Exercises - Chapter 1 - Chapter 2 (Correction)

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1 Université de Nice Sophia-Antipolis Master MathMods - Finite Elements - 28/29 Exercises - Chapter 1 - Chapter 2 Correction) Exercise 1. a) Let I =], l[, l R. Show that Cl) >, u C Ī) Cl) u H 1 I), u DĪ). 1) Let u DĪ). Let x,y Ī. The fondamental theorem of analysis gives ux) = uy) + x y u s)ds, which leads to l ux) uy) + u s) ds, By means of Cauchy-Schwarz inequality, one gets ux) uy) + l l u s) ds) Integrating the above inequality over [, l] in the y variable, and applying once again the Cauchy-Schwarz inequality, give or equivalently l ux) l ) 1 l uy) 2 2 l dy + l l u s) ds) 1 2 2, ux) max 1/ l, ) ) l u L 2 I) + u L 2 I). By using the the Cauchy-Schwarz inequality in R 2, denoting Cl) = 2 max 1/ l, ) l, and taking the supremum on x Ī, one obtains the result. Conclude that H 1 I) C Ī) in dimension 1. Let u H 1 I). Since DĪ) is dense in H1 I) for the norm H 1 I), there exists a sequence u n ) n DĪ) which converges to u in H1 I) for the norm H 1 I). Then the inequality 1) holds for each fonction u n : Cl) >, u n C Ī) Cl) u n H 1 I), n. 2) Since u n ) n is a Cauchy sequence in H 1 I), the sequence u n ) n is Cauchy in C Ī) for the uniform norm C Ī) by the inequality 2). The space C Ī) equipped with the norm C Ī) is complete: there exists w C Ī) such that u n) n converges to w in C Ī) for the norm C Ī). 1

2 On the one hand u n ) n converges to w in C Ī) for the norm C Ī) implies u n C Ī) w C Ī), and u n) n converges to u in H 1 I) for the norm H 1 I) leads u n H 1 I) u H 1 I). On the other hand u n ) n converges to u in H 1 I) for the norm H 1 I), leads to u n u a.e.. The sequence u n ) n converges to v in C Ī) for the norm C Ī), implies u n w in each point of Ī. This implies u = w a.e.. Then w is the continuous representant of the class u in H 1 I). Finally H 1 I) C Ī) in dimension 1. b) Let Ω = B,1/2) be the ball of radius 1/2 about the origine,) in R 2. Let v be the function defined on Ω by k vx) = ln x,k R. Study the continuity of v in the neighbourhood of the origine,), and then prove that for k < 1/2, v H 1 Ω). Conclude. Continuity lim ln x = + implies v is not bounded in the neighbourhood of the origine,) for x,) k >. Then v is continuous in Ω = B,1/2) for k. Regarding the belonging of v to H 1 Ω) The derivative of v with respect to x i, i = 1,2, is v x) = signln x )k x i x i x 2 k 1. ln x By the change of variables from cartesian coordinates to polar coordinates B, 1/2) ],1/2[ ],2π[, x = x 1,x 2 ) r,θ), one gets 2k 2 /2 v 2 H 1 I) = 2π 2k /2 ln r ln r r dr + 2πk 2 r 2 r dr. lim r = and for all α R,α, lim rln r) α =, imply the first integral is finite for all r r k R. The second integral is finite if only if 2k < i.e. k < 1/2. Then for k < 1/2, v H 1 Ω). Finally k < 1/2, v H 1 Ω) and v is not continuous in Ω. Exercise 2. Let be the following boundary value problem u x) = fx) on [,1], u) =, u 1) = α, 3) where f is a given function of L 2,1) and α R. a) Let V = {v H 1,1),v) = }. Prove that v 1,Ω = Ω = [,1], is a norm on V and V is a Hilbert space. ) 1 Ω v x) 2 dx 2, where 2

3 The critical point to prove that v 1,Ω = ) 1 Ω v x) 2 dx 2 is a norm, is to show that v 1,Ω = implies v = in H 1,1). Let v to be such a function. Then v = a.e. which implies v = constant a.e.. In dimension 1, v has a continuous representant, let call it w. Since v) =, one has w) = = constant and w =. Therefore v = w = a.e.. The norm v 1,Ω is equivalent to the norm v H 1,1)on H 1,1). v 1,Ω v 2 1,Ω + v 2 L 2,1)) 1 2 = v H 1,1). The converse inequality is obtained as follows. Let v V, then v H 1,1) in dimension 1 and v owns a continuous representant, let call it v. Then one has which implies vx) = v) + x v y)dy, x [,1], 1 vx) x2 v y) dy) 1 2 2, x [,1], Taking the square of the above inequality and then integrating in x over,1), one gets which leads to vx) 2 dx 1 2 v y) 2 dy, v 2 H 1,1) 3 2 v 2 1,Ω. 4) Let γ be the mapping γ : H 1,1) R, v v). Since H 1,1) C [,1]) in dimension 1, the mapping γ is well-defined. The mapping γ is linear and γv) = v) v C [,1]) C v H 1,1), where C >. This shows that γ is continuous. Then V = Ker γ is a closed subset of the Hilbert H 1,1). Therefore V is a Hilbert space. b) Give the variational problem and show that it has a unique solution. The variational problem Let v D[,1]). Multiplying the PDE 3) by v and integrating over [,1], one gets u x)vx)dx = fx)vx)dx. Integrating by part the left-hand side, one gets ) u x)v x)dx u 1)v1) u )v) = By taking v V et using u 1) = α, one obtains u x)v x)dx = α v1) + 3 fx)vx)dx. fx)vx)dx.

4 Let a : V V R, u,v) u x)v x)dx, L : V R, v αv1) + fx)vx)dx. The variational problem is the following: Find u V solution of v V, au,v) = Lv). 5) The solution of the variational problem The space V is a Hilbert space. The mapping a is a bilinear symmetric, continuous, coercive form on V : au,v) u 1,Ω v 1,Ω u,v V, av,v) = v 2 1,Ω v V. The mapping L is a linear continuous form on V : Lv) α v1) + f L 2,1) v L 2,1) α v C [,1]) + f L 2,1) v H 1,1) α C v H 1,1) + f L 2,1) v H 1,1) max α C, f L 2,1)) v H 1,1) β v 1,Ω, v V, 3 with β = 2 max α C, f L 2,1)). Finally thanks to Lax-Milgram theorem, there exists a unique solution u for the above variational problem. c) Recover formally the initial problem. Recovering the PDE Let u be the solution of the variational problem 5). Let one takes v D],1[) V in the variational problem 5). u x)v x)dx = α v1) + Since v D],1[), one has v1) = and u x)v x)dx = fx)vx)dx. fx)vx)dx. 6) One has also f L 2,1) D ],1[), u L 2,1) D ],1[). Then 6) rewrites as Then the PDE is u, v D D],1[) = f, v D, D],1[), or u, v D D],1[) = f, v D D],1[), or u f, v D D],1[) =. Since f L 2,1), the above PDE turns into u f = in D ],1[). u = f in L 2,1) a.e.. 7) 4

5 Recovering the boundary conditions Let one takes v V, multiplying 7) by v and integrating one gets u x)v x)dx = Then integrating by part and using v) =, one obtains fx)vx)dx. u x)v x)dx u 1)v1) = fx)vx)dx. The comparison of the above equation with the variational problem 5) gives u 1) = α. Obviously one has u) = since u V. Therefore the boundary value problem is u = f a.e. on [,1], u) =, u 1) = α, 8) where f is a given function of L 2,1) and α R. Exercise 3. a) Give the variational formulation of the boundary value problem u x) + ux) = fx) on [,1], u ) =, u 1) =, where f is a given function of L 2,1). Let V = H 1,1). Let a : V V R, u,v) u x)v x)dx + ux)vx)dx, L : V R, v fx)vx)dx. The variational problem of the PDE 3) is the following: Find u V solution of v V, au,v) = Lv). 9) b) Show that the variational problem has a unique solution. The solution of the variational problem The space V = H 1,1) is a Hilbert space. The mapping a is a bilinear symmetric, continuous, coercive form on V : au,v) u L 2,1) v L 2,1) + u L 2,1) v L 2,1), by Cauchy-Schwarz inequality in L 2,1) ) 1/2 1/2 u 2 L 2,1) + u 2 L 2,1) u 2 L 2,1) + u 2 L,1)), by Cauchy-Schwarz 2 = u H 1,1) v H 1,1) u,v V, av,v) = v 2 H 1,1) v V. inequality in R 2 5

6 The mapping L is a linear continuous form on V : Lv) f L 2,1) v L 2,1) f L 2,1) v H 1,1), v V. By Lax-Milgram theorem, there exists a unique solution u for the above variational problem. c) Recover formally the initial problem. Recovering the PDE Let u be the solution of the variational problem 9). Let one takes v D],1[) V in the variational problem 9). Since f L 2,1) D ],1[), u L 2,1) D ],1[), the variational problem 9) turns to Then the PDE is u, v D D],1[) + u, v D D],1[) = f, v D D],1[), or u, v D D],1[) + u, v D D],1[) = f, v D D],1[), or u + u f, v D D],1[) =. u + u f = in D ],1[). Since f L 2,1) and u L 2,1), one has u L 2,1) and the above PDE turns into u + u = f in L 2,1) a.e.. 1) Recovering the boundary conditions Let one takes v V = H 1,1), multiplying 9) by v and integrating one gets Then integrating by part gives u x)v x)dx u x)v x)dx = fx)vx)dx. ) u 1)v1) u )v) = fx)vx)dx. Then the comparison of the above equation with the variational problem 9) gives u ) = when taking v V = H 1,1) with v1) = and v). Then when taking v V = H 1,1) with v), one gets u ) =. Therefore the boundary value problem is u + u = f a.e. on [,1], u ) =, u 1) =, 11) where f is a given function of L 2,1). Exercise 4. Let Ω be a bounded subset of R n where n 1, with regular boundary Ω. Let be the following problem: Find u H 1 Ω) solution of v H 1 Ω), u v dx = Ω where f is a given function of L 2 Ω). Ω fv dx 12) 6

7 Show that this problem has a unique solution and give the associated initial boundary value problem. The solution of the variational problem Let a : H 1Ω) H1 Ω) R, u,v) Ω u v dx, L : H1 Ω) R, v Ω fv dx. The space V = H 1 Ω) is a Hilbert space. The trace mapping γ : H 1 Ω) L 2 Ω), u u Ω is linear continuous. Its kernel Ker γ = H 1Ω) is a closed subset of the Hilbert space H1 Ω). Therefore H 1 Ω) is a Hilbert space. Thanks to Poincaré theorem, the mapping u u 1,Ω = equivalent to the norm H 1 Ω) on H 1 Ω). ) /2 Ω ux) 2 dx is a norm The mapping a is a bilinear symmetric, continuous, coercive form on H 1 Ω): au,v) u L 2 Ω) v L 2 Ω) = u 1,Ω u 1,Ω, u,v H 1 Ω), av,v) = v 2 1,Ω v H1 Ω). The mapping L is a linear continuous form on H 1Ω): Lv) f L 2 Ω) v L 2 Ω) f L 2 Ω) v H 1 Ω) C f L 2 Ω) v 1,Ω, v H 1 Ω). By Lax-Milgram theorem, there exists a unique solution u for the above variational problem. Recovering the PDE Let u be the solution of the variational problem 12). Let one takes v DΩ) H 1 Ω) in the variational problem 12). Since f L 2 Ω) D Ω), u L 2 Ω) D Ω), the variational problem 12) turns to u, v D DΩ) = f, v D DΩ), or u, v D DΩ) = f, v D DΩ), or u f, v D DΩ) =. Then the PDE is u f = in D Ω). Since f L 2 Ω) and u L 2 Ω), one has u L 2 Ω) and the above PDE turns into u = f in L 2 Ω) a.e.. 13) Recovering the boundary conditions The solution of the the variational problem 12) u H 1 Ω), then u = on Ω. Therefore the boundary value problem is { u = f a.e. in Ω, u = on Ω, 14) where f is a given function of L 2 Ω). 7