lmost word-for-word the sme s our derivtion of formul (5.2) using the impulse response function. To estimte y(t) t some t >, divide the intervl [, t] into n equl prts of length h: = t < t < < t n = t. The impulse given to the system between t = t i nd t = t i is i t i f(t) dt, which is pproximtely f(t i )h. Insted of pplying the continuous force f(t), let s pply the force in discrete impulses t the times t i. Then our force will be n f h (t) = δ(t t i )f(t i )h. i= If h is smll, the force f h (t) should produce pproximtely the sme result s f(t). The response to the impulse δ(t t i )f(t i )h is of course h(t t i )f(t i )h, t t i. (The stimulus response function is shifted to t = t i nd multiplied by the pproprite constnt.) To get the response to f h (t) t the prticulr time t mentioned erlier, the responses to the little blows t ll the times t i must be dded. Therefore the response is n y h (t) = h(t t i )f(t i )h. i= This is Riemnn sum for the integrl h(t s)f(s) ds. Therefore, s h, y h(t) pproches h(t s)f(s) ds. This is therefore the response y(t) of the system to the continuous forcing function f(t). Actully, this is the response when the initil condition is y() =. It is therefore prticulr solution of (5.). To get the generl solution, dd the generl solution of the homogeneous problem, nd determine c nd c 2 from the initil condition.

9 STEP FUNCTIONS, DELTA FUNCTIONS 9 Exmple. Solve d 2 y dt + y = tn t, y() =, 2 y () =. Solution. According to Theorem 6., the solution is (6.3) y(t) = h(t s) tnsds, where h(t) is the impulse response function. The impulse response function h(t) is the solution of d 2 y dt + y =, 2 y() =, y () =. We esily find tht h(t) = sin t. From (6.3), Now y(t) = = = sin t = sin t sin(t s) tnsds (sin t coss costsin s) tnsds cos s tnsds cost sin s ds cost sin 2 s coss ds. sin s ds = coss] t = cost +. With the help of p. 96 in the text, we find tht Therefore (6.4) sin s tnsds sin 2 s cos s ds = ln sec s + tn s sin s]t = ln sec t + tnt sin t. y(t) = sin t ( cos t + ) cos t (ln sec t + tnt sin t) = sin t cos t ln sec t + tnt. A slightly more generl version of Theorem 6. is sometimes useful. Theorem 6.2. The solution of (6.5) my + by + ky = f(t), y() = y, y () = y is (6.6) y(t) = y c (t) + h(t s)f(s) ds. In this formul, y c (t) is the solution of the homogeneous initil vlue problem nd h(t) is the impulse response function. my + by + ky =, y() = y, y () = y,

10 SCHECTER 7. Using vrition of prmeters to find prticulr solution A consequence of Theorem 6. is tht prticulr solution of (7.) my + by + ky = f(t) is (7.2) y p (t) = h(t s)f(s) ds. In this formul, h(t) is the impulse response function, i.e., the solution of my + by + ky =, y() =, y () = m, nd is ny convenient number. This formul provides nother wy to find prticulr solution of (7.), in ddition to undetermined coefficients. Let s do n exmple from Chpter 4.6 this wy. Exmple. Find the generl solution on ( π 2, π 2 ) to d 2 y dt 2 + y = tnt. Solution. The generl solution of the homogeneous eqution is y = c cost + c 2 sin t. A prticulr solution is given by formul (7.2) with =. We lredy clculted this in formul (6.4): y p (t) = sin t costln sec t + tnt. However, sin t is solution of the homogeneous eqution, so if we subtrct it, we will get simpler prticulr solution: Therefore the generl solution is y p (t) = costln sec t + tn t. y(t) = c cost + c 2 sin t costln sec t + tn t. Solution 2. The generl solution of the homogeneous eqution is y = c cost + c 2 sin t. A prticulr solution is given by formul (7.2) with =. We clculted erlier tht h(t) = sin t, so y p (t) = = = sin t = sin t sin(t s) tnsds (sin t coss costsin s) tnsds cos s tnsds cost sin s ds cost sin 2 s coss ds. sin s tnsds

11 Now STEP FUNCTIONS, DELTA FUNCTIONS sin s ds = coss] t = cost + cos. This must be multiplied by sin t, which is solution of the homogeneous eqution. Therefore we ignore the term sin t cos nd just use sin t cost. With the help of p. 96 in the text, we find tht sin 2 s coss ds = ln sec s + tns sin s]t = ln sec t + tnt sin t (ln sec + tn sin ) This must be multiplied by cos t, which is solution of the homogeneous eqution. Therefore we ignore the term cos t(ln sec + tn sin ) nd just use cos t(ln sec t + tn t sin t). Therefore prticulr solution is nd the generl solution is y p (t) = sin t cost cos t(ln sec t + tn t sin t) = costln sec t + tn t, y(t) = c cost + c 2 sin t costln sec t + tn t. 8. From the impulse response function to Lplce trnsforms You wlk into lb nd see spring-mss system set up. You strike the mss with hmmer nd mesure the response h(t). Now you know the impulse response function. Hence if the spring-mss system is strted from rest nd subject to ny time-dependent force f(t), you cn predict the response! It will be y(t) = h(t s)f(s) ds. Tht s pretty good, especilly considering tht you didn t bother to mesure the mss m, the dmping coefficient b, or the spring constnt k. In effect, you re finding solutions of the differentil eqution without knowing wht the differentil eqution is. Let s be greedy nd try to use h(t), which we lredy know, to figure out wht the differentil opertor my + by + ky is. Then we ll know m, b, nd k without doing ny tedious experiments to mesure them. I will describe one wy to do this. Recll how we used undetermined coefficients to find prticulr to solution to the nonhomogeneous eqution (8.) my + by + ky = e st. (In this formul, s is constnt.) We ssumed there ws solution of the form y p = Ae st nd substituted into (8.): (ms 2 + bs + k)ae st = e st. We concluded tht A =. Therefore prticulr solution of (8.) is ms 2 +bs+k y p = ms 2 + bs + k est. This works s long s s is not root of the chrcteristic eqution.

12 2 SCHECTER ms 2 +bs+k The function H(s) = is clled the trnsfer function of the differentil opertor my +by +ky. If you knew the trnsfer function you could figure out the differentil opertor. Indeed, the chrcteristic polynomil of the differentil opertor is ms 2 + bs + k = H(s), nd once you know the chrcteristic polynomil, you know the differentil opertor. Cn we use the stimulus response function h(t) to find the trnsfer function H(s)? Let s give it try. Consider the differentil eqution (8.) with rest initil conditions. The solution is y(t) = h(t r)e sr dr. (I hve to use some letter other thn s for the vrible of integrtion. I chose to use r.) On the other hnd, ny solution of (8.) hs the form y = c e r t + c 2 e r 2t + H(s)e st, where r nd r 2 re the roots of the chrcteristic polynomil. Hence there must be constnts c nd c 2 such tht (8.2) c e r t + c 2 e r 2t + H(s)e st = h(t r)e sr dr. Now let s ssume tht s is bigger thn both r nd r 2. Multiply both sides of (8.2) by e st : c e (s r)t + c 2 e (s r2)t + H(s) = e st h(t r)e sr dr = = h(t r)e s(t r) dr h(u)e su du. The lst simplifiction comes from the chnge of vribles u = t r, du = dr. Finlly, let t on both sides of the previous eqution: H(s) = This eqution is usully written with t insted of u: (8.3) H(s) = h(u)e su du. h(t)e st dt. We re done! Eqution (8.3) tells us how to clculte the trnsfer function H(s) if we know the impulse response function h(t). Eqution (8.3) tkes the function h(t), defined for t, nd trnsforms it into new function H(s), defined for s greter thn the lrger of r nd r 2. It sys tht H(s) is the

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