FOURIER SERIES PART III: APPLICATIONS

FOURIER SERIES PART III: APPLICATIONS We extend the construction of Fourier series to functions with arbitrary eriods, then we associate to functions defined on an interval [, L] Fourier sine and Fourier cosine series and then aly these results to solve BVPs. 1. Fourier series with arbitrary eriods Let f : R R be a iecewise continuous function with eriod 2 ( > ). We would like to reresent f by a trigonometric series. We can reeat what we did in the revious note, when we had = π, and reach the sought reresentation. There is however a simle way of obtaining the same series by introducing the function g(s) = f ( s π ) ( f(x) = g ( πx )). Note that since f C(R), then g C(R) and that since f is 2-eriodic, then ( ) (s + 2π) ( s ) ( s ) g(s + 2π) = f = f π π + 2 = f = g(s). π That is, g is 2π-eriodic. We can therefore associate a Fourier series to g: g(s) a 2 + a n cos ns + b n sin ns. In terms of the function f, we have the association f(x) a 2 + a n cos nπx + b n sin nπx. π The coefficients are given by a = 1 π g(s)ds = 1 π ( s ) f ds = 1 f(x)dx π π π π π a n = 1 π g(s) cos nsds = 1 π ( s ) f cos(ns)ds = 1 π π π π π b n = 1 π g(s) sin nsds = 1 π ( s ) f sin(ns)ds = 1 π π π π ( ) nπx f(x) cos dx ( ) nπx f(x) sin dx The fundamental convergence theorem (Fourier theorem) states Theorem. Let f be a 2-eriodic and iecewise smooth function on R. Then f av (x) = f(x+ ) + f(x ) = a 2 2 + a n cos nπx + b n sin nπx, where the coefficients are given by a = 1 f(x)dx, Date: March 2, 216. 1

2 FOURIER SERIES PART III: APPLICATIONS and for n Z +, a n = 1 ( ) nπx f(x) cos dx, b n = 1 ( ) nπx f(x) sin dx. Again if f is continuous at x, then f(x ) is equal to its Fourier series: f(x ) = a 2 + a n cos nπx + b n sin nπx We also have uniform convergence with the additional condition that f is continuous on R. More recisely, Theorem. Let f be a 2-eriodic and iecewise smooth function on R. Suose that f is continuous in R, then f(x) = a 2 + a n cos nπx and the convergence is uniform. + b n sin nπx. x R Analogous results hold for termwise differentiation of Fourier series and termwise integration. Examle. Consider the function f(x) that is 2-eriodic and is given on the interval [, ] by 2Hx + H if x < /2 ; f(x) = 2Hx + H if /2 x < ; if /2 x where H is a ositive constant. H Figure 1. Grah of function of examle The function f is continuous on R and is iecewise smooth. It is also an even function (hence its b n Fourier coefficients are all zero). The Fourier coefficients of f are a = 2 f(x)dx = 2H /2 ( ) 2x + 1 dx = H 2

FOURIER SERIES PART III: APPLICATIONS 3 and for n = 1, 2, 3, we have a n = 2 f(x) cos nπx dx = 2H /2 ( ) 2x + 1 = 2H [( ) ] /2 2x nπx + 1 sin + 4H /2 nπ nπ = 4H (1 π 2 n 2 cos nπ ) 2 Since the function f is continuous and iecewise smooth, we have f(x) = H 4 + 4H 1 cos(nπ/2) π 2 n 2 cos nπx x R cos nπx dx sin nπx dx Furthermore, the convergence is uniform. We can aly termwise differentiation to obtain the Fourier series of the derivative f (where f (x) = 2H/ for < x < /2, f (x) = 2H/ for /2 < x < and f (x) = for /2 < x < ). We have f (x) 4H π 1 cos(nπ/2) n 2. Parseval s Identity sin nπx Parseval s identity is a sort of a generalized ythagorean theorem in the sace of functions. Theorem. (Parseval s Identity) Let f be 2-eriodic and iecewise continuous with Fourier series a 2 + a n cos nπx + b n sin nπx. Then a 2 4 + 1 2 (a 2 n + b 2 n) = 1 2 f(x) 2 dx. Proof. We will rove the identity when f is continuous and iecewise smooth. In this case the Fourier series (equals f),is uniformly convergent, and termwise integration is allowed. We have f(x) 2 = ( a 2 + a n cos nπx = a 2 f(x) + a n f(x) cos nπx ) + b n sin nπx f(x) + b nf(x) sin nπx We integrate from to and divide by 2. The term by term integration gives 1 2 f(x) 2 dx = a 1 f(x)dx+ 2 2 1 + a n f(x) cos nπx 2 dx + b 1 n 2 = a2 4 + 1 (a 2 n + b 2 2 n). f(x) sin nπx dx.

4 FOURIER SERIES PART III: APPLICATIONS The Parseval s identity is used to aroximate the average error when relacing a given function f by its N-th Fourier artial sum S N f. Recall that S N f(x) = a N 2 + a n cos nπx + b n sin nπx The mean square error, when relacing f by S N f, is defined as the number E N given by EN 2 = 1 (f(x) S N f(x)) 2 dx. If we use Parseval s identity to the function f S N f, we find that EN 2 = (a 2 n + b 2 n) n=n+1 Examle. Consider the 2π-eriodic function defined over [ π, π] by f(x) = 1 for < x < π and f(x) = 1 for π < x <. The fourier series of f is 4 sin(2j + 1)x. We would like to find N so that the aroximation π (2j + 1) j= f(x) 4 π 2j+1 N sin(2j + 1)x (2j + 1) guarantees that the mean square error is no more than.1. That is E N < 1 2. From the above discussion, we have EN 2 = 16 1 π 2 (2j + 1) 2. j>(n 1)/2 We can estimate the last series by using the integral test (see figure). We have. y=1/(2x+1) 2 M 1 M M+1 j=m Figure 2. Comarison of integral and sum 1 (2j + 1) 2 dx M 1 (2x + 1) 2 = 1 2(2M 1).

FOURIER SERIES PART III: APPLICATIONS 5 Hence, it follows from the above calculations that E 2 N 16 π 2 N. Therefore, in order to have E N <.1, it is enough to take N so that N > 16/π 2. That is N = 163. Parseval s identity can also be used to evaluate series. Examle. We have seen that Parseval s identity gives From this we get x = π 2 4 π π 2 4 + 8 π 2 j= j= cos(2j + 1)x (2j + 1) 2 x [ π, π]. 1 (2j + 1) 4 = 1 π x 2 dx = π2 2π π 3. 1 + 1 3 4 + 1 5 4 + 1 7 4 + 1 9 4 + = π4 96. 3. Even and Odd Periodic Extensions We would like to reresent a function f given only on a an interval [, L] by a trigonometric series. For this we extend f to the interval [ L, L] as either an even function or as an odd function then extend it to R as a eriodic function with eriod 2L. The Fourier series of this extension gives the sought reresentation of f. The even extension gives the Fourier cosine series of f and the odd extension gives the Fourier sine series of f. More recisely, let f be a iecewise smooth function on the interval [, L]. Let f even and f odd be, resectively, the even and the odd odd extensions of f to the interval [ L, L]. Now we extend f even to R as a 2L-eriodic function F even and grah of f odd grah of f grah of f even L L L L L we extend f odd to R as a 2L-eriodic function F odd. Note f(x) = F even (x) = F odd (x) x [, L]. The Fourier series of F even and F odd are F even (x) a 2 + a n cos nπx L and F odd (x) b n sin nπx L

6 FOURIER SERIES PART III: APPLICATIONS Even eriodic extension F even L L Odd eriodic extension F odd L L The Fourier coefficients are a n b n = 2 L = 2 L L L F even (x) cos nπx L dx = 2 L F odd (x) sin nπx L dx = 2 L L L f(x) cos nπx dx n =, 1, 2, L f(x) sin nπx dx n = 1, 2, 3, L This together with Fourier s Theorem give the following reresentations. Theorem Let f be a iecewise smooth function on the interval [, L]. Then f has the following Fourier cosine series reresentation: x (, L) f av (x) = a 2 + a n cos nπx L, where a n = 2 L L In articular at each oint x where f is continuous we have f(x) = a 2 + a n cos nπx L. f(x) cos nπx L dx. Theorem Let f be a iecewise smooth function on the interval [, L]. Then f has the following Fourier sine series reresentation: x (, L) f av (x) = b n sin nπx L, where b n = 2 L L In articular at each oint x where f is continuous we have f(x) = b n sin nπx L. f(x) sin nπx L dx. Examle 1. Let f(x) = x on the interval [, 1]. The 2-eriodic even extension of f is the triangular wave function and the 2-eriodic odd extension of f is the sawtooth function

FOURIER SERIES PART III: APPLICATIONS 7 Even eriodic extension of x 1 1 Odd eriodic extension of x 1 1 If we use the even extension we get the Fourier cosine reresentation of x with coefficients and for n 1, 1 a = 2 1 1 xdx = 1 a n = 2 x cos nπxdx = 2(( 1)n 1) 1 n 2 π 2 4 Since a 2j = and a 2j+1 = π 2, we get the Fourier cosine reresentation (2j + 1) 2 x over [, 1] as x = 1 2 4 π 2 j= cos[(2j + 1)πx] (2j + 1) 2. If we use the odd extension we get the Fourier sine reresentation of x with coefficients b n = 2 1 1 x sin nπxdx = 2( 1)n+1 nπ The Fourier sine reresentation of x over the interval [, 1) is x = 2 ( 1) n+1 sin(nπx). π n Examle 2. Find the Fourier cosine series of f(x) = sin x over the interval [, π]. We have a a 1 = 2 π = 2 π π π sin xdx = 4 π ; sin x cos xdx = 1 π To find a n with n 2, we use the identity π sin(2x)dx =. 2 sin x cos(nx) = sin(n + 1)x sin(n 1)x..

8 FOURIER SERIES PART III: APPLICATIONS a n π π = 2 sin x cos(nx)dx = 1 n π = 1 [ cos(n 1)x cos(n + 1)x π n 1 n + 1 = 2(( 1)n 1 1) π(n 2 1) We get the Fourier cosine of sin x on [, π] as sin x = 2 π + 2 π n=2 ( 1) n 1 1 n 2 1 (sin(n + 1)x sin(n 1)x) dx ] π cos(nx) = 2 π 4 π j=1 cos(2jx) (2j) 2 1. 4. Heat Conduction in a Rod Now we are in a osition to solve BVPs with more general nonhomogeneous terms than the ones considered in Note 4. Consider the following BVP for the temerature function u(x, t) in a rod of length L with initial temerature f(x) and with ends ket at temerature. u t = ku xx < x < L, t > u(, t) =, u(l, t) = t > u(x, ) = f(x) < x < L We can aly the method of searation of variables. The homogeneous art of the BVP is u t = ku xx, u(, t) =, u(l, t) = We have seen that solutions u(x, t) = X(x)T (t) (with searated variables) of the homogeneous art leads to the ODE roblems { X (x) + λx(x) = T (t) + kλt (t) =. X() = X(L) = The eigenvalues and eigenfunctions of the X-roblem (Sturm-Liouville roblem) are λ n = ν 2 n, X n (x) = sin(ν n x), where ν n = nπ L, n Z+ For each n Z +, the corresonding T -roblem has a solution T n (t) = e kν2 n t and a solution of the homogeneous art with searated variables is u n (x, t) = T n (t)x n (x). The rincile of suerosition imlies that any linear combination of these solutions is again a solution of the homogeneous art. Thus, u(x, t) = C n T n (t)x n (x) = C n e kν2 n t sin(ν n x) solves formally the homogeneous art of the BVP. At the oints (x, t) where the series converges and term by term differentiation (once in t and twice in x) is allowed, the function u(x, t) defined by the series is a true solution of the homogeneous art. This will be addressed shortly.

FOURIER SERIES PART III: APPLICATIONS 9 For now let us find the constants C n so that the formal solution solves also the nonhomogeneous condition u(x, ) = f(x). That is, we would like the constants C n so that u(x, ) = C n e kν2 n sin(ν n x) = f(x). Thus, after relacing ν n by nπ/l, we get f(x) = C n sin nπx L. This is the Fourier sine reresentation of the function f. Therefore the coefficients are given by C n = 2 L f(x) sin nπx L L dx, n Z+. Now we turn our attention to the series and verify that it indeed converges to a twice differentiable function u on < x < L and t > if f is iecewise smooth. For this we will use the Weierstrass M-test to rove uniform convergence. First, let M > be an uer bound of f (i.e. f(x) M for every x [, L]). We have C n 2 L L f(x) sin(ν n x) dx 2M. It follows that for a given t >, we have C n e kν2 n t sin(ν n x) 2Me kν2 n t, t t, x [, L] Since the numerical series n 2Me kν2 n t converges (use ratio or root tests), then it follows from the Weierstrass M-test that the series C n e kν2 n t sin(ν n x) converges uniformly on the set t t, x L. It follows at once that u is a continuous function. We can reeat the argument for the series giving u t and the series giving u xx. That is, the Weierstrass M-test shows that the series u t = ( kνn)c 2 n e kν2 n t sin(ν n x), and u xx = ( νn)c 2 n e kν2 n t sin(ν n x) converge uniformly on t t, x [, L]. We also have u t = ku xx. Consequently the function u(x, t) given by the above series satisfies the comlete BVP. Examle. Consider the BVP We have u t = u xx < x < π, t > u(, t) =, u(π, t) = t > u(x, ) = 1 < x < π π C n = 2 1 sin(nx)dx = 2(1 ( 1)n ) π πn The solution of the BVP is therefore u(x, t) = 2 1 ( 1) n e n2t sin(nx) π n

1 FOURIER SERIES PART III: APPLICATIONS 12 Temerature function u(x,t) at various times 1 t=.1 u axis (Temerature) 8 6 4 t=1 t=.5 t=.1 2 t=2.5 1 1.5 2 2.5 3 3.5 x axis (Rod) or equivalently u(x, t) = 4 π j= ex [ ((2j + 1) 2 t ] sin(2j + 1)x 2j + 1. 5. Wave Proagation in a String Consider the BVP for the vibrations of a string with fixed ends. u tt = c 2 u xx < x < L, t > u(, t) =, u(l, t) = t > u(x, ) = f(x) < x < L u t (x, ) = g(x) < x < L Thus u(x, t) reresents the vertical dislacement at time t of the oint x on the string. The initial osition and initial velocities of the string are given by the functions f(x) and g(x). The homogeneous art (HP) of the BVP is u tt = c 2 u xx, u(, t) =, u(l, t) = The solutions u(x, t) = X(x)T (t) (with searated variables) of the homogeneous art leads to the ODE roblems { X (x) + λx(x) = T (t) + c 2 λt (t) =. X() = X(L) = The eigenvalues and eigenfunctions of the X-roblem (SL roblem) are λ n = ν 2 n, X n (x) = sin(ν n x), where ν n = nπ L, n Z+ The corresonding T -roblem has two indeendent solutions T 1 n(t) = cos(cν n t) and T 2 n(t) = sin(cν n t). For each n Z +, we obtain solutions of (HP) with searated variables u 1 n(x, t) = Tn(t)X 1 n (x) = cos(cν n t) sin(ν n x) and u 2 n(x, t) = Tn(t)X 2 n (x) = sin(cν n t) sin(ν n x).

FOURIER SERIES PART III: APPLICATIONS 11 The rincile of suerosition imlies that any linear combination of these solutions is again a solution of (HP). Thus, u(x, t) = = A n Tn(t)X 1 n (x) + B n TnX 2 n (x) [A n cos(cν n t) + B n sin(cν n t)] sin(ν n x) is a formal solution of (HP). Now we use the nonhomogeneous conditions to find the coefficients A n and B n. First, we comute the (formal) derivative of u t u t (x, t) = [cν n B n cos(cν n t) cν n A n sin(cν n t)] sin(ν n x). The conditions u(x, ) = f(x) and u t (x, ) = g(x) lead to f(x) = A n sin(ν n x) = A n sin nπx and L cnπ g(x) = cν n B n sin(ν n x) = L B n sin nπx L. These are the Fourier series sine reresentations of f and g on the interval [, L]. Therefore A n = 2 L L f(x) sin nπx L dx and cnπ L B n = 2 L L g(x) sin nπx L dx By using criteria for uniform convergence of Fourier series (Proositions 1 and 2 of Note 7), it can be shown that if f is continuous and iecewise smooth and if g is iecewise smooth, then the series defining u is uniformly convergent and u(x, t) is a continuous function for t and x L. Moreover, we can show that if f, f, f, g, and g are continuous functions on [, L], the function u(x, t) defined by the infinite series is twice differentiable in (x, t) and term by term differentiations in the series are valid. This give u(x, t) as the (unique) solution of BVP. Remark 1. Many concrete roblems involve functions f that are only continuous and iecewise smooth. The series solution u is then only continuous. It is a continuous solution of the BVP. The roblem is understood in a more general sense: in the sense of distributions ( a notion of generalized functions that is beyond the scoe of this course). Remark 2. In concrete alication roblems, to overcome the lack of differentiability of the series solution u, we can to within any degree of accuracy ϵ, relace the functions f and g by their truncated Fourier series S N f and S N g so that f S N f < ϵ, g S N g < ϵ on [, L] The functions S N f and S N g are infinitely differentiable and the corresonding solution u N (the truncated series of u) is infinitely differentiable.

12 FOURIER SERIES PART III: APPLICATIONS Remark 3. By using the rincile of suerosition, this BVP could have been slit into two BVPs: BVP1 (lucked string) and BVP2 (struck string) v tt = c 2 v xx < x < L, t > v(, t) =, v(l, t) = t > v(x, ) = f(x) < x < L v t (x, ) = < x < L w tt = c 2 w xx < x < L, t > w(, t) =, w(l, t) = t > w(x, ) = < x < L w t (x, ) = g(x) < x < L The solutions to BVP1 and BVP2 are, resectively, v = w = A n cos(cν n t) sin(ν n x) B n sin(cν n t) sin(ν n x) The solution to the original BVP is u = v + w. Examle 1. (Plucked string) Consider the BVP u tt = 4v xx < x < 1, t > u(, t) =, u(l, 1) = t > u(x, ) = f(x) < x < 1 u t (x, ) = < x < 1 where 1 Initial osition of the string 5 1 f(x) = { x/5 if x 5 (1 x)/5 if5 x 1 For such an initial osition we have (B n = ) and A n = 2 1 1 5 = 1 25 = 8 π 2 n 2 sin nπ 2 f(x) sin nπx 1 dx x sin nπx 1 dx + 1 25 1 5 (1 x) sin nπx 1 dx

FOURIER SERIES PART III: APPLICATIONS 13 Hence, A 2j = and A 2j+1 = 8( 1)j π 2. The series solution is (2j + 1) 2 u(x, t) = 8 π 2 cos((2j + 1)t/5) ( 1)j sin((2j + 1)πx/1) (2j + 1) 2 j= = 8 ( cos(3t/5) sin(3x/1) π 2 cos(t/5) sin(x/1) + 9 cos(5t/5) sin(5x/1) cos(7t/5) sin(7x/1) + 25 49 ) + The individual comonents u n (x, t) = cos(nπt/5) sin(nπx/1) are called the harmonics or modes of vibrations. The function u n (x, t) is just a sine function in x being scaled by a cosine function in t with frequency n/1. First mode Second mode Third mode Figure 3. The first three modes of vibrations of the lucked string at various times. Examle 2. (Struck string) Consider the BVP u tt = 4v xx < x < 1, t > u(, t) =, u(l, 1) = t > u(x, ) = < x < 1 u t (x, ) = g(x) < x < 1 where g(x) = 1 if 4 < x < 6, and g(x) = elsewhere. This time A n =, and Thus B n = 5 π 2 n 2 The series solution is u(x, t) = 1 π 2 = 1 π 2 + 1 25 nπ 5 B n = 2 1 = 1 πn 1 ( g(x) sin nπx 1 dx cos 2nπ 5 cos 3nπ 5 ( cos 2nπ ) 3nπ cos = 1 5 5 π 2 n 2 sin nπ 2 1 n 2 sin nπ 2 ( sin π 1 sin πt 5 nπ nπt nπx sin sin sin 1 5 1 ) sin nπ 1 πx sin 1 1 3π 3πt 3πx sin sin sin 9 1 5 1 + 5π 5πt 5πx sin sin sin 1 5 1 1 7π 7πt 7πx sin sin sin 49 1 5 1 + )

14 FOURIER SERIES PART III: APPLICATIONS 6. Problems Dealing with the Lalace Equation Recall that the Dirichlet roblem in a rectangle is to find a harmonic function u inside the rectangle whose values on the boundary are given. That is u(x, y) = < x < L, < y < H u(x, ) = f 1 (x), u(x, H) = f 2 (x) < x < L u(, y) = g 1 (y), u(l, y) = g 2 (2) < y < H To solve this roblem, we use the rincile of suerosition to decomose it into four simler subroblems as in the figure. u 1 = u 1 = u 1 = u 2 =f 2 (x) u = 1 + u = 2 u = 2 + u = 2 u=f 2 (x) u 1 =f 1 (x) u 2 = u=g 1 (y) u= u=g 2 (y) = u=f 1 (x) u 3 = u 4 = u 3 =g 1 (y) u 3 = u 3 = + u 4 = u 4 = u 4 =g 2 (y) u 3 = u 4 = We can find the solution u(x, y) as u(x, y) = u 1 (x, y) + u 2 (x, y) + u 3 (x, y) + u 4 (x, y). Each of the subroblems can be solved by the method of searation of variables. Now we indicate how to find u 1 (x, y). The solutions with searated variables u 1 (x, y) = X(x)Y (y) of the homogeneous art leads to the ODE roblems { { X (x) + λx(x) = Y and (y) λy (y) = X() = X(L) = Y (H) = where λ is the searation constant. The X-roblem is an SL-roblem whose eigenvalues and eigenfunctions are λ n = ν 2 n, X n (x) = sin(ν n x), where ν n = nπ L, n Z+. For each λ n, the corresonding ODE for the Y -roblem has general solution Y n = A cosh(ν n y) + B sinh(ν n y). The boundary condition Y (H) =, imlies that (u to a multilicative constant), the solution of the Y -roblem is Y n (y) = sinh [ν n (H y)].

FOURIER SERIES PART III: APPLICATIONS 15 Hence, the solutions with searated variables of the homogeneous art of the u 1 - roblem are generated by u 1,n (x, y) = sinh [ν n (H y)] sin(ν n x) n Z +. A series solution of the u 1 -roblem is therefore u 1 (x, y) = C n sinh [ν n (H y)] sin(ν n x). Such a solution solves the nonhomogeneous condition u(x, ) = f 1 (x) if and only if f 1 (x) = C n sinh(ν n H) sin(ν n x) = C n sinh nπh nπx sin L L. Thus, C n sinh(ν n H) is the n-th Fourier sine coefficient of f 1 over [, L]: C n sinh nπh L = 2 L L f 1 (x) sin nπx L dx. The functions u 2, u 3, and u 4 can be found in a similar way. Examle 1. Consider the following BVP with mixed boundary conditions. u(x, y) = < x < π, < y < 2π u(x, ) = x, u(x, 2π) = < x < π u x (, y) =, u x (π, y) = 1 < y < 2π We decomose the roblem as shown in the figure (see next age). u= v= w= u x = u= u x =1 = v x = v= v x = + w x = w= w x =1 u=x v=x w= The method of searation of variables for the v-roblem leads to the ODE roblems { { X (x) + λx(x) = Y X () = X and (y) λy (y) = (π) = Y (2π) = where λ is the searation constant. The X-roblem is an SL-roblem whose eigenvalues and eigenfunctions are λ =, X (x) = 1 and for n Z + λ n = n 2, X n (x) = cos(nx). For λ =, the general solution of the ODE for the Y -roblem is Y (y) = A + By and in order to get Y (2π) =, we need A = 2Bπ. Thus Y (y) = (2π y) generates the solutions of the Y -roblem For λ n = n 2, the solutions of the Y -roblem are generated by Y n (y) = sinh [n(2π y)].

16 FOURIER SERIES PART III: APPLICATIONS The solutions with searated variables of the homogeneous art of the v-roblem are therefore v (x, y) = 2π y and v n (x, y) = sinh [n(2π y)] cos(nx) for n Z + The series solution is v(x, y) = C (2π y) + C n sinh [n(2π y)] cos(nx). In order for such a series to solve the nonhomogeneous condition v(x, ) = x, we need to have x = 2πC + C n sinh(2nπ) cos(nx). This is the Fourier cosine exansion of x over [, π]. Hence, and for n 1 This gives 2πC = 2 π sinh(2nπ)c n = 2 π π π C 2j = and C 2j+1 = The solution of the v-roblem is v(x, y) = 2π y 4 4 π j= xdx = π C = 1 2 x cos(nx)dx = 2 (( 1)n 1) πn 2. 4 π(2j + 1) 2 sinh [2π(2j + 1)]. sinh [(2j + 1)(2π y)] cos(2j + 1)x sinh [2(2j + 1)π] (2j + 1) 2. Now we solve the w-roblem. The searation of variables for the homogeneous art leads to the ODE roblems. { { X (x) λx(x) = Y X and (y) + λy (y) = () = Y () = Y (2π) = This time it is the Y -roblem that is a Sturm-Liouville roblem with eigenvalues and eigenfunctions λ n = n2 2 2, Y n(y) = sin ny 2, n Z+. For each n, a generator of the solutions of the X-roblem is X n (x) = cosh nx 2. The series solution of the w-roblem is therefore w(x, y) = C n cosh nx ny sin 2 2. To find the coefficients C n so that w x (π, y) 1, we need n w x (x, y) = 2 C n sinh nx ny sin 2 2

FOURIER SERIES PART III: APPLICATIONS 17 This gives 1 = n 2 C n sinh nπ 2 sin ny 2 (the Fourier sine exansion of 1 over the interval [, 2π]): Equivalently, n 2 C n sinh nπ 2 = 2 2π C 2j =, C 2j+1 = 2π sin ny 2 dy = 2 (1 ( 1)n ). nπ 8 π(2j + 1) 2 sinh [(2j + 1)π/2]. The solution of the w-roblem is therefore w(x, y) = 8 cosh [(2j + 1)x/2] sin [(2j + 1)y/2] π sinh [(2j + 1)π/2] (2j + 1) 2 j= The solution u of the original roblem is u(x, y) = v(x, y) + w(x, y). Examle 2. Consider the Dirichlet roblem in a disk (written in olar coordinates) We take the function f to be given by u(r, θ) = r < 1, θ [, 2π] u(1, θ) = f(θ) θ [, 2π] u(1,θ )=f(θ) u= f(θ) = { sin θ if θ π if π θ 2π Recall that the Lalace oerator in olar coordinates is = 2 r 2 + 1 r r + 1 2 r 2 θ 2 A solution with searated variables u(r, θ) = R(r)Θ(θ) leads to the ODEs Θ (θ) + λθ(θ) = and r 2 R (r) + rr (r) λr(r) = where λ is the searation constant. Note that since u(r, θ + 2π) = u(r, θ), then the function Θ and also Θ need to be 2π-eriodic. Thus to the ODE for Θ we need to add Θ() = Θ(2π) and Θ () = Θ (2π). Hence, the Θ-roblem is a eriodic SL-roblem whose eigenvalues and eigenfunctions are λ =, Θ (θ) = 1,

18 FOURIER SERIES PART III: APPLICATIONS and for n Z +, λ n = n 2, Θ 1 n(θ) = cos(nθ), Θ 2 n(θ) = sin(nθ). The ODE for the R-function is a Cauchy-Euler equation with characteristic equation m 2 λ =. For λ = λ =, the general solution of the R-equation are generated by R(r) 1 = 1 and R(r) 2 = ln r. For λ = λ n (we have m = ±n), the solutions of the R-equation are generated by R 1 n(r) = r n and R 2 n(r) = r n. The solutions with searated variables of the Lalace equation u = in the disk are therefore 1, ln r, r n cos(nθ), r n sin(nθ), r n cos(nθ), r n sin(nθ). Since we looking for solutions that are bounded in the disk and since ln r and r n cos(nθ) and r n sin(nθ) are not bounded, then we will discard then when forming u The series solution has the form u(r, θ) = A + (A n r n cos(nθ) + B n r n sin(nθ)) The initial condition u(1, θ) = f(θ) leads to f(θ) = A + A n cos(nθ) + B n sin(nθ). This is the Fourier series of f. I leave it as an exercise for you to verify that the Fourier series of f is f(θ) = 1 π + 1 2 sin θ 2 cos(2jθ) π 4j 2 1. The solution of the Dirichlet roblem is u(r, θ) = 1 π + r 2 sin θ 2 π j=1 j=1 7. Exercises r 2j cos(2jθ) 4j 2 1 Exercise 1. (a) Find the Fourier series of the function with eriod 4 that is defined over [ 2, 2] by f(x) = 4 x2. 2 (b) Use Parseval s equality to evaluate the series 1 n 4. (c) Use the integral test to estimate the mean square error E N when relacing f by its truncated Fourier series S N f. (d) Find N so that E N.1 and then find N so that E N.1 Exercise 2. (a) Find the Fourier series of the function with eriod 4 that is defined over [ 2, 2] by { 1 x if x 2 f(x) = 1 + x if 2 x

FOURIER SERIES PART III: APPLICATIONS 19 (b) Use Parseval s equality to evaluate the series j= 1 (2j + 1) 4. (c) Use the integral test to estimate the mean square error E N when relacing f by its truncated Fourier series S N f. (d) Find N so that E N.1 and then find N so that E N.1 Exercise 3. Find the Fourier sine series of f(x) = cos x over [, π] (What is the Fourier cosine series of cos x on [, π]?) Exercise 4. Find the Fourier cosine series of f(x) = sin x over [, π] (What is the Fourier sine series of sin x on [, π]?) Exercise 5. Find the Fourier cosine series of f(x) = x 2 over [, 1]. Exercise 6. Find the Fourier sine series of f(x) = x 2 over [, 1]. Exercise 7. Find the Fourier cosine series of f(x) = x sin x over [, π]. Exercise 8. Find the Fourier sine series of f(x) = x sin x over [, π]. Exercise 9. Solve the BVP u t = u xx, < x < 2, t > u(, t) = u(2, t) =, t > u(x, ) = f(x), < x < 2 where f(x) = { 1 if < x < 1 if 1 < x < 2 Exercise 1. Solve the BVP u t = u xx, < x < 2, t > u(, t) = u(2, t) =, t > u(x, ) = cos(πx), < x < 2 Exercise 11. Solve the BVP u t + u = (.1)u xx, < x < π, t > u x (, t) = u x (π, t) =, t > u(x, ) = sin x, < x < 2 Exercise 12. Consider the BVP modeling heat roagation in a rod where the end oints are ket at constant temeratures T 1 and T 2 : u t = ku xx, < x < L, t > u(, t) = T 1, u(l, t) = T 2, t > u(x, ) = f(x), < x < L Since T 1 and T 2 are not necessarily zero, we cannot aly directly the method of eigenfunctions exansion. To solve such a roblem, we can roceed as follows. 1. Find a function α(x) (indeendent on time t) so that α (x) =, α() = T 1 α(l) = T 2.

2 FOURIER SERIES PART III: APPLICATIONS 2. Let v(x, t) = u(x, t) α(x). Verify that if u(x, t) solves the given BVP, then v(x, t) solves the following roblem v t = kv xx, < x < L, t > v(, t) =, v(l, t) =, t > v(x, ) = f(x) α(x), < x < L The v-roblem can be solved by the method of searation of variables. The solution u of the original roblem is therefore u(x, t) = v(x, t) + α(x). Exercise 13. Aly the method of described in Exercise 12 to solve the roblem u t = u xx, < x < 2, t > u(, t) = T 1, u(2, t) = T 2, t > u(x, ) = f(x), < x < 2 in the following cases 1. T 1 = 1, T 2 =, f(x) =. 2. T 1 = 1, T 2 = 1, f(x) =. 3. T 1 =, T 2 = 1, f(x) = 5x. In roblems 14 to 16, solve the wave roagation roblem u tt = c 2 u xx, < x < L, t > u(, t) =, u(l, t) =, t > u(x, ) = f(x), u t (x, ) = g(x) < x < L { x if < x < 1 Exercise 14. c = 1, L = 2, f(x) =, g(x) = 2 x if 1 < x < 2 { x if < x < 1 Exercise 15. c = 1/π, L = 2, f(x) = sin x, g(x) = 2 x if 1 < x < 2 Exercise 16. c = 2, L = π, f(x) = x sin x, g(x) = sin(2x). In exercises 17 to 19, solve the wave roagation roblem with daming u tt + 2au t = c 2 u xx, < x < L, t > u(, t) =, u(l, t) =, t > u(x, ) = f(x), u t (x, ) = g(x) < x < L Exercise 17. c = 1, a =.5, L = π, f(x) =, g(x) = x Exercise 18. c = 4, a = π, L = 1, f(x) = x(1 x), g(x) =. Exercise 19. c = 1, a = π/6, L = 2, f(x) = x sin(πx), g(x) = 1. In exercises 2 to 22, solve the Lalace equation u(x, y) = inside the rectangle < x < L, < y < H subject the the given boundary conditions. Exercise 2. L = H = π, u(x, ) = x(π x), u(x, π) =, u(, y) = u(π, y) =. Exercise 21. L = π, H = 2π, u(x, ) =, u(x, 2π) = x, u x (, y) = sin y, u x (π, y) =. Exercise 22. L = H = 1, u(x, ) = u(x, 1) =, u(, y) = 1, u(1, y) = sin y. Exercise 23. Solve the Lalace equation u(r, θ) = inside the semicircle of radius 2 ( < r < 2, < θ < π ) subject to the boundary conditions u(r, ) = u(r, π) = ( < r < 2) and u(2, θ) = θ(π θ) ( < θ < π)

FOURIER SERIES PART III: APPLICATIONS 21 Exercise 24. Solve the Lalace equation u(r, θ) = inside the semicircle of radius 2 ( < r < 2, < θ < π ) subject to the boundary conditions u θ (r, ) = u θ (r, π) = ( < r < 2) and u(2, θ) = θ(π θ) ( < θ < π) Exercise 25. Solve the Lalace equation u(r, θ) = inside the quarter of a circle of radius 2 ( < r < 2, < θ < π/2 ) subject to the boundary conditions u(r, ) = u(r, π/2) = ( < r < 2) and u(2, θ) = θ ( < θ < π/2)