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=
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 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 = (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. (.)
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 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.
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 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 (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.
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.
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.
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)).
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.
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,
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.)
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
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)
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
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 > );
. 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 >.
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 > ); 5. 6. 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 >.
3. f(x) = e x. 7. 3. 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. 7. + 4π (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.
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