< 1. max x B(0,1) f. ν ds(y) Use Poisson s formula for the ball to prove. (r x ) x y n ds(y) (x B0 (0, r)). 1 nα(n)r n 1

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1 7 On the othe hand, u x solves { u n in U u on U, so which implies that x G(x, y) x n ng(x, y) < n (x B(, )). Theefoe u(x) fo all x B(, ). G ν ds(y) + max g G x ν ds(y) + C( max g + max f ) f(y)g(x, y) max x f G(x, y) Evans Ch. #6. Use Poisson s fomula fo the ball to pove n x ( + x ) n u() u(x) + x n u() ( x ) n wheneve u is positive and hamonic in B (, ). inequality. This is an explicit fom of Hanack s Solution: Poisson s fomula fo the ball gives an explicit smooth solution of Poisson s equation on B (, ). Recall that fo g C( B(, )) we take u(x) : x Futhemoe, since u is hamonic we have that u() ds(y) n x y n ds(y) (x B (, )). ds(y).

2 8 Note that u implies that g since g is assumed to be continuous. Theefoe, u(x) x x y n ds(y) n x nα n n x nα n x y n ds(y) ( x ) n ds(y) (Since x x y fo all y B(, ) and g.) n x ( x ) n nα n ds(y) n + x ( x ) n ds(y) n + x u(). ( x ) n The othe inequality is shown analogously by noting that x y + x. Evans Ch. #7. and fo we have (i) u C (B (, )), (ii) u in B (, ), and (iii) Pove Poisson s fomula fo the ball, which states that fo g C( B(, )) u : x x y n ds(y) (x B (, )) lim u(x) x x, x B (,) g(x ) fo each point x B(, ). (Hint: Since u solves { u in B (, ) u g on B(, ). fo g, the theoy automatically implies K(x, y) ds(y) fo each x B (, ).) Solution: Recall that G(x, y) : Φ(y x) φ x (y) whee Φ is the fundamental solution of Laplace s equation and φ x is a coecto function depending on the domain in question which is hamonic in that egion. Theefoe y G(x, y) is hamonic and since G(x, y) G(y, x) fo y x (a theoem), we have that x G(x, y) is also hamonic. Thus x G ν K(x, y) is hamonic fo x B (, ), y B(, ). A diect calculation shows that K(x, y) ds(y)

3 9 fo each x B (, ). Since g is continuous on a compact set it is bounded, and consequently u as defined is likewise bounded. Now, since x K(x, y) is a hamonic function, it is smooth fo x y. We easily veify that u C (B (, )), with u(x) x K(x, y), x B (, ). Now fix x B(, ), ɛ >. Choose δ > small enough to insue that Then if x x < δ, x B (, ), g(x ) < ɛ, if y x < δ, y B(, ). u(x) g(x ) B(x,δ) + : I + J. K(x, y)[ g(x ) B(x,δ) K(x, y) g(x ) K(x, y) g(x ) Now, since the integal of K ove the bounday is equal to (as in the hint), and g(x ) < ɛ, we have that I ɛ K(x, y) ɛ. Futhemoe if x x δ and y x δ, we have and so y x y x. Thus y x y x + δ y x + y x ; J g L B(x,δ) g L ( x ) C( x ) as x. K(x, y) B(x,δ) n y x n Evans Ch. #8. Let u be the solution of { u in n + u g on given by Poisson s fomula fo the half-space. Assume that g is bounded and g(x) x fo x, x. Show Du is not bounded nea x. (Hint: Estimate u(en) u().)

4 Solution: Recall that the fundamental solution of Laplace s equation with Diichlet bounday conditions in the uppe half-space is u(x) : x n x y n. Note that since x e n and y we have x y + y. Since we have extended n u continuously up to the bounday of +, u() g() in this case and u(e n ) u() u(e n) [ : (I + J) ( + y ) n/ ( + y ) n/ n \B n (,) ( + y ) n + B n (,) y ( + y ) n Now g is a bounded function so let s call this bound C. Then we see that I is a convegent integal since n \B n (,) ( + y ) C n/ C C n \B n (,) [ B n (,) ( + y ) n/ ds ( + y ) n/ (n )α(n ) n ( + ) n/ C(n ) α(n ) < n ( + ) n/ by the integal, limit compaison, and p-tests fo integals. Hence we conclude that I is a convegent integal. Howeve, y B n (,) ( + y ) n/ B n (,) ( + y ) n/ [ ds n/ B n (,) ( + ) n/ ( + ) [ B n (,) ds ( + ) n/ vol( B n (, ) ) ( + ) (n ) α(n n/ )n (n ) α(n ), n ( + ) n/

5 by the integal, limit compaison, and p-tests fo integals. Hence we conclude that J is a divegent integal fo all. Theefoe, u(e n ) u() Du (e n ) lim, so Du is unbounded, even though u is bounded. Evans Ch.9 #9. Let U + denote the open half-ball {x n x <, xn > }. Assume u C (Ū + ) is hamonic in U +, with u on U {x n }. Set { u(x) if x v(x) : n u(x,..., x n, x n ) if x n < fo x U B (, ). Pove v is hamonic in U. Solution: Clealy v is hamonic on B (, )\{x n } since it is equal to eithe of the hamonic functions ±u. By the symmety of v, it s clea that on the set {x n }, v satisfies the Mean Value popety, and is theefoe hamonic thee as well. Evans Ch. #. Suppose that u is smooth and solves u t u in n (, ). (i) Show u (x, t) : u(x, t) also solves the heat equation fo each. (ii) Use (i) to show v(x, t) : x Du(x, t) + tu t (x, t) solves the heat equation as well. Solution: (i) We have d dt u u d dt u(x, t) u(x, t) u t (x, t) u t (x, t) ( u t (y, z) u(y, z) ). (afte the change of vaiables y x, z t) (ii) We can show that v solves the heat equation without using (i) by poceeding via diect calculation. We intoduce the heat opeato h : t to abbeviate notation. Fist note that if h u then h u xi h u t as well, a fact elying on the equality of mixed patial deivatives: h u t (u t ) t u t n (u t ) t (u t ) t i n i u txix i u xix it ( n ) (u t ) t u xix i (u t u) t () t. i t