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IMEX method convergence for a parabolic equation

Although implicit-explicit (IMEX) methods for approximating solutions to semilinear parabolic equations are relatively standard, most recent works examine the case of a fully discretized model. We show that by discretizing time only, one can obtain an elementary convergence result for an implicit-explicit method. This convergence result is strong enough to imply existence and uniqueness of solutions to a class of semilinear parabolic equations.


Introduction
The use of implicit-explicit (IMEX) methods for approximating semilinear parabolic equations is well-established [1]. Many of the recent works on these methods employ discretizations in both space and time. These fully discrete approximations can be computed directly by a computer. However, one can obtain a stronger condition for convergence of the approximation if only the time dimension is discretized [2]. We show how an even stronger condition for convergence is met by the Cauchy problem for (1) ∂u(x, t) ∂t = ∆u(x, t) where a i ∈ L 1 (R n ) ∩ L ∞ (R n ), and how convergence of this method provides an elementary proof of existence and uniqueness of solutions. Existence and uniqueness of solutions for (1) under reasonable initial conditions have been known for some time. For instance, [4] and [6] contain straightforward proofs using semigroup methods. The purpose of this paper is to show how a more elementary proof can be obtained from a sequence of explicitly computed discrete-time approximations. The Cauchy problem for (1) arises in a variety of settings. Notably, some reaction-diffusion equations are of this form [3]. Another application is the special case ∂u(x, t) ∂t where a 0 is a nonzero function of x. This situation corresponds to a spatiallydependent logistic equation with a diffusion term, which can be thought of as a toy model of population growth with migration. Following [2], the approximation to be used is which is obtained by inverting the linear portion of a discrete version of (1). For brevity, we shall call (2) the implicit-explicit method. (In the summary paper [1], this is called an SBDF method, to distinguish it from other implicit-explicit methods.) One can compute the operator (I − h∆) −1 explicitly using Fourier transform methods, and obtain a proof of the numerical stability of the iteration as a whole.

A version of the fundamental inequality
In order to simplify the algebraic expressions, we make the following definitions.
Definition 2. Define the analytic functions Since we do not discretize the spatial dimension, we can employ some of the theory of ordinary differential equations. We therefore first prove a variant of the fundamental inequality for (1) as is done in [5]. The fundamental inequality gives a sufficient condition for approximate solutions to converge. A slightly weaker version of Lemma 3 was obtained in Theorem 3.1 of [2], where the existence of solutions was required.
Proof. Let i, j > 0 be given. Let Notice that the fourth condition in the hypothesis gives η(0) = 0.
But, d dt u i (t)−F (u i (t)) ∞ < ǫ i is equivalent to the statement that for each t ∈ [0, T ] and x ∈ R n , which allows and L 2 is complete, Lemma 3 gives conditions for existence and uniqueness of a short-time solution to (1).
wherever the limit exists.
Hence, by differentiating in t, x) for almost every x.

The implicit-explicit approximation
In this section, we consider the case of a 1-dimensional spatial domain, that is, x ∈ R. There is no obstruction to extending any of these results to higher dimensions, though it complicates the exposition unnecessarily.
As is usual, the first task is to define the function spaces to be used. Initial conditions will be drawn from a subspace of L 1 (R)∩L ∞ (R), as suggested by Lemma 3, and the first four spatial derivatives will be prescribed, for use in Lemma 10.
|f has bounded partial derivatives up to fourth order}.
For the remainder of this paper, we consider the case where each of the coefficients a i ∈ W . Then let X = {f ∈ W |g 1 ( f 1 ) < ∞ and g ∞ ( f ∞ ) < ∞}. We consider the case where the initial condition is drawn from X.
An approximate solution given by the implicit-explicit iteration will be the piecewise linear interpolation through the iterates computed by (2). A smoother approximation will prove to be unnecessary, as will be shown in Lemma 11.
Definition 7. Suppose f 0 and h > 0 are given. Put The function where n(t) = ⌊ t h ⌋, is called the implicit-explicit iteration of size h beginning at f 0 .
Using the method of residues, this can be simplified to give Calculation 9. Bounds on the L 1 and L ∞ operator norms of (I − h∆) −1 are now computed. First, let f ∈ L ∞ (R). Then Now, let f ∈ L 1 (R). So then which means (I − h∆) −1 1 ≤ 1. The third condition of Lemma 3 is a control on the slope error of the approximation. A bound on this error may be established for the implicit-explicit iteration as follows.
Proof. Recall every function in X will have bounded partial derivatives up to fourth order from Definition 6.
Now, using the fact that (I − h∆) −1 − I = (I − h∆) −1 (h∆), converges pointwise to a function in t. The limit function is piecewise differentiable in t.
Proof. Let u i be the implicit-explicit iteration of size h i . By Lemma 10, the slope error is bounded: Then, since X ⊂ C 2 (R n ), Lemma 3 applies, giving a pointwise limit function u(t). Finally, since the slope error uniformly vanishes, Lemma 5 implies that the solution is piecewise differentiable.

"A priori estimates" for the approximate solutions
Now we demonstrate that the implicit-explicit method converges for all initial conditions in X. Specifically, for each f 0 ∈ X, there exist A, B > 0 such that for each i and t ∈ [0, T ], u i (t) 1 ≤ A and u i (t) ∞ ≤ B, given sufficiently small T . We begin by recalling that from Calculation 9, the L ∞ -norm of (I − h∆) −1 is less than one. This means that for the implicit-explicit iteration, Hence the norm of each step of the implicit-explicit iteration will be controlled by the behavior of the recursion for f n , h, a > 0. Since we are only concerned with short-time existence and uniqueness, we look specifically at h = T /N and 0 ≤ n ≤ N , for fixed T > 0 and N ∈ N.
Remark 12. The recursion defined by (12) is an Euler solver for This equation is separable, and g ∞ is analytic near f 0 , so there exists a unique solution for the initial value problem (13) for sufficiently short time. Also, whenever y(t) > 0 d 2 y dt 2 = g ′ ∞ (y(t)) > 0, the function y(t) is concave up. As a result, the exact solution to (13) provides an upper bound for the recursion (12). More precisely, we have the following result.
Proof. Since the right side of (13) is strictly positive, the maximum of y is attained at T on any interval [0, T ] where y is continuous. Furthermore, since y(0) > 0, it follows from Remark 12 that y is concave up on all of [0, T ]. Therefore, y is a convex function on [0, T ]. Hence Euler's method, (12), will always underestimate the true value of y. Another way of stating this is that f n ≤ y(nh) ≤ y(T ).
Using Lemma 13, the growth of iterates to (12) may be controlled independently of the step size. This provides a uniform bound on the sequence of implicit-explicit approximations.
Then there exists a B > 0 such that for each i and t ∈ [0, T ], we have u i (t) ∞ ≤ B for sufficiently small T > 0.
Proof. Suppose f in is the n-th step of the implicit-explicit iteration of size h i . If we let y(0) = f 0 ∞ , Lemma 13 implies that for any i and any 0 ≤ n ≤ i f in ∞ ≤ y(T ) for sufficiently small T. Hence by (9) and the triangle inequality, u i (t) ∞ ≤ B for all i and t ∈ [0, T ].
With the bound on the suprema of the approximations, we can obtain a bound on the 1-norms.
Lemma 15. Suppose 0 < h i = T /i for i ∈ N. Let u i be the implicit-explicit iteration of size h i beginning at f 0 ∈ X on t ∈ [0, T ]. Then there exists an A > 0 such that for each i and t ∈ [0, T ], we have u i (t) 1 ≤ A for sufficiently small T > 0.
Proof. First, notice that Lemma 14 implies that there is a B > 0 such that for each i and t ∈ [0, T ], we have u i (t) ∞ ≤ A for sufficiently small T > 0. Again suppose f in is the n-th step of the implicit-explicit iteration of size h i . Then we compute This recurence leads to Once again, by referring to (9) and using the triangle inequality, it follows that u i (t) 1 ≤ B for all i and t ∈ [0, T ].
Theorem 16. Suppose 0 < h i = T /i for i ∈ N. Let u i be the implicit-explicit iteration of size h i beginning at f 0 ∈ X on t ∈ [0, T ]. Then, for sufficiently small T > 0, the sequence {u i (t)} ∞ i=1 converges pointwise to a function in t. The limit function is piecewise differentiable in t.
Proof. This compiles the results of Lemma 11, Lemma 14, and Lemma 15. where G is as in (4). If the operator L satisfies converges for whenever f ∈ X.

Conclusions
The convergence proof for the implicit-explicit method presented here has a number of advantages. First of all, like all implicit-explicit methods, each approximation to the solution is computed explicitly. As a result, a fully discretized version (as is standard in the literature) is easy to program on a computer. Theorem 16 therefore assures the convergence of these fully discrete methods.
However, since the implicit-explicit method presented here is discretized only in time, the convergence proof actually shows the existence of a semigroup of solutions. As a result, the convergence proof forms a bridge between the functional-analytic viewpoint of differential equations, namely that of semigroups, and the numerical methods used to approximate solutions. While the existence and uniqueness of solutions for (1) has been known via semigroup methods, the proof provided here gives a more elementary explanation of how this occurs. In particular, it approximates the semigroup action directly.