LECTURE 21. DISCUSSION OF MIDTERM EXAM. θ [0, 2π). f(θ) = π θ 2

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1 LECTURE. DISCUSSION OF MIDTERM EXAM FOURIER ANALYSIS (.443) PROF. QIAO ZHANG Problem. Cosider the itegrable -periodic fuctio f(θ) = θ θ [, ). () Compute the Fourier series for f(θ). () Discuss the covergece of the above Fourier series. More precisely, determie at which poit(s) θ the Fourier series coverges to f(θ), at which poit(s) it does ot, ad over which iterval(s) this covergece is uiform. Justify your aswers. Solutio. By defiitio, we have ˆf() If =, the we have ˆf() 4 If, the we have ˆf() = θ e iθ dθ 4 dθ e iθ dθ θe iθ dθ. θ dθ = () =. θe iθ dθ [ θe iθ i i () i. Hece we have the Fourier series f(θ) ˆf()e iθ = e iθ i = = si(θ). ] e iθ dθ We ote f(θ) is ot cotiuous at the poits Z ad is of class C at ay other poits. Further, the Fourier series evaluated at θ obviously is equal to, while f(θ) =. Hece the Fourier series for f(θ) coverges to f(θ) for every θ Z, ad this covergece is uiform over every closed iterval disjoit with Z. Remark.. A commo mistake i Problem (a) is to overlook the differet treatmets for = ad. This ca happe to eve experts, ad ca lead to disastrous results, so be careful!

2 FOURIER ANALYSIS (.443) PROF. QIAO ZHANG Problem. Keep cosiderig the itegrable -periodic fuctio f(θ) = θ Also, cosider the itegrable -periodic fuctio g(θ) = θ θ [, ). θ [, ). () Compute the covolutio h(θ) = (f g)(θ). () Compute the Fourier series for h(θ). Solutio. By periodicity, it suffices for us to compute h(θ) for θ [, ). Sice g(θ) = g(θ + ) = g(θ) = g(θ ) = for θ < we have h(θ) θ+ θ+ = 3θ, 4 ad for θ < we have Hece h(θ) ĥ() = θ θ (θ + ) (θ ) f(x) = + θ = 3 θ + θ + x dx f(x) + θ [ 3, ), θ [, 3), θ+ 3 θ + x dx + = 3θ θ +. 4 h(θ)e iθ dθ 3θ 4, if = ; ( ) 4, if, e iθ dθ θ+ + θ x dx θ θ h(θ)e iθ dθ + θ + x dx h(θ)e iθ dθ 3θ θ + e iθ dθ 4

3 LECTURE. DISCUSSION OF MIDTERM EXAM 3 ad we have h(θ) ( ) 4 eiθ = ( ) = cos(θ). Remark.. Alteratively, we ca compute ĥ() without determiig h(θ) first. I fact, sice h = f g, we have ĥ() = ˆf()ĝ(). Here ˆf() has bee obtaied i Problem (a), ad we have (this ca be also obtaied by observig g(θ) = f(θ + ) + ; how?) ĝ() θ e iθ dθ =, if = ; ( ) i, if. Hece ĥ() = ˆf()ĝ(), if = ; = ( ) 4, if. You ca stop here, or you may observe that ˆf() <, so the Fourier series ĥ()e iθ = ( ) 4 eiθ = ( ) cos(θ) defies a cotiuous fuctio ad shares the same Fourier coefficiet with h. Sice h, as a covolutio, is also cotiuous everywhere, they must be equal to each other, ad this gives h(θ) = ( ) 4 eiθ = = ( ) cos(θ). Problem 3. Isist o cosiderig the itegrable -periodic fuctio f(θ) = θ θ [, ). () State the requiremets for good kerels, ad verify whether the family of fuctios f(θ), f(θ), f(3θ),... } is a family of good kerels. ( ) () Compute the Abel sum for the ifiite series. Solutio. A family of itegrable -periodic fuctios K } is called a family of good kerels if () () K (x) dx ; = = K (x) dx is uiformly bouded for ; (3) for every δ > we have lim δ< x < K (x) dx =. Yes, this is a facy way for sayig that there exists a absolute costat such that K (x) dx < M.

4 4 FOURIER ANALYSIS (.443) PROF. QIAO ZHANG The family f(θ), f(θ),... } is ot a family of good kerels, as As for the Abel sum, write f(x) dx A(r) = = ( ) r. dx =. Obviously A(r) is absolutely coverget for r <, ad A (r) = ( ) r = ( r) = + r, = = A(r) = r A (t) dt + A() = Hece the Abel sum for ( ) = lim A r = r is lim r r dt = l( + r). + t l( + r) = l. Remark.3. How about the Abel sum for i my mid. = ( )? This is what was origially + Problem 4. Use Fourier series to solve the followig wave equatio u t = u x, u(x, ) = si(x), u(x, ) = si(x) ( x ). t Solutio. We recall that our classical discussio for the wave equatio is always for the period, so we have to make a chage of variables X = x, T = t to get U T = U X, U(X, ) = si(x), U(X, ) si(4x) ( X ). T Hece U(X, T ) = ( c cos(t ) + d ) si(t ) si(x), where c ad d are the Fourier coefficiets (i terms of the sie fuctio) for si(x) ad si(4x) respectively, so, if = ; c = d =, if = 4;, if,, if 4. Hece U(X, T ) = cos(t ) si(x) + u(x, t) = cos(t) si(x) + si(4t ) si(4x), si(t) si(x).

5 LECTURE. DISCUSSION OF MIDTERM EXAM 5 Remark.4. The aswer may be double checked applyig d Alembert s formula si((x + t)) + si((x t)) u(x, t) = + x+t x t si(y) dy cos((x + t)) cos((x t)) = cos(t) si(x) 8 si(t) si(x) = cos(t) si(x) +. However, sice the problem has specified to use Fourier series to solve the followig wave equatio, we caot use this approach as the official solutio to the problem. Problem 5. Let f be a itegrable -periodic fuctio. Show that Proof. We have f + f(x) dx f = f S N (f) + S N (f). f(x) S N (f)(f)(x) dx + (f(x) S N (f)(x)) + S N (f)(x) dx (f(x) S N (f)(f)(x))s N (f)(x) dx = f S N (f) + S N (f) + = f S N (f) + N = N N = N S N (f)(x) dx ( ˆf() f(x)e ix dx ) S N (f)(x)e ix dx ˆf(). Remark.5. Some studets applied the abstract theory of ier products ad orthoormal bases. This approach is also fie, but the they showed some isufficiet uderstadig i their argumets, just copyig certai seteces ad termiologies from memory, ad this is ot fie. We should always kow very clearly what we are talkig about!