75042 Partial differential equations

Problem set #2 — suggested solutions

The first two problems this week were picked from the notes entitled “Elements of analysis”.

Number 24: To show that a function f is not uniformly continuous, we must show that for some ε > 0 and any δ > 0 there exists x and y with |y − x| < δ but |f(y) − f(x)| > ε.

First, y^{2} − x^{2} = (x + y)(y − x). Given ε > 0 and δ > 0, let x = ε/δ and y = x + δ/2. Then y^{2} − x^{2} > 2x · δ/2 = ε.

Similarly, 1/x − 1/y = (y − x)/(xy). Given ε > 0 and δ > 0, let y = 2x where x > 0 is yet to be determined. Then y − x = x and 1/x − 1/y = 1/(2x). If 0 < x < min(δ, 1/(2ε)) we get y − x < δ and 1/x − 1/y > ε.

For any differentiable real function f on some interval, the secant theorem states f(y) − f(x) = f′(ξ) · (y − x) for some ξ between x and y. In particular |f(y) − f(x)| ≤ sup_{ξ} |f′(ξ)| · |y − x|. When f(x) = arctan x we have |f′(ξ)| = 1/(1 + ξ^{2}) ≤ 1, so | arctan y − arctan x| ≤ |y − x|. Thus arctan is actually Lipschitz continuous, and therefore certainly uniformly continuous.

Number 27: Assume that, for every ε > 0, there is some δ > 0 so that

|\frac{∂f}{∂x} (x, y) − \frac{∂f}{∂x} (x_{0}, y)| < εg(y)

where ∫_{−∞}^{∞} g(x, y) dy < ∞. Let F (x) = ∫_{−∞}^{∞} f(x, y) dy. Then as in the proof of Proposition 26, we get

|\frac{1}{x − x_{0}} (F (x) − F (x_{0})) − ∫_{−∞}^{∞} \frac{∂f}{∂x} (x_{0}, y) dy| = |\frac{1}{x − x_{0}} ∫_{−∞}^{∞} ∫_{x_{0}}^{x} (\frac{∂f}{∂x} (ξ, y) − \frac{∂f}{∂x} (x_{0}, y)) dξ dy|

≤ \frac{1}{|x − x_{0}|} ∫_{−∞}^{∞} ∫_{x_{0}}^{x} |\frac{∂f}{∂x} (ξ, y) − \frac{∂f}{∂x} (x_{0}, y)| dξ dy

< \frac{1}{|x − x_{0}|} ∫_{−∞}^{∞} ∫_{x_{0}}^{x} εg(y) dξ dy

= ε ∫_{−∞}^{∞} g(y) dy

and since the integral is just a finite constant, the desired differentiability result follows. Of course, in order for the above calculations to be correct, we also had to assume that the improper integrals ∫_{−∞}^{∞} f(x, y) dy and ∫_{−∞}^{∞} f_{y}(x, y) dy both converge.

Laplacian in the disk: An unusual but quick way to derive the Laplacian in polar coordinates, is the following: Apply the divergence theorem to Du in the region r_{1} < r < r_{2}, θ_{1} < θ < θ_{2}:

∫∫ _{R} Δu dxdy = ∫_{r_{1}}^{r_{2}} \frac{−u_{θ}(r, θ_{1})}{r} dr + ∫_{θ_{1}}^{θ_{2}} u_{r}(r_{2}, θ) r_{2}dθ + ∫_{r_{1}}^{r_{2}} \frac{u_{θ}(r, θ_{2})}{r} dr + ∫_{θ_{1}}^{θ_{2}} −u_{r}(r_{1}, θ) r_{1}dθ

= ∫_{r_{1}}^{r_{2}} ∫_{θ_{1}}^{θ_{2}} ((ru_{r})_{r} + \frac{u_{θθ}}{r} ) dθ dr

which, when compared to the formula ∫∫_{R} Δu dxdy = ∫_{r_{1}}^{r_{2}} ∫_{θ_{1}}^{θ_{2}} Δu(r, θ) dθ rdr yields the inevitable conclusion

Δu = \frac{(ru_{r})_{r}}{r} + \frac{u_{θθ}}{r^{2}} .

Attempting u(r, θ) = R(r)Θ(θ) we immediately arrive at (rR′)′Θ/r + RΘ′′/r^{2} = 0 which we multiply by r^{2}/(RΘ) to get r(rR′)′/R+Θ′′/Θ = 0, so that r(rR′)′/R = −Θ′′/Θ is constant. Since Θ must be periodic with period 2π this constant must be of the form n^{2} with n an integer, and we may write Θ = e^{inθ}. The R equation may be written r(rR′)′ = n^{2}R, which has the solutions (linear combinations of) R = r^{±n}. We want our solutions to be continuous at the origin, so from these two we pick only R = r^{|n|}. (For n = 0 the general solution is R = A + B ln r, and again we must throw away the part that is singular at the origin, leaving R = 1 = r^{|0|}.)

75042 PDE/1996: Problem set #2 Page 2

Taking linear combinations of the product solutions, we are therefore led to trying a series

u(x, y) = ∑_{n=−∞}^{∞} c_{n}r^{|n|}e^{inθ}

where x = r cos θ and y = r sin θ as usual. Clearly, each term in the series is harmonic. When we set r = 1 in the series and use the boundary conditions u(cos θ, sin θ) = g(θ), we simply get the Fourier series for g. Thus we should have

c_{n} = \frac{1}{2π} ∫_{−π}^{π} g(φ)e^{−inφ} dφ.

We insert this back into the series for u, and interchange the order of summation and integration to get

u(x, y) = ∫_{−π}^{π} g(φ)P (r, θ − φ) dφ

where P is the famous Poisson kernel:

P (r, θ) = \frac{1}{2π} ∑_{n=−∞}^{∞} r^{|n|}e^{inθ} = \frac{1}{2π} (∑_{n=0}^{∞} r^{n}e^{inθ} + ∑_{n=0}^{∞} r^{n}e^{−inθ} − 1)

= \frac{1}{2π} (\frac{1}{1 − re^{iθ}} + \frac{1}{1 − re^{−iθ}} − 1) = \frac{1}{2π} \frac{1 − r^{2}}{1 + r^{2} − 2r cos θ} .

As long as g is integrable (∫_{−π}^{π} |g(θ)| dθ < ∞) we may differentiate under the integral sign in the integral formula for u above, yielding

Δu = ∫_{−π}^{π} g(φ) (\frac{(rP_{r}(r, θ − φ))_{r}}{r} − \frac{P_{θθ}(r, θ − φ)}{r^{2}}) dφ = 0.

(I verified that the integrand is zero using Maple; there is a worksheet available from my Web page which you may download and load into xmaple if you’re curious.)

As for the boundary conditions, note that P (r, θ) → 0 uniformly in ε < θ ≤ π for every ε > 0 as r → 1, and also that ∫_{−π}^{π} P (rθ) dθ = 1 for all r (most easily seen from the series representation for P). These are properties typical of an approximate delta function; we’ll get back to those.

Problem set #3 — for next week

#3.1: Evans problem 1 (p. 80).

#3.2: If f is continuous at x, prove that − ∫_{∂B(x,ε)} f dx → f(x) when ε ↘ 0.

#3.3: Evans problem 3 (p. 80).

#3.4: Let u be a C^{2} function on some open set. Prove that u cannot have a local maximum at any point x where Δu(x) > 0.

#3.5: Let Ω be a bounded open set, and let u ∈ C^{2}(Ω) ∩ C(\overline{Ω}) satisfy the equation Δu = u^{3} − u in Ω. Assume that u = 0 on ∂Ω. Show that −1 ≤ u ≤ 1 throughout Ω.

#3.6: Evans problem 4 (p. 81).