FOURIER TRANSFORMS. 1. Fourier series 1.1. The trigonometric system. The sequence of functions
|
|
- Curtis Palmer
- 5 years ago
- Views:
Transcription
1 FOURIER TRANSFORMS. Fourier series.. The trigonometric system. The sequence of functions, cos x, sin x,..., cos nx, sin nx,... is called the trigonometric system. These functions have period π. The trigonometric system is orthogonal, that is, the following equalities hold (m, n N): dx = π, cos nx dx = cos mx cos nx dx = sin nx dx = sin mx sin nx dx = A trigonometric series is called a series of the form a sin mx cos nx dx =,, m n, π, m = n. + (a n cos nx + b n sin nx), (.) n= where a n, b n are real numbers (coefficients). By R[, π] we denote the class of all π periodic functions integrable in [, π]. Definition.. Let f R[, π] be a π-periodic function. Then the numbers a n = π b n = π f(x) cos nx dx (n =,,... ), (.) f(x) sin nx dx (n =,,... ) (.3) are called Fourier coefficients of the function f. The series (.), where a n and b n are Fourier coefficients of the function f, is called the Fourier series of f. We shall write f(x) a + (a n cos nx + b n sin nx). n=
2 Here the symbol denotes correspondence, it means only that to the function f it is assigned its Fourier series. Example.. Find the Fourier series for π-periodic extension of the function, < x < π, f(x) = f(kπ) = (k Z)., < x <, For this, evaluate the Fourier coefficients a n = π b n = π a = π f(x) cos nx dx = π f(x) dx = π f(x) sin nx dx = π cos nx dx = πn dx =, sin nx dx = π sin nx π = (n =,,... ), cos nx n = ( ) + ( ) n /(πn), n = k, = (n =,,... ). πn, n = k Thus, the given function has the following Fourier series f(x) + b n sin nx = + sin(k )x. π k n=.. The complex form of the Fourier series. By Euler formulas, Let cos ϕ = eiϕ + e iϕ k=, sin ϕ = eiϕ e iϕ. i f(x) a + (a n cos nx + b n sin nx). (.4) n= Rewrite the nth partial sum of this series as S n (x) = a n + (a k cos kx + b k sin kx) = a + k= n ( ak e ikx + a k e ikx ib k e ikx + ib k e ikx) k= = a + n k= ( ak ib k e ikx + a ) k + ib k e ikx. π
3 3 Set c = a, c k = a k ib k, c k = a k + ib k (k =,,..., n). (.5) We obtain that n ( S n (x) = c + ck e ikx + c k e ikx) = n c k e ikx. k= k= n Taking into account (.5) and formulas for the Fourier coefficients, we have that for any integer k c k = π f(x) cos kx dx i π Similarly, for any negative integer k c k = π f(x) sin kx dx = π f(x)e ikx dx. f(x)e ikx dx. Thus, the series (.4), assigned to the function f, can be rewritten in the form f(x) c n e inx, where c n = f(x)e inx dx. (.6) π n= We observe that c n = c n, where the over-line means the complex conjugation. The Fourier series in the complex form can be defined for any complex-valued π-periodic function f = u + iv integrable on [, π], that is, such that u, v R[, π]. Definition.3. The system of functions e inϕ }, where n runs over the set Z of all integers, and ϕ [, π], is called the exponential system. Proposition.4. The exponential system is orthogonal; more exactly,, m n, e inx e imx dx = π, m = n. Proof. Let m, n Z. Then, using the orthogonality of the trigonometric system, we obtain e inx e imx dx = (cos nx + i sin nx)(cos mx i sin mx) dx
4 4 = (cos(n m)x + i sin(n m)x) dx =, m n, π, m = n. Example.5. Find the complex Fourier series of the function f(x) = x ( < x < π). Evaluate the Fourier coefficients: if an integer n, then c n = π = Thus, πin xe inx c = π xe inx dx = π π + πin x i n Z\} x dx = ; [ u = x du = dx dv = e inx dx v = in e inx ] π e inx dx = ( e iπn + e iπn) = i ( )n in n. ( ) n n e inx ( x < π).. Definition of Fourier transform We say that a complex-valued function f defined on R is absolutely integrable on R if f is Riemann integrable in each bounded interval [a, b] and the integral f(x) dx R converges. We denote by A(R) the class of all complex-valued absolutely integrable functions on R. Definition.. For a function f A(R), its Fourier transform f is defined by f(ξ) = f(x)e iπξx dx. (.)
5 Since e iπξx =, the integral (.) is absolutely convergent for any ξ R. For a π periodic function integrable on [, π] we defined its Fourier coefficients a n, b n by formulas (.) and (.3) (and the complex Fourier coefficients c n by (.6)). Clearly, the Fourier transform is an analogue of Fourier coefficients which is suitable for functions defined on R (nonperiodic). Whereas the Fourier coefficients depend on discretely varying index n, the Fourier transform is a function of a variable ξ which ranges in the whole line R (that is, f and f have the same domain R). First we observe that the following simple properties hold. Theorem.. Let f A(R). Then: (i) if the function f is even, then f also is an even function, and f(ξ) = (ii) if f is odd, then f also is odd, and f(ξ) = i f(x) cos πξx dx; (.) f(x) sin πξx dx. (.3) Proof. We have e iπξx = cos πξx i sin πξx. Assume that f is even. Then f(x) sin πξx is an odd function of x and thus f(x) sin πξx dx =. On the other hand, f(x) cos πξx is an even function of x and thus f(x) cos πξx dx = f(x) cos πξx dx. 5 These observations imply (.). Similarly, we obtain (.3). We consider some important examples. Example.3. Let Then Π(ξ) = Π(x) = sin πξ πξ, x /,, x > /. (ξ ), Π() =. (.4)
6 6 Indeed, the function Π is even and thus by (.), Π(ξ) = Π(x) cos πξx dx = for any ξ. For ξ = we have Π() = / f(x) dx =. cos πξx dx = sin πξ πξ The function (.4) is called the Dirichlet kernel. Since sin t lim t t =, the function (.4) is continuous on the whole real line R. Example.4. Let Then Λ(x) = x, x,, x >. Λ(ξ) = sin πξ π ξ (ξ ), Λ() =. (.5) Indeed, taking into account that Λ is even, and applying integration by parts, we obtain Λ(ξ) = Λ(x) cos πξx dx = ( x) cos πξx dx = sin πξx dx = πξ for any ξ. For ξ = we have Λ() = cos πξ π ξ Λ(x) dx =. = sin πξ π ξ The function (.5) is called the Fejér kernel. Similarly to the Dirichlet kernel, the Fejér kernel is continuous on the whole real line R.
7 Example.5. Let f(x) = e π x. Then f(ξ) = Indeed, the function f is even and thus f(ξ) = f(x) cos πξx dx = π ( + ξ ). (.6) e πx cos πξx dx = π 7 e t cos ξt dt. Two consecutive integrations by parts give (.6). The function g(x) = e x is called the gaussian. This function plays an extremely important role in different areas of mathematics. We shall use the following equality which is well-known from the course of Mathematical Analysis e x dx =. (.7) Example.6. Let g(x) = e x. Then First we observe that the function Φ(s) = ĝ(ξ) = e ξ. (.8) e (x+is) dx, < s <, is constant. It can be proved, for example, with the use of Cauchy s Integral Theorem; also, one can apply differentiation with respect to s under the sign of the integral to show that Φ (s) =. We have ĝ(ξ) = Using completion of square we obtain e x e iπxξ dx. x + ixξ = (x + iξ) + ξ, ĝ(ξ) = e ξ e π(x+iξ) dx = e ξ e πx dx. Using (.7, we obtain (.8).
8 8 3. Basic properties of Fourier transforms In this section we summarize some basic formulas concerning Fourier transforms. Let a function ϕ be defined on R. For any h R, denote τ h ϕ(x) = ϕ(x h). The operation τ h is called translation or shifting. Further, for any λ >, set δ λ ϕ(x) = ϕ(λx). The operation δ λ is called dilation. Theorem 3.. Let f, g be complex-valued functions, absolutely integrable over R. Then the following properties hold. (i) Linearity. (f + g)(ξ) = f(ξ) + ĝ(ξ) and for any complex number α (ii) Shifting. For any h R (αf)(ξ) = α f(ξ). τ h f(ξ) = e πihξ f(ξ). (iii) Change of scale (dilation). For any λ > δ λ f(ξ) = ( ) λ f ξ. λ (iv) Modulation (shifting in Fourier transform). Let f η (x) = f(x)e iπηx, where η R. Then f η (ξ) = f(ξ η) = τ η f(ξ). Moreover, for g η (x) = f(x) cos πηx and h η (x) = f(x) sin πηx, we have ĝ η (ξ) = [ ] f(ξ η) + f(ξ + η), ĥ η (ξ) = i [ ] f(ξ + η) f(ξ η). Proof. (i) By the linearity of integration, we have (f + g)(ξ) = [f(x) + g(x)]e iπξx dx = f(x)e iπξx dx + g(x)e iπuξx dx = f(ξ) + ĝ(ξ), and αf(ξ) = αf(x)e iπξx dx = α f(x)e iπξx dx = α f(ξ).
9 9 (ii) Setting x h = y, we obtain τ h f(ξ) = f(x h)e iπx dx = f(y)e iπξ(y+h) dy = e iπξh f(y)e iπξy dy = e iπξh f(ξ). (iii) Similarly, substitution λx = y leads to the equality δ λ f(ξ) = (iv) First, we have f(λx)e iπξx dx = λ f(y)e iπξy/λ dy = λ f ( ) ξ. λ f η (ξ) = f(x)e iπ(ξ η)x dx = f(ξ η). Further, cos(πηx) = ( e iπηx + e iπηx), sin(πηx) = i ( e iπηx e iπηx). Thus, formulas for ĝ η and ĥη follow by linearity. We remind some definitions. Let a function f be defined on an open bounded interval (a, b). Definition 3.. The function f is said to be piecewise continuous on (a, b) if it has at most a finite number of points of discontinuity and, in addition, at each point of discontinuity x (a, b) the one-sided limits f (x +) = lim f(x), f (x ) = lim f(x) x x + x x exist and are finite. A function f defined on [a, b] is said to be piecewise continuous on [a, b], if it is piecewise continuous on (a, b) and one-sided limits f(a+) and f(b ) exist and are finite. We shall call a function f defined on R piecewise continuous on R if it is piecewise continuous on any bounded interval. The class of all such functions we denote by PC. If f (x +) f (x ), then we say that f has a jump equal to f (x +) f (x ). If a function f PC, then it is integrable on any bounded interval.
10 Definition 3.3. A function f is called piecewise smooth on a bounded interval (a, b) if: (i) f is piecewise continuous on (a, b); (ii) the derivative f exists and is continuous everywhere on (a, b), with a possible exception of a finite number of points; (iii) f has finite one-sided limits at every point x (a, b). We say that f is piecewise smooth on a closed interval [a, b] if it is piecewise continuous on [a, b] and there exist finite f (a+) and f (b ). We shall call f piecewise smooth on R if it is piecewise smooth on any bounded interval. The class of all such functions we denote by PS. We observe that a piecewise smooth function may have discontinuities. Further, if f PS, then, giving to the derivative f arbitrary values at the points where it does not exist, we obtain that f PC. The next theorem gives the Fourier transform of the derivative. Theorem 3.4. Let f be a continuous, absolutely integrable, and piecewise smooth function on R. Assume that f also is absolutely integrable. Then (f )(ξ) = πiξ f(ξ). (3.) Proof. By the Fundamental Theorem of Calculus, we have Since both the integrals f(x) f() = x f (t) dt and f (t) dt (x R). f (t) dt converge, there exist finite limits lim f(x) and lim f(x). x + x Since f is absolutely integrable, these limits are equal to zero. Integrating by parts, we get f (ξ) = f (x)e πixξ dx = +πiξ lim x + f(x)e πixξ lim x f(x)e πixξ f(x)e πixξ dx = πiξ f(ξ). Using induction, we have the following corollary.
11 Corollary 3.5. Let f, f,..., f (n ) be continuous and absolutely integrable over R. Assume also that f (n ) is piecewise smooth and f (n) is absolutely integrable over R. Then (f (n) )(ξ) = (πi) n ξ n f(ξ). It is possible to prove that for any function f A(R) its Fourier transform f is continuous on R. Furthermore, if f has a good rate of decay at infinity, then f ( ) is differentiable, and f is obtained by the differentiation with respect to the parameter ξ under the sign of the integral in (.). Theorem 3.6. Assume that the functions f and xf(x) are absolutely integrable over R. Then the Fourier transform f is continuously differentiable in R. Moreover, ( ) f (ξ) = πi xf(x)e πixξ dx. (3.) We shall not give the proof of this theorem (it requires some theorems from the theory of integrals depending on a parameter). Applying Theorem 3.6 and induction, we obtain the following corollary. Corollary 3.7. Let m N. Assume that the functions f and g(x) = x m f(x) are absolutely integrable over R. Then the Fourier transform ( ) (m) f has the continuous derivative f on R, and ( ) (m) f (ξ) = ( πi)mĝ(ξ). (3.3) 4. Convolutions In this section we introduce the notion of convolution. We consider some of its basic properties; the main result is that the operation of convolution is equivalent to the operation of multiplication of Fourier transforms. Let functions f and g be defined on R. Then their convolution is the function f g, defined by f g(x) = f(x y)g(y) dy, (4.) provided the integral exists.
12 First, we have the following property of symmetry. Proposition 4.. Let functions f and g be defined on R. Then f g = g f provided one of them exists. Proof. Assume that the integral (4.) exists. Setting x y = u, we obtain that y = x u and f g(x) = f(x y)g(y) dy = f(u)g(x u) du. By the definition, the latter integral is the convolution g f. Various conditions can be imposed on f and g to insure that the integral (4.) is absolutely convergent for all x R. In particular, the following proposition holds. Proposition 4.. Assume that f and g are defined on R and integrable in every bounded interval. Then each of the following conditions implies that the integral f(x y)g(y) dy (4.) converges uniformly with respect to x R: (i) functions f and g are integrable on R; (ii) one of functions f, g belongs to the class A(R), and the other is bounded on R; (iii) one of functions f, g vanishes outside of a bounded interval [a, b]. In the sequel we study convolutions f g under the assumption that functions f, g satisfy one of conditions (i)-(iii). Although these functions may be discontinuous at some points, their convolution is continuous. Proposition 4.3. Let functions f and g be defined on R and integrable in every bounded interval. Assume that one of conditions (i)-(iii) of Proposition 4. holds. Then the convolution ϕ = f g is continuous on R. The main result in this section is the following Convolution Theorem. To prove this theorem we apply interchanges of the order of integrations; these interchanges are based on one theorem from Mathematical Analysis. We do not formulate this theorem; nevertheless, we show that its main condition holds (see (4.4)).
13 Theorem 4.4. Let functions f and g belong to A(R) and satisfy one of conditions (i), (ii) or (iii). Then f g A(R) and Proof. First we show that f g A(R). Let I(x) = f g(ξ) = f(ξ)ĝ(ξ). (4.3) f(x y)g(y) dy. By Proposition 4.3, the function I = f g is continuous on R and therefore it is integrable in each bounded interval. Further, let J(y) = f(x y)g(y) dx. 3 Then J(y) = g(y) f(x y) dx = g(y) f(x) dx. Thus, J(y) is integrable in each bounded interval; moreover, J(y) dy = f(x) dx g(y) dy. (4.4) According to the general theorem, this enables us to interchange the order of integrations. Applying this interchange, we obtain I(x) dx = f(x y)g(y) dy dx = f(x y)g(y) dx dy = J(y) dy <. Thus, the integral I(x) dx converges. Since f g(x) I(x), it follows that f g A(R). Now we consider the Fourier transform of the convolution f g(ξ) = f g(x)e iπxξ dx = f(x y)g(y) dy e iπxξ dx.
14 4 As above, we interchange the order of integrations in the last iterated integral. Thus, we obtain f g(ξ) = g(y) f(x y)e iπxξ dx dy = g(y)e iπyξ dy f(u)e iπuξ du = ĝ(ξ) f(ξ). 5. Fourier inversion We shall consider the following problem: given f, recover f from f. The main result in this section is the following Fourier inversion theorem. Theorem 5.. Assume that f is continuous and absolutely integrable on R. If f is absolutely integrable on R, then f(ξ)e iπxξ dξ = f(x) (x R). (5.) As a corollary, we obtain the following uniqueness theorem for Fourier transform. Theorem 5.. Let f and g be continuous, absolutely integrable functions on R. If f = ĝ, then f = g. This statement follows immediately from Theorem 5. applied to the function f g (we have f g = ). Theorem 5. can be applied to evaluate some integrals. Example 5.3. Let f(x) = e π x. As it was shown in Example.5, f(ξ) = π ( + ξ ). Functions f and f are absolutely integrable on R, and f is continuous on R. Thus, the conditions of Theorem 5. hold. Applying this theorem, we obtain that for any x R e iπxξ π ( + ξ ) dξ = e π x,
15 5 or, equivalently, for any x (Laplace integral). cos πxξ dξ = π ( + ξ e πx ) Example 5.4. Let x /α, x α, f(x) =, x > α By Example.4 and Theorem 3. (iii), f(ξ) = sin απξ. α (πξ) (α > ). Since the conditions of Theorem 5. hold, we have that α sin απξ e iπξx dξ = f(x) (πξ) for all x R and all α >. In particular, if x =, α =, then, setting πξ = t, we obtain sin t t dt = π. Theorem 5. shows that a continuous function f A(R) can be obtained by equality (5.), provided its Fourier transform f also is absolutely integrable over R. However, the latter condition may not hold, and the integral at the left-hand side of (5.) may not exist. The following Dirichlet s Theorem states that for functions satisfying some good smoothness conditions, this integral may converge in the sense of principal value, that is, as the limit p.v. f(ξ)e iπxξ dξ = lim A A A f(ξ)e iπξx dξ. Theorem 5.5. Let a function f be absolutely integrable on R. If f is differentiable at a point x, then p.v. f(ξ)e iπxξ dξ = f(x). (5.)
16 6 6. The heat equation on the real line The one-dimensional homogeneous heat equation is u t = ku xx < x <, < t <. (6.) Here u is a function of x and t, k is a positive constant. The variable x represents a point on the real line and t is a time. The real line is a model for a rod of infinite length made of some material of uniform density. We assume that the rod is insulated. The value u(x, t) is the temperature of the rod at the point x at the time t. In order to determine a particular solution it is necessary to know the temperature at every point of the rod at some fixed time. Thus, we assume that we are given an initial temperature u(x, ) = f(x), < x <. (6.) We assume that u(x, t) for any fixed t and f(x) are absolutely integrable functions of x on R. We take the Fourier transform of u with respect to x: U(ξ, t) = u(x, t)e iπxξ dx. Differentiate U(ξ, t) with respect to t (under certain conditions the differentiation can be done under the sign of the integral): U t (ξ, t) = u t (x, t)e iπxξ dx. By the heat equation (6.) and Corollary 3.5, we get U t (ξ, t) = k Thus, we obtain the equation u xx (x, t)e iπxξ dx = kû xx e ıπxξ dx = 4π kξ U(ξ, t). U t + 4π kξ U =. This equation involves only the derivative with respect to t. Separating variables, we have U t U = 4π kξ. The general solution to this equation is U(ξ, t) = A(ξ)e 4π kξ t
17 7 (A(ξ) is an arbitrary function of ξ). For t = we get A(ξ) = U(ξ, ) = Using the initial condition (6.), we have Finally, A(ξ) = u(x, )e iπxξ dx. f(x)e iπxξ dx = f(ξ). U(ξ, t) = f(ξ)e 4π kξ t. (6.3) We shall use the Convolution Theorem (Theorem 4.4) to find u(x, t). We know that the Fourier transform of the gaussian g(x) = e x is equal to e ξ. Thus, by the dilation property (see Theorem 3. (iii)), the function e (ξ/λ) /λ is the Fourier transform of g(λx) = e (λx). In our case we determine λ from (6.3): ξ λ = 4πkξ t, λ = 4πkt. Thus, e 4π kξ t is the Fourier transform of the function K t (x) = 4πkt e x /(4kt). Therefore, applying (6.3) and the Convolution Theorem, we obtain that u(x, t) = f K t (x), that is, u(x, t) = 4πkt f(y)e (x y) /(4kt) dy (6.4) (t > ). We set also u(x, ) = f(x). It can be proved that f K t (x) f(x) as t. Now we can verify directly that (6.4) is the solution of (6.) satisfying the initial condition (6.). It is easy to see that the function K t (x) satisfies the heat equation. This function is called the heat kernel. We observe that the solution u(x, t) can be also obtained from (6.3) with the use of Fourier inversion theorem if f is known. Indeed, the right-hand side of (6.3) is absolutely integrable function of ξ on R. Thus, by Theorem 5. u(x, t) = f(ξ)e 4π kξ t e iπxξ dξ. (6.5)
18 8 Of course, it is easy to derive (6.4) from (6.5). Indeed, interchanging the order of integrations, we have from (6.5) ( ) u(x, t) = f(y)e iπyξ dy e 4π kξ t dξ ( ) = f(y) e iπ(y x)ξ e 4π kξ t dξ dy. The interior integral is the Fourier transform of the function ϕ t (ξ) = e 4π kξ t obtained from the gaussian by dilation. This Fourier transform is equal to K t (y x), and we again obtain (6.4). Example 6.. Find the solution of the equation u t = u xx, satisfying the initial condition u(x, ) = e π x. Solution. For the function f(x) = e π x we have (See Example.5) f(ξ) = π( + ξ ). Thus, by (6.5), the solution is u(x, t) = e 4π ξ t π + ξ eiπξx dξ e s t = 8π cos sx ds. 4π + s Also, by (6.4), we have u(x, t) = e πy e (x y) /(4t) dy. 4πt
19 EXERCISES Evaluate Fourier transforms of the following functions by the definition., < x <,. f(x) =, < x <,, x > ;. f(x) = 3. f(x) = 4. f(x) = x, x a,, x > a; cos x, x π,, x > π; sin x, x π,, x > π; 9 5. f(x) = x e x ; 6. f(x) = e x, x <, e x, x. Applying basic properties of Fourier transforms, evaluate Fourier transforms of the following functions. ). 7. f(x) = e x cos πx; ( 8. f(x) = Π x ) ; 9. f(x) =, x < c,, x > c;. f(x) = x e x ;. f(x) = e a x (a > );
20 . f(x) = e 4x 4x ; 3. f(x) = xe x. Let f A(R). functions Evaluate the Fourier transforms of the following 4. f( x); 5. f (x x ) (x R); 6. f(x)e iξ x (ξ R); 7. f(x) sin ξ x; 8. f(3x)e ix ; 9. f(x). We shall write f(x) f(u) to denote that f(u) is the Fourier transform of a function f(x). Prove that if f and f are absolutely integrable, and f is continuous, then f(x) f( u). Using this result, verify that.. + x πe π u ; sin πx π x Λ(u).. Let f A(R) be a continuous function. Assume that Find f. f(ξ) = ξ, ξ,, ξ >.
21 3. Let f A(R) be a continuous function. Let Find f. f(ξ) = ξ, ξ,, ξ >. Using the Fourier inversion formula, prove the following equalities 4. cos ξx a + ξ dξ = π a e a x (x R, a > ); π e σ ξ cos ξx dξ = σ π x e σ (x R, σ > ); sin ξ x, x, cos (ξx) dξ = ξ, x >. 7. Using the definition of a convolution, prove that Π Π(x) = Λ(x). Using Fourier transforms, solve the equations 8. f(t)f(x t) dt = e 4πx ; 9. f(t)f(x t) dt = + x. Applying (6.4) and (6.4), find the solution of the heat equation u t = u xx, satisfying the initial condition u(x, ) = f(x), where: 3. f(x) = + x ; 3. f(x) = x if x and if x >.
22 3. f(x) = e x f 4. f ANSWERS. f() =, f(ξ) = i sin πξ πξ sin πaξ. f() = a, f(ξ) = πξ ( ± ) 4πξ = π, f(ξ) = π 4π ξ sin π ξ (ξ ). (ξ ). ( ± ) i = πi, f(ξ) = π 4π ξ sin π ξ 5. 4π ξ ( + 4π ξ ). 6. 4πiξ + 4π ξ π (ξ + ) + + 4π (ξ ). ( ξ ± ). π ( 8. f() =, f(ξ) ) = e πiξ (ξ ). πiξ 9. f() = c, f(ξ) = sin πcξ πξ. a a + 4π ξ.. (ξ ).. ( ξ ± ). π ( ) π ξ e ξ. π π eπiξ 4 ξ. 3. π 3 iξe ξ. 4. f( ξ). 5. e πiξx f(ξ). 6. f (ξ ξ /(π)). [ ( f ξ ξ ) ( i π f ξ + ξ )]. 8. π. f() = 4 3, f(x) = π ( sin πx πx 3 f ) cos πx ( πξ 3. f() =, f(x) = sin πx π x (x ). 6π ). 9. f (x ). ( ) ξ.
23 3 8. ± e 8πx. 9. ± π + 4x. 3.u(x, t) = e u ut cos(ux) du; u(x, t) = + /(4t) 4πt + y e (x y) dy; 3.u(x, t) = sin u e 4ut cos(ux) du; π u u(x, t) = + y e (x y) /(4t) dy; 4πt 3.u(x, t) = e u/π e ut cos(ux) du; π u(x, t) = + e y e (x y) /(4t) dy. 4πt
1.5 Approximate Identities
38 1 The Fourier Transform on L 1 (R) which are dense subspaces of L p (R). On these domains, P : D P L p (R) and M : D M L p (R). Show, however, that P and M are unbounded even when restricted to these
More informationBernstein s inequality and Nikolsky s inequality for R d
Bernstein s inequality and Nikolsky s inequality for d Jordan Bell jordan.bell@gmail.com Department of athematics University of Toronto February 6 25 Complex Borel measures and the Fourier transform Let
More informationTOPICS IN FOURIER ANALYSIS-III. Contents
TOPICS IN FOUIE ANALYSIS-III M.T. NAI Contents 1. Fourier transform: Basic properties 2 2. On surjectivity 7 3. Inversion theorem 9 4. Proof of inversion theorem 10 5. The Banach algebra L 1 ( 12 6. Fourier-Plancheral
More informationFourier Series. ,..., e ixn ). Conversely, each 2π-periodic function φ : R n C induces a unique φ : T n C for which φ(e ix 1
Fourier Series Let {e j : 1 j n} be the standard basis in R n. We say f : R n C is π-periodic in each variable if f(x + πe j ) = f(x) x R n, 1 j n. We can identify π-periodic functions with functions on
More informationFourier Series. 1. Review of Linear Algebra
Fourier Series In this section we give a short introduction to Fourier Analysis. If you are interested in Fourier analysis and would like to know more detail, I highly recommend the following book: Fourier
More information3: THE SHANNON SAMPLING THEOREM
3: THE SHANNON SAMPLING THEOEM STEVEN HEILMAN 1. Introduction In this final set of notes, we would like to end where we began. In the first set of notes, we mentioned that mathematics could be used to
More information1 Fourier Integrals on L 2 (R) and L 1 (R).
18.103 Fall 2013 1 Fourier Integrals on L 2 () and L 1 (). The first part of these notes cover 3.5 of AG, without proofs. When we get to things not covered in the book, we will start giving proofs. The
More informationReal Analysis Chapter 8 Solutions Jonathan Conder. = lim n. = lim. f(z) dz dy. m(b r (y)) B r(y) f(z + x y) dz. B r(y) τ y x f(z) f(z) dz
3. (a Note that η ( (t e /t for all t (,, where is a polynomial of degree. Given k N, suppose that η (k (t P k (/te /t for all t (,, where P k (x is some polynomial of degree (k. By the product rule and
More informationLECTURE Fourier Transform theory
LECTURE 3 3. Fourier Transform theory In section (2.2) we introduced the Discrete-time Fourier transform and developed its most important properties. Recall that the DTFT maps a finite-energy sequence
More informationTopics in Fourier analysis - Lecture 2.
Topics in Fourier analysis - Lecture 2. Akos Magyar 1 Infinite Fourier series. In this section we develop the basic theory of Fourier series of periodic functions of one variable, but only to the extent
More informationHarmonic Analysis: from Fourier to Haar. María Cristina Pereyra Lesley A. Ward
Harmonic Analysis: from Fourier to Haar María Cristina Pereyra Lesley A. Ward Department of Mathematics and Statistics, MSC03 2150, 1 University of New Mexico, Albuquerque, NM 87131-0001, USA E-mail address:
More informationbe the set of complex valued 2π-periodic functions f on R such that
. Fourier series. Definition.. Given a real number P, we say a complex valued function f on R is P -periodic if f(x + P ) f(x) for all x R. We let be the set of complex valued -periodic functions f on
More informationStarting from Heat Equation
Department of Applied Mathematics National Chiao Tung University Hsin-Chu 30010, TAIWAN 20th August 2009 Analytical Theory of Heat The differential equations of the propagation of heat express the most
More informationMath 115 ( ) Yum-Tong Siu 1. Derivation of the Poisson Kernel by Fourier Series and Convolution
Math 5 (006-007 Yum-Tong Siu. Derivation of the Poisson Kernel by Fourier Series and Convolution We are going to give a second derivation of the Poisson kernel by using Fourier series and convolution.
More informationINTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES
INTRODUCTION TO REAL ANALYSIS II MATH 433 BLECHER NOTES. As in earlier classnotes. As in earlier classnotes (Fourier series) 3. Fourier series (continued) (NOTE: UNDERGRADS IN THE CLASS ARE NOT RESPONSIBLE
More informationCHAPTER VI APPLICATIONS TO ANALYSIS
CHAPTER VI APPLICATIONS TO ANALYSIS We include in this chapter several subjects from classical analysis to which the notions of functional analysis can be applied. Some of these subjects are essential
More informationMATH 5640: Fourier Series
MATH 564: Fourier Series Hung Phan, UMass Lowell September, 8 Power Series A power series in the variable x is a series of the form a + a x + a x + = where the coefficients a, a,... are real or complex
More informationn 1 f = f m, m=0 n 1 k=0 Note that if n = 2, we essentially get example (i) (but for complex functions).
Chapter 4 Fourier Analysis 4.0.1 Intuition This discussion is borrowed from [T. Tao, Fourier Transform]. Fourier transform/series can be viewed as a way to decompose a function from some given space V
More informationMATH 220 solution to homework 5
MATH 220 solution to homework 5 Problem. (i Define E(t = k(t + p(t = then E (t = 2 = 2 = 2 u t u tt + u x u xt dx u 2 t + u 2 x dx, u t u xx + u x u xt dx x [u tu x ] dx. Because f and g are compactly
More informationFOURIER METHODS AND DISTRIBUTIONS: SOLUTIONS
Centre for Mathematical Sciences Mathematics, Faculty of Science FOURIER METHODS AND DISTRIBUTIONS: SOLUTIONS. We make the Ansatz u(x, y) = ϕ(x)ψ(y) and look for a solution which satisfies the boundary
More informationFolland: Real Analysis, Chapter 8 Sébastien Picard
Folland: Real Analysis, Chapter 8 Sébastien Picard Problem 8.3 Let η(t) = e /t for t >, η(t) = for t. a. For k N and t >, η (k) (t) = P k (/t)e /t where P k is a polynomial of degree 2k. b. η (k) () exists
More informationu t = u p (t)q(x) = p(t) q(x) p (t) p(t) for some λ. = λ = q(x) q(x)
. Separation of variables / Fourier series The following notes were authored initially by Jason Murphy, and subsequently edited and expanded by Mihaela Ifrim, Daniel Tataru, and myself. We turn to the
More informationAverage theorem, Restriction theorem and Strichartz estimates
Average theorem, Restriction theorem and trichartz estimates 2 August 27 Abstract We provide the details of the proof of the average theorem and the restriction theorem. Emphasis has been placed on the
More informationChapter 5: Bases in Hilbert Spaces
Orthogonal bases, general theory The Fourier basis in L 2 (T) Applications of Fourier series Chapter 5: Bases in Hilbert Spaces I-Liang Chern Department of Applied Mathematics National Chiao Tung University
More informationMath 489AB A Very Brief Intro to Fourier Series Fall 2008
Math 489AB A Very Brief Intro to Fourier Series Fall 8 Contents Fourier Series. The coefficients........................................ Convergence......................................... 4.3 Convergence
More informationMath 121A: Homework 6 solutions
Math A: Homework 6 solutions. (a) The coefficients of the Fourier sine series are given by b n = π f (x) sin nx dx = x(π x) sin nx dx π = (π x) cos nx dx nπ nπ [x(π x) cos nx]π = n ( )(sin nx) dx + π n
More informationMath 4263 Homework Set 1
Homework Set 1 1. Solve the following PDE/BVP 2. Solve the following PDE/BVP 2u t + 3u x = 0 u (x, 0) = sin (x) u x + e x u y = 0 u (0, y) = y 2 3. (a) Find the curves γ : t (x (t), y (t)) such that that
More informationII. FOURIER TRANSFORM ON L 1 (R)
II. FOURIER TRANSFORM ON L 1 (R) In this chapter we will discuss the Fourier transform of Lebesgue integrable functions defined on R. To fix the notation, we denote L 1 (R) = {f : R C f(t) dt < }. The
More informationMATH 220 solution to homework 4
MATH 22 solution to homework 4 Problem. Define v(t, x) = u(t, x + bt), then v t (t, x) a(x + u bt) 2 (t, x) =, t >, x R, x2 v(, x) = f(x). It suffices to show that v(t, x) F = max y R f(y). We consider
More informationElliptic Problems for Pseudo Differential Equations in a Polyhedral Cone
Advances in Dynamical Systems and Applications ISSN 0973-5321, Volume 9, Number 2, pp. 227 237 (2014) http://campus.mst.edu/adsa Elliptic Problems for Pseudo Differential Equations in a Polyhedral Cone
More informationTopics in Harmonic Analysis Lecture 1: The Fourier transform
Topics in Harmonic Analysis Lecture 1: The Fourier transform Po-Lam Yung The Chinese University of Hong Kong Outline Fourier series on T: L 2 theory Convolutions The Dirichlet and Fejer kernels Pointwise
More informationFourier transforms, I
(November 28, 2016) Fourier transforms, I Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/ garrett/m/fun/notes 2016-17/Fourier transforms I.pdf]
More informationFourier Analysis, Stein and Shakarchi The Fourier Transform on R
Fourier Analysis, Stein and Shakarchi Chapter 5 The Fourier Transform on Yung-Hsiang Huang 8.3.7 Abstract After Problem 6, we contain a proof for Widder s uniqueness theorem in the class of nonnegative
More informationu xx + u yy = 0. (5.1)
Chapter 5 Laplace Equation The following equation is called Laplace equation in two independent variables x, y: The non-homogeneous problem u xx + u yy =. (5.1) u xx + u yy = F, (5.) where F is a function
More informationMath 172 Problem Set 8 Solutions
Math 72 Problem Set 8 Solutions Problem. (i We have (Fχ [ a,a] (ξ = χ [ a,a] e ixξ dx = a a e ixξ dx = iξ (e iax e iax = 2 sin aξ. ξ (ii We have (Fχ [, e ax (ξ = e ax e ixξ dx = e x(a+iξ dx = a + iξ where
More informationConvergence for periodic Fourier series
Chapter 8 Convergence for periodic Fourier series We are now in a position to address the Fourier series hypothesis that functions can realized as the infinite sum of trigonometric functions discussed
More information1.3.1 Definition and Basic Properties of Convolution
1.3 Convolution 15 1.3 Convolution Since L 1 (R) is a Banach space, we know that it has many useful properties. In particular the operations of addition and scalar multiplication are continuous. However,
More informationMAGIC058 & MATH64062: Partial Differential Equations 1
MAGIC58 & MATH6462: Partial Differential Equations Section 6 Fourier transforms 6. The Fourier integral formula We have seen from section 4 that, if a function f(x) satisfies the Dirichlet conditions,
More information2 Infinite products and existence of compactly supported φ
415 Wavelets 1 Infinite products and existence of compactly supported φ Infinite products.1 Infinite products a n need to be defined via limits. But we do not simply say that a n = lim a n N whenever the
More information13. Fourier transforms
(December 16, 2017) 13. Fourier transforms Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/ garrett/m/real/notes 2017-18/13 Fourier transforms.pdf]
More informationChapter 3 Second Order Linear Equations
Partial Differential Equations (Math 3303) A Ë@ Õæ Aë áöß @. X. @ 2015-2014 ú GA JË@ É Ë@ Chapter 3 Second Order Linear Equations Second-order partial differential equations for an known function u(x,
More informationDangerous and Illegal Operations in Calculus Do we avoid differentiating discontinuous functions because it s impossible, unwise, or simply out of
Dangerous and Illegal Operations in Calculus Do we avoid differentiating discontinuous functions because it s impossible, unwise, or simply out of ignorance and fear? Despite the risks, many natural phenomena
More information5. Some theorems on continuous functions
5. Some theorems on continuous functions The results of section 3 were largely concerned with continuity of functions at a single point (usually called x 0 ). In this section, we present some consequences
More informationPartial differential equation for temperature u(x, t) in a heat conducting insulated rod along the x-axis is given by the Heat equation:
Chapter 7 Heat Equation Partial differential equation for temperature u(x, t) in a heat conducting insulated rod along the x-axis is given by the Heat equation: u t = ku x x, x, t > (7.1) Here k is a constant
More information17 Source Problems for Heat and Wave IB- VPs
17 Source Problems for Heat and Wave IB- VPs We have mostly dealt with homogeneous equations, homogeneous b.c.s in this course so far. Recall that if we have non-homogeneous b.c.s, then we want to first
More informationIn this chapter we study elliptical PDEs. That is, PDEs of the form. 2 u = lots,
Chapter 8 Elliptic PDEs In this chapter we study elliptical PDEs. That is, PDEs of the form 2 u = lots, where lots means lower-order terms (u x, u y,..., u, f). Here are some ways to think about the physical
More informationMath 46, Applied Math (Spring 2009): Final
Math 46, Applied Math (Spring 2009): Final 3 hours, 80 points total, 9 questions worth varying numbers of points 1. [8 points] Find an approximate solution to the following initial-value problem which
More informationAdvanced Real Analysis
Advanced Real Analysis Edited and typeset by David McCormick Based upon lectures by José Luis Rodrigo University of Warwick Autumn Preface These notes are primarily based on the lectures for the course
More informationSome Aspects of Solutions of Partial Differential Equations
Some Aspects of Solutions of Partial Differential Equations K. Sakthivel Department of Mathematics Indian Institute of Space Science & Technology(IIST) Trivandrum - 695 547, Kerala Sakthivel@iist.ac.in
More informationMATH FALL 2014
MATH 126 - FALL 2014 JASON MURPHY Abstract. These notes are meant to supplement the lectures for Math 126 (Introduction to PDE) in the Fall of 2014 at the University of California, Berkeley. Contents 1.
More information7: FOURIER SERIES STEVEN HEILMAN
7: FOURIER SERIES STEVE HEILMA Contents 1. Review 1 2. Introduction 1 3. Periodic Functions 2 4. Inner Products on Periodic Functions 3 5. Trigonometric Polynomials 5 6. Periodic Convolutions 7 7. Fourier
More informationMath 342 Partial Differential Equations «Viktor Grigoryan
Math 342 Partial Differential Equations «Viktor Grigoryan 15 Heat with a source So far we considered homogeneous wave and heat equations and the associated initial value problems on the whole line, as
More information1 Review of di erential calculus
Review of di erential calculus This chapter presents the main elements of di erential calculus needed in probability theory. Often, students taking a course on probability theory have problems with concepts
More informationFourier Series and Integrals
Fourier Series and Integrals Fourier Series et f(x) beapiece-wiselinearfunctionon[, ] (Thismeansthatf(x) maypossessa finite number of finite discontinuities on the interval). Then f(x) canbeexpandedina
More informationMath 5588 Final Exam Solutions
Math 5588 Final Exam Solutions Prof. Jeff Calder May 9, 2017 1. Find the function u : [0, 1] R that minimizes I(u) = subject to u(0) = 0 and u(1) = 1. 1 0 e u(x) u (x) + u (x) 2 dx, Solution. Since the
More informationSloshing problem in a half-plane covered by a dock with two equal gaps
Sloshing prolem in a half-plane covered y a dock with two equal gaps O. V. Motygin N. G. Kuznetsov Institute of Prolems in Mech Engineering Russian Academy of Sciences St.Petersurg, Russia STATEMENT OF
More informationWe start with a simple result from Fourier analysis. Given a function f : [0, 1] C, we define the Fourier coefficients of f by
Chapter 9 The functional equation for the Riemann zeta function We will eventually deduce a functional equation, relating ζ(s to ζ( s. There are various methods to derive this functional equation, see
More informationMath 341: Probability Seventeenth Lecture (11/10/09)
Math 341: Probability Seventeenth Lecture (11/10/09) Steven J Miller Williams College Steven.J.Miller@williams.edu http://www.williams.edu/go/math/sjmiller/ public html/341/ Bronfman Science Center Williams
More informationLecture notes: Introduction to Partial Differential Equations
Lecture notes: Introduction to Partial Differential Equations Sergei V. Shabanov Department of Mathematics, University of Florida, Gainesville, FL 32611 USA CHAPTER 1 Classification of Partial Differential
More information1 The Complex Fourier Series
The Complex Fourier Series Adding Complexity to life The complex Fourier series is in some ways a superior product, at least for those people who are not terrified by complex numbers. Suppose fx) is a
More informationFinite-dimensional spaces. C n is the space of n-tuples x = (x 1,..., x n ) of complex numbers. It is a Hilbert space with the inner product
Chapter 4 Hilbert Spaces 4.1 Inner Product Spaces Inner Product Space. A complex vector space E is called an inner product space (or a pre-hilbert space, or a unitary space) if there is a mapping (, )
More informationPartial Differential Equations for Engineering Math 312, Fall 2012
Partial Differential Equations for Engineering Math 312, Fall 2012 Jens Lorenz July 17, 2012 Contents Department of Mathematics and Statistics, UNM, Albuquerque, NM 87131 1 Second Order ODEs with Constant
More informationSmoothing Effects for Linear Partial Differential Equations
Smoothing Effects for Linear Partial Differential Equations Derek L. Smith SIAM Seminar - Winter 2015 University of California, Santa Barbara January 21, 2015 Table of Contents Preliminaries Smoothing
More informationComplex Analysis, Stein and Shakarchi The Fourier Transform
Complex Analysis, Stein and Shakarchi Chapter 4 The Fourier Transform Yung-Hsiang Huang 2017.11.05 1 Exercises 1. Suppose f L 1 (), and f 0. Show that f 0. emark 1. This proof is observed by Newmann (published
More informationFRAMES AND TIME-FREQUENCY ANALYSIS
FRAMES AND TIME-FREQUENCY ANALYSIS LECTURE 5: MODULATION SPACES AND APPLICATIONS Christopher Heil Georgia Tech heil@math.gatech.edu http://www.math.gatech.edu/ heil READING For background on Banach spaces,
More informationMATH-UA 263 Partial Differential Equations Recitation Summary
MATH-UA 263 Partial Differential Equations Recitation Summary Yuanxun (Bill) Bao Office Hour: Wednesday 2-4pm, WWH 1003 Email: yxb201@nyu.edu 1 February 2, 2018 Topics: verifying solution to a PDE, dispersion
More informationA glimpse of Fourier analysis
Chapter 7 A glimpse of Fourier analysis 7.1 Fourier series In the middle of the 18th century, mathematicians and physicists started to study the motion of a vibrating string (think of the strings of a
More information1 Separation of Variables
Jim ambers ENERGY 281 Spring Quarter 27-8 ecture 2 Notes 1 Separation of Variables In the previous lecture, we learned how to derive a PDE that describes fluid flow. Now, we will learn a number of analytical
More informationMath 172 Problem Set 5 Solutions
Math 172 Problem Set 5 Solutions 2.4 Let E = {(t, x : < x b, x t b}. To prove integrability of g, first observe that b b b f(t b b g(x dx = dt t dx f(t t dtdx. x Next note that f(t/t χ E is a measurable
More informationChapter 9. Parabolic Equations. 9.1 Introduction. 9.2 Some Results from Functional Analysis
Chapter 9 Parabolic Equations 9.1 Introduction If u(x, t) represents the temperature in a body Ω R n at a point x Ω at time t, then the propagation (or diffusion) of heat is (approximately) described by
More informationJUHA KINNUNEN. Partial Differential Equations
JUHA KINNUNEN Partial Differential Equations Department of Mathematics and Systems Analysis, Aalto University 207 Contents INTRODUCTION 2 FOURIER SERIES AND PDES 5 2. Periodic functions*.............................
More informationYAO LIU. f(x t) + f(x + t)
DUNKL WAVE EQUATION & REPRESENTATION THEORY OF SL() YAO LIU The classical wave equation u tt = u in n + 1 dimensions, and its various modifications, have been studied for centuries, and one of the most
More information1.1 Appearance of Fourier series
Chapter Fourier series. Appearance of Fourier series The birth of Fourier series can be traced back to the solutions of wave equation in the work of Bernoulli and the heat equation in the work of Fourier.
More informationVectors in Function Spaces
Jim Lambers MAT 66 Spring Semester 15-16 Lecture 18 Notes These notes correspond to Section 6.3 in the text. Vectors in Function Spaces We begin with some necessary terminology. A vector space V, also
More informationNONLOCAL DIFFUSION EQUATIONS
NONLOCAL DIFFUSION EQUATIONS JULIO D. ROSSI (ALICANTE, SPAIN AND BUENOS AIRES, ARGENTINA) jrossi@dm.uba.ar http://mate.dm.uba.ar/ jrossi 2011 Non-local diffusion. The function J. Let J : R N R, nonnegative,
More informationContinuity. Chapter 4
Chapter 4 Continuity Throughout this chapter D is a nonempty subset of the real numbers. We recall the definition of a function. Definition 4.1. A function from D into R, denoted f : D R, is a subset of
More informationON THE BEHAVIOR OF THE SOLUTION OF THE WAVE EQUATION. 1. Introduction. = u. x 2 j
ON THE BEHAVIO OF THE SOLUTION OF THE WAVE EQUATION HENDA GUNAWAN AND WONO SETYA BUDHI Abstract. We shall here study some properties of the Laplace operator through its imaginary powers, and apply the
More informationPHYS 502 Lecture 3: Fourier Series
PHYS 52 Lecture 3: Fourier Series Fourier Series Introduction In mathematics, a Fourier series decomposes periodic functions or periodic signals into the sum of a (possibly infinite) set of simple oscillating
More informationDiffusion on the half-line. The Dirichlet problem
Diffusion on the half-line The Dirichlet problem Consider the initial boundary value problem (IBVP) on the half line (, ): v t kv xx = v(x, ) = φ(x) v(, t) =. The solution will be obtained by the reflection
More informationNotes on Fourier Series and Integrals Fourier Series
Notes on Fourier Series and Integrals Fourier Series et f(x) be a piecewise linear function on [, ] (This means that f(x) may possess a finite number of finite discontinuities on the interval). Then f(x)
More informationA review: The Laplacian and the d Alembertian. j=1
Chapter One A review: The Laplacian and the d Alembertian 1.1 THE LAPLACIAN One of the main goals of this course is to understand well the solution of wave equation both in Euclidean space and on manifolds
More informationOutline of Fourier Series: Math 201B
Outline of Fourier Series: Math 201B February 24, 2011 1 Functions and convolutions 1.1 Periodic functions Periodic functions. Let = R/(2πZ) denote the circle, or onedimensional torus. A function f : C
More informationSingular Integrals. 1 Calderon-Zygmund decomposition
Singular Integrals Analysis III Calderon-Zygmund decomposition Let f be an integrable function f dx 0, f = g + b with g Cα almost everywhere, with b
More informationG: Uniform Convergence of Fourier Series
G: Uniform Convergence of Fourier Series From previous work on the prototypical problem (and other problems) u t = Du xx 0 < x < l, t > 0 u(0, t) = 0 = u(l, t) t > 0 u(x, 0) = f(x) 0 < x < l () we developed
More informationA Bowl of Kernels. By Nuriye Atasever, Cesar Alvarado, and Patrick Doherty. December 03, 2013
December 03, 203 Introduction When we studied Fourier series we improved convergence by convolving with the Fejer and Poisson kernels on T. The analogous Fejer and Poisson kernels on the real line help
More information1 Distributions (due January 22, 2009)
Distributions (due January 22, 29). The distribution derivative of the locally integrable function ln( x ) is the principal value distribution /x. We know that, φ = lim φ(x) dx. x ɛ x Show that x, φ =
More informationFourier Transform & Sobolev Spaces
Fourier Transform & Sobolev Spaces Michael Reiter, Arthur Schuster Summer Term 2008 Abstract We introduce the concept of weak derivative that allows us to define new interesting Hilbert spaces the Sobolev
More informationANALYTIC SEMIGROUPS AND APPLICATIONS. 1. Introduction
ANALYTIC SEMIGROUPS AND APPLICATIONS KELLER VANDEBOGERT. Introduction Consider a Banach space X and let f : D X and u : G X, where D and G are real intervals. A is a bounded or unbounded linear operator
More information7.1. Calculus of inverse functions. Text Section 7.1 Exercise:
Contents 7. Inverse functions 1 7.1. Calculus of inverse functions 2 7.2. Derivatives of exponential function 4 7.3. Logarithmic function 6 7.4. Derivatives of logarithmic functions 7 7.5. Exponential
More informationHyperbolic PDEs. Chapter 6
Chapter 6 Hyperbolic PDEs In this chapter we will prove existence, uniqueness, and continuous dependence of solutions to hyperbolic PDEs in a variety of domains. To get a feel for what we might expect,
More informationPartial Differential Equations
Part II Partial Differential Equations Year 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2015 Paper 4, Section II 29E Partial Differential Equations 72 (a) Show that the Cauchy problem for u(x,
More informationMATH 126 FINAL EXAM. Name:
MATH 126 FINAL EXAM Name: Exam policies: Closed book, closed notes, no external resources, individual work. Please write your name on the exam and on each page you detach. Unless stated otherwise, you
More informationIndeed, the family is still orthogonal if we consider a complex valued inner product ( or an inner product on complex vector space)
Fourier series of complex valued functions Suppose now f is a piecewise continuous complex valued function on [, π], that is f(x) = u(x)+iv(x) such that both u and v are real valued piecewise continuous
More informationGreen s Functions and Distributions
CHAPTER 9 Green s Functions and Distributions 9.1. Boundary Value Problems We would like to study, and solve if possible, boundary value problems such as the following: (1.1) u = f in U u = g on U, where
More informationExamples of the Fourier Theorem (Sect. 10.3). The Fourier Theorem: Continuous case.
s of the Fourier Theorem (Sect. 1.3. The Fourier Theorem: Continuous case. : Using the Fourier Theorem. The Fourier Theorem: Piecewise continuous case. : Using the Fourier Theorem. The Fourier Theorem:
More informationSINC PACK, and Separation of Variables
SINC PACK, and Separation of Variables Frank Stenger Abstract This talk consists of a proof of part of Stenger s SINC-PACK computer package (an approx. 400-page tutorial + about 250 Matlab programs) that
More informationLet R be the line parameterized by x. Let f be a complex function on R that is integrable. The Fourier transform ˆf = F f is. e ikx f(x) dx. (1.
Chapter 1 Fourier transforms 1.1 Introduction Let R be the line parameterized by x. Let f be a complex function on R that is integrable. The Fourier transform ˆf = F f is ˆf(k) = e ikx f(x) dx. (1.1) It
More informationd(x n, x) d(x n, x nk ) + d(x nk, x) where we chose any fixed k > N
Problem 1. Let f : A R R have the property that for every x A, there exists ɛ > 0 such that f(t) > ɛ if t (x ɛ, x + ɛ) A. If the set A is compact, prove there exists c > 0 such that f(x) > c for all x
More informationSolutions to Exercises 8.1
Section 8. Partial Differential Equations in Physics and Engineering 67 Solutions to Exercises 8.. u xx +u xy u is a second order, linear, and homogeneous partial differential equation. u x (,y) is linear
More informationarxiv: v1 [math.ca] 31 Dec 2018
arxiv:181.1173v1 [math.ca] 31 Dec 18 Some trigonometric integrals and the Fourier transform of a spherically symmetric exponential function Hideshi YAMANE Department of Mathematical Sciences, Kwansei Gakuin
More information