Math 220B Final Exam Solutions March 18, 2002

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1 Math 0B Fial Exam Solutios March 18, (1 poits) (a) (6 poits) Fid the Gree s fuctio for the tilted half-plae {(x 1, x ) R : x 1 + x > 0}. For x (x 1, x ), y (y 1, y ), express your Gree s fuctio G(x, y) i terms of x 1, x, y 1 ad y Aswer: G(x, y) 1 π l (y 1, y ) (x 1, x ) + 1 π l (y 1, y ) + (x, x 1 ) (b) (6 poits) Use the Gree s fuctio from part (a) to write the solutio formula for { u 0 x u g Simplify your aswer as much as possible. Aswer: u(x) x 1 + x π x. g(y) y x ds(y).. (8 poits) Fid a Neuma fuctio for the iterval [a, b] R. That is, fid a fuctio N(x, y) such that for each x (a, b), { y N(x, y) δ x a < y < b N ν (x, y) 1 y a, b Aswer: N(x, y) 1 y x. 3. (10 poits) Let be the triagle bouded by x 0, x 1 x 1 ad x 1 + x 1. Let λ i () be the i th eigevalue of { u λu x u 0 x (a) Prove that 5 4 π λ 1 (). Aswer: Let 1 {(x 1, x ) : 1 < x 1 < 1, 0 < x < 1}. We kow 1 λ i () λ i ( 1 ). Now ( π ) λ i ( 1 ) + (mπ). λ 1 ( 1 ) 5 4 π λ 1 (). 1

2 (b) Prove that 5 π λ (). Aswer: Let be the square with vertices at (1, 0), (0, 1), ( 1, 0) ad (0, 1). This square has side legths, ad cotais. The eigevalues of are give by ( ) ( ) π mπ λ i ( ) +. λ ( ) π + π 5 π λ (). 4. (1 poits) Cosider the eigevalue problem with Neuma boudary coditios, { u λu x ( ) 0 x. ν (a) (8 poits) Let X {w C (); w 0, w, v i 0 for i 1,..., 1} where the v i are the first 1 eigefuctios. Prove that the th eigevalue of (*) satisfies λ mi w X w L w L. Aswer: Suppose u X is the miimizer of this quotiet over all w X. Pick ay v X ad let i(t) (u + tv) L () u + tv L. If u is a miimizer, the i has a local miimum at t 0, ad, therefore, i (0) 0. By a straightforward calculatio, we see that i (0) ( u )( u v) ( u )( uv) ( u ). We see that i (0) 0 uv dx + for all v X. [ u + mu]v dx ν v ds(x) m uv dx. ν v ds(x)

3 Now let v j be oe of the first 1 eigefuctios for this problem. By assumptio, u X implies that u is orthogoal to v j. we see that [ u + mu]v j dx uv j dx u v j dx + ν v j ds(x) u v j ν ds(x) λ j uv j dx + ν v j ds(x). ν v j ds(x). Now let h be a arbitrary trial fuctio (C fuctio which vaishes o the boudary of ). Defie 1 v h c i v i where First, we ote that we coclude that v X. [ u + mu]h dx c i h, v i v i, v i. 1 v, v j h c i v i 0. { } 1 [ u + mu] v + c i v i dx 1 [ u + mu]v dx + c i 1 ν v ds(x) + c i h ds(x) 0, ν v i dx ν v i ds(x) usig the assumptio that h vaishes o. we coclude that u + mu 0, ad, thus, u is a eigefuctio with correspodig eigevalue m. Next, we will show that m is the th eigevalue. First, we ote that X X 1 X.... m λ 1 λ.... To show that m λ, let v j be 3

4 a eigefuctio with correspodig eigevalue λ j for j. Now v j X. w m mi () v j L () j v j v j λ j v j λ w X w j. L () v j L () (b) (4 poits) Use the results of part (a) to give a estimate o the secod eigevalue for (*) (eigevalue problem of the Laplacia with Neuma boudary coditios) i the case whe is the triagle from the previous problem (the triagle bouded by x 0, x 1 x 1 ad x 1 + x 1 ). Aswer: We eed to fid a fuctio i X, the space of fuctios which are orthogoal to the first eigefuctio. As the first eigevalue for the Laplacia with Neuma boudary coditios is zero, we kow the first eigefuctio is the costat fuctio. X cosists of C fuctios which are orthogoal to costat fuctios; that is, fuctios which satisfy v dx 0. Here is the triagle with vertices at (1, 0), (0, 1) ad ( 1, 0). Oe fuctio i X would be v(x, y) x. Usig this test fuctio, we get the followig approximatio for the secod eigevalue, λ dx 1 dx 1 x 1 dx 1 dx 1/ (15 poits) Determie whether the followig statemets are true or false. Briefly explai your aswer. (a) Let be a ope, bouded, coected subset of R. Cosider the exterior Neuma problem, { u 0 x c v j g x. ν A ecessary coditio for solvability is g(y) ds(y) 0. Aswer: False. For example, let B(0, 1) i R 3 ad cosider u(x) 1. x We see that u is a solutio of u 0 x c ν 1 x but 1 ds(x) 0. 4 v j v j

5 (b) Let be a ope, bouded subset of R. Let a(x) 0. All eigevalues of { u λu x + a(x)u 0 ν x are o-egative. Aswer: True. λ u dx u u dx u dx u dx + a(x)u dx 0. λ 0. u ν ds(x) (c) Let {(x 1, x ) R : 0 < x 1 < π, 0 < x < π}. Suppose u ad v are liearly idepedet eigefuctios of { u λu x u 0 x. The u ad v are orthogoal. Aswer: False. The eigevalues are give by λ m m +. We see that λ 5 is a eigevalue with multiplicity. it has a two-dimesioal eigespace, ad cosequetly the eigefuctios eed ot be orthogoal. (d) Let {(x 1, x ) R : 0 < x 1 < 1, 0 < x < π}. Suppose u ad v are liearly idepedet eigefuctios of { u λu x u 0 x. The u ad v are orthogoal. Aswer: True. The eigevalues are give by λ m (mπ) +. All eigevalues are distict. We kow that for symmetric boudary coditios, eigefuctios correspodig to distict eigevalues are orthogoal. all liearly idepedet eigefuctios must be orthogoal. (e) Let be a ope, bouded set i R. Assume u C () C() is a harmoic fuctio o. If there exists a poit x 0 such that the u(x) costat. u(x 0 ) max u(x), Aswer: False. We eed the domai to be coected for the strog maximum priciple to hold. 6. (7 poits) Aswer the followig short aswer questios. 5

6 (a) (4 poits) State the miimax priciple for the th eigevalue of { u λu x R u 0 x Aswer: Let w 1,..., w be a set of liearly idepedet trial fuctios (C fuctios which vaish o ). Let { } w λ L (w 1,..., w ) max () : w c c 0 w i w i. L () The λ mi λ (w 1,..., w ), where the miimum is take over all possible sets of liearly idepedet trial fuctios. (b) (3 poits) State Liouville s Theorem. Aswer: If u is a bouded, harmoic fuctio o R, the u must be costat. 7. (10 poits) Let be a ope, bouded subset of R. Let {v i, λ i } be the eigefuctios ad eigevalues of { u λu x u 0 x Solve the followig iitial/boudary value problem, u t k u 0 u(x, 0) C 1 v 1 (x) + C v (x) u(x, t) 0 x x. Aswer: Look for a solutio of the form u(x, t) X(x)T (t). The we see that u (x, t) A v e kλt is a solutio of the heat equatio which satisfies the boudary coditio for each. Now, we wat u(x, 0) C 1 v 1 (x) + C v (x). by lettig A 1 C 1, A C ad A i 0 for i 3, we arrive at the solutio u(x, t) C 1 v 1 (x)e kλ 1t + C v (x)e kλ t. 8. (16 poits) 6

7 (a) (4 poits) Let B 3 (0, 1) be the ball of radius 1 about the origi i R 3. Use the fudametal solutio of Laplace s equatio to costruct a solutio of { u 0 x (B3 (0, 1)) ( ) c u 1 x B 3 (0, 1) which decays to zero as x +. Aswer: u(x) 1 x. (b) (4 poits) Prove uiqueess of solutios of (*) which decay to zero as x +. Aswer: Suppose u ad v are two solutios of (*) which decay to zero as x +. Fix C > 1. Let C (B 3 (0, 1)) c B 3 (0, C). The u ad v are harmoic o C, ad w u v satisfies w 0 x C w 0 x B 3 (0, 1) w ɛ x B 3 (0, C) for some ɛ ɛ(c). By the maximum priciple, max C w(x) max w(x) max{0, ɛ}. C w max{0, ɛ}. This is true for all ɛ > 0 by choosig C sufficietly large. we coclude that w u v 0 o (B 3 (0, 1)) c. By a similar aalysis with w v u, we coclude that w v u 0 o (B 3 (0, 1)) c. we coclude that u v. (c) (4 poits) Give the defiitio of a double-layer potetial with momet h. If we ca write the solutio of (*) as a double-layer potetial, what equatio must h satisfy? Aswer: where u(x) h(y) Φ (x y) ds(y) ν y g(x) 1 h(x) h(y) Φ ν y (x y) ds(y). (d) (4 poits) Usig parts (a)-(c), explai why we caot write the solutio of (*) as a double-layer potetial. Aswer: Suppose we ca write the solutio as a double-layer potetial. By parts (a) ad (b), we kow the uique solutio (which decays to zero) is give by u(x) 1. (Note: A solutio give by the double-layer potetial will decay x 7

8 to zero.) By part (c), we kow that if the solutio is give by a double-layer potetial, the there must exist a cotiuous fuctio h such that 1 x h(y) Φ (x y) ds(y) ν y which satisfies B 3 (0,1) 1 1 h(x) + 1 x for all x B 3 (0, 1). we eed 1 1 h(x) + 1, which implies h 0. But, we ca see this caot be satisfied. 9. (10 poits) Let f (x) 4π e x /4. Prove that f coverges weakly to δ 0 i the sese of distributios as +. Aswer: Let F f be the distributio such that We eed to show that (F f, φ) f (x)φ(x) dx. (F f, φ) (δ 0, φ) φ(0) as +. That is, we eed to show that for all ɛ > 0, there exists a N such that for N. Usig the fact that (F f, φ) φ(0) < ɛ 4π e x /4 dx 1, we write (F f, φ) φ(0) /4 4π e x [φ(x) φ(0)] dx /4 B(0,δ) 4π e x [φ(x) φ(0)] dx + /4 4π e x [φ(x) φ(0)] dx I + J R B(0,δ) 8

9 for δ to be determied below. Now for term I, we boud as follows. We write I /4 4π e x [φ(x) φ(0)] dx B(0,δ) φ(x) φ(0) L (B(0,δ)) < ɛ by choosig δ sufficietly small, usig the fact that φ is cotiuous. Now for that choice of δ, term J is bouded as follows. We write J /4 R B(0,δ) 4π e x [φ(x) φ(0)] dx C e x /8 dx C 4π e δ /8 4π e δ /8 < ɛ R B(0,δ) by choosig sufficietly large. our claim is prove. 9