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9 Pages

### sol3

Course: AM 223, Fall 2009
School: Sanford-Brown Institute
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Word Count: 2618

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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 &gt; 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...

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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 ...

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Microsymposium 36, MS002, 2002ABLATION OF METEORITES V.A. Alexeev, Vernadsky Institute of Geochemistry and Analytical Chemistry, Russian Academy of Sciences, 117975, Moscow, Russia (aval@icp.ac.ru) The interaction of galactic cosmic-ray particles w
Sanford-Brown Institute - MICRO - 36
Microsymposium 36, MS001, 2002ANALYSIS OF VENUSIAN CORONAE, ARACHNOIDS AND NOVAE ON THE BASIS OF THEIR GEOLOGICAL ENVIRONMENT. M. Aittola and V.-P. Kostama, 1 Astronomy, Dept. of Physical Sciences, Univ. of Oulu, P.O. BOX 3000, FIN-90014, Finland (
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Microsymposium 36, MS073, 2002LANDSAT TM CLASSIFICATION OF THE ARCTIC KARA IMPACT CRATER TUNDRA, RUSSIA. K. Ojala 1, D. Badjukov 2, J. Raitala1(jouko.raitala@oulu.fi), T. hman 1 and C. Lorenz2, 1Physics Dept., Univ. of Oulu, Finland; 2Vernadsky Ins
Sanford-Brown Institute - MICRO - 36
Microsymposium 36, MS063, 2002SEASONAL VARIATIONS OF SUBSURFACE HYDROGEN AS SEEN BY HIGH ENERGY NEUTRON DETECTOR, MARS ODISSEY. M. L. Litvak1, I. G. Mitrofanov1, A.S. Kozyrev1, A.B. Sanin1, W. Boynton 2, C. Shinohara2, D. Hamara2, R. S. Saunders3; 1
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Microsymposium 36, MS051, 2002DISTRIBUTION AND CLASSIFICATION OF VOLCANO-TECTONIC FEATURES OF VENUS. V.-P. Kostama 1 and M. Aittola 1, 1Astronomy, Dept. of Physical Sciences, University of Oulu, P.O. BOX 3000, FIN90014, Finland (&lt;petri.kostama@oulu
Sanford-Brown Institute - MICRO - 36
Microsymposium 36, MS062, 2002DISTRIBUTION OF CHEMICALLY BOUND WATER IN SURFACE LAYER OF MARS BASED ON DATA ACQUIRED BY HIGH ENERGY NEUTRON SPECTROMETER, MARS ODYSSEY. M. L. Litvak1, A. T. Basilevsky 2, I. G. Mitrofanov 1, W. Boynton 3, R. S. Saunde