v( x) u( y) dy for any r > 0, B r ( x) Ω, or equivalently u( w) ds for any r > 0, B r ( x) Ω, or ( not really) equivalently if v exists, v 0.
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1 Sep. 26 The Perron Method In this lecture we show that one can show existence of solutions using maximum principle alone.. The Perron method. Recall in the last lecture we have shown the existence of solutions to the Laplace equation u = 0 in Ω, u = g on Ω ( ) for the special case of Ω = B R by writing down the explicit formula for the solution. Unfortunately, for general Ω this is not possible. The Perron method show the existence of the general problem by combining the explicit formula and the maximum principle. First we recall the following property of subharmonic functions. Let Ω R n and f: Ω [, ) be upper semicontinuous in Ω and f. The function f is called subharmonic in Ω if for all Ω Ω, the following property holds: If u is harmonic in Ω, and f u on Ω, then f u in Ω. or equivalently, v( x) B r v( x) B r B r( x ) B r( x ) if v exists, v 0. u( y) dy for any r > 0, B r ( x) Ω, or equivalently u( w) ds for any r > 0, B r ( x) Ω, or ( not really) equivalently The idea of the Perron method can be best illustrated via the D example: solving u = 0 in ( 0, ) ; u( 0) = u 0, u( ) = u. ( 2) The solution is a straight line connecting ( 0, u 0 ) and (, u ). Now let U 0 be any function connecting these two points. And we modify U 0 as follows. Take any subinterval ( a, b), and replace U 0 ( a, b) by the solution of u = 0 in ( a, b) ; u( a) = U 0 ( a), u( b) = U 0 ( b). ( 3) Doing this again and again, we see that the sequence approaches the straight line, which is the solution we need. It is hard to make this argument rigorous for general U 0. But if we take U 0 to be subharmonic, or roughly speaking U 0 0, then an important observation is that every modification makes the function larger. On the other hand they are all bounded from above by max ( u 0, u ). Therefore there must be an upper bound and this upper bound is the solution. This argument can be generalized to higher dimensional problems. We see that we need to guarantee. Modifying any subharmonic function by patching it with harmonic functions; 2. The modified function is still subharmonic; 3. Subharmonic functions is upper-bounded by its boundary values. They are indeed true. Lemma. We have i. If v C 0 ( Ω) is subharmonic and B R ( y) Ω, then the harmonic replacement v of v, defiend by v( x) x Ω\B R ( y) v ( x) R 2 2 x y v( z) n ds z x B R ( y) d w d R z x B R( y) ( 4) is subharmonic in Ω ; Furthermore v v in Ω.
2 ii. Let v be subharmonic in Ω, if there is x 0 Ω with v( x 0 ) = sup Ω v( x), then v is constant. In particular, if v C 0 ( Ω), then v( x) max Ω v( y) for all x Ω. iii. If v,, v k are subharmonic, so is v max v,, v k }. Proof. i. It is clear that v v in Ω. Now we argue that v is be subharmonic. Take arbitrary Ω Ω, and let u be harmonic in Ω with v u on Ω. Since v v in Ω, v u on Ω. Since v is subharmonic, v u in the whole Ω ( this implies v u in Ω \B R ( y) ). In particular, v u on ( Ω B R ( y) ). Since both u and v are harmonic in Ω B R ( y), we have v u in there too. ii. Left as exercise. iii. Left as exercise. Now we turn to the existence of solutions to the problem u = 0 in Ω, u = g on Ω. ( 5) Let S g be the set of all subharmonic functions which are C 0 ( Ω) and are below g on the boundary. We claim that the upper bound is in fact the solution. First we show that the upper bound is harmonic. Theorem 2. Let then u is harmonic in Ω. u( x) sup v S g v( x), ( 6) Proof. We devide the argument into several steps.. u is well-defined. To see this, first note that v min Ω g is subharmonic, so S ϕ is not empty; Next note that any v S g, we have v( x) sup Ω g for any x Ω. Therefore the supreme is well-defined at every point. 2. Now consider any x Ω. Let v n be such that lim v n ( x) = u( x). By replacing v k by max v,, v k, inf Ω g}, we can assume that v k } is monotonically increasing and bounded from below. Denote by B R the ball B R ( x) Ω. Now let v k be the harmonic replacements of v k. It is easy to see that v k is also monotonically increasing. Since v k v k, we must have lim k v k( x) = u( x). ( 7) 3. We show that the limit v = lim k v k is harmonic in. To see this, note that v k v l is harmonic in B R for any k, l. Thus Harnack inequality implies the existence of C such that v k( y) v l( y) C v k( x) v l( x) ( 8) for any y. As a consequence, the convergence v k v is uniform in and therefore v must be harmonic. 4. We need to show that v = u in. Since u is the supreme, v u. Now if there is x such that u( x ) > v( x ), since u = sup v Sg v, there is ṽ subharmonic such that u( x ) > ṽ ( x ) > v( x ). Set w k = max v k, ṽ }. Then the harmonic replacements w k convergens to a harmonic function w in satisfying which implies v w u, ( 9) v( x 0 ) = w( x 0 ). ( 0). To see this, recall that v is harmonic when it has the mean value property, which is kept by uniform convergence.
3 On the other hand we have v( x 0 ) = u( x 0 ) which gives w( y) dy v( y) dy = v( x 0 ) = w( x 0 ) = w( y) dy. ( ) The only possibility is v w in. But this gives a contradiction as by assumption w( x ) ṽ ( x ) > v( x ). ( 2) Thus ends the proof. The next task is to show that u satisfies the boundary condition. That is lim u( x) = g( x 0). ( 3) x x 0 Ω We use the concept of a barrier. Recall that a function f is superharmonic if f is subharmonic. Note that we can replace all subharmonic in the above construction by superharmonic and obtain u( x) = inf v where the infimum is over all superharmonic functions whose boundary value g. Definition 3. a ) Let ξ Ω. A function β C 0 ( Ω) is called a barrier at ξ with respect to Ω if i. β > 0 in Ω\ ξ}, β( ξ) = 0. ii. β is superharmonic in Ω. b ) ξ Ω is called regular if there existes a barrier β at ξ with respect to Ω. Remark 4. If β is a local barrier at ξ with respect to U Ω, then for any B ρ ( ξ) U, m inf β x Ω \B ρ ( ξ) β U \ B ρ( ξ) min ( m, β) x Ω B ρ ( ξ) ( 4) is a barrier at ξ with respect to Ω. To see this, take any Ω Ω, and let u be harmonic in Ω with boundary value β. We need to show that u β in Ω. Since β m, u m in Ω. Next we show u β in Ω B ρ ( ξ). At the boundary, we have min ( m, β) β on Ω B ρ u = β m = inf β on B ρ Ω u β in Ω B ρ ( ξ). ( 5) U \ B ρ( ξ) Lemma 5. If ξ is a regular point of Ω and g is continuous at ξ, then u takes g as its boundary value at ξ, that is lim u( x) = g( ξ). ( 6) x ξ Proof. The idea is the construct super and subharmonic function from g and the barrier β, which gives lower and upper bounds of u. For any ε > 0, there is δ > 0 such that Now consider We have g( x) g( ξ) < ε for all x ξ < δ; c β( x) 2 M 2 sup ϕ for all x ξ δ. ( 7) Ω v g( ξ) + ε + c β( x), v g( ξ) ε c β( x). ( 8) v u v ( 9)
4 which gives the convergence estimate. Summarizing, we obtain Theorem 6. Let Ω R n be bounded. The Dirichlet problem u = 0 in Ω, u = g on Ω ( 20) is solvable for all continuous boundary values g if and only if all points ξ Ω are regular. Proof. We have already done the if part. For the only if part, we take g 0 be such that g( ξ) = 0 while g( x) > 0 for all other x Ω. The solution u 0 is a barrier due to strong maximum principle. 2. The alternating method of H. A. Schwartz. It is important from the numerical point of view and is the foundation of domain decomposition method. Theorem 7. Let Ω and Ω 2 be bounded domains all of whose boundary points are regular for the Dirichlet problem. Suppose that Ω Ω 2 φ and that Ω and Ω 2 are of class C in some neighborhood of Ω Ω 2, and that they intersect at a nonzero angle. Then the Dirichlet problem fo the Laplace equation on Ω Ω Ω 2 is solvable for any continuous boundary values. Proof. We denote γ = Ω Ω 2, γ 2 = Ω 2 Ω, Γ = Ω \ γ, Γ 2 = Ω 2 \ γ 2.. Let M sup Ω g and m inf Ω g. We solve the Dirichlet problem 2 and obtain u, defined on Ω. Next we solve u = 0 in Ω, u = u = 0 in Ω 2, u = and obtain u 2. This process can be continued to obtain u 3, u 4, Ω 2. g Γ M γ. ( 2 ) g Γ2 u γ 2. ( 22) with u 2 k + defined on Ω and u 2 k defined on 2. By strong maximum principle we have u 2 M and in particular u 2 u on γ 2. This implies u 2 u on Ω Ω 2. ( 23) Next comparing the boundary values of u, and u 3 and using maximum principle, we have Comparing the values of u 2 and u 3 on γ γ 2 we have Comparing the values of u 2 and u 4 on Ω 2 = Γ 2 γ 2 we obtain 3. In general, we have u 3 u on Ω. ( 24) u 3 u 2 on Ω Ω 2. ( 25) u 4 u 2 on Ω 2. ( 26) u u 3 on Ω ; u 2 u 4 on Ω 2 ; u u 2 on Ω Ω 2. ( 27) Therefore there is u on Ω Ω 2 such that lim u2 u = k + lim u 2 k Ω. Ω 2 ( 28) 2. Note that the boundary value here may not be continuous. But it is still bounded and the Perron construction still works with convergence to boundary values only fail at discontinuous points.
5 Using Harnack inequality we can conclude that the convergence is locally uniform which means u is harmonic. 4. Next we show that u is continuous up to the boundary except for those points in Ω Ω 2. We can start from g Γ v = 0 in Ω, v = ( 29) m γ and obtain a increasing sequence v n }. It is clear that v n u n and therefore u is between each pair of ( v n, u n ). The continuity of v n, u n up to boundary then gives the continuity of u, except for those points in Ω Ω To show continuity at Ω Ω 2, one has to modify the iteration process. Instead of using M on γ to obtain u, we use a continuous extension of g. This way we can still obtain u n } but the monotonicity relation does not hold anymore. Instead, we show that ( u 2 n+ 3 u 2 n+ ) max γ u 2 n+ 2 u 2 n q <, on γ 2 ( 30) for a uniform q. This, combined with the maximum principle, implies u n+ u n converges to 0 as fast as the geometric series q n, which in turn gives uniform convergence of the sequence. For details of the last step, see 3. 3, 3. 4 of J. Jost Partial Differential Equations, GTM Further readings. J. Jost, Partial Differential Equations, Chap. 3. Exercise. Exercise. Let v be subharmonic in Ω, if there is x 0 Ω with v( x 0) = sup Ω v( x), then v is constant ( If you are not comfortable with upper semicontinuity, just assume v to be continuous). In particular, if v C 0 ( Ω), then v( x) max Ω v( y) for all x Ω. Exercise 2. Prove: If v,, v k are subharmonic in Ω, so is v max v,, v k }. 3. The key lemma is the following: There exists 0 < q < depending on Ω, Ω 2, such that if w: Ω and continuous on the closure Ω, with w = 0 on Γ, w on γ, then R is harmonic in Ω, A corresponding result holds if Ω is replaced by Ω 2. w q on γ 2. ( 3 )
u( x) = g( y) ds y ( 1 ) U solves u = 0 in U; u = 0 on U. ( 3)
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