Introduction. f(t) dt. lim. f(t) dt = lim. 1 (1+t)1 p. dt; here p>0. h = (1+h)1 p 1 p. dt = (1+t) p 1 p

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1 Introduction Method of Laplace transform is very useful in solving differential equations. Its main feature is to convert a differential equation into an algebric equation, which allows us to obtain the Laplace transform of the solution. There immediately lies the main difficulty to obtain solution from its Laplace transform (Inversion problem) is not easy at all except for some special forms. We shall start with a brief review of improper integral. Improper Integral and Laplace Transform Suppose f(t) is a function on inteval t< (we can regard t as time). Improper Integral: If the limit of integral over finite intervals h lim h f(t) dt exists, then we say improper integral f(t) dt converges and define it by f(t) dt = lim h Otherwise, we say the improper integral diverges. h f(t) dt. Example: Evaluate the improper integral dt; here p> (+t) p Solution: First assume p. Forh>, we have h (+t) p dt = (+t) p p h = (+h) p p p, which implies that h { lim h (+t) p dt = p ; p> ; p< Therefore the improper integral converges for p>, but diverges for p<. In case of p =, we have h dt =log(+h) (+t) as h,

2 and the improper integral diverges. In one word, the improper integral diverges for p, and convergess for p>withvalue (+t) p dt = for p>. p Example: Evaluate improper integral +t 2 dt Solution: It follows that h +t 2 dt =arctant h =arctanh π, as h. 2 Therefore, the improper integral converges with value +t 2 dt = π 2 We state without proof a theorem that is very useful in determining the convergence of improper integral. Theorem: If f(t) g(t) for all t and improper integral g(t) dt converges, then improper integral f(t) dt also converges. On the other hand, if f(t) g(t) and g(t) dt diverges, then f(t) dt diverges. Example: Prove that the improper integral converges. Actually, it can be shown that e t2 dt e t2 dt = Solution: It is easy to see that e t2 e t for t, and e t2 for t<. Hence e t2 g(t) with { g(t) = ; t< e t ; t But g(t) dt =+ e converges. Therefore e t2 dt converges. To calculate the improper integral, it suffices to show that However, we have A 2 = A = π 2. e t2 dt = π e x2 dx e y2 dy = e (x2 +y 2) dxdy. 2

3 Using Polar coordinates, x = r cos θ, y = r sin θ, withr, θ<2π, wehavedxdy = rdrdθ and A 2 = 2π This completes the proof. e r2 rdrdθ=2π re r2 dr == π e r2 = π. Exercise: Determine whether the following improper integrals converge or diverge.. t t dt. (Hint: Diverge, since the integrand goes to as t ) e st sin tdt.heres>. (Hint: Converge, since e st sin t e st ) 3. +t +t dt. (Hint: Diverge, since +t 2 +t 2 +t ) Definition of Laplace Transform: Suppose f(t) is defined for t <. The Laplace transform of f is defined as the improper integral L{f} = F (s) = for all s such that the improper integral converges. Remark: The Laplace transform of f(t) is a function of s. e st f(t) dt The following theorem will give a satisfying description regarding the domain on which the Laplace transform is well-defined. Theorem: If improper integral e st f(t) dt converges for some s = s,thenitconvergesforall s>s. Example: Find Laplace transform of f(t) =e at.hereais a constant. Solution: The Laplace transform L { e at} = e st e at = s a, for s>a. Example: Find Laplace transform of f(t) = t. Solution: The Laplace transform is { } L = e st dt = t t 2 e x2 dx = s π s. Here the second equality follows from change of variable t = x2 s otherwise the Laplace transform is clearly ). (we can assume s>, Example: Find Laplace transform of f(t) =sinat, wherea is a constant. Solution: One way to find its Laplace transform is using formula sin at = eiat e iat. 2i 3

4 However, for s>, We have, for s>, L { e iat} = L{sin at} = 2i e (s ia)t dt = s ia. ( s ia ) = s + ia a s 2 + a 2 Exercise: Find the Laplace transform of f(t) = cos at. (Answer: s s 2 +a 2 ) There are many theorems that facilitate the computations of Laplace transforms of more complicated functions. We shall state below a relatively simple one without proof. Theorem: Suppose Laplace transform F (s) =L{f(t)} is well-defined for s>s.wehave L{tf(t)} = F (s), L { t 2 f(t) } = F (s),, L{t n f(t)} =( ) n F (n) (s), for s>s. Intuition: We have F (s) = e st f(t) dt. If we can exchange the order of differentiation and integration, then df ds = d ds e st f(t) dt = d ds ( e st f(t) ) dt = e st tf(t) dt = L {tf(t)}. The derivation of F (s), follows similarly.. Let us use this theorem to calculate several examples. Example: Find Laplace transform of f(t) =t n e at.heren is an integer. Solution: We know F (s) = L { e at} = s a. Hence L { t n e at} =( ) n F (n) (s) =( ) n ( ) n n! (s a) n+ = n! (s a) n+. Example: Find Laplace transform of f(t) =t n 2 for all integer n. { } Solution: We know F (s) = L. Hence t { } L t n 2 =( ) n F (n) 3 5 (2n ) (s) = 2 n 2 Properties of Laplace Transform Laplace transform is a linear operator, in the sense that for all constants c,c 2.Moreprecisely,wehave L{c f + c 2 f 2 } = c L{f } + c 2 L{f 2 } π s n+ 2. 4

5 Theorem: If Laplace transforms of f (t) andf 2 (t) convergesfors>s and s>s 2 respectively. Then for s>max{s,s 2 }, where c,c 2 are constants. Proof: Exercise. L{c f + c 2 f 2 } = c L{f } + c 2 L{f 2 }, Another extremely important property of Laplace transform is the one-to-one respondence between functions and their Laplace transforms. More precisely, we have Theorem: If f = f 2,thenL{f } = L{f 2 }.Conversely,ifL{f } = L{f 2 },thenf = f 2 provided that f,f 2 are both continuous. The proof of this theorem is beyond the scope of this text. More details can be found in D.V. Widder The Laplace Transform. This theorem enable us to define the inverse Laplace transform. Definition: Suppose F is the Laplace transform of a continuous funtion f, that is F = L{f}. then the inverse Laplace transform of F, written as L {F }, isf. Inanotherword f = L {F }. Remark: The inverse is well-defined and unambiguous by the previous theorem. Example: Find the inverse Laplace transform of s 2. Solution: Since L{t} = s 2,wehave L { s 2 } = t. Exercise: Find the inverse Laplace transform of s.(answer: πt ). Theorem: The inverse Laplace transform is a linear opeartor. Thatis L {c F + c 2 F 2 } = c L {F } + c L {F 2 } for all constants c,c 2. Proof: Let F = L{f } and F 2 = L{f 2 },wheref,f 2 are continuous functions. We have By definition, we have L{c f + c 2 f 2 } = c L{f } + c 2 L{f 2 } = c F + c 2 F 2 L {c F + c 2 F 2 } = c f + c 2 f 2 = c L {F } + c L {F 2 }. This completes the proof. 5

6 Remark: It follows easily that L {c F + c 2 F c n F n } = c L {F } + c L {F 2 } + + c n L {F n }. for all n and constants c,c 2,,c n. The linear property of inverse Laplace transform helps to deal with more complicated Laplace transforms. Before doing some examples, we shall briefly review Partial Fraction Expansion, which is also very useful in solving initial value problems. 2. Partial Fraction Expansion Aquotientofform P (x) Q(x),whereP(x),Q(x) are both polynomials and P (x) isofdegreeless than Q(x), can be expanded into partial fractions. For example, we can expand quotient 3x x 2 = 3x 2 (x +)(x ) into the sum of two partial fractions A x + + B x, where A, B are two constants yet to be determined. It is easy to see that which implies that A x + + B (A + B)x +(B A) = x x 2 A + B =3, B A = A =, B =2 3x x 2 = x x and we complete the partial fraction expansion. In general, the form of each partial fraction in the expansion depends only on Q(x). It shall be clear from the following example. Suppose Q(x) =(x + a)(x 2 + b)(x 2 + c) 2 (x + d) 3 and P (x) is some polynomial with degree less than Q(x). The partial fraction expansion of P (x) Q(x) will take form P (x) Q(x) = A x + a + Bx + C x 2 + b + Dx + E x 2 + c + Fx+ G (x 2 + c) 2 + H x + d + The general rule can be concluded as I (x + d) 2 + H (x + d) 3. Term such as (x + a) =(x + a) has exponent outside the parenthesis. Hence term (x + d) will once in the denominators of the expansion as (x + a). 2. Term such as (x 2 + b) =(x 2 + b) has exponent outside the parenthesis. Hence term (x + d) will once in the denominators of the expansion as (x 2 + b). 6

7 3. Term such as (x + d) 3 has exponent 3 outside the parenthesis. Hence term (x + d) will appear three times in the denominators of the expansion as (x + d), (x + d) 2,and(x + d) 3 respectively. 4. Term such as (x 2 + c) 2 has exponent 2 outside the parenthesis. Hence term (x 2 + c) will appear twice in the denominators of the expansion as (x 2 + c) and(x 2 + c) 2 respectively. 5. Each numerator is a polynomial of degree one less than that of the term inside the parenthesis of its denominator. Let us study the following examples. Example: With the aid of partial fraction expansion, find the Inverse Laplace Transforms of the following functions.. 3s+2 s(s+)(s+2) 2. s s 2 +2s+ 3. 3s 2 +s+ s(s 2 +) Solution: From the general rule of determining the form of each term in partial fraction expansion, it follows that 2s 3 s(s +)(s +2) s +2 s 2 +2s + 3s 2 + s + s(s 2 +) = A s + B s + + C A = B =,C = 2 s +2 s +2 = (s +) 2 = A s + + B (s +) 2 A =,B = = A s + Bs + C s 2 + A =,B =2,C = Therefore, the inverse Laplace transform are () + e t 2e 2t (2) e t te t (3) + 2 cos t +sint respcetively. 3 Solutions to Initial Value Problems withconstant Coefficients Consider the following linear differential equation L[y] = a y (n) + a y (n ) + + a n y + a n y = f(t), where a,a,,a n are constants with a n. An advantange of Laplace transform method is that non-homogeneous (g(t) ) and non-homogeneous (g(t) ) equations are handled exactly the same way. Taking Laplace transform on both sides, we have { a L y (n)} { + a L y (n )} + + a n L { y } + a n L{y} = L{f(t)} := F (s) Our next task is to evaluate L { y (n)} in terms of Y (s) = L{y}. We have the following important theorem. 7

8 Theorem: Let Y (s) = L{y(t)}. Wehave,forn, { L y (n)} = s n Y (s) s n y() sy (n 2) () y (n ) () under some reguarlity conditions. Proof: We first assume n =. Inthiscase, L{y } = e st y (t) dt = e st dy(t) = e st y(t) + y(t)se st dt = sy (s) y(). (In the last equality, we implicitly assume that lim t e st y(t) = ). Similarly L{y } = sl{y } y () = s [ sy (s) y() ] y () = s 2 Y (s) sy() y (). Here we implicitly assume lim t e st y (t) = for the first equality and lim t e st y(t) = for the second equality. In general, it can be shown that { L y (n)} = s n Y (s) s n y() sy (n 2) () y (n ) () under condition lim t e st y (k) (t) =fork =,,,n. Remark: The condition that ensure the theorem to hold is lim y t! (k) (t) =, k =,,,n. e st This condition usually holds, especially, it always holds for a solution obtained by Laplace transform method; see D. Widder, Advanced Calculus (96), for details. In the following examples, we will omit the details of verifying this condition for the solutions we obtained. Example: Use the Laplace transform method to solve initial value problem y +2y = e x, y() = Solution: Observe that F (s) = L{e x } = s+. Taking Laplace transform on both sides, and letting Y (s) = L{y}, wehave L{y } +2Y (s) = s + sy (s)+2y (s) = s + Y (s) = (s +)(s +2) To determine the inverse Laplace transform of follows that Y (s) = (s+)(s+2), we expand it in partial fraction. It (s +)(s +2) = A s + + B A(s +2)+B(s +) = s +2 (s +)(s +2) 8

9 which yileds that or But L { s a A + B =, 2A + B = A =, B = Y (s) = s + s +2 y(t) =L { Y (s) } { = L s + } = e at,wehave y(t) =e t e 2t. Exercise: Solve initial value problem y +2y = e x, y() = 4 } { } L s +2 Solution: It is very similar to the preceding example the only difference is that L{y } = sy (s) y() = sy (s) 4. We obtain that (s +2)Y (s) 4= s + Y (s) = 4s +5 (s +)(s +2) = s s +2 We have y(t) =e t +3e 2t. Example: Use Laplace transform method to solve initial value problem Solution: Let Y (s) = L{y(t)}. It follows that y 2y + y = e t, y() =, y () = L{y } = s 2 Y (s) sy() y () = s 2 Y (s) s L{y } = sy (s) y() = sy (s) Taking Laplace transform on both sides, we have which implies that Y (s) = s 2 Y (s) s 2sY (s)+2+y (s) = s (s ) 3 + s 2 (s ) 2 = (s ) + (s ) 3 (s ) 2 = Therefore y(t) =(t 2 t +)e t (exercise!). Exercise: Solve initial value problem (s ) 3 (s ) 2 + s y (4) y =, y() =, y () =, y () = 2, y () =. Solution: Let Y (s) = L{y(t)}, wehave s 4 Y (s) s 3 2s Y (s) = Y (s) = s3 +2s s 4 = s 3 +2s (s +)(s )(s 2 +). 9

10 Usinf partial fraction, we have Y (s) = which implies that A s + + B s + Cs + D s 2 + y(t) = ( e t + e t) + 3 cos t. 4 2 Example (Resonance): Consider initial value problem A = B = 4,C = 3,D = 2 y + ω 2 y = F m cos ωt, y() =, y () = Solution: Taking Laplace transform on both sides, we have In case ω ω,wehave where A = B = Y (s) = F s m (s 2 + ω 2)(s2 + ω 2 ) Y (s) = F. This implies that m(ω 2 ω2 ) y(t) = In case that ω = ω (Resonance), we have As s 2 + ω 2 Bs s 2 + ω 2, F m(ω 2 ω2 ) (cos ωt cos ω t) Y (s) = F s m (s 2 + ω 2 = F (s) )2 where Therefore, We recover the same results. F (s) = F 2m s 2 + ω 2 y(t) = F 2mω t sin ω t { } F = L sin ω t 2mω Example: (linear system) Solve the following system of first order equations: z = 2z +4z 2 z 2 = z 3z 2 with initial condition z () =, z 2 () =. Solution: Let Z i (s) =L{z i (t)} for i =, 2. We have L{z } = sz (s) z () = sz, L{z 2 } = sz 2(s) z 2 () = sz 2.

11 Therefore sz = 2Z +4Z 2 sz 2 = Z 3Z 2 or Z = 4 (s +2)(s ) = s s, Z 2 = 4 s 2 (s +2)(s ) = 3 s s, which implies z = 4 ( e t e 2t), z 2 = ( e t +4e 2t) 3 3 The main difficulty lies in the fact that the inverse Laplace transform is not that easy to calculate, contrary to what we have shown in examples. We have to resort to numerical schemes to obtain the inverse in most cases. Remark: In above discussions, the initial conditions are evaluated at t =. There is no loss of generality, indeed. For example, if the initial value problem takes form L[y] =f(t), y(t )=y,y (t )=y,, y(n ) (t )=y (n ) We can define ỹ(t) = y(t + t ), we have an equivalent initial value problem L[ỹ](t) =f(t + t ), ỹ() = y, ỹ () = y,, ỹ (n ) () = y (n ). For example, consider initial value problem y +4y = 3sint, y(π) =, y (π) =. Let ỹ(t) = y(t + π). It follows that ỹ +4ỹ = 3sin(t + π) =3sint, ỹ() =, ỹ () =. It is not difficult to find the solution as ỹ(t) =sint (exercise) Hence, y(t) =ỹ(t π) = sin t. Exercise: To find Ỹ (s) =L{ỹ}, show that Ỹ (s) = s 2 +, which implies that ỹ =sint.

12 4 Miscellaneous Results of Laplace Transform In this section we collect some useful results about Laplace transform. following lemma. Lemma: Suppose F (s) =L{f(t)}, orf(t) =L {F (s)}. Wehave Let us start with the L { e ct f(t) } = F (s c) or L { F (s c) } = e ct f(t) Proof: Use definition. Example: Solve the following initial value problem Solution: Let Y (s) = L{y}. Wehave y 2y +5y =, y() =, y () =. L{y } = s 2 Y (s) s, L{y } = sy (s). It follows that s 2 Y (s) = s 2 2s +5 = s 2 (s ) 2 = F (s ) +4 where F (s) = s s 2 +4 L {F } =cos2t sin 2t 2 Hence y(t) =e t ( cos 2t 2 sin 2t). Remark: We can also use partial fraction expansion to solve the inverse problem in this example with the aid of complex variable. Indeed Y (s) = It follows that Therefore, s 2 (s ) 2 +4 = s 2 ( )( ) = s (+2i) s ( 2i) A s (+2i) + B s ( 2i) A + B =, A( 2i)+B(+2i) =2 A = i, B = 2 4 i. y(t) = Ae (+2i)t + Be ( 2i)t = = ( e t cos 2t ) 2 sin 2t ( 2 + ) ( 4 i e t (cos 2t + i sin 2t)+ 2 ) 4 i e t (cos 2t i sin 2t) Exercise: Solve the linear system z = 3z 5z 2 z 2 = z + z 2 with initial condition z () =, z 2 () =. (Answer: z = e t (cos t 7sint), z 2 = e t (cos t + 3sint).) 2

13 4. Convolution and Integral Equation The Faltung Theorem, which is very useful helping evaluating integrals, involves the convolution of two functions. Let us first give the definition. Definition of Convolution: The convolution of two functions f and g, denoted by f g, isa function whose value at t is determined by (f g)(t) = This integral is called convolution integral. f(t s)g(s) ds Example: Let f(t) = t, g(t) =cost. Determine f g. Solution: By definition, we have (f g)(t) = (t s)cossds=(t s)sins t + sin sds= cos t Here we state several properties of convolution product, which resemble those of ordinary product. Lemma: For any functions f,g,h, wehave. (Commutative Law) f g g f. 2. (Distributive Law) f (g + h) f g + f h. 3. (Associative Law) (f g) h f (g h). Proof: We will only give the proof of Associative Law. The Commutative and Distributive Laws are left as exercises. By definition, we have ( ) t [ u ] (f g) h (t) = (f g)(t u)h(u) du = f(t u w)g(w)h(u) dw du For the integral in the bracket, make change of variable w = s u. Wehave ( ) t [ ] (f g) h (t) = f(t s)g(s u)h(u) ds du This multiple integral is carried out over the region {(s, u); u s t} as depicted by shaded region in the following graph. u Change the order of integration, we have ( ) t [ s ] (f g) h (t) = f(t s)g(s u)h(u) du ds = f(t s)(g h)(s) ds = ( f (g h) ) (t) This completes the proof. 3

14 The importance of convolution shows in the following theorem. Faltung theorem: Let F (s) =L{f(t)}, G(s) =L{g(t)}. Wehave Proof: By definition, L{f g}(s) = F G = L{f g} or L {F G} = f g = L {F } L {G} e st [ ] f(t u)g(u) du dt = The multiple integral is carried out over region {(t, u); u t} as depicted by the shaded region in the following graph. [ ] e st f(t u)g(u) du dt Changing the order of integration, we have [ ] L{f g}(s) = e st f(t u)g(u) dt du = u [ ] = e su g(u) e sw f(w) dw du = = F (s) This completes the proof. e su g(u) du = F (s) G(s) [ s su g(u) u e su g(u) F (s) du ] e s(t u) f(t u) dt du An immediate application of Faltung theorem is finding a particular solution of the nonhomogeneous linear equation. We will illustrate with a second-order linear equation, but same methodology works for higher order linear equations too. Consider non-homogeneous equation ay + by + cy = g(t). We wish to find a particular solution. Actually, it suffices to solve the following initial value problem ay + by + cy = g(t), y() =, y () =. Suppose the solution to this initial value problem is y p (t). Let Y p (s) =L{y p (t)}. Wehave Letting h(t) = L { as 2 +bs+c G(s) Y p (s) = as 2 + bs + c, },wehavethat y p (t) =(h g)(t) = here G(s) =L{g(t)} h(t s)g(s) ds. 4

15 Remark: Function h(t) is easy to obtain. Let r,2 stand for the two roots of the characteristic equation as 2 + bs + c =. It follows that as 2 + bs + c = a(s r )(s r 2 ).. r r 2 are both real: Using partial fraction expansion, we have ( as 2 + bs + c = ) a(r r 2 ) s r s r 2 or ( h(t) = e r t e r 2t ). a(r r 2 ) 2. r = r 2 := r are both real: Inthiscase as 2 + bs + c = a (s r) 2 = F (s) where F (s) = a s r. Therefore, h(t) = a tert. 3. r r 2 are complex: Writer,2 = λ ± iµ. Using partial expansion, we still have that h(t) = Example: Write the solution of the initial value problem ( e r t e r 2t ) = a(r r 2 ) aµ eλt sin µt. y +4y +5y = f(t), t ; y() =, y () =, in term of a definite integral. Solution: Let F = L{f} and Y = L{y}. Wehave and L{y } = sy (s), L{y } = s 2 Y (s) L{y } +4L{y } +5L{y} = L{f} s 2 Y (s) +4sY (s)+5y (s) =F (s). Hence Note that It follows that Y (s) = +F (s) s 2 +4s +5 = (s +2) (s +2) 2 + F (s) { } L (s +2) 2 = e 2t sin t. + y(t) =e 2t sin t + e 2(t s) sin(t s)f(s) ds 5

16 Example (Beta Functions): Show that for all non-negative integers m, n. ( t) m t n dt = Proof: Let f(x) = x m and g(x) = x n. It follows that h(x) =(f g)(x) = However, Faltung theorem yields that x H(s) = L{h(x)} = L{f(x)} L{g(x)} = m! s m+ m!n! (m + n +)! f(x t)g(t) dt = n! s n+ = x m!n! s m+n+2 = (x t) m t n dt. m!n! (m + n +)! (m + n +)! s m+n+2 Therefore, for all x, In particular, This completes the proof. h(x) =L {H(s)} = h() = ( t) m t n dt = m!n! (m + n +)! tm+n+. m!n! (m + n +)!. 4.2 Elementary Volterra Integral Equation Convolution is very useful in the study of Volterra integral equations, which take form y(t) = y(t s)g(s) ds + y (t), t. Here g(t),y (t) defined on t are given functions. We shall assume throughout this section that g(t) andy (t) arebothnon-negative functions, which implies the solution y(t) isalsoanon-negative function. Population Model (with age structure): Suppose in a population, the fraction of individuals that will survive to age s is ρ(s), while each individual of age s will produce offspring at rate β(s). We are interested in the dynamics of y(t), the overall birth rate at time t for the population. That is, in a small time interval [t, t + dt), the total popaultion will produce y(t) dt new offsprings. Now at time t, the number of individuals aged from s to s + ds (where s t) is y(t s)ρ(s) ds. These individuals will produce offspring y(t s)ρ(s) ds β(s) dt in time interval [t, t + dt). Let y (t) denote the birth rate of individuals with age greater than t (or born before time ). We have y(t) dt = y(t s)ρ(s) ds β(s) dt + y (t) dt 6

17 which implies that where g(s) = ρ(s) β(s). The integral can be rewritten as y(t) = y(t s)g(s) ds + y (t) y(t) =(y g)(t)+y (t). Taking Laplace transform on both sides, we obtain Y (s) =Y (s)g(s)+y (s) Y (s) = Y (s) G(s) Here Y (s) =L{y(t)}, G(s) =L{g(t)} and Y (s) = L{y (t)}. We will first take a look of a few examples that are explicitly solvable. Example: If y (t), then Y (s),whichinturnimplesthaty (s), or y(t). Example: Solve the integral equation y(t) = sin(t s)y(s) ds +. Solution: Taking Laplace transform on both sides, we have Therefore Y (s) = s 2 + Y (s)+ s Y (s) = s2 + s 3 = s + s 3 y(t) =+ 2 t2. Example: Solve the following integrodifferential equation y (t) 2 sin(t s)y(s) ds =; t with initial condition y() =,y () =. Solution: Let Y (s) =L{y(t)}. Wehave L{y } = s 2 Y (s). Hence, s 2 Y (s) 2 s 2 + Y (s) = Y (s) = s2 + s 4 + s 2 2 = s 2 + (s 2 +2)(s +)(s ). Partial fraction expansion yields Y (s) = 3 s s 3 s + Therefore, y(t) = 3 2 sin 2t + 3 (et e t ). 7

18 In general it is impossible to find explicit inverse Laplace transform. However, we can obtain an asymptotic result as t using Laplace transform. From now on, we assume that function g is non-negative and Y = L{y } is bounded. It can be shown that y(t) Ce rt for some constant C and r, ast.inanotherword, lim t y(t) = Cert for some constants C > andr. This asymptotic result shows that y(t) increases (or decay) exponentially with exponent r. The question now is how to determine C and r using Laplace transform. Actually, we have r is the unique solution to equation G(r) =,whilec = Y (r) G (r). We shall illustrate the main idea without being rigorous. The following lemma plays a key role in the development, which is also of its own interest. Lemma: Let F (s) =L{f(t)}. Iff(t) c as t,thensf (s) c as s. Proof: For any ( ɛ>, there exists T> ) such that f(t) c <ɛfor all t T. It follows that, ɛ for all s<min, R T sf (s) c = log( ɛ/ c ) f(t) dt, T T s e st f(t) dt + s s T ɛ + ɛ s f(t) dt + s T ɛ + ɛ + ɛ =3ɛ. T T e st f(t) dt c e st f(t) c dt + s e st cdt c e st dt + ( e st ) c T This completes the proof. Now suppose y(t) Ce rt or lim t e rt y(t) =C. It follows from the preceding lemma that sl { e rt y(t) } = sy (s + r) = sy (s + r) G(s + r) C as s. However, since sy (s + r), we have G(s + r) ass, or G(r) =. ButG is a decreasing function. It follows that r is uniquely determined. By L Hospital rule, we have This completes our discussion. C = lim s sy (s + r) G(s + r) = lim s Y (s + r)+sy (s + r) G = Y (r) (s + r) G (r) 8

19 Example: Suppose that k(t) is positive and continuous, with k(t) dt <. Let y(t) beasmooth function that solve the Volterra Equation y(t) =+ k(t s)y(s) ds, t Show that y(t) is non-decreasing and lim t y(t) = ( k(s) ). Proof: We have y(t) =+ y(t s)k(s) ds. Taking derivatives on both sides, we obtain y (t) =y()k(t)+ However, y() =, therefore, with φ(t) = y (t), we have φ(t) =k(t)+ y (t s)k(s) ds φ(t s)k(s) ds φ() = k() >. It follows that φ(t) > for all t (why?) andy(t) is increasing. Hence, lim t y(t) must exist, say c (could be + ). We have c = lim sl{y}(s) := lim sy (s) s s However, taking Laplace transform on both sides of the integral equation, Y (s) = s + K(s)Y (s) sy (s) = K(s) c = By definition K(s) = e st k(t) dt lim K(s) = lim s s e st k(t) dt = This completes the proof. Example (Abel Equation): Formally solve the Abel equation x y(s) ϕ(x) = ds, t. x s lim s K(s) k(t) dt. Here ϕ is a given non-negative, smooth function with ϕ() =. See the following graph: Solution: Note the equation is indeed ϕ = y we have, with Φ = L{ϕ}, that Φ(s) =Y (s) x. Taking Laplace transform on both sides, π s Y (s) = π Φ(s) s 9

20 But there is no function with Laplace transform as s. However, note L{ϕ } = sφ(s) sϕ() = sφ(s). It follows that which implies that Y (s) = π sφ(s) π s = π L{ϕ } L{ x }, y(t) = π ϕ = x ϕ (s) ds. x π x s 2

21 5 Discontinuous and Impulse forcing funcitons The same idea of Laplace transform can be carried out to solve differential equations with discontinuous non-homogeneous terms. Example: Consider the following initial value problem { y +2y 5 ; t< +5y = g(t) = ; t. with initial condition y() =, y () =. Solution: Let Y (s) = L{y}. It follows that Therefore, we have which implies that Y (s) = Ly = sy y() = sy ; Ly = s 2 Y sy() y () = s 2 Y. (s 2 +2s +5)Y = L{g} = e st g(t) dt = 5e st dt = 5 ( e s ), s 5 ( e s ) s(s 2 = H(s) H(s)e s y(t) =L {H(s)} L { H(s)e s}. +2s +5) We first compute L {H}, which is straightforward since or H(s) = 5 s(s 2 +2s +5) = s s +2 s 2 +2s +5 = (s +)+ s (s +) 2 +4 h(t) =L {H} = e t ( cos 2t + 2 sin 2t ). It remains to compute L {e s H} (to be continued). In general, suppose L{f} = F,or For any constant c, assume { g(t) = e st f(t) dt = F (s). ; t<c f(t c) ; t c in other words, g is a translation of f; see the following graph., 2

22 The Laplace transform of g is L{g} = e st g(t) dt = Notation: From now on, we will denote u c (t) = c e st f(t c) dt = e cs F (s). { ; t<c ; t c Furthermore, it follows from the preceding discussion that Example (continued): We have and the solution L{u c (t)f(t c)} = e cs F (s), or L { e cs F (s) } = u c (t)f(t c). L { e s H } = u (t)h(t ), y(t) =h(t) u (t)h(t ); t. Example: Find the Laplace inverse of the following functions (). e cs s ; ( 2). e πs s 2 + Solution: () The Laplace inverse is u c (t). (2) The Laplace inverse is u π (t)sin(t π) = u π (t)sint. Example: Solve the initial value problem y +2y +5y =5 ( ) n u n (t); t n= with initial condition y() =, y () =. Solution: The function n= ( )n u n (t) looks like Letting Y = L{y}, wehave (s 2 +2s +5)Y =5 n= ( ) n e ns s. 22

23 or or Y (s) = ( ) n e ns 5 s(s 2 +2s +5) n= y(t) = ( ) n u n (t)h(t n). n= Exercise: Solve the above equation with a different initial condition y() = 4, y () = 2. Hint: The solution is y(t) =e t (4cos2t +3sin2t)+ ( ) n u n (t)h(t n). n= Example: Find the Laplace inverse of the following function and graph the inverse function (roughly). Solution: Using the formula F (s) = s 2 ; s> + e πs a =+a + a2 + a 3 +, a <, we have e πs =+e πs + e 2πs + = { } Since L =sint, wehave s 2 + Below are the figures. f(t) =L {F } = n= e nπs. n= u nπ (t)sin(t nπ) =sint ( ) n u nπ (t) n= 23

24 5. Impulse functions The impulse function is not a function in the usual sense indeed, it is a generalized derivative of step funcitons. One way to define the unit impluse function δ(t) is as the intuitive limit of a sequence of functions { δ(t) = lim f ɛ (t) := lim 2ɛ ; t <ɛ ɛ ɛ ; t ɛ. The function δ has the following property:. δ(t) =ast. 2. δ(t) dt =. 3. δ(t)g(t) dt = g() for all functions g(t) continuous at t =. These properties are straightforward; e.g., ɛ δ(t)g(t) dt = lim g(t) dt = g() ɛ 2ɛ ɛ Intuitively, the impulse function δ(t) can be understood as a force of a large magnitude acting for a very short amount of time around t =, hence the name impusle. In general, one can define impulse function δ(t t ) (sometimes denoted by δ t (t)), which satisfies. δ(t t )=ast t. 2. δ(t t ) dt =. 3. δ(t t )g(t) dt = g(t ) for all functions g(t) continuous at t = t. It is easy to see that the Laplace transform of δ(t t ), t is L{δ(t t )} = e st δ(t t ) dt = e st δ(t t ) dt = e st. Remark: The other way to define the impulse function δ(t) is to understand it as the generalized derivative of the step function { h(t) = ; t< ; t. It roughly goes as follows. For any continuously differentiable function f, its derivative f can be uniquely characterized by the following identity f(t)g (t) dt = f (t)g(t),dt; g C (IR). Here C (IR) is the set of all smooth functions with compact support (i.e., g(t) = for all t outside some interval). This definition of derivatives can be easily extended to obtained the generalized derivative of h(t). In other words, h (t) is defined so that h(t)g (t) dt = 24 h (t)g(t),dt; g C (IR).

25 But h(t)g (t) dt = g (t) dt = g() h (t)g(t) dt = g(). for all g C (IR). It is easy to see that h (t) also satisfies the property 3 above. Indeed, h (t) here is the true meaning of δ(t). Example: Solve the IVP Solution: Let Y = L{y}. Wehave y +2y +5y = δ(t ); y() =, y () =. (s 2 +2s +5)Y = e s Y = (s +) 2 +4 e s Hence y(t) =u (t)f(t ), where f(t) = 2 sin 2t e t. Remark: There is another way to find the Laplace inverse of function of form e cs F (s). Indeed, by Faltung theorem, However, (δ c f)(t) = L { e cs F (s) } = L { e cs} L {F } = δ c f f(t s)δ c (s) ds = { f(t s)δ(s c) ds = ; t<c f(t c) ; t c } = u c (t)f(t c). 25