u(e iθ )K(e iθ,z)dθ, z (1). (126) h(e iθ )K(e iθ,z)dθ, z (1),

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1 2 Dirichlet Problem Dirichlet problem Dirichlet s problem is as follows: h C ( (1)), unique u C ( (1)) Harmonic( (1))) such that { u, (125) u (1) h. Recall thatgiven aharmonicfunction uinaneighborhoodof (1), we knew fromsection 2 that u(z) 1 u(e iθ )K(e iθ,z)dθ, z (1). (126) By this approach, we define u(z) 1 h(e iθ )K(e iθ,z)dθ, z (1), which is harmonic in (1). If we can prove that such u(z) is continuous function on (1), then u h holds on the boundary D, i.e., Dirichlet problem is proved. In next theorem, this continuity property is proved. Theorem 2.1 Let h be a continuous real-valued function defined on (1). Then the function defined by u(z) 1 h(e iθ )K(e iθ,z)dθ, z (1), is harmonic in (1), and e it (1), we have Remark: lim u(z) h(e it ). z e it,z (1) 1. Dirichlet ( ), German mathematician, has been hired by Göttingen in 1855 as Gauss s successor. After his death, Riemann became full professor. 2. Dirichlet s principle is a major theorem in the potential theory. The term potential theory arises from the fact that, in 19th-century physics, the fundamental forces of nature were believed to be derived from potentials which satisfied Laplace s equation. It is still the case that the Laplace equation is used in applications in several areas of physics like heat conduction and electrostatics. 194

2 . Since the Laplace operator appears in the heat equation, one physical interpretation of this problem is as follows: fix the temperature on the boundary of the domain according to the given specification of the boundary condition. Allow heat to flow until a stationary state is reached in which the temperature at each point on the domain doesn t change anymore. The temperature distribution in the interior will then be given by the solution to the corresponding Dirichlet problem. 4. The Dirichlet problem is named after Peter Gustav Lejeune Dirichlet, who proposed a solution by a variational method which became known as Dirichlet s principle. The existence of a unique solution is very plausible by the physical argument : any charge distribution on the boundary should, by the laws of electrostatics, determine an electrical potential as solution. However, Karl Weierstrass found a flaw in Dirichlet s argument, and a rigorous proof of existence was found only in 19 by David Hilbert. It turns out that the existence of a solution depends delicately on the smoothness of the boundary and the prescribed data. 5. The well-known Riemann mapping theorem states that if D is a non-empty simply connected open subset of C which is not all of C, then there exists a biholomorphic (bijective and holomorphic) mapping f from D onto the open unit disk (1). The theorem was stated (under the assumption that the boundary of U is piecewise smooth) by Bernhard Riemann in 1851 in his PhD thesis. Riemann s proof depended on the Dirichlet principle (which was named by Riemann himself), which was considered sound at the time. However, Karl Weierstrass found that this principle was not universally valid. Later, David Hilbert was able to prove that, to a large extent, the Dirichlet principle is valid under the hypothesis that Riemann was working with. However, in order to be valid, the Dirichlet principle needs certain hypotheses concerning the boundary of U which are not valid for simply connected domains in general. Lars Ahlfors wrote once, concerning the original formulation of the theorem, that it was ultimately formulated in terms which would defy any attempt of proof, even with modern methods. Simply connected domains with arbitrary boundaries were first treated by William Fogg Osgood (19) without using the Dirichlet principle. 195

3 Proof of Theorem 2.1: Step 1. To show: u is harmonic in (1) We knew that the function f(z) : 1 h(e iθ ) eiθ +z dθ, z (1), z (1), e iθ z is holomorphic. Recall Re ( e iθ +z e iθ z) K(e iθ,z). Since u is a real-valued function, by taking the real part, it implies that is harmonic. u(z) 1 h(e iθ )K(e iθ,z)dθ, z (1), Step 2. To show: u(z) h(e it ) as z e it Taking u 1 in (126), we get 1 1 K(e iθ,z)dθ. Then h(e it ) 1 h(e it )K(e iθ,z)dθ so that u(z) h(e it ) 1 h(e iθ )K(e iθ,z)dθ 1 1 [ h(e iθ ) h(e it ) ] K(e iθ,z)dθ 1 h(e iθ ) h(e it ) K(e iθ,z)dθ. h(e it )K(e iθ,z)dθ (1) e it We have to estimate K(e iθ,z) Is K(e iθ,z) bounded for all θ? Recall here that K(ζ,z) ζ 2 z 2 ζ z 2 > and K(e iθ,z) 1 z 2 e iθ z

4 Step. Estimate for K(e iθ,z) K(e iθ,z) 1 z 2 e iθ z 2 (cos θ +isin θ z)(cos θ isin θ z) cos 2 θ icos θsin θ zcos θ+isin θcos θ+sin 2 θ izsin θ zcos θ +izsin θ+ z 2 1+ z 2 zcos θ zcos θ izsin θ +izsin θ (1 z ) 2 +2 z z(cos θ isin θ) z(cos θ+isin θ) (1 z ) 2 + z (2 e i arg z e iθ e i arg z e iθ ) (use z z e i arg z and z z e i arg z ) (1 z ) 2 + z (2 2cos(θ arg z)). (1 z ) 2 +4 z sin 2 θ arg z 2 From the above calculation, we see that if z 1 and θ arg z, the function K(e iθ,z) is unbounded. On the other hand, if z 1 but θ arg z b >, then K(e iθ,z) 1 z 2 4 z sin 2 b. (127) Step 4. To show: u(z) h(e it ) as z e it To show: ǫ >, δ > such that In fact, let us consider u(z) h(e it ) < ǫ, z (1) with z e it < δ. (128) u(z) h(e it ) 1 1 t+π 1 ( t+δ + t δ t π t δ t π h(e iθ ) h(e it ) K(e iθ,z)dθ h(e iθ ) h(e it ) K(e iθ,z)dθ t+π ) + h(e iθ ) h(e it ) K(e iθ,z)dθ t+δ 197

5 where δ > is to be determined. We need to estimate three integrals t+δ, t δ and t+π t δ t π respectively, as follows. Consider the first integral Recall t is the angle of the fixed point e it. Here since θ and argz could be arbitrarily close, by the formula in Step, K(e iθ,z) is out of control, namely, unbounded. However, the term h(e iθ ) h(e it ) is under control, namely, for the given ǫ, there exists δ > such that t+δ, h(e iθ ) h(e it ) < ǫ, θ [t δ,t+δ] (i.e., θ t < δ) (129) 1 t+δ so that the first integral satisfies t δ 1 t+δ h(e iθ ) h(e it ) K(e iθ,z)dθ ǫ 1 t δ Now we fix the δ. t+δ Consider the second and third integrals t δ t π and t+π t+δ t δ K(e iθ,z)dθ ǫ 1 K(e iθ,z)dθ ǫ. In this case we have opposite inequality θ t δ (compare (129)). We choose δ > such that if z e it < δ, it must satisfy so that t arg z < δ 2, and 1 2 max h 1 z 2 z sin 2 δ 4 < 2ǫ θ arg z θ t t arg z δ δ 2 δ 2 > δ 4. (1) Then we estimate the second and the third integrals: 1 ( t δ t+π ) + h(e iθ ) h(e it ) K(e iθ,z)dθ t π t+δ 2 max h ( t δ 4 z sin 2 δ 4 t π 2 max h ( 2δ) 4 z sin 2 δ 4 < 1 1 z 2 max h 2 z sin 2 δ 4 < 2ǫ. (by (1)) t+π ) + dθ (use (127) and t+δ 198 f f )

6 Then (128) has been proved. Dirichlet problem for general domains Let D R n be a bounded domain with C 2 -boundary. For every φ C ( D), there is a unique u C (D) such that { u in D, The general solution is given by u(x) u φ on D. D ν(s) G(x,s) ds n where G(x, y) is the Green s function for the partial differential equation, and G(x, s) n n s G(x,s) i n i G(x,s) s i is the derivative of the Green s function along the inward-pointing unit normal vector n. The integration is performed on the boundary, with measure ds. The function ν(s) is given by the unique solution to the Fredholm integral equation of the second kind, φ(x) ν(x) 2 + D ν(s) G(x,s) ds. n 199