75042 Partielle differensialligninger

Oppgavesett #8 (til 1997–10–28 ) — løsningsforslag

Evans 2.5–9: Extend u to a function v on the full ball by antisymmetry in the x_{n} variable, as explained in the problem. The function

w(x) = ∫_{U} K(x, y)v(y) dS(y)

is then continuous in \overline{U} (v is certainly continuous on ∂U, since u = 0 on x_{n} = 0) and harmonic in U. The Poisson kernel K(x, y) is clearly symmetric in y_{n} when x_{n} = 0 (see its formula!) and therefore, when x_{n} = 0 the integrand above is antisymmetric in y_{n}. Thus w(x) = 0 when x_{n} = 0. Since w is harmonic in U^{+} with the same boundary values as u, w = u in all of U^{+}. Further, w(x) is antisymmetric in x_{n} and therefore w = v. This antisymmetry can be seen directly (note that K(x, y) in unchanged if x_{n} and y_{n} both are replaced by −x_{n} and −y_{n} respectively).

Alternatively, one can note that the function −w(x_{1}, …, x_{n−1}, −x_{n}) is harmonic with the same boundary values as w to get the antisymmetry (and in particular the vanishing of w(x) when x_{n} = 0).

Evans 2.5–11: That u_{λ}(x, t) = u(λx, λ^{2}t) solves u_{t} − Δu = 0 if u does so, is trivial, since the chain rule yields u_{λ,t} − Δu_{λ} = λ^{2}(u_{t} − Δu).

(In cases like this where not all subscripts mean differentiation, is is handy to introduce a comma to separate those subscripts that indicate a derivative from those that do not.)

Now differentiate u_{λ} with respect to λ. Equality of mixed partials, i.e., u_{λ,λt} = u_{λ,tλ} and Δu_{λ,λ} = (Δu_{λ})_{λ} guarantee that u_{λ,λ} satisfies the heat equation. Put λ = 1 to obtain the solution v = x · Du + 2tu_{t}.

#8–1: 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|}.)

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.

75042 PDE 1997: Oppgavesett #8 Side 2

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)K(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} .

The general Poisson formula for n = 2 (Evans page 43) is

u(x) = ∫_{∂B(0,1)} K(x, y)g(y) dS(y), K(x, y) = \frac{1 − |x|^{2}}{2π|x − y|^{2}} (x, y ∈ ℝ^{2}).

In this formula, write x = re^{iφ} and y = e^{i(φ+θ)} and note that then |x − y|^{2} = |r − e^{iθ}|^{2} = 1 + r^{2} − 2 cos θ, so this P is really K expressed in polar coordinates.

Oppgavesett #9 (til 1997–11–04 )

Fra Evans side 82–83: Oppgave 13,14,16. (Det er en tyrkleif i oppgave 14: Du må legge til h(t) i formelen for u(x, t).)

#9–1: [Oppgaven er gjenbrukt fra i fjor, derfor på engelsk.] We say a continuous function u is subharmonic in an open set Ω if u(x) ≤ − ∫_{∂B(x,r)} u dS whenever B(x, r) ⊂ Ω. Note that we do not require u to be differentiable. Prove that, if u is subharmonic and v is harmonic in Ω and both functions are continuous on \overline{Ω} with u ≤ v on ∂Ω, then u ≤ v in Ω. Also prove that, if u ∈ C^{2}(Ω), then u is subharmonic in the present sense if and only if −Δu ≤ 0 in Ω.