Math 5520 Homework 2 Solutions

Transcription

1 Math 552 Homework 2 Solutions March, 26. Consider the function fx) = 2x ) 3 if x, 3x ) 2 if < x 2. Determine for which k there holds f H k, 2). Find D α f for α k. Solution. We show that k = 2. The formulas are 6x ) 2 if x, Dfx) = and D 2 fx) = 6x ) if < x 2 2x ) if x <, 6 if < x 2. Note first that f L 2. It suffices to show that D 2 f is the weak derivative of Df, since Df is continuous and clearly has a finite L 2 -norm. Then k 3 is immediate, since D 2 f is discontinuous and the Sobolev Imbedding Theorem says that D 2 f must be continuous for k = 3 to hold. Alternatively, one may show directly that the formula for D 3 f must involve a δ-type distribution, which is not an L 2 -function, and thus there is no third weak derivative for f. In the latter case, one should be careful not to simply say that the intuitive piece-wise definition of D 3 f does not satisfy the definition; you must argue that there is no such L 2 function. To show that D 2 f is the weak derivative of Df, we check that it satisfies the definition. Let φ C, 2) be chosen arbitrarily. If follows that Df)Dφ) dx = 6x ) 2 Dφ) dx + 6x )Dφ) dx = 6x ) 2 φx) ) x= x= + 6x )φx)) x=2 x= = = D 6x ) 2) φ dx 2x )φ dx 6φ dx D 6x )) φ dx D 2 f ) φ dx D 2 f ) φ dx = D 2 f ) φ dx,

2 as required by the definition. Clearly, D 2 f L 2, 2) and so we have shown that f H 2, 2). 2. Prove that in a Hilbert space,, ) /2 defines a norm. Solution. The scalar field is not specified but let us assume it to be the real numbers, since this is the field always used in the course. Results shown with other fields are fine too. For notational purposes, let, ) and let H denote the Hilbert space. Claim : u for all u H and u = u =. The claim follows immediately from the properties of the inner product: u, u), u, u) = u =. Claim 2: ku = k u for all u H and all k R. Indeed, via the properties of the inner product over the reals, ku 2 = ku, ku) = k 2 u, u) ku = k u. Claim 3: u + v u + v for all u, v H. It is fine to cite the Cauchy-Schwarz inequality, since it can be proved without using the fact that the inner product induces a norm. Then u + v 2 = u + v, u + v) = u 2 + u, v) + v, u) + v 2 u u, v) + v 2 u u v + v 2 = u + v ) 2, thus proving the desired result. 3. Let H be a Hilbert space. Prove the parallelogram law x + y 2 + x y 2 = 2 x y 2, x, y H. Solution. The result follows directly from the properties of the underlying inner-product for the Hilbert space, say, ): x + y 2 + x y 2 = x + y, x + y) + x y, x y) = 2x, x) + 2y, y) + x, y) + y, x) x, y) y, x) = 2 x y Let X and Y be normed vector spaces. Prove that a linear operator T : X Y is bounded if and only if it is continuous. Solution. There is some room for debate about the way continuous is even defined in this context, but I will provide an example that is common, arguably. Let X and Y denote the norms for spaces X and Y, respectively. Continuity shall mean that there exists a number M < such that T x ) T x 2 ) Y M x x 2 X, 2

3 independent of the choices of x X and x 2 X. Boundedness shall mean that there exists a number T < such that T x) Y T x X, independent of the choice of x X. Proof that boundedness implies continuity. Well, the spaces are linear and T is linear, so for arbitrary choices of x X and x 2 X, also x x 2 X and T x ) T x 2 ) Y = T x x 2 ) Y T x x 2 X. Then take M T. Proof that continuity implies boundedness. Since X is linear, X and since T is linear, T ) =. Let x X be arbitrary. Then Then take T M. T x) Y = T x) T ) Y M x X = M x X. 5. Let H be a Hilbert space and L H. Prove that the null space of L, N L, is a closed subspace of H. Prove also that NL is a closed subspace of H. Solution. The norm on H is denoted by. Let x n N L H for n =, 2,... be a sequence such that x n x as n for some x H. Then since L H, L is continuous and we have Lx) = lim n Lx n) =, since Lx n ) = for all n. It follows that x N L and hence N L is closed. Furthermore, L is linear, so that if u and v are arbitrary elements of N L and α, β are arbitrary scalars, then Lαu + βv) = αlu) + βlv) = and αu + βv N L. Since L is linear, L) = and N L as well. Then N L is a closed) linear subspace of H. Similarly, for arbitrary u N L, v, w NL and scalars α and β, we have αv + βw, u) = αv, u) + βw, u) =, so that αv + βw NL. Clearly, N L. Then N L is a linear subspace of H. Now let x n NL H for n =, 2,... be a sequence such that x n x as n for some x H. Given arbitrary y N L, y, x n ) = holds for all n and thus y, x) = y, x) y, x n ) = y, x x n ) y x x n, as n. Thus y, x) = and x N L ; N L is a closed linear subspace of H. 6. Let H be a Hilbert space and let G be a closed subspace of H. Prove that the orthogonal projection operator P G has the following properties: a) P G is a linear operator on H b) P 2 G = P G c) P G = d) I P G is the orthogonal projection operator onto G 3

4 Solution. There are many valid approaches. Here is one. First, recall that in class we showed H = G G. Algebraically, given any x H we have x = P G x + I P G )x, and P G x G by definition of P G. Since x = P G x + x for some unique x G, x = I P G )x and hence I P G ) : H G must therefore be the orthogonal projection operator onto G, using the definition of the orthogonal projection operator from class. Now, let me explain another way. Note that G is a closed subspace of H details as in Problem 5 above) and G = G, trivially. Thus, if we define S = G it follows that H = S S is an orthogonal decomposition of H. The above arguments show that I P G ) : H S is the orthogonal projection operator onto S = G. Thus d) holds. Next, note that if x, y H are arbitrary and α, β are arbitrary scalars, then αx + βy = α P G x + I P G )x) + β P G y + I P G )y) = αp G x + βp G y) + αi P G )x + βi P G )y). Here, αp G x + βp G y G and αi P G )x + βi P G )y G. But also αx + βy = P G αx + βy) + I P G ) αx + βy). Since the decompositions are unique, we have that P G αx + βy) = αp G x + βp G y and also that I P G ) αx + βy) = αi P G )x + βi P G )y, proving a), as well as an analogous result for I P G. The result b) holds by noting simply that P G x = P G x + = P G P G x) + I P G )P G x), so that by uniqueness of the decomposition, P 2 G = P G and also I P G )P G = ). Finally, for arbitrary x H we have x 2 = P G x+i P G )x, P G x+i P G )x) = P G x 2 + I P G )x 2 P G x 2, since P G x, I P G )x) =, hence P G x x. Therefore, P G x P G sup. x x In case G = H, there is a problem because in fact P G =, but if G is nontrivial which I should have specified), then there exists x with P G x. Thus, for y = P G x we have This proves c). P G y y = P 2 G x P G x = P Gx P G x =. 7. Modify your code from Homework to solve the problem with mixed boundary conditions u) = α, a)u ) = β. 4

5 Perform the tasks as specified in Homework, part 2c), except for the plots. Instead, just plot the error field ux) u h x) for the coarsest grid, where u h x) is your FE approximation. The error plot with elements is shown in Figure. The norms of the error are displayed in Table. Your results should be quite close to what is shown, except your plot only requires error at the mesh nodes. In Figure, the error plot is not just the error at the mesh nodes; I have shown what happens in-between the mesh nodes with a higher-resolution plot. Also, note that at x = the error is not zero, since we no longer have a Dirichlet condition here like for homework. Figure : Error plot with elements. Note the behavior of the error at x = and in-between mesh nodes. In terms of convergence rates, we observe in Table precisely the optimal-order results that one expects for our smooth solution. Results could vary somewhat depending on how you coded your methods and calculated the errors. But the convergence rates should be optimal. u u h L 2 Rate u u h H Rate E E.27E E E E E E E E E E 3. Table : The convergence rates are optimal in both norms. 5