1. Quiz 4
(1) Solve the following Dirichlet boundary value problems.
(1.1) ∆f = 0 on B(0, 1)
f |S1 = g
Here g is the following given periodic functions on R.
(a) g(θ) = 1 + 2 cos θ − 3 sin θ − 5 cos 2θ + sin 2θ
(b) g(θ) = 1 + a1cos θ + a2sin θ + a3cos θ sin θ + a4sin2θ + a5cos2θ + a6sin3θ + a7sin3θ.
(c) g(θ) = a0+PN
n=1(ancosnθ + bnsinnθ). (You do not need to solve this one in the discussion session. Think!)
Express f as a linear combination of {1, Cn(x, y), Sn(x, y), n ≥ 1}, where Cn(x, y) = Re(x + iy)n and Sn(x, y) = Im(x + iy)n.
(2) Let f : B(0, R) → R be a function. Here B(0, R) = {(x, y) ∈ R2: x2+ y2 < R2} for R > 0. Suppose that f ∈ C2(B(0, R)) and (x, y) ∈ B(0, R).
(a) Define a function
F (t) = f (tx, ty), |t| < 1.
Compute F0(t).
(b) Show that
f (x, y) = f (0, 0) + x Z 1
0
fx(tx, ty)dt + y Z 1
0
fy(tx, ty)dt.
(c) Show that there exist θ, ξ ∈ [0, 1] such that
f (x, y) = f (0, 0) + fx(θx, θy)x + fy(ξx, ξy)y.
(d) Show that
fx(tx, ty) = fx(0, 0) + tx Z 1
0
fxx(tsx, tsy)ds + ty Z 1
0
fxy(tsx, tsy)ds fy(tx, ty) = fy(0, 0) + tx
Z 1 0
fyx(tsx, tsy)ds + ty Z 1
0
fyy(tsx, tsy)ds.
(e) Using fxy = fyx on U, to prove that
f (x, y) = f (0, 0) + fx(0, 0)x + fy(0, 0)y + x2
Z 1 0
Z 1 0
fxx(tsx, tsy)tdtds + 2xy
Z 1 0
Z 1 0
fxy(tsx, tsy)dtds + y2
Z 1 0
Z 1 0
fyy(tsx, tsy)tdtds.
(f) Assume that f ∈ C3(B(0, R)). We may do the above process further. For example,
fxx(tsx, tsy) = fxx(0, 0) + tsx Z 1
0
fxxx(tsux)du + tsy Z 1
0
fxxy(tsux, tsuy)du.
1
2
Notice that
Z 1 0
Z 1 0
tdtds = 1 2!
and so forth. Find R2(f )(x, y) such that f (x, y) = f (0, 0) + 1
1!(fx(0, 0)x + fy(0, 0)y) + 1
2!(fxx(0, 0)x2+ 2fxy(0, 0)xy + fyy(0, 0)y2) + R2(f )(x, y).
(g) Let
H0(f )(x, y) = f (0, 0)
H1(f )(x, y) = fx(0, 0)x + fy(0, 0)y
H2(f )(x, y) = fxx(0, 0)x2+ 2fxy(0, 0)xy + fyy(0, 0)y2. If f ∈ C3, we can find a continuous function R2(f ) such that
f (x, y) =
2
X
j=0
1
j!Hj(f )(x, y) + R2(f )(x, y).
If f ∈ C4, using the same precess, we obtain a polynomial H3(f )(x, y) of degree 3 and a function R3(f )(x, y) such that
f (x, y) =
3
X
j=0
1
j!Hj(f )(x, y) + R3(f )(x, y).
Find H3(f )(x, y). (You do not need to do the process again. You only need to guess the formula.) In general, for f ∈ Ck+1, find (guess) the formula for Hk(f )(x, y) such that
f (x, y) =
k
X
j=0
1
j!Hj(f )(x, y) + Rk(f )(x, y).
Remark. The function Hk(f )(x, y) in your answer must be a (homogeneous) polynomial of degree k. A polynomial H(x, y) is said to be homogenous of degree k if H(λx, λy) = λkH(x, y) for any λ > 0.
