Suggested Solution to Assignment 6

Transcription

1 MATH 4 (6-7) partial diferential equations Suggested Solution to Assignment 6 Exercise 6.. Note that in the spherical coordinates (r, θ, φ), Thus, Let u = v/r, we get 3 = r + r r + r sin θ θ sin θ θ + r sin θ u rr + r u r = 3 u = k u. φ. u r = v r r v r, u rr = v rr r v r r + v r 3. by the equation of u, v rr = k v, which implies v(r) = Ae kr + Be kr, A, B are constants. Therefore, u(r) = A r e kr + B r ekr, A, B are constants. 4. We have known that c r + c is a solution, c and c satisfy the equation: c a + c = A, c b + c = B. u(x, t) = ab A B b a r + A + b B A b a, r = x + y + z, is a solution. Therefore, it is the unique solution by the Uniqueness Theorem of the Dirichlet problem for the Laplace s equation. 6. Firstly, we find a solution depending only on r. Let u(r), r = x + y, is a solution. As before, we have u = 4 r + c ln r + c, c, c are constants. By the boundary conditions, we get 4 a + c ln a + c =, is the unique solution by the Uniqueness Theorem. 4 b + c ln b + c =, u(x, y) = 4 (r a ) b a (ln r ln a), 4(ln b ln a) 7. Firstly, we look for a solution depending only on r = x + y + z. Let u(r) be a solution, then as before, u rr + r u r =, from which we have u = 6 r + c r + c, c, c are constants. Thus, by the boundary conditions, we get 6 a + c a + c =, 6 b + c b + c =. u(x, y) = 6 (r a ) + ab a + b 6 ( r a ), is the unique solution by the Uniqueness Theorem.

2 MATH 4 (6-7) partial diferential equations 9. (a) Firstly, we look for a solution depending only on r = x + y + z. Let u(r) be a solution, then as before, u rr + r u r =, from which we have u r = c + d, c, d are constants. Thus, by the boundary conditions, we r have c + d =, c = 4γ. Therefore, u = 4γ r + 4γ. u is unique by the maximal principle. (b) The hottest temperature is C, the coldest is γ. (c) By assumption, we have γ =, therefore, γ = 4.. Integrating the equation u = f and using the divergence theorem, u f dxdydz = u dxdydz = n ds = D D bdy(d) bdy(d) g ds. there is no solution unless f dxdydz = g ds. D bdy(d) Exercise 6.. By the boundary conditions, we can guess u x (x, y) = x a and u y (x, y) = y + b. Luckily these also satisfy the equation. u(x, y) = x y ax + by + c, c is any constant, are solutions. Actually, we can prove that they are all solutions by the Hopf maximum principle.. Let (m, n) (m, n ), then =( (sin my sin nz)(sin m y sin n z)dydz sin my sin m ydy)( sin nz sin n zdz) =, so the eigenfunctions {sin my sin nz} are orthogonal on the squre { < y < π, < z < π}. 3. Let u(x, y) = X(x)Y (y), then Therefore, X X = Y Y = λ, X() = Y () = Y (π) =. λ n = n, Y n (y) = cos(ny), X = x, X n+ = sinh[(n + )x], n =,,,... u(x, y) = A x + A n sinh(nx) cos(ny).

3 MATH 4 (6-7) partial diferential equations By the inhomogeneous boundary condition, we get which implies Therefore, A π + A = π, A = A n sinh() cos(ny) = ( + cos y), sinh(π), A n =, if n,. u(x, y) = x π + sinh(x) cos(y). sinh(π) 4. Let u satisfies u =, in the squre { < x <, < y < }, u (x, ) = x, u (x, ) = u,x (, y) = u,x (, y) =, and u satisfies u =, in the squre { < x <, < y < }, u (x, ) = u (x, ) = u,x (, y) =, u,x (, y) = y, then u = u + u is a harmonic function which we want to find. By the method of separate variables, And Therefore, u = A (y ) + A n cos(x)[cosh(y) coth() sinh(y)], A =, A n = u = B n = sinh() = sinh() x cos(x)dx = n π [( )n ], n =,,.... B n cosh(x) sin(y), y sin(y)dy { ( ) n+ n π + n 4 π 4 [( )n ] }, n =,,.... u = (y ) + A n cos(x)[cosh(y) coth() sinh(y)] + A n = n π [( )n ], B n = B n cosh(x) sin(y), { ( ) n+ sinh() n π + } n 4 π 4 [( )n ], n =,,.... 3

4 MATH 4 (6-7) partial diferential equations 6. Let u(x, y, z) = X(x)Y (y)z(z), then and Therefore, X X + Y Y + Z Z =, X () = X () = Y () = Y () = Z () =. X m (x) = cos(mπx), Y n (y) = cos(y), m, n =,,,..., Z = (m + n )π Z, Z () =. u(x, y, z) = 4 A + A m cos(mπx) cosh(mπz) + A n cos(y) cosh(z) m= n= + A mn cos(mπx) cos(y) cosh( m + n πz). m= n= Finally, by the inhomogeneous boundary condition, we get g(x, y) = A m mπ sinh(mπ) cos(mπx) cosh(mπz) + which implies + m= m= n= A n sinh() cos(y) cosh(z) A mn m + n π sinh( m + n π) cos(mπx) cos(y) cosh( m + n πz), n= A mn = 4 m + n )π sinh( m + n π) g(x, y) cos(mπx) cos(y)dxdy, m + n, and A is any constant. We can prove that they are all solutions by the Hopf maximum. 7(a). Let u(x, y) = X(x)Y (y), then Thus, X X + Y Y =, X() = X(π) =. X n (x) = sin(), n =,,..., and Y = n Y, lim y Y (y) =. u(x, y) = A n sin()e ny. Finally, by the inhomogeneous condition h(x) = A n sin(nx), we have And the solution is u(x, y) = A n = π h(x) sin(nx)dx. ( π ) h(x) sin(nx)dx sin(nx)e ny. π 4

5 MATH 4 (6-7) partial diferential equations Exercise 6.3. (a) By the Maximum Principle, max D u = max D u = max (3 sin θ + ) = 4. θ (b) By the Mean Value property, u(, ) = π. By the formula ()-() in the textbook, Since h(θ) = + 3 sin θ, we get u(x, y) = A + (3 sin θ + )dθ =. r n (A n cos nθ + B n sin nθ), A n = πa n h(θ) cos nθdθ, B n = πa n h(θ) sin nθdθ. A =, A n = (n > ), B = 3 a, B m = (m > ). 3. As before, since u(r, θ) = + 3r a sin θ. h(θ) = sin 3 θ = 3 4 sin θ sin 3θ, 4 we get A n = (n N), B = 3 4a, B 3 = 4a 3, B m = (m, 3). Use the same way, the solution should be u(r, θ) = 3 r3 r sin θ sin 3θ. 4a 4a3 Problem 4. By Poisson s formula, u(x, y) = u(r, θ) = ( r ) u(, φ) dφ r cos(θ φ) + r π since u, cos(θ φ) and u has the Mean-Value Property. Similarly, u(x, y) = u(r, θ) = ( r ) u(, φ) r cos(θ φ) + r dφ π since u, cos(θ φ) and u has the Mean-Value Property. Problem 5. (a)use the Strong Maximum Principle. (b)by Problem 4(r = /), 3 = / +/ +/ u(x, y) / = 3 r ( r) u(, φ) dφ r ( + r) u(, φ) dφ π = + r u(, ), r π = r u(, ), + r 5

6 MATH 4 (6-7) partial diferential equations Problem 6. Since u is a harmonic function in B () \ (, ), u(x, y) is smooth in B () \ (, ). Define v(x, y) = v(r, θ) = /4 r π h(φ) dφ, for r < /, /4 cos(θ φ) + r h(φ) = u(/, φ), and w := u v, then v(x, y) is a harmonic function in B / (), w is a harmonic function in B / ()\(, ), w = on B / () and w is bounded in B / (). Now, it suffices to show that w in B / () \ (, ): For any fixed (x, y ) B / () \ (, ), r := x + y. ε >, define v ε(r) := ε log(r), which is harmonic in B / () \ (, ). Since v ε = on B / () and lim r + v ε = +, we can choose r small enough such that < r < r and v ε (r) > sup B/ \(,) w on r = r. Thus, by the Maximum Principle on A := {(x, y) : r < x + y < /}, we get w(x, y ) ε log(r ). Let ε +, we get w(x, y ). Similarly, for w we get w(x, y ). Therefore, w(x, y ) =. Exercise 6.4. Since the only difference between the formulas of harmonic function in the interior and exterior of a disk is that r and a are replaced by r and a. Therefore, by the result in the exercise 6.4. u(r, θ) = + 3a r sin θ. 6. Using the separation of variables technique, we have Θ + λθ =, r R + rr λr =. So the homogenous conditions lead to Θ + λθ =, Θ() = Θ(π) =. λ n = n, Θ(θ) = sin nθ, n =,,..., and then R n (r) = r n, n =,,.... u(r, θ) = A n r n sin nθ. Finally, the inhomogeneous boundary condition requires that π sin θ sin θ = A n sin nθ, which implies So the solution is A = π, A =, A n = (n, ). u(r, θ) = πr sin θ r sin θ.. By the example in the textbook Section 6.4 (Please do it again) and letting β = π/, h(θ) =, we have u(r, θ) = A n r n sin nθ, 6

7 MATH 4 (6-7) partial diferential equations The first two nonzero terms are A n = a n / aπ r sin θ, sin(nθ)dθ = a n n π [ ( )n ]. 9a 5 π r6 sin 6θ.. Multiplying u in the both sides of equation and using the divergence theorem, u u n u =. Using Robin boundary condition, a u u =, D D which implies u = in the D and u = on D since a >. So u in D. D 3. It is similiar to the Example in Section 6.4 in the textbook. Here we only give the result and leave the details to you. For the eigenvalue problem of Θ(θ), we have λ n = ( ), Θ n (θ) = sin For the eigenvalue problem of R(r), we have So the solution is D, n =,,.... R n (r) = A n r β α + B n r β α, n =,,.... u(r, θ) = (A n r β α + B n r β α ) sin. By setting r = a and r = b the coefficients A n and B n should satisfy A n a β α + B n a β α = β g(θ) sin dθ then A n b β α + B n b β α = α β α h(θ) sin Aa β α Bb β α A n = a β α b β α B n = Aa β α Bb β α a β α b β α dθ A = β g(θ) sin α dθ, B = β h(θ) sin α dθ.,, Problem 7. Write u(r, θ) = R(r)Θ(θ), then r R R + r R R = Θ Θ =: λ and Θ () = = Θ(π), λ = is nontrivial. We can write λ = β, β > and Θ + β Θ =, β n = / + n, Θ n = cos(β n θ), n =,,,... Θ Θ > since u 7

8 MATH 4 (6-7) partial diferential equations u(r, θ) = n= (c nr βn + d n r βn ) cos(β n θ) By the boundary conditions, u(, θ) = cos 3 (θ/) = /4 cos(3θ/) + 3/4 cos(θ/) and u(, θ) = 4 cos(5θ/) c + d = /4; c + d = 3/4; c n + d n = if n, ; c β + d β = 4; c n βn + d n βn = if n c n = d n = ifn,, ; c = 3/4, d = 3/, c = /8, d = /7, c = 6 3 = d. 8