LECTURE # 0 BASIC NOTATIONS AND CONCEPTS IN THE THEORY OF PARTIAL DIFFERENTIAL EQUATIONS (PDES)

LECTURE # 0 BASIC NOTATIONS AND CONCEPTS IN THE THEORY OF PARTIAL DIFFERENTIAL EQUATIONS (PDES) RAYTCHO LAZAROV 1 Notations and Basic Functional Spaces Scalar function in R d, d 1 will be denoted by u, v, z, f etc Vector functions in R d will be denoted by u, b etc with components u i and b i, i = 1,, d, correspondingly Also we shall use the standard notation for a dot (scalar) product of two vector functions u and b: u b = u 1 b 1 + u 2 b 2 + + u d b d and the operator = ( x 1,, x d ) We shall use the following widely accepted notations for the differential operators used in this class (see, eg [1, 3]): (1) Gradient of a scalar function u: ( ) u :=,, u = (u x1,, u xd ), (obviously this is a vector); x 1 x d Here an further u xi denotes u x i (2) Divergence of a vector function u = (u 1,, u d ) u := u 1,x1 + + u d,xd ; divergence of a vector function is a scalar function (3) Laplace s operator u = ( u) = u x1 x 1 + + u xd x d Further, in we shall consider various function spaces: (1) C() = {u : R u is continuous on }; (2) C k () = {u : R u is k-times continuously differentiable in }; (3) C () = {u : R u is infinitely differentiable in }; (4) C 0 (), C0 k(), C 0 () the above classes of function with compact support in (5) For 1 p < L p () = {u : R u is Lebesgue measurable u L p () < }, where ( ) 1 u L p () = u p p dx We say that u L 1 loc () if u L 1 (D) = D u dx < for any subdomain D Further, we introduce the notion of generalized (weak) derivative of u We say that xi u = u x i L 1 loc () is a weak derivative of u if xi uφdx = u xi φ dx, φ C0 () Date: April 2, 2013 1

2 RAYTCHO LAZAROV Example: Show that for = (0, 1) { 0 x < 1 u(x) = 2 x 1 2 x > 1 2 has weak derivative u (x) = { 0 x < 1 2 1 x > 1 2 Now we define the Sobolev space W 1,p () as set of functions u L p () having their weak derivatives of first order in L p () We shall work with the Hilbert space H 1 () := W 1,2 () Then the norm in H 1 () is defined by Further, we shall use the space u 2 H 1 = u 2 L 2 () + u 2 L 2 () H 1 0 () = {u H 1 () : u(x) = 0, x } Obviously, H 1 () is a subspace of L 2 () We shall further use two important properties for these two spaces (a) Poincaré-Friedrichs inequality: there is a constnat c 0 > 0, dependent only on the domain, such that u H 1 0 () : c 0 u 2 L 2 () u 2 L 2 (), u H 1 () : c 0 u 2 L 2 () u 2 L 2 () + ( udx) 2 (b) H 1 () is compactly embedded in L 2 (): this means that from any bounded in L 2 () sequence one can extract a convergent in H0 1 () subsequence; In a similar way one can define higher order weak derivatives and Sobolev spaces of any integer order k To introduce these we need some notations for higher order partial derivatives Denote by α = (α 1,, α d ) a multi-index and α i 0 integer numbers, α = α 1 + + α d Then D α α u 2 u u := x α 1 1, eg = D α u, α = (1, 1) xα d d x 1 x 2 Then the weak α derivative of u is defined as D α uφdx = ( 1) α u D α φ dx, φ C0 () Now we define the Sobolev space W k,p () as set of functions u L p () having their weak derivatives of order k in L p () We shall work with the Hilbert space H k () := W k,2 () Then the norm in H k () is defined by u 2 H = D α u 2 k L 2 () α k We shall use the following basic result, called Sobolev lemma: Lemma 1 (Sobolev s lemma, [5, more general form on page 107]) Assume that the region has Lipshitz boundary and k > d/2 Then the elements of H k () are bounded continuous functions on and H k () is continuously embedded in C( )

LECTURE # 0: BASIC NOTATIONS AND CONCEPTS 3 In the modern numerical analysis related to PDEs important role play the notion of weak solutions of the differential equation Now we formulated various PDEs in their strong form Usually, this assumes that all derivatives that are involved in the PDE are continuous (or piecewise continuous) functions in the domain 2 Basic Differential Equations (in strong form) We shall consider various problems related to the following PDEs in a bounded domain R d A list of all these equations and many more could be found in the book of L Evans [1, p 3-6], which I recommend as a basic advanced textbook on PDEs (1) Poisson Equation for u = u(x): u = f, x Appropriate boundary conditions for this equations are: Dirichlet: u = g on the boundary ; Neumann: u n = g on the boundary ; Robin type: u n + u = g on the boundary Here n is the outer unit normal vector to and g is a given function on (2) Helmholtz Equation for u = u(x): u ω 2 u = f, x, ω a real number, with appropriate boundary conditions (see, eg Poisson equation) (3) Second Order Elliptic Equation (diffusion-reaction): (K(x) u) + c(x)u = f, x with appropriate boundary conditions Here K(x) is a symmetric matrix that is uniformly positive definite in, ie there is a constant α 0 > 0 such that ξ T K(x)ξ α 0 ξ 2 for all ξ R d and x and c(x) 0 in The functions K(x), c(x) and f(x) are given on Remark: The one-dimensional variant (ie all functions depend only on the single variable x) of this equation is (k(x)u ) + c(x)u = f(x), x (0, 1) with relevant boundary conditions (4) Mixed Form of the Diffusion-Reaction Equation: Introduce the flux dependent variable q = K(x) u so that the diffusion-reaction equation reads as q + c(x)u = f This first order system is called mixed form of the equation (5) Diffusion-Convection-Reaction Equation: (K(x) u + bu) + c(x)u = f, x Proper boundary conditions for this equation are in general more complicated One simple BC is Dirichlet BC (6) Transient (Time-Dependent) Heat Equation: Here the unknown function depends on x and time t > 0, ie u(x, t): u t u = f, x, t > 0 Together with proper BC we give also an initial condition (IC): u(x, 0) = u 0 (x), where u 0 (x) is a given function in

4 RAYTCHO LAZAROV (7) Second Order Hyperbolic Equation for u = u(x, t): u tt u = f, x, t > 0, This equation describes the vibrations of a thin membrane under the transverse load f(x, t) and the initial displacement u(x, 0) = u 0 (x) and initial velocity u t (x, 0) = u 1 (x) for x Appropriate BC have to be specified as well (8) Transport Equation for u = u(x, t): u t + b u + c(x)u = f, x, t > 0 with standard initial condition u(x, 0) = u 0 (x) and BC given on the inflow part of the boundary, Γ in, namely, u n = g, x Γ in, t > 0, Γ in := {x : b(x) n(x) < 0} Sometimes it is better to have this in a conservative form u t + (b u) + c(x)u = f, x, t > 0 with the steady state case (b u) + c(x)u = f (9) Nonlinear Conservation Law for u = u(x, t): u t + b(u) = f, x R, t > 0, where b(u) is a convex function of u, eg Burger s equation for d = 1, b(u) = u 2 (10) Nonlinear Poisson Equation for u = u(x): (11) p-laplacian Equation for u = u(x): u = f(u), x ( u p 1 u) = 0, x with appropriate boundary conditions (12) Minimal Surface Equation for u = u(x): ( ) u = 0, x (1 + u 2 ) 1 2 (13) Stokes System: This is a system for the fluid velocity u and pressure p in a bounded domain : u + p = f, u = 0, with Dirichlet boundary condition for u and no BC for p (14) Equations of linear elasticity The primary dependent variable is the displacement vector u(x) = (u 1,, u d ) that describes the deformations of an elastic body occupying a volume The gradient of u is the tensor u 1 x 1 u := u d x 1 u 1 x d u d x d, ɛ = 1 2 ( u + ( u)t ), σ = Kɛ

LECTURE # 0: BASIC NOTATIONS AND CONCEPTS 5 Here ɛ is the strain tensor, σ is the stress tensor and K is a matrix giving the relation between the strains and the stresses (elements of the matrix depend on the Young and Poisson moduli) The equation then is σ = f x, is taken row-wise from σ 3 Boundary Value Problems Is is well known that in order to select a unique solution of a given PDE we need to specify some boundary conditions (BC) or/and initial conditions (IC) Eg the equation u = 1 has infinitely many solutions u(x) = 1 2 x2 + c 1 x + c 2 depending on two arbitrary constants c 1 and c 2 However, the boundary value problem u = 1, x (0, 1), u(0) = u(1) = 0 has unique solution u(x) = 1 2x(x 1) One of the most common boundary conditions are Dirichlet BC u(x) = g(x), and/or Neumann BC, u n = g(x), x Γ, where g(x) is a given function and n is the outer normal unit vector to the boundary Γ Another possibility is to have Dirichlet BC on Γ D and Neumann BC on Γ N where Γ = Γ D Γ N We shall consider differential equations and boundary conditions that constitute a well posed problem Loosely speaking, this is a problem that has unique solution and the solution depends continuously on the data in certain sense (1) 4 The notion of weak form of a boundary value problem (BVP) (a) We shall consider the following boundary value problem: find u(x) such that (K(x) u) + c(x)u = f for x and u(x) = 0 for x Γ We multiply this equation by a test function v(x) (a smooth function with compact support in the domain ) and integrate over the domain to get: ( (K(x) u) + c(x)u)v dx = fv dx After applying the Stokes theorem in the first term on the left and assuming v = 0 on Γ we get ( (K(x) u) v dx = K(x) u v dx K(x) u vds Γ = K(x) u v dx Thus, the solution u(x) satisfies the following integral identity: ( ) K(x) u v + c(x)uv dx = fv dx v H0 1 () Now if K(x) is smooth (in fact it is enough to be measurable and bounded in ) then the above integral identity is well defined for u, v H 1 0 () To simplify the notation and to reduce some extensive writing the weak formulation of the BVP can be written as (2) Find u H 1 0 () such that a(u, v) = l(v) v H 1 0 (), where the bilinear form a(u, v) and the linear from l(v) are defined by ( ) (3) a(u, v) := K(x) u v + c(x)u v dx and l(v) := fv dx

6 RAYTCHO LAZAROV Note, that u H0 1 () implies that the solution u(x) = 0 on the boundary Γ This boundary condition is often called essential boundary condition (b) Consider the same problem with nonhomegeneous Dirichlet data in a mixed form: q + K(x) u = 0, q + c(x)u = f, x, u(x) = g(x), x Γ Rewrite the first equation in the form K 1 (x)q + u = 0, multiply the last one by a test vector-function r, integrate over, apply the Stokes theorem to the second term and take into account the boundary condition to get K 1 (x)q rdx u r dx + g r nds = 0 Further, we multiply the second equation by a test function v and integrate over to get q vdx + c(x)uv dx = fv dx Obviously, these integral identities make sense for vector fields such that the divergence q exists in a weak sense and belongs to L 2 () Now we define the space This space is endowed with the norm H(div; ) = {r L 2 () d : r L 2 ()} r 2 H div = r 2 L 2 + r 2 L 2 Then the weak form of the mixed system is: find q H(div) and u L 2 such that the integral identity (K 1 (x)q, r) (u, r) ( q, v) (cu, v) = g, r n (f, v) is satisfied for all r H(div) and v L 2 Here we have used the shorthand notation K 1 (x)q rdx = (K 1 (x)q, r), fv dx = (f, v) and gvds = g, v Note that the solutions q and u are not required to satisfy any boundary conditions The boundary conditions have become part of the weak formulation, ie they are included in the weak form Such boundary conditions are called natural BC for the setting 5 Linear Partial Differential Equations of Second Order All equations considered above are linear This means that if we introduce the operator, for example Lu := (K(x) u + bu) + c(x)u then it satisfies the relation L(u + v) = Lu + Lv In more general cases we consider the linear differential operator d 2 u d u Lu := K ij (x) + b i, x R d x i x j x i i,j=1 with given coefficients K ij and b In analogy with the classification of conic sections, at point x the differential equation is called elliptic, if the eigenvalues of the matrix K(x) with entries K ij are not zero and have the same sign; i=1 Γ Γ

LECTURE # 0: BASIC NOTATIONS AND CONCEPTS 7 hyperbolic, if one eigenvalue is positive and the others are all negative (or if one eigenvalue is negative and the others are all positive); parabolic, if exactly one eigenvalue is equal to 0 6 Exercises: Problem 1: (see [4]) Decide whether the operator is linear or nonlinear Lu := u x1 x 1 + x 1 u x2, u = u(x), x R 2, Lu := u tt u xx + u 2, u = u(x, t), x R, Lu := u t + u xx + 1 + u, u = u(x, t), x R Problem 2: (see [4]) Determine the type of the differential operator (x R 2 ): Lu := u x1 x 1 u x1 x 2 + 2u x2 + u x2 x 2 3u x2 x 1 + 4u; Lu := 3u x2 + u x1 x 2 Lu := 9u x1 x 1 + 6u x1 x 2 + u x2 x 2 Problem 3: Consider the following two-points boundary value second order problem in 1-D: Find a function u defined ae in ]0, 1[ such that (4) ( xk(x)u (x) ) + xc(x)u(x) = xf(x) ae in ]0, 1[, ( xu (x) ) = 0 and K(1)u (1) + u(1) = 0, lim x 0 where K C 1 ([0, 1]), q C 0 ([0, 1]) and f L 2 (0, 1) are given functions Assume that there exists a constant κ 0 > 0 such that K(x) κ 0 and c(x) 0 for all x [0, 1] (1) Assume that the Sobolev space V is obtained by completion of the continuously differentiable functions in ]0, 1[, C 1 ([0, 1]), with respect to the norm ( v V = 1/2 xv 2 L 2 (0,1) + xv 2 L (0,1)) 2 Derive the weak form of the problem (4) in the space V (2) Prove that the corresponding bilinear from of the weak form of (4) is coercive in V Problem 4: Consider the boundary value problem u (4) = f(x), 0 < x < 1, u(0) = 0, u (0) = 0, u (1) + u (1) = β, u (1) = γ, where f(x) is a given function on (0, 1) and β and γ are given constants (1) Give the weak formulation of this problem in an appropriate space V and characterize V (2) Show that the corresponding bilinear form is coercive and continuous in V and the linear form is continuous in V (3) Set up a finite dimensional space V h V of piece-wise polynomial functions over a uniform partition of (0, 1) Define the nodal basis in terms of the degrees of freedom (4) Introduce the Galerkin method for the problem (??) for V h ; state the error estimate in the V -norm assuming smooth solution u(x) Problem 5: Consider the following boundary value problem for the biharmonic equation: find u(x) that satisfies the differential equation (5) 4 1111u + 2 4 1122u + 4 2222u = f(x), x,

8 RAYTCHO LAZAROV and homogeneous Dirichlet conditions on the boundary Γ, (6) u(x) = 0 and u(x) n = 0, x Γ Here is the unit square in R 2, Γ is its boundary, and f(x) L 2 () is given Here 1 u denotes the derivative x1 u and 2 u denotes the derivative x2 u The problem (5)-(6) describes the transverse deformation of homogeneous isotropic plate with constant thickness and clamped boundary due to external force f(x) (1) Derive a weak formulation of this problem which gives rise to a symmetric bilinear form on V = H 2 0 () {φ H2 1 () : φ = 1φ = 2 φ = 0 on Γ} (2) Show that the bilinear from is coercive on V (Hint: you may use the Poincaré inequality on H 1 0 ()) (3) Let T h be a triangulation of into triangles Consider the FE (K, P K, Σ), where: (α) K T h is a triangle determined by its three vertexes z 1, z 2, z 3, (β) P K = P 5 (K) is the set of polynomials of degree 5 over K, and (γ) Σ = {v(z i ), 1 v(z i ), 2 v(z i ), 2 11 v(z i), 2 12 v(z i), 2 22 v(z i), ν v(m i ), i = 1, 2, 3}, with m i denoting the midpoints of the triangle sides Show that the finite dimensional space V h based on the finite element (K, P K, Σ) is a subspace of H 2 () Problem 6: Let = (0, 1) and u be the solution of the boundary value problem u (4) (k(x)u ) + c(x)u = f(x) x, u(0) = u (0) = 0, u(1) = 0, u (1) + βu (1) = γ, where k(x) 0, c(x) 0, f(x), γ, and β > 0 are given data Derive the weak formulation of this problem Specify the appropriate Sobolev spaces and show that the corresponding bilinear form is coercive Problem 7: Consider the variational problem: find u H 1 (), such that a(u, v) = l(v) for all v H 1 (), where = (0, 1) (0, 1), Γ is its boundary, and 1 a(u, v) = u v dx + u(x, 0)v(x, 0) dx and l(v) = gvds (a) Derive the strong form to this problem (b) Prove that the bilinear form a(u, v) is coercive in H 1 () Problem 8: Derive weak formulation of the Stokes system 0 References [1] L C Evans, Partial Differential Equations, Graduate Studies in Mathematics, vol 19, AMS, 1998 [2] Ch Grossmann, H-O Ross, and M Stynes, Numerical Treatment of Partial Differential Equations, Springer, Berlin, 2005 [3] M Renardy and R Rogers, An Introduction to Partial Differential Equations, Texts in Applied Mathematics, Springer-Verlag, 1993 [4] P Knabner and L Angermann, Numerical Methods for Elliptic and Parabolic PDEs, Springer-Verlag, New Yrok Inc, 2003 [5] J Wloka, Partial Differential Equations, Cambridge University Press, 1992 Γ