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Energy Methods in Parabolic PDE Theory

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Energy Methods in Parabolic PDE Theory Math 951 Lecture Notes Chapter 3 { Energy Methods in Parabolic PDE Theory Mathew A. Johnson 1 Department of Mathematics, University of Kansas [email protected] Contents 1 Introduction1 2 Autonomous, Symmetric Equations3 3 Review of the Method: Galerkin Approximations 10 4 Extension to Non-Autonomous and Non-Symmetric Diffusion 11 5 Final Thoughts 15 6 Exercises 16 1 Introduction Previously, we have studied the existence and regularity of solutions of uniformly elliptic PDE. Such equations are generalizations of the famous Poisson equation ∆u = f; x 2 Ω n posed on some domain Ω ⊂ R , equipped with an appropriate set of boundary conditions. We now turn our eye to study a class of \evolution equations" which generalize the well- known heat equation ut = ∆u; x 2 Ω; t > 0: (1) Usually the heat equation is posed with some initial condition u(x; 0) prescribing the state of the system at time t = 0, along with appropriate boundary conditions on @Ω. Note that (1) describes the evolution (in time) of a function of x. That is, for each time t the solution of (1) will be a function of x, and (1) describes how that function of x will change over time. Consequently, instead of thinking of a solution of (1) as being a function u :Ω × [0; 1) ! R; 1Copyright c 2020 by Mathew A. Johnson ([email protected]). This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 1 u(x, 0.1) ϕ(x) 1.0 1.0 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 x x -4 -2 2 4 -4 -2 0 2 4 u(x, 0.5) u(x, 1) 1.0 1.0 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 x x -4 -2 0 2 4 -4 -2 0 2 4 Figure 1: Time evolution of the solution of the heat equation posed on R with initial data u(x; 0) = φ(x), where φ(x) = 1 if jxj < 1 and φ(x) = 0 for x > 1. Thus, for each time t > 0 we get a function u(t): R ! R. it is more natural to think of solutions of such evolution equations as being maps u : [0; 1) ! X; where here X is some appropriate function space on Ω, possibly encoding the boundary conditions: for example, if (1) is equipped with homogeneous Dirichlet boundary conditions, 1 then X = H0 (Ω). Note this is also consistent with how one would think of numerically simulating the solutions: for each time, you would plot the solution as a function of x. See Figure1. With the above change in perspective, we now have two scales of regularity to track: the regularity of the map u : [0; 1) ! X and the regularity of each u(t):Ω ! R. For example, if the map u : [0; 1) ! X is continuous, we write u 2 C([0; 1); X) noting that the space C([0; 1); X) equipped with the norm kuk := sup ku(t)kX t2[0;1) 2 is a normed linear vector space space. Note above we are essentially equipping C([0; 1); X) with the L1([0; 1); X) norm. The next result, which we will use several times, implies that C([0; 1); X) is in fact a Banach space. Lemma 1. If X is a Banach space, then the space C([0; 1); X) equipped with the above 1 norm is complete and hence itself is a Banach space. That is, if fukgk=1 is a Cauchy sequence in C([0; 1); X), then there exists a u 2 C([0; 1); X) such that lim sup kuk(t) − u(t)kX = 0: k!1 t≥0 1 Proof. If fukgk=1 is Cauchy then given any > 0 there exists a K > 0 such that kuj(t) − uk(t)kX ≤ (2) for all j; k ≥ N and all t 2 [0; 1). In particular, for each fixed t ≥ 0 the sequence fuk(t)g is Cauchy in X and hence, since X is a Banach space, there exists a function u(t) 2 X such that uk(t) ! u(t) in X for each t ≥ 0. To see that the function u : [0; 1) ! X is continuous, we use the fact that the estimate (2) is independent of time and apply a standard /3-argument. Specifically, note by the triangle inequality that for all j 2 N we have ku(t1) − u(t2)kX ≤ ku(t1) − uj(t1)kX + kuj(t1) − uj(t2)kX + kuj(t2) − u(t2)kX : Thus, if j > K, then for all t1; t2 ≥ 0 we have ku(t1) − u(t2)kX ≤ kuj(t1) − uj(t2)kX + 2. Since uj 2 C([0; 1); X) by assumption, we know there exists a δ > 0 such that kuj(t) − uj(s)kX < 8 t; s ≥ 0; jt − sj < δ: It follows that if jt1 − t2j < δ then ku(t1) − u(t2)kX ≤ 3, which implies that the function u : [0; 1) ! X is in fact uniformly continuous. 2 Autonomous, Symmetric Equations n Given an open, bounded domain Ω ⊂ R with smooth boundary, lets consider the following 2 evolution problem: given u0 2 L (Ω), find a function u : Ω × [0; 1) ! R such that 8 ut + Lu = 0; x 2 Ω; t > 0 <> u(x; t) = 0; x 2 @Ω; t > 0: (3) > : u(x; 0) = u0(x); x 2 Ω 3 where here n X i;j Lu := − a ux i xj i;j=1 is a time-independent, uniformly elliptic differential operator on Ω. Specifically, here we are making the important assumption that the evolution equation (3) is autonomous, i.e. its coefficients do not depend on time2 More precisely, we assume that ai;j = aj;i and that ai;j 2 L1(Ω). The goal of this section is to essentially show the problem (3) can be solved (weakly) by classical separation of variables. To derive the appropriate weak formulation, note that 1 if v 2 Cc (Ω) then Z Z n X i;j ut(x; t)v(x)dx = − a uxi (x; t)vxj (x)dx: Ω Ω i;j=1 Further, under reasonable assumptions on the regularity of u(t) in time3, the above could be written as d Z u(x; t)v(x)dx = −B[u(t); v]; dt Ω 1 1 where here B : H0 (Ω) × H0 (Ω) ! R is the standard bilinear form associated to L. This motivates the following. Weak Formulation: Find u : [0; 1) ! L2(Ω) such that 2 (a) u 2 C([0; 1); L (Ω)) and u(0) = u0. 1 (b) u(t) 2 H0 (Ω) for all t > 0. 1 (c) for all t > 0 and v 2 H0 (Ω) we have d Z u(x; t)v(x)dx = −B[u(t); v]; dt Ω Remark 1. Note that condition (a) ensures that the initial condition makes sense, while 1 condition (b) ensures that B[u(t); v] makes sense for each v 2 H0 (Ω). Further, note that since the initial data is only L2(Ω) it is only required that the solution approach the initial data as t ! 0 in L2(Ω). 2Recall a finite dimensional dynamical system is said to be autonomous if its of the form x0 = f(x), while it is non-autonomous if the vector field f depends on t, i.e. is of the form x0 = f(x; t). 3 1 1 Technically, I believe H ([0; 1); H0 (Ω)) is sufficient. Note weak differentiation of the map u : [0; 1) ! 1 H0 (Ω) is defined as you might expect. 4 To state the main result for this section recall the First Existence Theorem for Uniformly 2 1 Elliptic PDE guarantees that for each f 2 L (Ω) there exists a unique u = S(f) 2 H0 (Ω) that weakly satisfies the BVP ( Lu = f in Ω u = 0 on @Ω 2 1 Specifically, there exists a weak solution operator S : L (Ω) ! H0 (Ω) such that Z 1 B[S(f); v] = fv dx 8v 2 H0 (Ω): Ω Further, recall that since Ω is bounded that the operator S : L2(Ω) ! L2(Ω) is compact. 1 We may thus let f(λk; φk)gk=1 denote eigenvalue and eigenfunction pairs of S, where the 2 eigenfunctions fφkg are chosen to form an O.N.B. of L (Ω) as well as an orthogonal basis 1 1 of H0 (Ω) with respect to the inner product B[·; ·]. It follows that f(µk; φk)gk=1 are the 1 eigenvalue and eigenfunction pairs associated to L, where here µk = is an increasing λk sequence satisfying 0 < µ1 < µ2 ≤ µ3 ≤ ::: % 1: With this setup, we now state the main result for this section. n Theorem 1. Let Ω ⊂ R be bounded, and consider the linear diffusion problem (3) with ij ji 1 2 a = a 2 L (Ω) and u0 2 L (Ω). Then the unique weak solution of (3) is given by 1 X −µkt u(t) = ake φk (4) k=1 where here ak := hu0; φkiL2(Ω). Before proving this result, I want to emphasize this is EXACTLY what you should expect by performing a formal separation of variables argument to the IVBVP (3). Indeed, in an undergraduate PDE class one might attempt to solve this problem by seeking solutions of the form u(x; t) = X(x)T (t) which, when substituted into the PDE in (3) yields the equation (assume here X; T 6= 0) X00(x) T 0(t) = : X(x) T (t) Clearly, the only way a function of x can be identically equal to a function of t is if both functions are constant, say −λ 2 R.
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