TD M1 EDP 2018 no 2 Elliptic equations: regularity, maximum principle

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1 TD M EDP 08 no Elliptic equations: regularity, maximum principle Estimates in the sup-norm I Let be an open bounded subset of R d of class C. Let A = (a ij ) be a symmetric matrix of functions of class C on satisfying the ellipticity condition A(x)ξ ξ θ ξ, x, ξ R d, (.) where θ is a positive constant. Let f L () and let u H 0 () be the weak solution to the Dirichlet problem (we use the convention of summation over repeated indices) i (a ij j u) = f in, (.) u = 0 on. (.3). We assume d 3. Show that there exists a constant C 0 depending on and d such that u L () C( f L () + u L ()). (.4). We assume = B(0, R): the open ball of center 0 and radius R, with R <. (a) Compute v when v(x) = ψ( x ) is a radial function. (b) By considering the function u(x) = ln ln x and the case A(x) = I d, discuss the validity of (.4) when d 4. Answer: by the result of regularity for (.)-(.3), we have u H () C ( f L () + u L ()), (.5) where C depends on and d only. By Sobolev Injection Theorem, we have u L () if 4 > d and, thus, u L () C u H (), (.6) where C depends on and d only, for d 3. Combining (.5) and (.6) gives (.4) with C = C C. To compute the Laplacian of radial functions, we use the formulas v = div( v), div(λφ) = λ div Φ + λ Φ, v, λ: R, Φ: R d. (.7) Let us introduce r : x x. By the chain-rule formula, we have v(x) = ψ ( x ) x r(x). To compute r(x), we note that r(x) = x, hence r(x) =. This gives v(x) = x r(x)

2 ψ ( x ) x r(x). Then we use repeatedly the second formula in (.7) and the fact that x x = r(x), to obtain [ ] v(x) = [ψ x x ( x )] r(x) + ψ ( x ) div r(x) = ψ ( x ) x [ r(x) x div(x) r(x) + ψ ( x ) x ] r(x) r(x) r(x) [ d = ψ ( x ) + ψ ( x ) r(x) x ] r(x) x r(x) = ψ ( x ) + d r(x) ψ ( x ), i.e. [ v(x) = ψ (r) + d ] ψ (r). (.8) r r= x For u(x) = ln(ln( x )), we compute therefore, for x 0, [ x u(x) = x ln( x ) x, u(x) = r ln(r) + ] d r. ln(r) r= x For radial functions, we have an integrand r d when integrating on B(0, R). By the usual results on Bertrand s integral, we see that: u H (), u C ( \ {0}), u = f in \ {0}, where f L (). We have also u ū H0 (), where ū = ln ln(r). Therefore, considering u ū, we will have a counterexample to (.4), provided we show that u is a weak solution to (.)-(.3). We want to establish that u vdx = fvdx, (.9) for all v H0 (). Since u H0 (), it is sufficient to prove (.9) for v Cc () smooth. Note then that we have (.9), by integration by parts, for all v Cc ( \ {0}). To get (.9) for v Cc (), we will use a cut-off function ω ε defined as follows: ω ε (x) = θ(ε x ), θ C (R + ), where θ(r) = 0 for r, θ(r) = for r. We apply (.9) to vω ε. This gives u vω ε dx + u ω ε vdx = fvω ε dx. (.0) The first and third term in (.0) converge to both sides of (.9) by the dominated convergence theorem. To conclude, we have to show that the second term in (.0) tends to zero. We estimate it, in absolute value, by v L () ε ε which gives the desired result. r ln(r) ε θ (ε r) r d dr v L () ε ε ln(ε) ε θ L (ε) d d v L () θ L ln(ε), ε d

3 Hölder regularity I In this exercise (in two parts), we show a gain of (almost) two derivatives in Hölder spaces for the solution u to the Laplace equation u = f. Let f C(R 3 ) be a function with compact support. Let G(x) = 4π x. We admit that the function u defined by u(x) = G f(x) = G(x y)f(y)dy = R 3 G(y)f(x y)dy, R 3 (.) is solution to the Poisson equation u = f in R 3. We assume f C α (R 3 ) for a given α (0, ). Therefore f Ċα (R 3 ) := sup f(x) f(z) x z R 3 x z α (.) is finite. Let K be a compact of R 3.. Show that u C α (K), u Ċα (K) C f Ċα (R 3 ).. Show that u C α (K), u Ċα (K) C f Ċα (R 3 ). To prove the existence of u, you may introduce a cut-off function ω ε, ω ε (x) = θ(ε x ), θ C (R + ), where θ(r) = 0 for r, θ(r) = for r, 0 θ (r), and consider the approximation u ε = (Gω ε ) f. In Question and, the constants C, C depend on K, d and α only. 3 A semi-linear elliptic equation Let be an open bounded subset of R d of class C. Let A = (a ij ) be a symmetric matrix of functions of class C on satisfying the ellipticity condition (.). We consider the Dirichlet Problem i (a ij j u) + u = G( u) + f in, (3.) u = 0 on, (3.) where f L () and G: R d R d is a Lipschitz continuous function (G( u) is the function defined pointwise as x G( u(x))). We recall the Leray-Schauder fixed-point theorem (also called Schaefer s fixed-point Theorem): Theorem 3.. Let T : X X be a continuous and compact mapping of the Banach space X. Assume that the set {u X; u = λt (u), λ [0, ]} is bounded. Then T has a fixed-point in X. Show that, if Lip(G) is small enough compared to θ, then (3.)-(3.) has a unique weak solution in H H 0 (). You may consider the mapping v = T (u) defined on X = H 0 () by the fact that v is the weak solution to the Dirichlet problem for the equation i (a ij j v) + v = f u := G( u) + f. (3.3) It is usually deduced from the Schauder fixed-point theorem, which is a generalization in infinite dimension of the Brouwer fixed-point theorem 3

4 4 Estimates in the sup-norm II We go on Exercise. We assume now d 4 and f L p (), where p > d. Note that f L () thus. We will show an estimate somehow analogous to (.4): u L () C( f Lp () + u L ()), (4.) for a constant C that depends on d,, p only. We recall the following Sobolev inequality: and also the interpolation inequality for all r < s < t and all v L t ().. Let q satisfy Show that for all v L (). v L () C S v L (), = d, v H 0 (), (4.) v L s () v θ L r () v θ L t (), s = θ r + θ, (4.3) t < q < q + <. (4.4) v L q+ () v θ L q () v θ L (), q + = θ q + θ, (4.5). Let q be the conjugate exponent to p. Show that q satisfies (4.4). 3. Let u be the weak solution to (.)-(.3). Let ϕ: R R be a Lipschitz function such that ϕ(0) = 0 and ϕ 0 a.e. Show that ζ(u) L () θ C S f Lp () ϕ(u) Lq (), ζ(u) := u 0 ϕ (s)ds. (4.6) 4. By considering ϕ k (u) = (min(u, k) z) + for 0 z k, show that (u z) + L () and that (u z) + L () θ C S f Lp () (u z) + Lq (). (4.7) 5. Show that (u z) + L q+ () [ θ ] C S f θ L p () (u z) + +θ L q (). (4.8) 6. We introduce the function F (z) = (u z) + q+ L q+ (). Show that F (z) (q + ) [ θ ] θ +θ C S f Lp () F (z) β, β = q q + + θ <. (4.9) 7. Prove that there exists z > 0 such that F (z) = 0 for z z. What does this imply on u? 8. Establish (4.). 4

5 5 Hölder regularity II We go on the exercise. We will show that, for every δ (0, α), where the constant C 3 depends on K, d, α, δ only. D u Ċδ (K) C 3 f Ċα (R 3 ), (5.). Show that u is twice differentiable, with x ix j u(x) = x ix j G(x y)(f(y) f(x))dy f(x) B(0,R ) B(0,R ) xi G(x y)ν j (y)dσ(y), (5.) where ν(y) is the outward unit normal to B(0, R ) at y B(0, R ) (you may use the same cut-off function as in Exercise, Question ).. Show (5.). 6 Comparison for the mean curvature equation Let A: R d R d be a function of class C. We set a ij (p) = ai p j (p) and we assume that for all R > 0, there exists θ(r) > 0 such that a ij (p)ξ i ξ j θ(r) ξ, ξ R d, p B(0, R). (6.) i,j Let be an open bounded subset of R d. Let u, v Lip( ) H () satisfy the inequalities in the following weak sense: for all ϕ H 0 (), and. Show that u v a.e. div(a( u)) div(a( v)) in, u v on (6.) (A( u) A( v)) ϕdx 0, (6.3) (u v) + H 0 (). (6.4). We consider the case of the mean-curvature operator: A(p) = d. + p (a) Show that (6.) is satisfied. (b) Let u Lip( ) H () be a weak solution to the mean-curvature equation ( ) d div u = 0. (6.5) + u Show that sup u = sup u. (c) We assume = B(0, ) and let r >. Let u Lip( ) H () be a weak solution to the mean-curvature equation ( ) d div u = + u r. (6.6) Show that u(x) sup u+ r x for all x. You will admit that the function w : x r x is solution to (6.6). 5

6 7 Maximum principle: estimates on the gradient Let be an open bounded subset of R d. Let A = (a ij ) i,j d be a symmetric definite positive d d matrix (note well that a ij is independent on x). Let f Lip( ). We will establish gradient estimates for solutions u to the equation Lu = f with Dirichlet homogeneous boundary condition, where L is the elliptic operator d Lu = a ij iju, i,j= under the assumption that there exists a function ψ Lip() C () such that Lψ f in and ψ = 0 on. For simplicity, we will consider the case f = Cst > 0 only.. Let ω. Let u, v C (ω) C( ω) satisfy Lu Lv in ω. Show that. Let u C () C( ) satisfy Lu = f in. sup(u v) sup(u v). (7.) ω (a) Show that { } { } u(x) u(y) u(x) u(y) sup ; x, y, x y = sup ; x, y. (7.) x y x y ω Hint: given x, x with τ := x x 0, compare u and u τ : x u(x + τ) in ω := ( τ + ). (b) We assume furthermore u = 0 on. Show that Lip(u) Lip(ψ). 3. A ψ as above is called a barrier function. Construct a barrier function in the case = B(0, ) (you may consider ψ(x) = γ x + C for some given constants γ > 0, C R). 6