Analysis III Solutions - Serie 12

Transcription

1 .. Necessary condition Let us consider the following problem for < x, y < π, u =, for < x, y < π, u y (x, π) = x a, for < x < π, u y (x, ) = a x, for < x < π, u x (, y) = u x (π, y) =, for < y < π. Find all the values of a R for which the problem admits a solution. Sol. We recall that, since u is harmonic in the square < x, y < π, then, from the Green identities, in order to have a solution the flux of such solution must be equal to zero. Namely, if we denote Q our domain, and Q is its boundary, then the problem will have a solution if n u ds =, Q where n u denotes the outward normal derivative of u. (See equation (7.9) from the book.) In this case, we should have = = = = Q n u ds u y (x, π) dx + (x a) dx + (x a) dx = 3 π3 aπ. u x (π, y) dy (a x ) dx u y (x, ) dx u x (, y) dy Thus, we must have a = 3 π for the problem to have a solution. (In fact, once the problem has a solution, adding any constant gives another solution, so the problem has infinitely many solutions.).. Separation of variables Find the solution to the following problem, posed for < x < π and < y <. u =, for < x < π, < y <, u(x, ) =, for x π, u(x, ) = + cos(x), for x π, u x (, y) = u x (π, y) =, for < y <. December 3, 8 /7

2 In order to do that, first find nontrivial solutions w(x, y) = X(x)Y (y) to the following problem, { w =, for < x < π, < y <, w x (, y) = w x (π, y) =, for < y <, and use them as a basis to generate the solution to the previous problem. Sol. Let us proceed by separation of variables. Since w =, then X (x)y (y) + X(x)Y (y) = X (x) X(x) = Y (y) Y (y) = λ, for some λ R. Thus, we know X (x) + λx(x) =, for < x < π, Y (y) λy (y) =, for < y <. On the other hand, from the boundary condition we must have that X () = X (π) =. As it is already standard, the solutions to the eigenvalue problem for X are given by ( ) ( ) nx n X n (x) = cos, λ n =, for n =,,,.... Similarly, solving for Y using the values of λ n we have just found, we reach that Y (y) = α y + β, and for n, ( ) ( ) n(y + ) n(y ) Y n (y) = α n sinh + β n sinh. Notice that we have chosen the basis into which express Y n so that, when trying to impose the conditions at y =,, we get a simpler expression. That is, instead of considering the standard basis given by sinh(ny/) and cosh(ny/) we have shifted it so to make is simpler in the following steps. (See the previous exercise sheet for a discussion on why this can be done.) Thus, our solutions w n (x, y) are of the form: w (x, y) = α y + β, ( ) ( ( ) ( )) nx n(y + ) n(y ) w n (x, y) = cos α n sinh + β n sinh, /7 December 3, 8

3 for n =,, 3,.... We now use the superposition principle, to say that we are looking for a solution to our problem of the form u(x, y) = w (x, y) + w (x, y) + w (x, y) +... = α y + β + ( ) ( ( ) ( )) nx n(y + ) n(y ) cos α n sinh + β n sinh. n Notice that, from the way we have built the functions w n, such solution already fulfils the Neumann boundary conditions u x (, y) = u x (π, y) = for < y <. Moreover, since each w n is harmonic, u is also harmonic. Thus, we just need to impose the other boundary conditions (the Dirichlet boundary conditions for this problem). Notice that, the special base we have chosen to express w n, will be extremely useful in this step. Let us start imposing the boundary condition at y = : u(x, ) = α + β + n cos ( ) nx β n sinh ( n) =, so that, since sinh( n), we get β n = for all n, and β = α. On the other hand, imposing the boundary condition at y =, u(x, ) = α + β + n cos ( ) nx α n sinh (n) = + cos(x). That is, on the one hand we get that α + β =, so that, recalling that α = β we get that α = β =. For n 4 we get that α n sinh(n) = (and therefore, α n = if n 4), and α 4 sinh(4) =. That is, α 4 = sinh(4). Putting all together, we reach that our solution is u(x, y) = y + + cos(x) sinh ((y + )). sinh(4).3. Neumann problem Consider the Neumann boundary problem for the Laplace equation for < x, y < π: u =, for < x, y < π, u x (, y) =, for < y < π, u x (π, y) = sin(y), for < y < π, u y (x, ) =, for < x < π, u y (x, π) = sin(x), for < x < π. December 3, 8 3/7

4 (a) Show that this problem admits a solution. Sol. As in exercise., we need to check that the outward flux of the solution (the integral of the normal derivative) vanishes: Q n u ds = = =. u y (x, π) dx + sin(x) dx + u x (π, y) dy sin(y) dy u y (x, ) dx u x (, y) dy Hence, the problem admits a solution (infinitely many up to adding a constant). (b) We now want to split the problem into two different problems such that the Neumann condition is zero in opposite sides. In order to do that, though, we need to make sure that the arising problems can still be solved (namely, the outward flux must be zero). For that, let us consider the function v = u + a(x y ), for some a R to be determined. That is, we have subtracted an harmonic polynomial such that the flow of the normal derivative in opposite sides is non-zero. What is the problem solved by v? Write it in terms of the constant a R. Sol. Notice that v x (, y) = u x (, y) + a(x y ) x x= = + ax x= =. Similarly, we have that v x (π, y) = u x (π, y) + ax x=π = sin(y) + aπ, v y (x, ) = and v y (x, π) = sin(x) aπ. Thus, v solves the problem v =, for < x, y < π, v x (, y) =, for < y < π, v x (π, y) = sin(y) + aπ, for < y < π, v y (x, ) =, for < x < π, v y (x, π) = sin(x) aπ, for < x < π. (c) Split the problem for v into two different problems with zero Neumann conditions on opposite sides of the domain. Determine the value of a R for which such problems can be solved. Sol. We split v = v + v, where v and v solve the problems (v ) =, for < x, y < π, (v ) x (, y) =, for < y < π, (v ) x (π, y) = sin(y) + aπ, for < y < π, (v ) y (x, ) =, for < x < π, (v ) y (x, π) =, for < x < π. 4/7 December 3, 8

5 and (v ) =, for < x, y < π, (v ) x (, y) =, for < y < π, (v ) x (π, y) =, for < y < π, (v ) y (x, ) =, for < x < π, (v ) y (x, π) = sin(x) aπ, for < x < π. The flux for each problem must be. In particular, we must have that for v, (sin(y) + aπ) dy = aπ = [ cos(y)] π = a = π. We get the same result for v, either by the same computation, or by symmetry. Thus, each of the previous two problems can be solved if a = π. December 3, 8 5/7

6 .4. Practice Exercise: Exam Summer 8 Let u solve the Laplace equation with Dirichlet boundary conditions (a) Find the explicit formula for u. u = in [, π] [, π], u(x, ) = sin x for x π, u(x, π) = sin x + sin(x) for x π, u(, y) = u(π, y) =, for y π. (b) Is there a point (x, y) (, π) (, π) such that u(x, y) =? Hint: if possible, try use the strong maximum principle. If you do not like this option, then use the explicit formula found in point (a). Solution of (a). We look for a solution of the form u(x, y) = n sin(nx) [ A n sinh(ny) + B n sinh ( n(y π) )]. Note that this formula guarantees that u = and u(, y) = u(π, y) =, so it is enough to impose the boundary conditions at y = and y = π. From the boundary conditions we get sin x = u(x, ) = B n sin(nx) sinh( nπ) = B n sin(nx) sinh(nπ), n n sin x + sin(x) = u(x, π) = n A n sin(nx) sinh(nπ), therefore A = B = sinh(π), B n = n, sinh(π), A = sinh(π) A n = n 3. Hence u(x, y) = sinh(π) sin x sinh y + sin(x) sinh(y) sin x sinh(y π). sinh(π) sinh(π) Solution of (b): first possible solution. Since sin(x) = sin x cos x sin x for x [, π], 6/7 December 3, 8

7 we see that, for x [, π], u(x, ) = sin x and u(x, π) = sin x + sin(x) = sin x + sin x cos x. This proves that u on the boundary of [, π] [, π], hence u inside [, π] [, π] by the maximum principle. In addition, by the strong maximum principle, if u(x, y) = for some point (x, y) (, π) (, π), then u would be identically zero, which is in contradiction to the boundary data. This proves that there is no point (x, y) (, π) (, π) such that u(x, y) =. Solution of (b): second possible solution. We note that ey + e y e π + e π y [, π] sinh y sinh π sinh(y) sinh(π) y [, π] and that thus sin(x) = sin x cos x sin x x [, π], sinh(π) sin x sinh y + sin(x) sinh(y) (x, y) [, π] [, π]. sinh(π) Since sinh(π) sin x sinh(y π) = sin x sinh(π y) > sinh(π) (x, y) (, π) (, π), this proves that there is no point (x, y) (, π) (, π) such that u(x, y) =. December 3, 8 7/7