MATH 220 solution to homework 4

Transcription

1 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 the domain U T,L = {(t, x) : t T, L x L} where T and L are some positive number and the auxiliary function w(t, x) = v(t, x) F M L 2 (2āt + x2 ), where ā = > a(x), F = sup y R f(y), and M = sup t,x R v(t, x) + F (since u(t, x) is a bounded solution, M is a well defined real number). We have w t (t, x) a(x + w bt) 2 x 2 (t, x) = 2M (ā a(x + bt)) <, L2 (t, w(, x), x ( L, L), w(t, ±L), t [, T ]. x) U T,L, Now we show that w(t, x) must attain its maximum over U T,L on the parabolic boundary Γ T,L = {(t, x) : t = L < x < L t T x = ±L} (This is similar to Theorem.6 in the notes): Suppose w(t, x) attains its maximum at (t, x ) U T,L \ Γ T,L, then w t (t, x ), = 2 w x 2 (t, x ), w t (t, x ) a(x + bt ) 2 w x 2 (t, x ), which contradicts w t (t, x ) a(x + bt ) 2 w x 2 (t, x ) <. Thus w must attain its maximum on Γ T,L. We also note that w(t, x) for x Γ T,L. Thus we have w(t, x), x U T,L. In other words, v(t, x) F + M L 2 (2āt + x2 ), t [, T ], x [ L, L].

2 Now, for any point (t, x ) R + R, we want to show that v(t, x ) F. Fist we set T = t, and then we choose L > x. Then we have v(t, x ) F + M L 2 (2āt + x 2 ). Now, note that the inequality holds for any L > x. Let L and we have v(t, x ) F. Problem 2. (i) By Fourier transform we have û t (t, k) + 4π2 k 2 û(t, k) =, t >, k R, û(, k) = ˆf(k). Solve the equation we have û(t, k) = e 4π2 k 2 t ˆf(k), u(t, x) = û(t, k)e 2πikx dk = e 2πikx 4π2 k 2 t ˆf(k) dk. (ii) u(t, x) = e 2πikx 4π2 k 2 t ˆf(k) dk = e 2πikx 4π2 k 2 t e 2πiky f(y) dy dk ( ) = e 2πik(x y) 4π2 k 2t dk f(y) dy = ( ) e πξ2 e 2πiξ((x y)/ 4πt) dξ f(y) dy 4πt = e π((x y)/ 4πt) 2 f(y) dy 4πt = G(t, x y)f(y) dy (ξ = 4πtk) Problem 3. (i) ˆf = ˆf n = ( x) dx =. 2 e 2πinx ( 2 x) dx = ( e 2πinx 4πin 2 + xe 2πinx 2πin 4π 2 n 2 ) e 2πinx x= x= =, n. 2πin

3 (ii) N (S N f) (x) = ( k= N N = e 2πikx k= N = D N (x), N ˆf k e 2πikx ) = (2πik ˆf k )e 2πikx k= N and S N f() = N ˆf k= N k =, thus S N f(x) = (D N (t) ) dt = sin((2n + )πt) sin(πt) dt x. (iii) S N f(x) S N f(x) = Define g(z) = sin(z), we have z ( sin(πt) ) sin((2n + )πt) dt. πt g(z) = sin(z) z = z z 3 /3! + z 5 /5! z 7 /7! +... z = z2 /3! z 4 /5! + z 6 /7! +... z z 3 /3! + z 5 /5! z 7 /7! +... = z/3! z3 /5! + z 5 /7! +... z 2 /3! + z 4 /5! z 6 /7! +... so z = is a removable singular point. We define g() =, then g(z) is a C function on [, π/4], then then h(t) := sin(πt) πt C [, /4], x [, /4], ( sin(πt) ) sin((2n + )πt) dt πt = h(t) sin((2n + )πt) dt t=x cos((2n + )πt) x = h(t) h (t) cos((2n + )πt) (2N + )π + dt (2N + )π C + C 2 (2N + )π, t= 3

4 where C = 2 sup t [,/4] h(t), C 2 = /4 h (t) dt. Define C = C + C 2, then h(t) sin((2n + )πt) dt C/N, x [, /4]. (iv) /(2N) S N f(x N ) = /(2N) + = /(2N) + π = lim S N f(x N ) N ( = lim N = π π (2N+)π/(2N) /(2N) + π sin(y) y (v) Consider the function r(x) := π x π dy. sin((2n + )πt) πt sin(y) y (2N+)π/(2N) sin(x), p(x) =: x. What we want to prove is that π π Obviously, we have p(x) dx > /2. π π dy sin(y) y dt (y = (2N + )πt) dy ) r(x) dx = /2 we just need to prove p(x) r(x) (which is obvious) and p(x) r(x) ( x [, π]): First, we have p(x) r(x) = sin(x) x π x π = π sin(x) x(π x). πx Now we consider the function q(x) := π sin(x) x(π x). We need to show that q(x), x [, π]. Because of the symmetry of q(x) (which means that q(x) = q(π x), x [, π]), we 4

5 just need to show q(x), x x [, π/2]. Note that q() =, q (x) = π cos(x) π + 2x, q (x) is concave on [, π/2], and q () = q (π/2) =, we have q (x), x [, π/2], q(x), x [, π/2], q(x), x [, π], p(x) r(x), x [, π], S > /2. This means, as N goes to infinity, there exists x N = /(2N), which goes to, such that S N f(x N ) goes to some constant which is strictly greater than /2. But we want to use S N f(x) to approximate f(x), and we know that f(x) /2. So the sequence S N f(x) does not converge to f(x) uniformly. This is mainly because f(x) is not continuous at x = (after we extend f periodically). This is the Gibbs phenomenon. If we want to use S N f(x) as an approximation of f(x), then near x =, we need N to be large to get a relatively good approximation. Problem 4. (i) From Problem 3.(ii) we know S N f(x) = N k= N a k e 2π ikx sin((2n + )πt) = x + dt. sin(πt) It s obvious that the first term x is bounded for x [, ]. We only need to show that the second term is bounded. sin((2n + )πt) We note that the function h(t) := is even w.r.t. x = /2, i.e., h(x) = h( x) sin(πt) for any x [, ], thus it suffices to show that sin((2n + )πt) dt sin(πt) M, x [, /2]. To show this, we notice that the denominator of h(t) is a positive increasing function on (, /2] and the numerator of h(t) is a periodic function which oscillates with period O(/N). If we split the interval [, x] into smaller chunks: [, /(2N + )], [/(2N + ), 2/(2N + )],..., [k/(2n + ), x], and we compute the integral on each chunk as b j = j/(2n+) (j )/(2N+) h(t) dt, then the integral over [, x] can be written as sin((2n + )πt) sin(πt) dt = b b 2 + b 3 b ( ) k b k + ( ) k b k+, 5

6 where b j is a non-negative sequence and b b 2 b 3 b k b k+. The point of doing this is to rewrite the integral as an alternating sum of the sequence {b j }. Then we see that the integral can be bounded by b due to the alternating property. We also have that b = /(2N+) h(t) dt. Thus the integral is bounded. (ii) We define g k (x) := e 2π in kx B k(x) k 2, then from (i) we know that g k (x) M/k 2. Since each g k (x) is continuous and the dominating series k= M/k2 has finite sum, we know that f(x) is continuous. For the Fourier coefficients, we expand the expression for f(x) as f(x) = k= m k j= m k a j k 2 e2π i(n k+j)x. Since N k + m k < N k+ m k+, we see that the terms for different k in the above expansion do not share the same exponentials. If we write the Fourier expansion of f(x) as f(x) = l= c le 2π ilx, then by comparing the coefficients we have a j c l = k 2, if l can be written as l = N k + j for some k N + and m k j m k,, otherwise. (iii) For S Nk f(x), we only need to keep the terms in the Fourier expansion where N k l N k. That gives k S Nk f(x) = e 2π inpx B p(x) p 2 p= Note that B p () = and we have S Nk f() = k 2 a j = k 2 j= m k + e2π inkx k 2 2π ij = j= m k j= m k a j e 2π ijx. i 2πk 2 Now the harmonic series m k j= /j is bounded below by C log(m k) for some C >, thus S Nk f() C log(m k) k 2. 6 m k j= j.

7 (iv) We set m k = e k3 and we define N k recursively by N = m +, N k+ = m k + N k + m k+ +, k =, 2,..., Then N k+ m k+ > N k + m k is satisfied and log(m k )/k 2 k, whose sum k= k diverges to, which means the Fourier series of f(x) at x = diverges. Problem 5. We have v xx + v( v 2 ) =. Multiply by v x on both sides we have v x v xx + v( v 2 )v x =, (v 2 x/2) x + (v 2 /2) x (v 4 /4) x =, v 2 x + v 2 v 4 /2 = C, v 2 x = ( v 2 ) 2 /2 + C 2, where C 2 = C /2. Note that v( ) =, v(+ ) = and < v(x) <, we have lim v x(x) 2 = C 2. x But v(+ ) = exits, so C 2 =, and v x = ±( v 2 )/ 2. Since v(+ ) > v( ), we should choose the branch v x = ( v 2 )/ 2. Then we have dv v 2 = dx/ 2, log + v v = 2(x C 3 ), v(x) = e 2(x C 3) e 2(x C 3) + = tanh where C 3 is an arbitrary number. If the equation is εv xx + v( v 2 ) =, then by a similar calculation we have ( ) x C3 v(x) = tanh. 2ε ( ) x C3, 2 7

8 As ε, we have, x < C 3, v(x) sgn(x C 3 ) =, if x = C 3,, x > C 3. This means, as ε, we have a jump at the point x = C 3. 8