Suggested Solution to Assignment 7

MATH 422 (25-6) partial diferential equations Suggested Solution to Assignment 7 Exercise 7.. Suppose there exists one non-constant harmonic function u in, which attains its maximum M at x. Then by the mean value property, we have u(x ) uds, B(x, r) B(x,r) for all B(x, r). maximum. Thus u(x) M for all x B(x, r) since u is continuous and M is its Now since u is not a constant, there exists x such that u(x ) u(x ). Since is a region, we can find a continuous a curve γ(t) ( t ) such that γ() x and γ() x. Set E : { t u(γ(t)) M}, then E is closed since u and γ are both continuous. Let t E, then by the above result, u(x) M for all x B(γ(t ), r). So E is open. Hence, E [, since E and [, is connected. But this contradicts to u(x ) u(x ) and then we complete the proof. 2. Suppose u and u 2 are solutions of the problem. Then u u u 2 satisfies u in, on. Thus, by the Green s first identity for v u, we have u ds u 2 dx + Hence, Therefore, u and then u(x) is a constant.. u 2 dx. u udx. 3. Suppose u and u 2 are solutions of the problem. Then u u u 2 satisfies u in, + au on. Thus, by the Green s first identity for v u, we have u ds u 2 dx + u udx. Hence, u 2 dx au 2 ds. Since a(x) >, u and then u(x) is a constant C. So by the Robin boundary conditions, we have Ca(x). This shows that C and u.. 4. Suppose u and u 2 are solutions of the problem. Then u u u 2 satisfies u t k u in (, ), u on (, ), u(x, ) in. Thus, by the Green s first identity for v u, we have u ds u 2 dx + u udx.

MATH 422 (25-6) partial diferential equations Therefore, by the diffusion equation u 2 dx + uu t dx k u 2 dx + d u 2 dx. 2k dt efine E(t) : u2 dx, then the above equality implies d dte(t). Note that E() and E(t), we have E(t). So we obtain u and the uniqueness for the diffusion equation with irichlet boundary conditions is proved.. 5. Suppose u(x) minimizes the energy. Let v(x) be any function and ɛ be any constant, then E[u E[u + ɛv E[u + ɛ u vdx ɛ hvds + ɛ 2 v 2 dx. Hence, by calculus for any function v. By Green s first identity, u vdx + u vdx hvds v ds hvds. Noting v is an arbitrary function, let v vanish on firstly and then the above equality implies u in. So the above equality changes to v ds hvds. Since v is arbitrary, h on and then u(x) is a solution of the following Neumann problem u in, h on bdy. Note that the Neumann problem has a unique solution up to constant and the functional E[w does not change if a constatn is added to w, thus it is the only funtion that can minimize the energy up to a constant. Note that here we assume functions u, v and the domain are smooth enough, at least under which the Green s first identity can be hold. 6. (a) Suppose u and u 2 are solutions of the problem. Then u u u 2 tends to zero at infinity, and is constant on A and constant on B, and satisfies ds ds. A Suppose that u, then without loss of generality, we assume that there exists x such that u(x ) >. Since u tends to zero at infinity, so there exists R such that u(x) u(x ) 2 if x R. Then u is a harmonic function in B(, R) and u. Thus, the B max u u(x ) 2 u(x ) max B(,R) B(,R) maximum is attained on A or B by the Maximum Principle. WOLG, we assume that u attain its maximum on A and then by any point of A is a maximum point since u is constant on A. Hence, by the Hopf maximum principle, / > on A, but this contradicts to the condition Therefore, u and then the solution is unique. A 2 ds.

MATH 422 (25-6) partial diferential equations (b) If not, then u(x) have a negative minimum. As above, by the Minimum Principle, we get that u attains its minimum on A or B. WOLG, we assume that u attains its minimum on A, then < by the Hopf principle since u is not a constant. But this contradicts to A ds Q >. So u in, (c) Suppose that there exists x such that u(x ). Then by (b) and the Strong Minimum Principle, u is a constant in. But this contradicts to A ds Q >. So u > in. (Actually, we can prove that u > on A and B by the Hopf principle again as in (b)). 7. (Extra Problem ) Suppose u and u 2 are solutions of the problem. Then (u u 2 ) u 3 u 3 2 (u u 2 )(u 2 + u u 2 + u 2 2) in Thus, (u u 2 ) + a(x)(u u 2 ) on (u u 2 ) 2 (u 2 + u u 2 + u 2 2) (u u 2 ) (u u 2 ) (u u 2 ) (u u 2 ) u u 2 2 a(x)(u u 2 ) 2 u u 2 2 since a(x). (u u 2 ) 2 (u 2 + u u 2 + u 2 2 ) u u 2 in 8. (Extra Problem 2) (a) E[u : 2 ( u 2 + b(x)u 2 ) + f(x)u (b)if u is a solution, then v C 2(), v u + b(x)uv + f(x)v u v + b(x)uv + f(x)v w C 2 (), w h on, take v w u, E[w 2 v 2 + 2 u 2 + v u + 2 b(x)v2 + 2 b(x)u2 + b(x)uv + f(x)v + f(x)u 2 v 2 + 2 b(x)v2 + E[u since b(x). E[u Conversely, v C 2(), t R, w : u + tv C2, w u h on d dt E[u + tv t u v + b(x)uv + f(x)v v( u + b(x)u + f(x)), v C 2 () u + b(x)u + f(x) in 3

MATH 422 (25-6) partial diferential equations Exercise 7.2 2. Let r x and ɛ : {x ɛ < r < R}, where R is large enough such that φ outside r < R/2. Using the Green s second identity, [ ɛ x φ(x)dx x φ φ ds x x R x ɛ ɛ [ x φ φ x [ ɛ φ + φ c 2 ds + x ɛ ds 4φ 4ɛ φ, where φ denotes the average value of φ on the sphere {r ɛ}, and φ this sphere. Since φ is continuous and φ is bounded, Hence, 4φ 4ɛ φ B(x,R) φ() x <R [ x φ φ 4φ() as ɛ. x φ(x)dx 4. x ds denotes the average value of φ on 3. Choosing B(x, R) in the representation formula () and using the divergence theorem, [ u(x ) u(x) ( x x ) + x x ds 4 x x R 4R 2 4R 2 [ R 2 u(x) + ds R 4 x x R x x R uds + 4R uds. x x <R 4. (Extra Problem) φ(x) Cc 2 (R 2 ), we need to show that φ() log x φ(x) dx R 2 2 udx proof: Let r x and ε : x : ε < r < R, where R is large enough such that φ if r R/2. Using the Green s identity, log x φ(x)dx [log x φ ε ε φ log x ds [log x φ x ε φ log x ds [log ε φ φ ε ds x ε 2 φ 2ε log ε φ 4

MATH 422 (25-6) partial diferential equations where φ, bounded, Exercise 7.3 φ φ denote the average value of φ, φ on the sphere x ε. Since φ is continuous and 2 φ φ 2ε log ε 2φ(), as ε. Suppose G and G 2 both are the Green s functions for the operator and the domain at the point x d. By (i) and (iii), we obtain that G (x) + 4 x x and G 2(x) + both are harmonic 4 x x functions in. Thus, by (ii) and the uniqueness theorem of harmonic function, we have G (x) + that is, the Green s function is unique. 3. In the textbook, 4 x x G 2(x) + 4 x x, A ɛ x a ɛ where u(x) G(x, a) and v(x) G(x, b). Note that (u v v )ds, u(x) G(x, a) + H(x, a) 4 x a, where H(x, a) is a harmonic function throughout the domain. What s more, v(x) G(x, b) is a harmonic function in { x a < ɛ}, then we have [ v A ɛ ( + H(x, a)) x a ɛ 4 x a v ( + H(x, a)) ds 4 x a [ v x a ɛ 4 x a + v ( 4 x a ) ds [ + H(x, a) v x a ɛ v H(x, a) ds v(a) [H(x, a) v v H(x, a) dx v(a). x a <ɛ is Here we use the representation formula (7.2.) and. So lim A ɛ v(a) G(a, b). ɛ Exercise 7.4. By (i), we know that G(x) is a linear function in [, x and [x, l. Since (ii) and G(x) is continuous at x, we have { kx, < x x ; G(x) kx x l (x l), x < x < l. Hence, G(x) + 2 x x is harmonic at x if and only if if and only if (G(x) + 2 x x ) x (G(x) + 2 x x ) x, + k l x. l 5

MATH 422 (25-6) partial diferential equations So the one-dimensional Green s function for the interval (, l) at the point x (, l) is G(x) { l x l x, < x x ; x l (x l), x < x < l. 2. Assume that h(x, y) is a continuous function that vanished outside {(x, y) x 2 +y 2 R 2 } and h(x, y) M. Then by the formula (3), we have u(x, y, z ) M [(x x ) 2 + (y y ) 2 + z 2 2 3 2 dxdy {(x,y) x 2 +y 2 R 2 } MR 2 2( x 2 + y2 +, when z2 R)3 x 2 + y2 + z2 > R. Therefore, u satisfies the condition at infinity: u(x), as x. 3. From (3), we have u(x, y, z ) z 2 z 2 [(x x ) 2 + (y y ) 2 + z 2 3 2 h(x, y)dxdy [x 2 + y 2 + z 2 3 2 h(x + x, y + y )dxdy. Now we change variables such that x z s cos θ, y z s sin θ. Then Since u(x, y, z ) z 2 2 2 2 (z 2 s 2 + z 2 ) 3 2 h(z s cos θ + x, z s sin θ + y )z 2 s dsdθ s(s 2 + ) 3 2 h(z s cos θ + x, z s sin θ + y ) dsdθ. s(s 2 + ) 3 2 ds (s 2 + ) 3 2, lim u(x, y, z ) h(x, y ), z if the limit can be taken in the integration, for example, when h(x, y) is bounded. 5. Here is one of suggested explains. Since the half-plane {y > } is not bounded, this only means that the solution is not unique under the (uncompleted) boundary condition or it is not a well-posed problem. But generally it will be a well-posed problem if we add another boundary condition, such as u(x), as x. 6. (a) Using the method of reflection, as in the dimension three, we have where x (x, y), x (x, y ), x (x, y ). G(x, x ) 2 log x x 2 log x x, 6

MATH 422 (25-6) partial diferential equations (b) Since G y y + y 2[(x x ) 2 + (y + y ) 2 y y 2[(x x ) 2 + (y y ) 2 y [(x x ) 2 + y 2 on y, the solution is given by u(x, y ) y h(x) (x x ) 2 + y 2 dx. (c) By (b), we have u(x, y ) y h(x) (x x ) 2 + y 2 dx x 2 dx. + 7. (a) If u(x, y) f( x ), then y (b) So f has to satisfy the following OE where a, b are contants. (c) Using the polar coordinate, u x y f ( x y ), u y x y 2 f ( x y ). u xx + u yy y 2 f ( x y ) + x2 y 4 f ( x y ) + 2x y 3 f ( x y ). f(t) ( + t 2 )f (t) + 2tf (t). t a ds + b a arctan t + b, + s2 x x + y y y f ( x y )x r x y 2 f ( x y )y r. v(r, θ) cθ + d, where c, d are constants. So in the (x, y)-coordinate, we have v(x, y) c arctan y x + d. (d) By (a), So u(x, y) f( x y ) a arctan x y + b. h(x) lim y u(x, y) 2 a + b, x > ; b, x ; a + b, x <. 2 (e) By (d), we see that h(x) is not continuous unless u is a constant. This agrees with the condition that h(x) is continuous in Exercise 6. 7

MATH 422 (25-6) partial diferential equations 9. Here the methof of reflection is used. Let x (x, y, z ) such that ax + by + cz >, then its reflection point x (x, y, z ) about the hyperplane {ax + by + cz } satisfies x 2a x a 2 + b 2 + c 2 (ax + by + cz ), y 2b y a 2 + b 2 + c 2 (ax + by + cz ), z 2c z a 2 + b 2 + c 2 (ax + by + cz ). So the Green s function for the tilted half-space {(x, y, z) : ax + by + cz > } is given by G(x, x ) 4 x x + 4 x x.. In case x, the formula for the Green s function is G(x, ) 4 x + 4a, since 4 x + 4a C2 and is harmonic in B a (), except at the point x ; [ 4 x + 4a x a ; and [ 4 x + 4a + 4 x is harmonic at. 3. Let x, then we have G(x, x ) 4 x x + 4 r x/a ax /r is the Green s function at x for the whole ball, where a is the radius and r x. Let x (x, y, z ), then G(x, x ) 4 x x + 4 r x/a ax /r is the Green s function at x for the whole ball. Note that G(x, x ) G(x, x ) on and G(x, x ) is a harmonic function on. Hence, G(x, x ) G(x, x ) is the Green s function for by the uniqueness of the Green s function. 5. (a) If v(x, y) is harmonic and u(x, y) v(x 2 y 2, 2xy), then Thus u x 2xv x + 2yv y, u y 2yv x + 2xv y. u xx + u yy 2v x + 4x 2 v xx + 4xyv xy + 4xyv yx + 4y 2 v yy 2v x + 4y 2 v xx 4xyv xy 4xyv yx + 4x 2 v yy 4(x 2 + y 2 )(v xx + v yy ). (b) Using the polar coordinates, the transformation si (r cos θ, r sin θ) (r 2 cos 2θ, r 2 sin 2θ). Therefore, it maps the first quadrant onto the half-plane {y > }. Note that the transformation is one-one which also maps the x-positive-axis to x-positive-axis and y-positive-axis to x-negative-axis. 8

MATH 422 (25-6) partial diferential equations 7. (a) Here the method of reflection is used and you can also use the result of Exercise 55 to find the Green s function for the quadrant Q. Let x (x, y ) Q, x y ( x, y ), x x (x, y ) and x ( x, y ). Then it is easy to verify that G(x, x ) 2 log x x 2 log x xy 2 log x xx + 2 log x x is the Green s function at x for the quadrant Q. (b) (Extra Problem 3) By formula () in Section 7.3 in the textbook, u(x ) u(x) G(x, x ) ds bdy x G G x,y> x x ( x x 2 x x ) 2 G G x>,y y y ( x x 2 x x ) 2 g(y)( x x 2 x x )dy + y 2 h(x)( x x 2 x x )dx 2 9. Consider the four-dimensional laplacian u u xx + u yy + u zz + u ww. Show that its fundamental solution is r 2, where r 2 x 2 + y 2 + z 2 + w 2. proof: irect computation according to its definition. 2. Using the conclusion in 7 and 2, we have the singular part of the Green function is 8ω 4 x x 2, where ω 4 is the volume of S 3, x (x, y, z, w) and x (x, y, z, w ). So the Green function is 8ω 4 x x 2 + 8ω 4 x x 2, where x (x, y, z, w ). 2. By the formula (2)(3) in Section7.3, we get u(x ) bdy (u N N) bdy N since N on the boundary. Thus we proved the following theorem: If N(x, x ) is the Neumann function, then the solution of the Neumann problem is given by the formula u(x ) 22. (Extra Problem 4)By the formula (4) in Section7.4, u(x ) z 2 z bdy N x 2 +y 2 + x x 3 ds. If the bounded condition is dropped, then we have v u + az where a is arbitrary real number is also a solution. 9