The Heat and Schrödinger Equations

The Heat and Schrödinger Equations

viii CHAPTER 5 The Heat and Schr¨odinger Equations The heat, or diffusion, equation is (5.1) ut = ∆u. Section 4.A derives (5.1) as a model of heat flow. Steady solutions of the heat equation satisfy Laplace’s equation. Using (2.4), we have for smooth functions that ∆u(x) = lim ∆u dx r→0+ − ZBr (x) n ∂ = lim u dS r→0+ r ∂r − "Z∂Br(x) # 2n = lim u dS u(x) . r→0+ r2 − − "Z∂Br(x) # Thus, if u is a solution of the heat equation, then the rate of change of u(x, t) with respect to t at a point x is proportional to the difference between the value of u at x and the average of u over nearby spheres centered at x. The solution decreases in time if its value at a point is greater than the nearby mean and increases if its value is less than the nearby averages. The heat equation therefore describes the evolution of a function towards its mean. As t solutions of the heat equation typically approach functions with the mean value→ ∞ property, which are solutions of Laplace’s equation. We will also consider the Schr¨odinger equation iu = ∆u. t − This PDE is a dispersive wave equation, which describes a complex wave-field that oscillates with a frequency proportional to the difference between the value of the function and its nearby means. 5.1. The initial value problem for the heat equation Consider the initial value problem for u(x, t) where x Rn ∈ n ut = ∆u for x R and t> 0, (5.2) ∈ u(x, 0) = f(x) for x Rn. ∈ We will solve (5.2) explicitly for smooth initial data by use of the Fourier transform, following the presentation in [34]. Some of the main qualitative features illustrated by this solution are the smoothing effect of the heat equation, the irreversibility of its semiflow, and the need to impose a growth condition as x in order to pick out a unique solution. | | → ∞ 127 128 5. THE HEAT AND SCHRODINGER¨ EQUATIONS 5.1.1. Schwartz solutions. Assume first that the initial data f : Rn R is a smooth, rapidly decreasing, real-valued Schwartz function f (see Section→ 5.6.2). The solution we construct is also a Schwartz function of x at∈ later S times t> 0, and we will regard it as a function of time with values in . This is analogous to the geometrical interpretation of a first-order system of ODEs,S in which the finite- dimensional phase space of the ODE is replaced by the infinite-dimensional function space ; we then think of a solution of the heat equation as a parametrized curve in theS vector space . A similar viewpoint is useful for many evolutionary PDEs, where the SchwartzS space may be replaced other function spaces (for example, Sobolev spaces). By a convenient abuse of notation, we use the same symbol u to denote the scalar-valued function u(x, t), where u : Rn [0, ) R, and the associated vector- valued function u(t), where u : [0, ) ×. We∞ write→ the vector-valued function corresponding to the associated scalar-valued∞ → S function as u(t)= u( ,t). · Definition 5.1. Suppose that (a,b) is an open interval in R. A function u : (a,b) is continuous at t (a,b) if → S ∈ u(t + h) u(t) in as h 0, → S → and differentiable at t (a,b) if there exists a function v such that ∈ ∈ S u(t + h) u(t) − v in as h 0. h → S → The derivative v of u at t is denoted by ut(t), and if u is differentiable for every t (a,b), then u : (a,b) denotes the map u : t u (t). ∈ t → S t 7→ t In other words, u is continuous at t if u(t)= -lim u(t + h), Sh→0 and u is differentiable at t with derivative ut(t) if u(t + h) u(t) ut(t)= -lim − . Sh→0 h We will refer to this derivative as a strong derivative if it is understood that we are considering -valued functions and we want to emphasize that the derivative is defined as the limitS of difference quotients in . We define spaces of differentiable Schwartz-valuedS functions in the natural way. For half-open or closed intervals, we make the obvious modifications to left or right limits at an endpoint. Definition 5.2. The space C ([a,b]; ) consists of the continuous functions S u :[a,b] . → S The space Ck (a,b; ) consists of functions u : (a,b) that are k-times strongly S → S j differentiable in (a,b) with continuous strong derivatives ∂t u C (a,b; ) for 0 j k, and C∞ (a,b; ) is the space of functions with continuous∈ strong derivativesS ≤ of≤ all orders. S Here we write C (a,b; ) rather than C ((a,b); ) when we consider functions defined on the open intervalS (a,b). The next propositionS describes the relationship between the C1-strong derivative and the pointwise time-derivative. 5.1. THE INITIAL VALUE PROBLEM FOR THE HEAT EQUATION 129 Proposition 5.3. Suppose that u C(a,b; ) where u(t) = u( ,t). Then u C1(a,b; ) if and only if: ∈ S · ∈ S n (1) the pointwise partial derivative ∂tu(x, t) exists for every x R and t (a,b); ∈ ∈ (2) ∂tu( ,t) for every t (a,b); (3) the map· ∈t S ∂ u( ,t) belongs∈ C (a,b; ). 7→ t · S Proof. The convergence of functions in implies uniform pointwise conver- gence. Thus, if u(t)= u( ,t) is strongly continuouslyS differentiable, then the point- wise partial derivative ∂ ·u(x, t) exists for every x Rn and ∂ u( ,t) = u (t) , t ∈ t · t ∈ S so ∂tu C (a,b; ). Conversely,∈ S if a pointwise partial derivative with the given properties exist, then for each x Rn ∈ u(x, t + h) u(x, t) 1 t+h − ∂ u(x, t)= [∂ u(x, s) ∂ u(x, t)] ds. h − t h s − t Zt Since the integrand is a smooth rapidly decreasing function, it follows from the dominated convergence theorem that we may differentiate under the integral sign with respect to x, to get u(x, t + h) u(x, t) 1 t+h xα∂β − = xα∂β [∂ u(x, s) ∂ u(x, t)] ds. h h s − t Zt Hence, if is a Schwartz seminorm (5.72), we have k·kα,β u(t + h) u(t) 1 t+h − ∂tu( ,t) ∂su( ,s) ∂tu( ,t) α,β ds h − · α,β ≤ h t k · − · k | | Z max ∂su( ,s) ∂tu( ,t) α,β , ≤ t≤s≤ t+h k · − · k and since ∂ u C (a,b; ) t ∈ S u(t + h) u(t) lim − ∂tu( ,t) =0. h→0 h − · α,β It follows that u(t + h) u(t) -lim − = ∂tu( ,t), Sh→0 h · so u is strongly differentiable and u = ∂ u C (a,b; ). t t ∈ S We interpret the initial value problem (5.2) for the heat equation as follows: A solution is a function u : [0, ) that is continuous for t 0, so that it makes sense to impose the initial condition∞ → S at t = 0, and continuously≥ differentiable for t > 0, so that it makes sense to impose the PDE pointwise in t. That is, for every t> 0, the strong derivative ut(t) is required to exist and equal ∆u(t) where ∆ : is the Laplacian operator. S → S Theorem 5.4. If f , there is a unique solution ∈ S (5.3) u C ([0, ); ) C1 (0, ; ) ∈ ∞ S ∩ ∞ S of (5.2). Furthermore, u C∞ ([0, ); ). The spatial Fourier transform of the solution is given by ∈ ∞ S 2 (5.4)u ˆ(k,t)= fˆ(k)e−t|k| , 130 5. THE HEAT AND SCHRODINGER¨ EQUATIONS and for t> 0 the solution is given by (5.5) u(x, t)= Γ(x y,t)f(y) dy Rn − Z where 1 2 (5.6) Γ(x, t)= e−|x| /4t. (4πt)n/2 Proof. Since the spatial Fourier transform is a continuous linear map on with continuous inverse, the time-derivative ofFu exists if and only if the time derivativeS ofu ˆ = u exists, and F (u ) = ( u) . F t F t Moreover, u C ([0, ); ) if and only ifu ˆ C ([0, ); ), and u Ck (0, ; ) if and only ifu ˆ∈ Ck (0∞, ;S ). ∈ ∞ S ∈ ∞ S Taking the∈ Fourier∞ transformS of (5.2) with respect to x, we find that u(x, t) is a solution with the regularity in (5.3) if and only ifu ˆ(k,t) satisfies (5.7)u ˆ = k 2u,ˆ uˆ(0) = f,ˆ uˆ C ([0, ); ) C1 (0, ; ) . t −| | ∈ ∞ S ∩ ∞ S Equation (5.7) has the unique solution (5.4). To show this in detail, suppose first thatu ˆ satisfies (5.7). Then, from Propo- sition 5.3, the scalar-valued functionu ˆ(k,t) is pointwise-differentiable with respect to t in t> 0 and continuous in t 0 for each fixed k Rn. Solving the ODE (5.7) with k as a parameter, we find that≥ u ˆ must be given∈ by (5.4). Conversely, we claim that the function defined by (5.4) is strongly differentiable with derivative 2 (5.8)u ˆ (k,t)= k 2fˆ(k)e−t|k| . t −| | To prove this claim, note that if α, β Nn are any multi-indices, the function ∈ 0 kα∂β [ˆu(k,t + h) uˆ(k,t)] − has the form |β|−1 2 2 2 aˆ(k,t) e−h|k| 1 e−t|k| + h hiˆb (k,t)e−(t+h)|k| − i i=0 h i X wherea ˆ( ,t), ˆb ( ,t) , so taking the supremum of this expression we see that · i · ∈ S uˆ(t + h) uˆ(t) 0 as h 0.

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