A sufficient condition for the existence of the Fourier transform of f : R C is. f(t) dt <. f(t) = 0 otherwise. dt =

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1 Fourier transform Definition.. Let f : R C. F [ft)] = ˆf : R C defined by The Fourier transform of f is the function F [ft)]ω) = ˆfω) := ft)e iωt dt. The inverse Fourier transform of f is the function F [ft)] = ˇf : R C defined by F [ft)]ω) = ˇfω) := ft)e iωt dt. π We shall interpret the integral over the real line in the sense of the principle value, i.e. R ft)e iωt dt = lim ft)e iωt dt. R R A sufficient condition for the existence of the Fourier transform of f : R C is ft) dt <. The set of all such functions will be denoted by L R), i.e. } L R) = f : R C ft) dt <. Example.. Suppose that a > and if t [ a, a], ft) = otherwise. Then F [ft)]ω) = a a [ e e iωt iωt dt = iω Example.3. Let a >. It can be shown that π F [e at ]ω) = ] a a ω a e 4a. Let us recall one useful result from complex analysis. = sinaω). ω Theorem.4. Let P and Q be polynomials and Qx) for all x R.

2 i) If + deg P < deg Q, then P x) dx = πi Qx) ii) If α > and deg P < deg Q, then P x) Qx) eiαx dx = πi iii) If α < and deg P < deg Q, then P x) Qx) eiαx dx = πi Example.5. Suppose that If ω, then Theorem.4 gives z z C Qz)=,Im z>} z z C Qz)=,Im z>} z z C Qz)=,Im z<} ft) = + t. e iωt + t dt = πires i ) e iωz = πe ω + z Similarly, if ω >, then ) e iωt e iωz + t dt = πires i = πe ω. + z ) P z) res z. Qz) ) P z) res z Qz) eiαz. ) P z) res z Qz) eiαz. Therefore, F [ft)]ω) = πe ω. Theorem.6. If f L R), then ˆf is continuous. Theorem.7. Suppose that f : R C and g : R C have the Fourier transform. i) If a, b C, then F [aft) + bgt)]ω) = af [ft)]ω) + bf [gt)]ω). ii) F [ft)]ω) = πf [ft)] ω). iii) F [fat)]ω) = a F [ft)] ω a ) whenever a is a nonzero real number. iv) If a R, then F [ft a)]ω) = e iωa F [ft)]ω). v) If a R, then F [e iat ft)]ω) = F [ft)]ω a). Proof. i) It follows directly from the linearity of the integral.

3 ii) F [ft)]ω) = iii) Let a. Then iv) v) F [fat)]ω) = ft)e iωt dt = π ft)e i ω)t dt = πf [ft)] ω). π fat)e iωt dt = fτ)e iω τ ω ) a dτ = a a F [ft)], a where we have used the substitution τ = at. F [ft a)]ω) = F [e iat ft)]ω) = ft a)e iωt dt = e iat ft)e iωt dt = fτ)e iωτ+a) dτ = e iωa F [ft)]ω). fτ)e iω a)t dt = F [ft)]ω a). Example.8. i) F [sint)e t ]ω) = F [ eit e it e t ]ω) = F i i [eit e t ]ω) F i [e it e t ]ω) = F i [e t ]ω ) F i [e t ]ω + ) = ) π e ω ) 4 e ω ) 4. i ii) Let F [ft)]ω) = F ω), a R \ } and b R. Then where gu) = u + b. Hence For example, [ F F [fat + b)]ω) = F [fgat))]ω), F [fat + b)]ω) = F [fgat))]ω) = ω ) a F [fgt))] a = ω ) a F [ft + b)] = a a e iωb ω ) a F [ft)] a = a e iωb ω ) a F. a + t 3) ] ω) = 3iω e [ ] ω ) F + t where we have used the result of Example.5. 3 = π 3iω e e ω,

4 Theorem.9. If f, f,..., f n) L R) are continuous, then F [f n) t)]ω) = iω) n F [ft)]ω). Example.. If a >, then π F [e at ) ]ω) = iωf [e at ]ω) = iω a e ω 4a. Theorem.. If ft) and tft) belong to L R), then Example.. F [tft)]ω) = i d dω F [ft)]ω). F [te t ]ω) = i d ω πe dω 4 = i π ω ωe 4. Definition.3. Let f, g L R). The convolution of f and g is the function h = f g given by the formula ht) = f g)t) := fs)gt s)ds. It is easy to see that f g = g f. Indeed, by the substitution t s = u, f g)t) = fs)gt s)ds = Example.4. Consider the function if t [, ], ft) = otherwise. Using the definition of the convolution, we have f f)t) = fs)ft s)ds = ft u)gu)du = g f)t). ft s)ds = t+ t fτ)dτ. The last integral is equal to the length of the interval [, ] [t, t + ] and so if t, t + if t, ], f f)t) = t if t, ], if t > a. 4

5 Theorem.5. If f, g L R), then F [f g)t)]ω) = F [ft)]ω)f [gt)]ω). Example.6. Let ft) = if t [, ], otherwise. It follows from Theorem.5 and the result of Example. that F [f f)t)]ω) = F [ft)]ω)f [ft)]ω) = sin ω ) = 4 sin ω. ω ω Example.7. Let a > and f a t) = e at if t, otherwise. Using the definition of the Fourier transform, [ e F [f a t)]ω) = e at e iωt a+iω)t dt = a + iω) ] = a + iω. Hence whenever b >. F [f a f b )t)]ω) = a + iω b + iω In the sequel, we shall use the following notation: ft+) = lim x t+ fx) and ft ) = lim x t fx) Theorem.8. Let f L R). i) If f is continuous on R and ˆf L R), then for all t R. ft) = F [ ˆfω)]t) ii) If f and f are piecewise continous on R, then ft+) + ft ) = F [ ˆfω)]t) for all t R. Corollary.9. Two continuous functions from L R) are equal if they have the same Fourier transform. 5

6 Proof. Let f, g L R) be continuous and F [ft)]ω) = F [gt)]ω) for all ω R. Then F [ft) gt)]ω) = F [ft)]ω) F [gt)]ω) =. Applying the inverse Fourier transform, we get ft) gt) = for all t R. Example.. Consider the differential equation where ft) = y t) 4yt) = ft), e t if t, otherwise. Applying the Fourier transform to both sides of the differential equation, we have iω) ŷω) 4ŷω) = + iω. This implies that i ŷω) = ω + 4)ω i). The function ŷω) has simple poles at points i and ±i. By Theorem.4, yt) = F πi [ res i [ŷω)]t) = πŷω)eiωt) + res i πŷω)eiωt)] if t, πires i πŷω)eiωt) if t <. Therefore, yt) = 3 e t + 4 e t if t, et if t <. Example.. Consider the differential equation Taking the Fourier transform, we get Therefore, Since we have [ ] F t) = ω + π F ŷω) = F [ e t ] ω)f y t) yt) = e t. iω) ŷω) ŷω) = F [e t ]ω). ŷω) = ω + F [e t ]ω). [ ] t) = ω + π πe t = e t, [e t] ω) = F [ e t e t] ω). Applying the inverse Fourier transform, we express the solution of the differential equation in the convolution form yt) = 6 e s e t s) ds.

7 Laplace transform Definition.. Let f : [, ) C. The Laplace transform of f is the complex function L [ft)] = F defined by L [ft)]s) = F s) := provided the integral converges for at least one s C. Example.. If ft) = for t, then ft)e st dt F s) = L []s) = for every s C such that Re s >. e st dt = s Example.3. Let a C. If ft) = e at for t, then F s) = L [e at ]s) = for every s C such that Re s > Re a. e at e st dt = e s a)t dt = s a Now we look at a sufficient condition for the existence of the Laplace transform. Definition.4. A function f : [, ) C is said to be of exponential order α R if there is a positive real number M such that ft) Me αt for all t [, ). A function f : [, ) C is of exponential order if there is α R such that f is of exponential order α. Theorem.5. If f is a piecewise continuous function of exponential order α, then the Laplace transform of f exists for all s C such that Re s > α. Definition.6. The set L consists of all piecewise continuous functions of exponential order. Let us introduce one useful convention. We will make no distinction between a function f : [, ) C and the function f : R C given by ft) if t, f = if t <. Definition.7. The function is called Heaviside step function. t) = Theorem.8. Suppose that f, g L. if t if t < i) If a, b C, then L [aft) + bgt)]s) = al [ft)]s) + bl [gt)]s). 7

8 ii) L [fat)]s) = a L [ft)] s a) whenever a is a positive real number. iii) If a C, then L [e at ft)]s) = L [ft)]s a). iv) If a R is positive, then L [ft a)t a)]s) = e as L [ft)]s). Proof. i) It follows directly from the linearity of the integral. ii) Let a >. Then L [fat)]s) = fat)e st dt = a fτ)e s τ s ) a dτ = a L [ft)]. a iii) L [e at ft)]s) = ft)e s a)t dt = L [ft)]s a). iv) Let a >. Then L [ft a)t a)]s) = = ft a)t a)e st dt = fτ)e sτ+a) dτ = e as L [ft)]s). a ft a)e st dt Example.9. i) Let ω be a real number. By the linearity of the Laplace transform and Example.3, we have [ ] e iωt e iωt L [sinωt)]s) = L s) = [ ] L e iωt s) L [ e iωt] s) ) i i = i s iω ) s + iω s iω) = s + iω is iω)s + iω) = s + ω for all s C satisfying Re s >. Similarly, one can show that for all s C satisfying Re s >. ii) Let a C. Then L [cosωt)]s) = s s + ω L [e at sin t]s) = L [sin t]s a) = for all s C satisfying Re s a) >. s a) + 8

9 iii) If a C, then it follows from Example. that iv) for all s C satisfying Re s >. L [t a)]s) = e as L []s) = e as s L [t )e t ]s) = L [t )e t + ]s) = e L [t )e t ]s) for all s C satisfying Re s >. = e s e L [e t ]s) = e s ) s Theorem.. If f, f,..., f n) L, then L [f n) t)]s) = s n L [ft)]s) s n f) s n f ) sf n ) ) f n ) ). Example.. Let ft) = t n. Then f n) t) = n! and f) = = f n ) ) =. Hence L [n!]s) = L [f n) t)]s) = s n L [ft)]s) = s n L [t n ]s). As L [n!]s) = n!l []s) = n!, we have s L [t n ]s) = n! s n+. Theorem.. If f L is of exponential order α R, then L [ft)]s) is holomorphic on the half-plane Re s > α and Example.3. d L [ft)]s) = L [tft)]s). ds L [t sin t]s) = d ds L [sin t]s) = d ds s + = s s + ). Definition.4. Let f, g L. The convolution of f and g is the function h = f g defined by ht) = f g)t) := t fs)gt s)ds. Example.5. Suppose that ft) = t and gt) =. Then f g)t) = t sds = t. This shows that gt) = is not the unit with respect to the convolution i.e. f g f). 9

10 Theorem.6. If f, g L, then L [f g)t)]s) = L [ft)]s)l [gt)]s). Example.7. L [ ]s) = L []s)l []s) = s. Example.8. L [t e t ]s) = L [t]s)l [e t ]s) = s s ). In applications of Laplace transform, we often need to find the inverse transform. More specifically, given a Laplace transform F of an unknown function f L we want to determine the function f. This means that we have to solve the integral equation F s) = ft)e st dt. A solution ft) is called an inverse Laplace transform. The following theorem shows that the inverse Laplace transform is uniquely determine if f is a continuous function. Theorem.9. Two continuous functions from L are equal if they have the same Laplace transform. The continuous inverse Laplace transform of F s) will be denoted by the symbol L [F s)]t). The inverse Laplace transform can be described by a line integral in complex plane in the following way. Theorem.. Let f L be a continuous function of exponential order α. Suppose that f is piecewise continuous on [, ). Then ft) = a+i L [ft)]s)e ts ds := lim L [ft)]s)e ts ds, πi a i b πi C a,b where a > α and C a,b is the line segment with the parametrization ϕt) = a + it, t [ b, b]. If L [ft)]s) is a rational function, then the evaluation of the line integral from the preceding theorem leads to the following result. Theorem.. Let F s) = P s) be a rational function such that deg P < deg Q. Qs) Assume that z,..., z n are poles of F s). Then there is a continuous function f L satisfying L [ft)]s) = F s) on a certain half-plane in the complex plane) given by the formula n ft) = res ) zk F s)e st for every t [, ). k=

11 Example.. Let F s) = s+3 s +. Then L [F s)]t) = res i F s)e st ) + res i F s)e st ) = i + 3 i = e it + 3 i eit + e it 3 i e it = cos t + 3 sin t. e it + i + 3 e it i Example.3. Let F s) = s s ). Then Since res F s)e st ) = e t and L [F s)]t) = res F s)e st ) + res F s)e st ). res ) ) F s)e st e st ) te st s ) e st = lim = lim = t, s s s s ) we have L [F s)]t) = e t t. As a by-product, we have found that the convolution of t and e t is e t t see Example.8). Example.4. Consider the initial value problem y t) + yt) =, y) = y ) =. Applying the Laplace transform to both sides of the differential equation, we obtain L [y t)]s) + L [yt)]s) = L []s). Denote the Laplace transform L [yt)]s) of yt) simply by Y s). Since L [y t)]s) = s Y s) sy) y ) and y) = y ) =, we get Thus s Y s) + Y s) = s. Y s) = ss + ) The function Y s) has simple poles at points and ±i. By Theorem., yt) = L [Y s)]t) = res Y s)e st ) + res i Y s)e st ) + res i Y s)e st ) = e it eit = cos t.

12 Example.5. Consider the initial value problem y t) + 4y t) + 3yt) = e t, y) = and y ) =. Applying the Laplace transform to both sides of the differential equation, we have s Y s) s + + 4sY s) 4 + 3Y s) = s +. Therefore, the Laplace transform of y is Y s) = s + )s + 4s + 3) + s + s + 4s + 3 = s + 3s + 3 s + ) s + 3). The function Y s) has a simple pole at -3 and a pole of order at. It is easy to see that res 3 Y s)e st ) = 3 4 e 3t, res ) Y s)e st s + 3s + 3)e st = lim s s + 3 Hence the solution is = lim s = t + e t. 4 ) s s t + 3st + 3t)e st s + 3) s + 3s + 3)e st s + 3) yt) = res ) 3 Y s)e st + res ) Y s)e st = 3 4 e 3t + t + e t. 4 3 Z transform Definition 3.. Let a n ) be a sequence of complex numbers. The Z transform of a n ) is the complex function Z [a n ) ] = F defined by Z [a n ) ]z) = F z) := provided the series converges on a neighbourhood of infinity. Example 3.. Let α C. Then Z [α n ) ]z) = α n z n = α z for every z C such that z > α. In particular, for every z C satisfying z >. Z [) ]z) = Z [ n ) ]z) = a n z n = z z α z z

13 Example 3.3. for every z C \ }. Z [ ) ] z) = n! n!z n = e z Definition 3.4. A sequence a n ) of complex numbers is said to be of exponential order if there are constants M > and α R such that a n Me αn for all n N. The next result characterizes sequences whose Z transform exists. Theorem 3.5. The series a n z n converges on a neighbourhood of infinity if and only if a n ) is of exponential order. Definition 3.6. The set Z consists of all sequences of complex numbers of exponential order. Theorem 3.7. Suppose that a n ), b n ) Z. i) If α, β C, then Z [αa n ) + βb n ) ]z) = αz [a n ) ]z) + βz [b n ) ]z). ii) If α, then Z [α n a n ) ]z) = Z [a n ) ] z α). Proof. i) It is obvious. ii) Clearly, Z [α n a n ) ]z) = α n a n z n = a n z z n = Z [a n ) α) ]. α) Example 3.8. i) Let ω C. It follows from Example 3. that Z [sinωn)) ] z) = i Z [ e iωn) Similarly, = i z z e iω Z [cosnω)) ] z) = ] z) i Z [ e iωn) ) z = z e iω z z cos ω z z cos ω +. ii) By the results of Example 3. and Example 3.3, ) ] [ Z [3 n z) = 3Z [) n n! ] z) Z n! = 3z [ ) ] z ) z Z n! 3 ] z) z sin ω z z cos ω +. ) ] z) = 3z z e z.

14 Theorem 3.9. If a n ) Z and k N, then Z [a n+k ) ]z) = z k Z [a n ) ]z) a z k a z k a k z. Proof. We obtain from the definition of Z transform that Z [a n+k ) ]z) = a n+k z n = z k a n+k = zk zn+k k = z k Z [a n ) ]z) a n z k n. a n z k a n n zk z n Example 3.. Using the result of Example 3.8, [ π )) ] [ Z sin n + ) z) = z Z sin nπ ) ] z) z sin z sin π = z3 z + z = z z +. Theorem 3.. If a n ) Z, then Proof. By the definition of Z transform, Z [na n ) ]z) = z d dz Z [a n) ]z). z d dz Z [a n) ]z) = z d dz a n z = z n na n z n+ = na n z n = Z [na n) ]z). Example 3.. Z [n) ]z) = z d dz Z [) ]z) = z ) z = z z z ). Definition 3.3. Let a n ), b n ) Z. The convolution of a n ) and b n ) is the sequence c n ) = a n ) b n ) defined by c n := n a k b n k k= for every n N. 4

15 Example 3.4. n ) ) = ) k= = n + ). Theorem 3.5. If a n ), b n ) Z, then Z [a n ) b n ) ]z) = Z [a n ) ]z)z [b n ) ]z). Example 3.6. Z [) ) ]z) = Z [) ]z)z [) ]z) = z z ). Theorem 3.7. The Z transform is a bijection from Z onto the set H of all holomorphic functions on a neighbourhood of infinity with the finite limit at infinity. Definition 3.8. The inverse Z transform of a function F H is a sequence a n ) such that Z [a n) ]z) = F z). We shall sometimes denote the nth element of the inverse Z transform of F by the symbol Z [F z)]n). The inverse Z transform of a function F H can be found by the expansion of F in Laurent series a n z n which converges on a neighbourhood of infinity U ). The coefficients a n may also be determined by the integral formula a n = F z)z n dz, πi C where C is an arbitrary positively oriented simple closed path that lies entirely in U ). This integral can sometimes be evaluated by Residue theorem. There are also explicit formulas for coefficients a n. In particular, the formula is very useful. Example 3.9. Let Then and F z) = a = lim z F z) z )z ). a = Z [F z)]) = lim z F z) = a n = Z [F z)]n) = res F z)z n ) + res F z)z n ) = + n for every n N. Note that a can also be computed using Residue theorem. In this case, we have a = res F z)z ) + res F z)z ) + res F z)z ) = + =. The formula follows immediately from the fact that lim z a n z n is a. 5

16 Example 3.. Consider difference equation y n+ 3y n+ + y n = with initial conditions y = and y =. Applying the Z transform to both sides of the difference equation, we obtain z Y z) z z 3zY z) + 3z + Y z) =, where Y z) is the Z transform of y n ). Thus Y z) = z z z 3z + = zz ) z )z ) = z z = z Therefore, the solution is y n = n for every n N. Example 3.. Consider difference equation y n+ y n+ 3y n = 3 n = n z n. with initial conditions y = y =. Taking the Z transform, we have z Y z) zy z) 3Y z) = z z 3, where Y z) is the Z transform of y n ). Hence z Y z) = z 3)z z 3) = z z 3)z z 3) = z z 3) z + ). Therefore, z n y n = res z 3) z + ) + res 3 z 3) z + ) = )n 6 = )n 6 nz n z + ) z n + lim = )n + n3n 3n z 3 z + ) z n ) z n + lim z 3 z + for every n N. The value y is already known from initial conditions.) Example 3.. Consider difference equation n y n+ + k y n k = k= with initial conditions y = y =. We see that n k= k y n k is the nth element of the sequence n ) y n ). Applying the Z transform to both sides of the difference equation, we obtain z Y z) + z z Y z) = z z, where Y z) is the Z transform of y n ). Hence Y z) = z z z z 3 z + z = z z ). 3 This implies that y n = res Y z)z n ) = lim z zn z n ) nn ) n )n ) = = n ) n ) for every n N. The value y is already known from initial conditions.) 6

17 4 Exercises. Find the Fourier transform of the function ) Result: ˆfω) = π sin ω + cos ω e ω. ft) = t Find the Fourier transform of the function ft) = t + ) t + ). Result: ˆfω) = π e ω π ω )e ω. 3. Use the Fourier transform to solve the differential equation y t) + 4y t) + 4yt) = e t. Result: yt) = 8 et if t, t 3 e t 4 9 e t + e t if t >.. 4. Use the Laplace transform to solve the initial value problem y t) + y t) + 5yt) =, y) = y ) =. Result: yt) = [ 5 e t cost) e t sint) ]. 5. Use the Laplace transform to solve the initial value problem y t) + 6yt) + 8yt) = e t e 3t, y) =, y ) =. Result: yt) = 4 e t + te t + e 3t 3 4 e 4t. 6. Use the Laplace transform to solve the initial value problem y t) + y t) = sin t, y) =, y ) =. Result: yt) = + e t cos t sin t. 7. Use Z transform to find the solution of difference equation satisfying the initial condition y =. Result: y n = )n + n 9 3 for n N. 9 y n+ + y n = n 7

18 8. Use Z transform to find the solution of difference equation y n+ y n+ 3y n = satisfying initial conditions y = and y =. Result: y n = 4 + 3n+ + ) n+ 8 for n N. 9. Use Z transform to find the solution of difference equation y n+ 3y n+ + y n = n satisfying initial conditions y = and y =. Result: y n = n n n + for n N. 8