Regularity Theory a Fourth Order PDE with Delta Right Hand Side

Transcription

1 Regularity Theory a Fourth Order PDE with Delta Right Hand Side Graham Hobbs Applied PDEs Seminar, 29th October 2013

2 Contents Problem and Weak Formulation Example - The Biharmonic Problem Regularity Theory Application - Biomembrane Deformation Further Work

3 Problem Formulation We will consider the following PDE κ 2 u σ u = δ X in where κ > 0, σ 0 and R 2 is a bounded, Lipschitz domain. The right hand side δ X is the delta function centred at the point X. We impose some appropriate homogeneous boundary conditions which will lead to the weak formulation.

4 Weak Formulation We use integration by parts to derive a weak formulation and the possible homogeneous boundary conditions. (κ 2 u σ u)v = + = + κ u v + σ u v ( u) ν v + u ν v κ u v + σ u v ( u) ν v + u v v + u ν ν

5 Weak Formulation - Boundary Conditions We thus pose the problem in H 2 () and impose boundary conditions by considering the problem in a subspace V H 2 (). We take the first boundary condition to be u = 0. The boundary integral now vanishes if we choose the second boundary condition as follows. Dirichlet boundary conditions: u = u ν = 0 on Navier boundary conditions: u = u = 0 on

6 Weak Formulation - Imposing Boundary Conditions For Dirichlet boundary conditions we take the test space to be: { V = v H 2 () v = v } = 0 on = H0 2 (). ν For Navier boundary conditions we take the test space to be: { } V = v H 2 () v = 0 on = H 2 H0 1 ().

7 Weak Formulation - Left Hand Side We have now constructed the left hand side of the weak formulation: a(u, v) := κ u v + σ u v. Notice that a : V V R is bilinear, bounded and coercive.

8 Weak Formulation - Right Hand Side For the right hand side we follow the same process and integrate the right side against a test function v V. δ X v = v(x )

9 Weak Formulation - Right Hand Side For the right hand side we follow the same process and integrate the right side against a test function v V. δ X v = v(x ) This doesn t make sense!

10 Weak Formulation - Right Hand Side For the right hand side we follow the same process and integrate the right side against a test function v V. δ X v = v(x ) This doesn t make sense! We must interpret δ X as an element of the dual space, V.

11 Weak Formulation - Right Hand Side By the Sobolev embedding theorem we have H 2 () C(). We then interpret δ X as l V such that l(v) = v(x ). Notice that l(v) v C() C v V, so indeed l V.

12 Weak Form We may thus state the weak form of our problem; Find u V s.t. a(u, v) = l(v) v V. By the Lax-Milgram theorem we have the existence of a unique solution u V.

13 An Example - The Biharmonic Problem To gain an idea of the regularity of solutions consider the problem 2 u = δ in = B(0; 1) Beginning with Navier boundary conditions, we use the Green s function of the laplacian to construct a radial solution. The Green s function is given by: Φ(r) = 1 2π ln(r). So we find u(r) such that 1 r r ( r u ) = Φ(r). r

14 An Example - The Biharmonic Problem We then conclude that u has the form u(r) = r 2 [ ] ln(r) 1 + a ln(r) + b. 8π If u H 2 () it is continuous and thus a = 0. We then enforce the zero boundary condition and so b = 1/8π. Notice that u C 1,γ () for all γ [0, 1). The same result holds for Dirichlet boundary conditions.

15 Regularity for General Problem In fact the same regularity result holds for the general problem. Theorem Suppose has C 3 boundary and u V is the weak solution of κ 2 u σ u = δ X Then u W 3,p () for all p (1, 2) and hence u C 1,γ ( ) for all γ [0, 1).

16 Proof Define T : L 2 () C() by Tf = v f such that a(v f, v) = (f, v) L 2 () v V (1) And thus define the adjoint operator T : C() L 2 (). Hence, for any f L 2 () Thus u = T (δ X ). (T (δ X ), f ) L 2 () = δ X [Tf ] = v f (X ) = a(u, v f ) = (u, f ) L 2 ()

17 Proof Now let ψ C0 () and find v H1 0 () such that κ v σv = ψ. By elliptic regularity, v is smooth, we will only require v H 4 (). uψ = = u ψ u[κ 2 v σ v] = (T δ X, κ 2 v σ v) = δ X [v] = v(x)

18 Proof = uψ v C(, p) ψ W 1,p () for p (1, 2) Where ψ W 1,p () is given by ψ[w] = ψw. C0 () is dense in W 1,p () thus u W 1,p (). By elliptic regularity u W 3,p (), hence u C 1,γ ( ).

19 Application - Biomembrane Deformation If we consider an elastic membrane which is deformed by point forces we are lead to the energy functional: R P + PN + E(u, X± ) = Ω κ2 u 2 + σ2 u 2 + α N j=1 u(xj ) β j=1 u(xj ) Elastic Energy I I A A A AU Coupling Energy Energy minimisation produces our fourth order equation. Gradient flow produces particle movement X ± u(x ± ). j j

20 Further Work - Eighth Order Equations Another model for biomembrane deformation considers the coupling energy: E C (u, X ± ) = α Du(X + j ) β Du(X j ). N + j=1 We add higher order terms to the elastic energy and arrive at an equation of the form: N j=1 κ 8 4 u κ 6 3 u + κ 2 u σ u = Dδ X