HW 1
PDE, 3 solutions
Problem 1. No. If a sequence of harmonic polynomials on [-1, 1]n converges uniformly to a limit f then f is harmonic. Problem 2. By definition Ur U for every r > 0. Suppose w is a barrier at y for U . Then the restriction of w to Ur is also a barrier. Therefore, if y is regular for U , it is also regular for Ur . Conversely, if w is a barrier at y for Ur it may be extended to a barrier on Ur as follows. Let Ar = {x Ur |r/2 |x - y| r}, and let M = maxxAr w(x). By the definition of a barrier, M < 0. For x Ur let w(x) = max{w(x), M }. This is the maximum of two subharmonic functions, ~ and is therefore subharmonic. Moreover, since w M on Ar it also follows that w(x) = M for every x Ar . We extend w to U by setting w = M < 0 ~ ~ ~ for all x U, |x - y| r. This extension is a barrier at y for U . First proof for Problem 3. I hope you did not struggle too much with the calculation. After some experimentation, I realized that the calculation is simpler if one works with = xn /|x| rather than as defined in the problem statement. Assume u = |x| h(). Then D is given in coordinates by ,i = - A nice cancellation is x D = xi ,i = - xn xi xi xn + = 0. |x|3 |x| xn xi ni + . |x|3 |x|
The derivatives of u are obtained by the chain rule Du = |x|-2 xh + |x| h D, or in coordinates u,i = xi |x|-2 h - h + |x|-1 h ni := (a) + (b). Evaluate each term in turn. By using the cancellation above we have (a),i = |x|-2 ( - 2 + n)(h - h ),
2 and (b),i = ( - 1)|x|-2 h + |x|-2 (1 - 2 )h . Summing up, we have the differential equation (1 - 2 )h + (1 - n)h + ( - 2 + n)h = 0. If we change variables according to cos = and g() = h() we have g + (n - 2) cot g + ( + n - 2)g = 0. The analysis is simplest (and unecessary) when n = 2. However, this is a good check that the calculation is OK. Here we have g + 2 g = 0, with an even solution g = cos(). This solution is positive for || < if we choose < /2. More generally, our task is to find a positive solution to (0.1) on any interval (-a, a) for 0 < a = cos < 1. One way to do this is to look at a classical reference on Legendre's differential equation or hypergeometric functions (Abramowitz and Stegun for example), but a crude analysis will also suffice. Fix cos < 1. Observe that g 1 is an even positive solution on (-a, a) when = 0. Now consider solutions with g(0) = 1 and g (0) = 0 for small > 0. The only problem is at 0, but here lim0 g +(n-2) cot g = (n-1)g (0) = -(+n-2)g(0) = -(+n-2). Therefore, for sufficiently small > 0 one may use a series expansion to construct an analytic solution g, convergent in the neighborhood (-, ) for all 0 < 1. In particular, g = 1 - ( + n - 2) 2 + o( 2 ), 2 (0.1)
and the perturbation is continuous in . For fixed > 0, the equation for g is regular on [, a] and one may use continuous dependence of the solution of an ODE on parameters to say that for sufficiently small > 0 there is a solution g that is positive such that g (x) 1 as 0. Second proof of Problem 3. I can see why the proof above may be unpalatable. Here is a classical argument that avoids the ODE, and relies only on a direct construction of a barrier. Let U denote the domain U = B(0, 1) C. Consider the Perron function for boundary data u = |x| on U . Then every point of U is regular except the origin. Since u(x) 0, x U we also have lim inf u(x) 0.
x0
3 We only need show c = lim supx0 u(x) = 0. It is clear that c 0. We use the scaling invariance of the cone: for any k > 1 consider the scaled function v(x) = u(kx). Then on all boundary points |x| = 0 we have v(x) = k|x| > u(x), and v(0) = u(0). Thus, there is a constant a < 1 such that u(x) av(x), x U, x = 0.
We claim that this inequality also holds at all points in the interior. If so, by taking x 0 we obtain, lim sup u(x) = c ac = a lim sup u(kx) < c.
x0 x0
Thus, c = 0 and u is a barrier. It remains to show u av for all x in the interior. We cannot use the maximum principle directly since we do not know a priori that u and v are continuous on the boundary. However, this is an easy problem to fix. For any 1 > > 0 consider the domain U = U {|x| > }. Consider a harmonic function on the annulus { |x| 1} with boundary values w = 0 on |x| = 1 and w = supxU,|x|(u - av) < 0 on |x| = . By the maximum principle, we then have u - av w 0 for all x U . Let 0 to find u av, x U as desired. Problem 4. Let denote the given measure 1|y|<a dy. The potential is u (x) =
R3
|x - y|-1 (dy).
Since is invariant under rotations, we may evaluate the integral by assuming x is along the x1 -axis. If is the polar angle measured from the axis, and we denote |x| = and |y| = r we have |x - y|2 = 2 - 2r cos + r 2 .
First consider the case where |x| > a. In this case,
a
u (x) =
0 0
(2 - 2r cos + r 2 )-1/2 (2r sin )r d dr.
Evaluate the inner integral via the substitution p = 2 - 2r cos + r 2 :
a (+r)2
u (x) = 2 = 2
0 a 0 (-r)2
p-1/2
dp r dr 2 2
a 0
(| + r| - | - r|) r dr =
2r 2 dr =
4a3 . 3
4 If |x| a, then we may use the same calculation to evaluate the integral on the shell 0 < r < to obtain a contribution 42 /3. However, on the shell < r < a we have | + r| - | - r| = 2 and we have the integral 2
a
2r dr = 2(a2 - 2 ).
Summing these two contributions we have u (x) = 2(a2 - 2 ). 3
The attraction F = Du is found by differentiating u (with left and right derivatives at = a that are equal). We have F (x) = - 4x , |x| < a, 3 3 - 4a3 x , |x| a 3|x|
If we define F as Du this calculation is entirely legitimate. On the other hand, if we define F as the integral F (x) = (2 - n) x-y (dy), |x - y|n
Rn
we must show this equals Du. This can be done using finite differences. To clarify ideas, let us do this in generality. Fix with || = 1. Let Dh u = u(x + h) - u(x) = h-1 h
Rn
|x + h - y|2-n - |x - y|2-n (dy).
If we know a priori that u is differentiable (for example, as in this problem) we may use the mean value theorem to obtain a number (0, 1) such that Dh u = (2 - n)
Rn
(x + h - y) (dy). |x + h - y|n (x + h - y) x-y - n |x + h - y| |x - y|n (dy).
We use the definition of F to obtain Dh u - F (x) = (2 - n)
Rn
As h 0, the integrand converges to 0 pointwise. All that is needed is to justify the interchange of limits. This is best done through the dominated convergence theorem. For any h = 0, the integrand is bounded by (n - 2) (|x + h - y|1-n + |x - y|1-n )(dy).
Rn
5 If this term is finite, we are done. We only need consider the case |x| a. Since is absolutely continuous with respect to Lebesgue measure, the singularity is integrable. Indeed,
Rn
|x - y|1-n (dy) =
B(0,a)
|x - y|1-n dy
B(0,2a)
|y|1-n dy = 2an .
Problem 5. First suppose x R. Fix p > 0 and g(x) let = e-x , x > 0 and g = 0, x 0. As you have shown, g is C . To construct a bump function, choose f (x) = g(1/2 + x)g(1/2 - x). Then supp(f ) (-1, 1). In Rn one may choose the product x1 x2 xn (x) = f ( )f ( ) . . . f ( ), n n n
-p
and obtain a normalized bump function by defining (x) = (x) . Rn (y)dy
The support of is compactly contained within (-n-1/2 , n-1/2 )n which is contained within B(0, 1). Now consider g as defined in the problem. Suppose dist(x, U ) > . Let V = Rn \U . We then have either B(x, ) U or B(x, ) V . To see this, suppose x is not in U . Observe that dist(x, U ) = dist(x, U ) since U = U \U and inf yU |x - y| cannot be attained in U . Therefore, dist(x, U ) > which implies B(x, ) V . All we have used here is that U is open. Since V is also open, if x U we have similarly B(x, ) U . In either case, 1U (x - y) = 1U (x) if |y| < . ~ Problem 6. Let Ul and Um be two sequences of increasing, bounded domains used in the definition of pF . That is, Ul Ul+1 U and Ul = U , l=1 ~ and similarly for Um . We define a sequence of Perron functions pl by solving the Dirichlet problem with boundary data pl = 1, x F , pl = 0 on the `outer boundary' of Ul (and similarly for pm ). We have shown that pl and ~ pm converge to functions harmonic on U = Rn \F . Denote these by pF and ~ pF respectively. We must show that pF = pF . ~ ~ ~ Fix l. Since Ul is a bounded domain, we must have Ul Uml for sufficiently large ml . By the maximum principle 1 > pml > pl on Ul . Passing to ~ the limit, we find pF pF , x U . A similar argument with Ul replaced by ~ ~ Um yields pF pF , x U . ~
6 Problem 7. pF is generated by a charge F with support in F pF (x) =
F
|x - y|2-n F (dy).
If x U , we have r(x) |x - y| R(x), and we have the desired estimate R(x)2-n cap(F ) pF (x) r(x)2-n cap(F ).
Remark 0.1. The virtue of the above estimate is that it gives us the leading order asymptotics of pF as x . Conversely, if we are able to determine pF (xn ) along some sequence xn , we may determine the capacity. For example, to compute the capacity of a planar disc in R3 we can use symmetry to determine pF exactly along the axis of the disk, and let x to find the capacity (one needs elliptic integrals away from the axis). Problem 8. First suppose F is smooth. In this case, observe that p (x) = pF (x) is the potential associated to F (use uniqueness of potentials here). Now change variable to say that cap(F ) = n-2 cap(F ) in this case. If F is not smooth, use the approximation theorem. I apologize for an error in equation (1.62), p.38 of the notes. I hope you were able to fix this. It should read
k=0
k(2-n) cap(Fk ) = .
The proof of the exterior cone condition using Wiener's criterion goes as follows. Suppose y U satisfies an exterior cone condition. Let = 1/2 and let Fk be the compact sets in Theorem 1.50. Observe that Fk contains a ball of radius ck for some c > 0 (independent of k). Therefore, using the scaling property just proved, and the fact that balls of radius R have capacity Rn-2 we find cap(Fk ) (ck )n-2 . Therefore, the sum in Wiener's criterion
k=0
k(2-n) cap(Fk ) cn-2
k=0
1 = .
7 Problem 9. In case, you were confused about the notation for |Df |, what was intended was the following uniform Lipschitz estimate |f (x + y) - f (x)| |y|, (0.2)
which is all that is needed in this problem. If f is C 1 this bound is derived as follows. By the fundamental theorem of calculus
1
f (x + y) - f (x) =
0
d f (x + ty) dt = dt
1
Df (x + ty)y dt.
0
If |Df | denotes the (Euclidean) norm of the matrix Df we obtain (0.2). The estimate cap(f (F )) cap(F ) is intuitively believable, but quite subtle to prove. It is easy to fall into the following trap: Since f contracts distances, f (F ) F and therefore cap(f (F )) cap(F ). The inclusion f (F ) F works for balls (after a suitable translation), but not in general. The key property of capacity needed here is the following. Consider a finite collection of balls B(ak , rk ), k = 1, . . . N . If we move the centers of ...