Zeitschrift für Analysis und ihre Anwendungen

Journal for Analysis and its Applications Volume 19 (2000), No. 1, 159–201

On a Class of Parabolic Integro-Differential Equations

W. Kohl

Abstract. Existence and uniqueness results for the integro-differential equation

u_{t}(x, t) − au_{xx}(x, t) = c(x, t)u(x, t) + ∫_{0}^{1} k(s, x)h(s, t, u(s, t)) ds + f(x, t) ((x, t) ∈ Q)

subject to the boundary condition

u(x, t) = φ(x, t) ((x, t) ∈ R)

and, especially, for the linear case h(s, t, u) = u are given. To this end, this equation is written as operator equation in a suitable Hölder space. The main tools are the calculation of the spectral radius in the linear case, and fixed point principles in the nonlinear case.

Keywords: Integro-differential equations, parabolic operators, multiplication operators, integral operators, Hölder spaces, heat potential, existence and uniqueness of solutions, Neumann series, fixed point principle

AMS subject classification: 47 G 20, 47 H 10, 47 H 30, 45 K 05, 35 K 99, 26 B 35

1. Introduction

In this paper we study existence and uniqueness results for the parabolic integrodifferential equation

u_{t}(x, t) − au_{xx}(x, t) = c(x, t)u(x, t) + ∫_{0}^{1} k(s, x)u(s, t) ds + f(x, t) ((x, t) ∈ Q) (1)

subject to the boundary condition

u(x, t) = φ(x, t)  ((x, t) ∈ R).  (2)

Here c : Q → ℝ, k : (0, 1) × (0, 1) → ℝ, f : Q → ℝ, and φ : R → ℝ are given functions, where Q = (0, 1) × (0, T] and R = \overline{Q} \ Q is its parabolic boundary; the parameter a is a real constant. Equations of this type occur in the mathematical modelling of

W. Kohl: Neue Vockenroter Steige 17, D-97877 Wertheim

ISSN 0232-2064 / $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag Berlin

various transport problems, e.g., describing the propagation of radiation through the atmospheres of planets and stars [4, 5], or the transfer of neutrons through thin plates and membranes in nuclear reactors [6]. In the case a = 0 this boundary value problem has been studied in the recent survey paper [1]. By means of a simple scaling argument we may suppose that a = 1.

If we introduce the differential operator

Lu(x, t) = u_{t}(x, t) − u_{xx}(x, t),  (3)

the multiplication operator

Cu(x, t) = c(x, t)u(x, t),  (4)

and the partial integral operator

Ku(x, t) = ∫_{0}^{1} k(s, x)u(s, t) ds,  (5)

we may write (1) as operator equation

Lu = (C + K)u + f.  (6)

Our strategy for proving existence (and sometimes also uniqueness) of solutions to the operator equation (6) with boundary condition (2) is standard: First we give conditions under which the classical parabolic boundary value problem

\begin{matrix}Lu & = f  in Q \\ u & = φ  on R \end{matrix} \}  (7)

has a unique solution for each f and φ in some suitable Banach space; this allows us to define the operator L^{−1} on this Banach space. Afterwards we pass from the operator equation (6) to the equivalent equation

u − L^{−1}(C + K)u = L^{−1}f  (8)

and try to find conditions under which the spectral radius of the operator L^{−1}(C + K) is less than 1, in order to apply the classical Neumann series. In fact it turns out that the spectral radius of the linear operator L^{−1}(C + K) is 0, if we take a Hölder space as underlying Banach space of the operator equation (8).

Apart from the linear equation (1), we will also be interested in the nonlinear equation

\begin{matrix}u_{t}(x, t) − au_{xx}(x, t) \\ = c(x, t)u(x, t) + ∫_{0}^{1} k(s, x)h(s, t, u(s, t)) ds + f(x, t) \end{matrix}  ((x, t) ∈ Q)  (9)

where h : Q × ℝ → ℝ is some Carathéodory function. Introducing the nonlinear Nemytskij operator

Hu(x, t) = h(x, t, u(x, t))  (10)

generated by the function h, we may write (9) again as operator equation

Lu = (C + KH)u + f.  (11)

If we suppose again that the parabolic operator L be invertible in some Banach space, we end up, analogously to (8), with the nonlinear operator equation

u − L^{−1}(C + KH)u = L^{−1}f,  (12)

which may be studied by several (classical and non-classical) fixed point principles.

The plan of this paper is as follows. First we introduce some special spaces of continuous functions in which the operator (3) and its inverse have particularly “nice” properties. In Lemma 1 and Lemma 2 we describe some features of the inverse operator by estimations which are not only useful for later functional analytic considerations. These estimations fill also a gap in the literature of the heat equation. So we aimed at thoroughness in proving them. Afterwards we give sufficient conditions under which the operators (4) and (5) are bounded in these spaces. It turns out that analogous results for the nonlinear operator (10) are much more involved. Finally, we show how our results give existence and uniqueness results for solutions of the linear boundary value problem (1)/(2) and the nonlinear boundary value problem (9)/(2).

2. The heat potential

Following the theory of the heat equation in the book of J. R. Cannon [2: Chapter 19] we know that the inhomogeneous heat equation (7) is invertible, if the data f is bounded and uniformly Hölder continuous on each compact subset of the domain under consideration. A detailed discussion of the inverse operator L^{−1} in the case of the infinite set (−∞, +∞) × (0, T] is given in this book. Because we could not find similiar investigations for the finite set Q in the literature, we turn now our attention to this case. The inverse L^{−1} corresponding to the rectangular set Q can be represented as a linear Volterra operator

L^{−1}f(x, t) = ∫_{0}^{t}∫_{0}^{1} Γ(x, t; ξ, τ)f(ξ, τ) dξdτ,  (13)

which is generated by the Green’s function Γ for the Dirichlet problem [3: p. 195]. This function can be expressed with the help of the θ-function

θ(x, t; ξ, τ) = ∑_{n=−∞}^{+∞} exp \frac{−n^{2} + n(x − ξ)}{t − τ} − ∑_{n=−∞}^{+∞} exp \frac{−n^{2} + n(x + ξ) − xξ}{t − τ}

and the heat kernel

γ(x, t) = \frac{e^{−x^{2}/4t}}{\sqrt{4πt}}

in the form

Γ(x, t; ξ, τ) = \{ \begin{matrix}γ(x − ξ, t − τ)θ(x, t; ξ, τ) & if x, ξ ∈ ℝ and τ < t \\ 0 & if x, ξ ∈ ℝ and τ ≥ t. \end{matrix}  (14)

Thus the function Γ is infinitely often continuously differentiable for all x, ξ ∈ ℝ and τ < t. For fixed (ξ, τ) ∈ ℝ^{2} it solves the heat equation for all x ∈ ℝ and t > τ, while for fixed (x, t) ∈ ℝ^{2} it is a solution of the adjoint heat equation for all ξ ∈ ℝ and τ < t. Moreover, we have the boundary properties

Γ(0, t; ξ, τ) = Γ(1, t; ξ, τ) = 0 (ξ ∈ ℝ, τ < t)  (15)
Γ_{xx}(0, t; ξ, τ) = Γ_{xx}(1, t; ξ, τ) = 0 (ξ ∈ ℝ, τ < t).  (16)

In order to investigate the operator L^{−1} we introduce for fixed ε > 0 the family of functions u_{h} with

u_{h}(x, t) = ∫_{0}^{t−h} ∫_{0}^{1} Γ(x, t; ξ, τ)f(ξ, τ) dξdτ ((x, t) ∈ ℝ × [ε, T], 0 < h < \frac{ε}{2}).  (17)

Now the singularity (x, t) of the Green’s function lies not in the domain of integration. So we conclude assuming f ∈ L^{∞}(Q) that each function u_{h} is infinitely often continuously differentiable with respect to x and differentiating under the integral sign is permitted

\frac{∂^{k}u_{h}}{∂x^{k}} (x, t) = ∫_{0}^{t−h} ∫_{0}^{1} \frac{∂^{k}Γ}{∂x^{k}}(x, t; ξ, τ)f(ξ, τ) dξdτ (k ∈ ℕ).

In the case f ∈ C^{0}(\overline{Q}) we may differentiate (17) with respect to t to yield identity (18) for all (x, t) ∈ ℝ × [ε, T], 0 < h < \frac{ε}{2}

\frac{∂u_{h}}{∂t} (x, t) = \frac{∂^{2}u_{h}}{∂x^{2}} (x, t) + ∫_{0}^{1} Γ(x, t; ξ, t − h)f(ξ, t − h) dξ.  (18)

The properties of functions L^{−1}f with f ∈ L^{∞}(Q) are summarized in the following lemma. Let C^{α,0}(\overline{Q}) denote, as usual, the set of all v ∈ C^{0}(\overline{Q}) such that there exists a c > 0 with

höl_{α}(v(·, t)) := sup_{x,y∈[0,1],x≠y} \frac{|v(x, t) − v(y, t)|}{|x − y|^{α}} ≤ c (t ∈ [0, T]).

Lemma 1. For f ∈ L^{∞}(Q), the heat potential

u(x, t) = L^{−1}f(x, t) = ∫_{0}^{t} ∫_{0}^{1} Γ(x, t; ξ, τ)f(ξ, τ) dξdτ  (19)

has the following properties:

(a) u|R = 0 and u ∈ C^{0}(\overline{Q}) with sup_{x∈[0,1]} |u(x, t)| ≤ c_{1}(T) ∥f∥_{∞} t where ∥f∥_{∞} = inf_{μ(N)=0} sup_{(x,t)∈Q\N} |f(x, t)| is a norm in L^{∞}(Q).

(b) u_{x} ∈ C^{0}(\overline{Q}) with u_{x}(x, t) = ∫_{0}^{t} ∫_{0}^{1} Γ_{x}(x, t; ξ, τ)f(ξ, τ) dξdτ, u_{x}(x, 0) ≡ 0, and sup_{x∈[0,1]} |u_{x}(x, t)| ≤ c_{2}(T)∥f∥_{∞}\sqrt{t}.

(c) u_{x} ∈ C^{\frac{1}{3} ,0}(\overline{Q}), i.e. |u_{x}(x+δ, t)−u_{x}(x, t)| ≤ c_{3}(T)∥f∥_{∞} |δ|^{\frac{1}{3}} (x+δ, x ∈ [0, 1], t ∈ [0, T]).

Proof. Part (a): Since the θ-function is bounded on the set

D = {(x, t; ξ, τ) ∈ ℝ^{4}| x, ξ ∈ [0, 1], t, τ ∈ [0, T], τ < t},

we may estimate the function u by

|u(x, t)| ≤ ∫_{0}^{t}∫_{0}^{1} |Γ(x, t; ξ, τ)| |f(ξ, τ)| dξdτ

≤ \underbrace{sup_{D} |θ(x, t; ξ, τ)|}_{=c_{1}(T)} ||f||_{∞} ∫_{0}^{t} ∫_{−∞}^{∞} γ(x − ξ, t − τ) dξdτ

≤ c_{1}(T) ||f||_{∞}t.

So the function u is well defined on \overline{Q} and satisfies the asserted inequality. Furthermore, we estimate the difference u − u_{h} by

\begin{aligned}|u(x, t) − u_{h}(x, t)| & ≤ ∫_{t−h}^{t} ∫_{0}^{1} |Γ(x, t; ξ, τ)| |f(ξ, τ)| dξdτ \\ & = c_{1}(T) ||f||_{∞}h. \end{aligned}

Taking a sequence (h_{n}) with lim_{n→∞} h_{n} = 0 the sequence (u_{h_{n}}) of continuous functions converges uniformly on [0, 1] × [ε, T] towards the function u for all ε > 0. Hence we have u ∈ C^{0}([0, 1] × (0, T]) and the function u(·, t) possesses zero boundary values

u(0, t) = lim_{n→∞} u_{h_{n}}(0, t) = lim_{n→∞} u_{h_{n}}(1, t) = u(1, t) (t ∈ (0, T]).

Moreover, the estimation |u(x, t)| ≤ c_{1}(T) ||f||_{∞}t shows that u(x, t) → 0 as t ↘ 0 uniformly for all x ∈ [0, 1], and we conclude u ∈ C^{0}(\overline{Q}) with u|R = 0.

Part (b): The existence of the first derivative u_{x} of the heat potential is based on the crucial inequality

∫_{0}^{1} |Γ_{x}(x, t; ξ, τ)| dξ ≤ c(T) \frac{1}{\sqrt{t − τ}} ,  (20)

which we prove first. The product rule and further estimation leads to

∫_{0}^{1} |Γ_{x}(x, t; ξ, τ)| dξ

≤ ∫_{0}^{1} |θ_{x}(x, t; ξ, τ)|γ(x − ξ, t − τ) dξ + ∫_{0}^{1} |θ(x, t; ξ, τ)| |γ_{x}(x − ξ, t − τ)| dξ

=: J_{1} + J_{2}. We estimate the integral J_{1} by the two integrals

∫_{0}^{1} |θ_{x}(x, t; ξ, τ)|γ(x − ξ, t − τ) dξ ≤ A + B

with

A = ∫_{0}^{1} ∑_{n=−∞}^{n=+∞} \frac{|n|}{t − τ} exp ( − \frac{n^{2}}{t − τ} + n\frac{x − ξ}{t − τ} )γ(x − ξ, t − τ) dξ

and

B = ∫_{0}^{1} ∑_{n=−∞}^{n=+∞} \frac{|n − ξ|}{t − τ} exp ( − \frac{n^{2}}{t − τ} + n\frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )γ(x − ξ, t − τ) dξ

and consider each integral separately using the convention of constants. We write A as a sum of integrals A = A_{1} + A_{2} + A_{3} and treat each integral separately as follows:

A_{1} = ∫_{0}^{1} ∑_{|n|≥2} \frac{|n|}{t − τ} exp ( − \frac{n^{2}}{t − τ} + n\frac{x − ξ}{t − τ} )γ(x − ξ, t − τ) dξ

≤ \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} ∫_{0}^{1} ∑_{|n|≥2} \frac{|n|}{t − τ} exp (\frac{1 − |n|}{t − τ})dξ

≤ \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} \frac{2}{t − τ} exp \{ − \frac{1}{t − τ} ∑_{n=0}^{∞}(n + 2)( exp −\frac{1}{T} )^{n}\}

≤ \frac{c(T)}{\sqrt{t − τ}} ,

n = 1 : A_{2} = ∫_{0}^{1} \frac{1}{t − τ} exp ( − \frac{1}{t − τ} + \frac{x − ξ}{t − τ} )\frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

= \frac{1}{t − τ} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} ∫_{0}^{1} exp −\frac{(2 − (x − ξ))^{2}}{4(t − τ)} dξ

≤ \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} \frac{1}{t − τ} ∫_{0}^{1} exp −\frac{1}{4(t − τ)} dξ

≤ \frac{c(T)}{\sqrt{t − τ}} ,

n = −1 : A_{3} = ∫_{0}^{1} \frac{1}{t − τ} exp ( − \frac{1}{t − τ} − \frac{x − ξ}{t − τ} )\frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

= \frac{1}{t − τ} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} ∫_{0}^{1} exp −\frac{(2 + (x − ξ))^{2}}{4(t − τ)} dξ

≤ \frac{c(T)}{\sqrt{t − τ}} .

Next we turn to the integral B = B_{1} + B_{2} + B_{3} + B_{4}, where we look at

B_{1} = ∫_{0}^{1} ∑_{|n|≥2} \frac{|n − ξ|}{t − τ} exp ( − \frac{n^{2}}{t − τ} + n\frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )γ(x − ξ, t − τ) dξ

≤ ∫_{0}^{1} 2 ∑_{n=2}^{∞} \frac{n + 1}{t − τ} exp (\frac{1 − n}{t − τ})\frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

≤ ∫_{0}^{1} \underbrace{\frac{2(4π)^{− \frac{1}{2}}}{t − τ} exp −\frac{1}{t − τ} ∑_{n=0}^{∞}(n + 3)( exp −\frac{1}{T} )^{n}}_{≤c(T)} \frac{1}{\sqrt{t − τ}} \underbrace{exp −\frac{(x − ξ)^{2}}{4(t − τ)}}_{≤1} dξ

≤ \frac{c(T)}{\sqrt{t − τ}} .

Then we estimate the integral B_{2}

\begin{aligned}n = 0 : B_{2} & = ∫_{0}^{1} \frac{ξ}{t − τ} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{(x + ξ)^{2}}{4(t − τ)} dξ \\ & ≤ ∫_{0}^{1} \frac{ξ}{t − τ} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{ξ^{2}}{4(t − τ)} dξ \end{aligned}

166 W. Kohl

and substitute by φ(ξ) = ξ\sqrt{4(t − τ)} to gain the desired inequality

B_{2} ≤ \frac{4(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} ∫0^{\frac{1}{\sqrt{4(t−τ)}}} ξ exp −ξ^{2}dξ ≤ \frac{1}{\sqrt{t − τ}} \frac{2}{\sqrt{π}} ∫_{0}^{∞}ξ exp −ξ^{2}dξ.

For the integral B_{3} we obtain

n = 1 : B_{3} = \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} ∫_{0}^{1} \frac{1 − ξ}{t − τ} exp −\frac{((1 − x) + (1 − ξ))^{2}}{4(t − τ)} dξ

≤ \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} ∫_{0}^{1} \frac{1 − ξ}{t − τ} exp −\frac{(1 − ξ)^{2}}{4(t − τ)} dξ

≤ ∫_{− \frac{1}{\sqrt{4(t−τ)}}}^{0} −2ξ exp −ξ^{2}dξ \frac{π^{− \frac{1}{2}}}{\sqrt{t − τ}}

≤ ∫_{−∞}^{0} −2ξ exp −ξ^{2}dξ \frac{π^{− \frac{1}{2}}}{\sqrt{t − τ}} ,

where we applied the substitution ψ(ξ) = 1 + ξ\sqrt{4(t − τ)}. Finally, the asserted estimation holds for the integral B_{4}

n = −1 : B_{4} = ∫_{0}^{1} \frac{1 + ξ}{t − τ} exp −\frac{(1 + x)(1 + ξ)}{t − τ} γ(x − ξ, t − τ) dξ

≤ \frac{2}{t − τ} exp −\frac{1}{t − τ} ∫_{0}^{1} γ(x − ξ, t − τ) dξ

≤ c(T) \frac{1}{\sqrt{t − τ}} .

Considering the integral J_{2} we see that the boundedness of the θ-function and the substitution by φ(ξ) = x + ξ\sqrt{4(t − τ)} yield

J_{2} ≤ c_{1}(T) ∫_{0}^{1} \frac{|x − ξ|}{2(t − τ)} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

≤ c_{1}(T) ∫ \frac{1−x}{\sqrt{4(t−τ)}} ; \frac{−x}{\sqrt{4(t−τ)}} |ξ| exp −ξ^{2}dξ \frac{π^{− \frac{1}{2}}}{\sqrt{t − τ}}

≤ c_{1}(T) ∫_{−∞}^{+∞} |ξ| exp −ξ^{2}dξ \frac{π^{− \frac{1}{2}}}{\sqrt{t − τ}}

= c_{1}(T) \frac{π^{− \frac{1}{2}}}{\sqrt{t − τ}} .

Now we are able to estimate the function q, where

q(x, t) = ∫_{0}^{t}∫_{0}^{1} \frac{∂Γ}{∂x}(x, t; ξ, τ)f(ξ, τ) dξdτ ((x, t) ∈ [0, 1] × [0, T])

with the help of the just derived inequality as

|q(x, t)| ≤ ∫_{0}^{t} ∫_{0}^{1} |\frac{∂Γ}{∂x}(x, t; ξ, τ)| |f(ξ, τ)| dξdτ

≤ c(T) ||f||_{∞} ∫_{0}^{t} \frac{1}{\sqrt{t − τ}} dτ

= 2c(T) ||f||_{∞}\sqrt{t}

= c_{2}(T) ||f||_{∞}\sqrt{t}.

Obviously, the function q(·, t) is uniformly bounded on [0, 1] for each t and q(x, t) → 0 uniformly on [0, 1] as t ↘ 0. Looking at the difference q − \frac{∂u_{h}}{∂x}, we get

|q(x, t) − \frac{∂u_{h}}{∂x} (x, t)| ≤ c_{2}(T) ||f||_{∞}\sqrt{h} ((x, t) ∈ [0, 1] × [ε, T]).

We conclude like in part (a) q ∈ C^{0}(\overline{Q}) with q(x, 0) = 0 for all x ∈ [0, 1]. For each t the functions u_{h}(·, t) are continuously differentiable on [0, 1] and satisfy the equation u_{h}(0, t) = 0; after the fundamental theorem of calculus the identity

u_{h}(x, t) = ∫_{0}^{x} \frac{∂u_{h}}{∂x} (ξ, t) dξ (t ∈ [ε, T])

holds, and we gain applying the uniform convergence of the functions u_{h} and \frac{∂u_{h}}{∂x} as h ↘ 0 on [0, 1] × [ε, T] the equation

u(x, t) = ∫_{0}^{x} q(ξ, t) dξ ((x, t) ∈ [0, 1] × (0, T]).

By the uniform convergence of q(x, t) as t ↘ 0 this relationship is also true for t = 0. Differentiating with respect to x leads to u_{x}(x, t) = q(x, t) on \overline{Q}.

Part (c): In order to show the claimed inequality, we proof first that an estimate of the type

∫_{0}^{1} |Γ_{xx}(x, t; ξ, τ)| dξ ≤ c(T) \frac{1}{t − τ}  (x, ξ ∈ [0, 1], 0 ≤ τ < t ≤ T )  (21)

holds. For this sake we apply the product rule and obtain

\begin{aligned}∫_{0}^{1} |Γ_{xx}(x, t; ξ, τ)| dξ & ≤ ∫_{0}^{1} |θ(x, t; ξ, τ)| |γ_{xx}(x − ξ, t − τ)| dξ \\ & + 2 ∫_{0}^{1} |θ_{x}(x, t; ξ, τ)| |γ_{x}(x − ξ, t − τ)| dξ \end{aligned}

\begin{matrix}+ ∫_{0}^{1} |θ_{xx}(x, t; ξ, τ)| |γ(x − ξ, t − τ)| dξ \\ =: J_{1} + 2J_{2} + J_{3} \end{matrix}

where each integral J_{1}, J_{2} and J_{3} will be investigated separately.

With regard to the integral J_{1} we employ the boundedness of the θ-function

\begin{aligned}J_{1} & ≤ c_{1}(T) ∫_{0}^{1} |γ_{xx}(x − ξ, t − τ)| dξ \\ & ≤ c_{1}(T) ∫_{0}^{1} (\frac{1}{2(t − τ)} + \frac{(x − ξ)^{2}}{4(t − τ)^{2}} )\frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ \end{aligned}

and substitute by φ(ξ) = x + ξ\sqrt{4(t − τ)} to get

J_{1} ≤ c_{1}(T)π^{− \frac{1}{2}} ∫ \frac{1−x}{\sqrt{4(t−τ)}} ; \frac{−x}{\sqrt{4(t−τ)}} (\frac{1}{2(t − τ)} + \frac{ξ^{2}}{t − τ} ) exp −ξ^{2}dξ

≤ c_{1}(T)π^{− \frac{1}{2}} \frac{1}{(t − τ)} ∫_{−∞}^{+∞} (\frac{1}{2} + ξ^{2}) exp −ξ^{2}dξ

≤ c(T)\frac{1}{t − τ} .

Next we estimate the absolute value of the integral J_{2} by the sum A + B of the two integrals

A = ∫_{0}^{1} ∑_{n=−∞}^{n=+∞} \frac{|n|}{t − τ} exp ( − \frac{n^{2}}{t − τ} + n\frac{x − ξ}{t − τ} )\frac{|x − ξ|}{2(t − τ)} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

and

\begin{aligned}B & = ∫_{0}^{1} ∑_{n=−∞}^{n=+∞} \frac{|n − ξ|}{t − τ} exp ( − \frac{n^{2}}{t − τ} + n\frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} ) \\ & × \frac{|x − ξ|}{2(t − τ)} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ. \end{aligned}

On a Class of Integro-Differential Equations 169

Similiar estimations as in the proof of Part (b) lead to

A ≤ \frac{1}{t − τ} ∫_{0}^{1} 2 exp ( − \frac{1}{t − τ} ) ∑_{n=0}^{∞}(n + 2)( exp −\frac{1}{T} )^{n} \frac{|x − ξ|}{2(t − τ)} \frac{1}{\sqrt{4π(t − τ)}} dξ

+ \frac{1}{t − τ} ∫_{0}^{1} \frac{|x − ξ|}{2(t − τ)} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(2 + (x − ξ))^{2}}{4(t − τ)} dξ

+ \frac{1}{t − τ} ∫_{0}^{1} \frac{|x − ξ|}{2(t − τ)} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(2 − (x − ξ))^{2}}{4(t − τ)} dξ

≤ c(T) \frac{1}{t − τ} .

Let us write B = B_{1} + B_{2} + B_{3} + B_{4}. We estimate B_{1} via

B_{1} ≤ ∫_{0}^{1} ∑_{|n|≥2} \frac{|n − ξ|}{t − τ} exp ( − \frac{n^{2}}{t − τ} + n\frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )

× \frac{|x − ξ|}{2(t − τ)} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

≤ \frac{1}{t − τ} ∫_{0}^{1} \underbrace{∑_{n=2}^{∞} 2(n + 1) exp (\frac{1 − n}{t − τ})\frac{|x − ξ|}{2(t − τ)} \frac{1}{\sqrt{4π(t − τ)}}}_{≤c(T)} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

≤ c(T) \frac{1}{t − τ} .

Then we consider B_{2}:

B_{2} ≤ ∫_{0}^{1} \frac{ξ}{t − τ} exp ( − \frac{xξ}{t − τ} )\frac{|x − ξ|}{2(t − τ)} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

= \frac{1}{4(π)^{\frac{1}{2}}} ∫_{0}^{1} \frac{1}{(t − τ)^{\frac{5}{2}}} ξ|x − ξ| exp −\frac{(x + ξ)^{2}}{4(t − τ)} dξ

≤ \frac{1}{4(π)^{\frac{1}{2}}} \frac{1}{t − τ} ∫_{0}^{1} \frac{ξ}{t − τ} exp ( − \frac{(x + ξ)^{2}}{8(t − τ)} ) \underbrace{\frac{x + ξ}{\sqrt{t − τ}} exp −\frac{(x + ξ)^{2}}{8(t − τ)}}_{≤C} dξ

≤ C \frac{1}{t − τ} ∫_{0}^{1} \frac{ξ}{t − τ} exp −\frac{ξ^{2}}{8(t − τ)} dξ

≤ c(T) \frac{1}{t − τ} .

170 W. Kohl

For the integral B_{3} we obtain

B_{3} ≤ ∫_{0}^{1} \frac{|1 − ξ|}{t − τ} exp ( − \frac{1}{t − τ} + \frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )\frac{|x − ξ|}{2(t − τ)} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

= ∫_{0}^{1} \frac{1 − ξ}{t − τ} exp ( − \frac{((1 − x) + (1 − ξ))^{2}}{4(t − τ)} )\frac{|x − ξ|}{2(t − τ)} \frac{1}{\sqrt{4π(t − τ)}} dξ.

Using the estimation

|x − ξ| ≤ |x − 1| + |1 − ξ| = 2 − (x + ξ) (x, ξ ∈ [0, 1])

we calculate further

B_{3} ≤ ∫_{0}^{1} \frac{1 − ξ}{(t − τ)^{2}} exp ( − \frac{((1 − x) + (1 − ξ))^{2}}{8(t − τ)} ) \underbrace{\frac{2 − (x + ξ)}{4\sqrt{π}\sqrt{t − τ}} exp −\frac{(2 − (x + ξ))^{2}}{8(t − τ)}}_{≤C} dξ

≤ \frac{C}{t − τ} ∫_{0}^{1} \frac{1 − ξ}{t − τ} exp −\frac{(1 − ξ)^{2}}{8(t − τ)} dξ

≤ c(T) \frac{1}{t − τ} .

At last, we proceed with the integral B_{4} to get

B_{4} ≤ ∫_{0}^{1} \frac{1 + ξ}{t − τ} exp ( − \frac{1}{t − τ} − \frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )\frac{|x − ξ|}{2(t − τ)} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

= ∫_{0}^{1} \frac{1 + ξ}{t − τ} exp ( − \frac{(1 + x)(1 + ξ)}{t − τ})\frac{|x − ξ|}{2(t − τ)} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

≤ ∫_{0}^{1} \frac{2}{t − τ} exp ( − \frac{1}{t − τ} )\frac{1}{2(t − τ)} \frac{1}{\sqrt{4π(t − τ)}} dξ

≤ c(T) \frac{1}{t − τ} .

Finally, it remains to investigate the integral J_{3}, which we estimate by

J_{3} ≤ ∫_{0}^{1} ∑_{n=−∞}^{n=+∞} \frac{n^{2}}{(t − τ)^{2}} exp ( − \frac{n^{2}}{t − τ} + n\frac{x − ξ}{t − τ} )\frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

+ ∫_{0}^{1} ∑_{n=−∞}^{n=+∞} (\frac{n − ξ}{t − τ} )^{2} exp ( − \frac{n^{2}}{t − τ} + n\frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )

× \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

=: C + D.

On a Class of Integro-Differential Equations 171

We treat the integral C in a similiar manner as the integral A above and derive without difficulties

C ≤ c(T) \frac{1}{t − τ} .

For the integral D_{1} we obtain

D_{1} ≤ ∫_{0}^{1} ∑_{|n|≥2} (\frac{n − ξ}{t − τ} )^{2} exp ( − \frac{n^{2}}{t − τ} + n\frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )\frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

≤ ∫_{0}^{1} ∑_{n=2}^{∞} 2(n + 1)(t − τ)^{2} exp (\frac{1 − n}{t − τ})\frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

≤ \frac{1}{t − τ} ∫_{0}^{1} ∑_{n=0}^{∞} 2(n + 3)( exp −\frac{1}{T} )^{n} \frac{1}{t − τ}

× exp ( − \frac{1}{t − τ} )\frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

≤ c(T) \frac{1}{t − τ} .

Next we go on estimating

D_{2} = ∫_{0}^{1} \frac{ξ^{2}}{(t − τ)^{2}} exp ( − \frac{xξ}{t − τ} )\frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

= ∫_{0}^{1} \frac{ξ^{2}}{(t − τ)^{2}} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x + ξ)^{2}}{4(t − τ)} dξ

≤ \frac{1}{t − τ} ∫_{0}^{1} \frac{ξ}{t − τ} exp ( − \frac{ξ^{2}}{8(t − τ)}) \underbrace{\frac{ξ}{\sqrt{4π(t − τ)}} exp −\frac{ξ^{2}}{8(t − τ)}}_{≤C} dξ

≤ C \frac{1}{t − τ} ∫_{0}^{1} \frac{ξ}{t − τ} exp −\frac{ξ^{2}}{8(t − τ)} dξ

≤ c(T) \frac{1}{t − τ} .

The integral D_{3} will be estimated by

D_{3} = ∫_{0}^{1} \frac{(1 − ξ)^{2}}{(t − τ)^{2}} exp ( − \frac{1}{t − τ} + \frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )\frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

= ∫_{0}^{1} \frac{(1 − ξ)^{2}}{(t − τ)^{2}} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{((1 − x) + (1 − ξ))^{2}}{4(t − τ)} dξ

≤ \frac{1}{t − τ} ∫_{0}^{1} \frac{1 − ξ}{t − τ} exp ( − \frac{(1 − ξ)^{2}}{8(t − τ)} ) \underbrace{\frac{1 − ξ}{\sqrt{4π(t − τ)}} exp −\frac{(1 − ξ)^{2}}{8(t − τ)}}_{≤C} dξ

≤ \frac{C}{t − τ} ∫_{0}^{1} \frac{1 − ξ}{t − τ} exp −\frac{(1 − ξ)^{2}}{8(t − τ)} dξ

≤ c(T) \frac{1}{t − τ} .

At last we calculate for the integral D_{4} without difficulties

D_{4} = ∫_{0}^{1} \frac{(1 + ξ)^{2}}{(t − τ)^{2}} exp ( − \frac{1}{t − τ} − \frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )\frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

= ∫_{0}^{1} \frac{(1 + ξ)^{2}}{(t − τ)^{2}} exp ( − \frac{(1 + x)(1 + ξ)}{t − τ})\frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ

≤ ∫_{0}^{1} \frac{4}{(t − τ)^{2}} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{1}{t − τ} dξ

≤ c(T) \frac{1}{t − τ} .

Now we turn to the asserted inequality of this lemma and obtain for a positive parameter η

|u_{x}(x + δ, t) − u_{x}(x, t)| ≤ ∫_{0}^{t} ∫_{0}^{1} |Γ_{x}(x + δ, t; ξ, τ) − Γ_{x}(x, t; ξ, τ)| |f(ξ, τ)| dξdτ

≤ ∫_{0}^{t−η}∫_{0}^{1} |Γ_{x}(x + δ, t; ξ, τ) − Γ_{x}(x, t; ξ, τ)| |f(ξ, τ)| dξdτ

+ ∫_{t−η}^{t} ∫_{0}^{1} |Γ_{x}(x + δ, t; ξ, τ)| |f(ξ, τ)| dξdτ

+ ∫_{t−η}^{t} ∫_{0}^{1} |Γ_{x}(x, t; ξ, τ)| |f(ξ, τ)|dξdτ

=: I_{1} + I_{2} + I_{3}.

We already know by the result of part (b) that I_{2} + I_{3} ≤ 2c_{2}(T) ||f||_{∞}η^{\frac{1}{2}} is true. Moreover, we obtain by the mean value theorem

I_{1} = ∫_{0}^{t−η}∫_{0}^{1} |Γ_{xx}(y, t; ξ, τ)| |f(ξ, τ)| dξdτ δ

On a Class of Integro-Differential Equations 173

for y between x and x + δ. We calculate further applying the inequality above

I_{1} ≤ c(T) ||f||_{∞} ∫_{0}^{t−η}\frac{1}{t − τ} dτ δ

= c(T) ||f||_{∞}δ ln \frac{t}{η}

≤ c(T) ||f||_{∞}δ \frac{t}{η} .

Setting η = δ^{\frac{2}{3}} we have for all t ∈ [δ^{\frac{2}{3}} , T] the estimation

|u_{x}(x + δ, t) − u_{x}(x, t)| ≤ c(T) ||f||_{∞}tδ^{\frac{1}{3}} + 2c_{2}(T) ||f||_{∞}δ^{\frac{1}{3}}
≤ (c(T)T + 2c_{2}(T))||f||_{∞}δ^{\frac{1}{3}}
= c_{3}(T) ||f||_{∞}δ^{\frac{1}{3}}

whereas in the case t ∈ [0, δ^{\frac{2}{3}} ] the inequality

|u_{x}(x + δ, t) − u_{x}(x, t)| ≤ 2 sup_{x∈[0,1]} |u_{x}(x, t)|
≤ 2c_{2}(T) ||f||_{∞}t^{\frac{1}{2}}
≤ 2c_{2}(T) ||f||_{∞}δ^{\frac{1}{3}}

holds

Equipped with the norm

∥v∥_{C^{α,0}(\overline{Q})} = ||v||_{∞} + sup_{t∈[0,T]} höl_{α}(v(·, t), [0, 1]),

C^{α,0}(\overline{Q}) is a Banach space. The subspace C_{0}^{α,0}(\overline{Q}) consisting of all v ∈ C^{α,0}(\overline{Q}) with

v(0, t) = v(1, t) = 0 (t ∈ [0, T]) (22)

is a closed subspace of C^{α,0}(\overline{Q}), hence also a Banach space. We point out that the norm

∥v∥_{C_{0}^{α,0}(\overline{Q})} = sup_{t∈[0,T]} höl_{α}(v(·, t), [0, 1])

is equivalent to the norm ∥ · ∥_{C^{α,0}(\overline{Q})} on C_{0}^{α,0}(\overline{Q}). In the Banach space C_{0}^{α,0}(\overline{Q}) we obtain the following

Lemma 2. For f ∈ C_{0}^{α,0}(\overline{Q}), the heat potential (19) has the following properties:

(a) u ∈ C^{0}(\overline{Q}) with sup_{x∈[0,1]} |u(x, t) − ∫_{0}^{t} f(x, τ) dτ| ≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} t^{1+ \frac{α}{2}} .

(b) u|R = 0 and sup_{x∈[0,1]} |u(x, t)| ≤ c_{1}(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} t.

(c) u_{x} ∈ C^{0}(\overline{Q}) with u_{x}(x, 0) ≡ 0 and sup_{x∈[0,1]} |u_{x}(x, t)| ≤ c_{2}(α, T) ∥f∥_{C_{0}^{α,0}(\overline{Q})} t^{\frac{1+α}{2}}.

(d) u_{xx} ∈ C^{0}(\overline{Q}) with u_{xx}|R = 0, u_{xx}(x, t) = ∫_{0}^{t} ∫_{0}^{1} Γ_{xx}(x, t; ξ, τ)f(ξ, τ) dξdτ and sup_{x∈[0,1]} |u_{xx}(x, t)| ≤ c_{3}(α, T) ∥f∥_{C_{0}^{α,0}(\overline{Q})} t^{\frac{α}{2}} .

(e) u_{xx} ∈ C_{0}^{α,0}(\overline{Q}), i.e. |u_{xx}(x + δ, t) − u_{xx}(x, t)| ≤ c_{4}(α, T) ∥f∥_{C_{0}^{α,0}(\overline{Q})}|δ|^{α}.

(f) u_{t} ∈ C_{0}^{α,0}(\overline{Q}) with u_{t}(x, t) = u_{xx}(x, t) + f(x, t) on \overline{Q}, u_{t}(x, 0) = f(x, 0), and sup_{x∈[0,1]} |u_{t}(x, t)| ≤ ||f||_{C_{0}^{α,0}(\overline{Q})}(1 + c_{3}(α, T)t^{\frac{α}{2}} ).

Proof. Parts (a) and (b): By the continuity of the imbedding C_{0}^{α,0}(\overline{Q}) ⊆ L^{∞}(Q) we conclude u ∈ C^{0}(\overline{Q}) with u|R = 0 and sup_{x∈[0,1]} |u(x, t)| ≤ c_{1}(T) ||f||_{C_{0}^{α,0}(\overline{Q})} t. In view of the asserted inequality in statement (a) we estimate

|u(x, t) − ∫_{0}^{t} f(x, τ) dτ| ≤ ∫_{0}^{t} | ∫_{0}^{1} (Γ(x, t; ξ, τ) − γ(x − ξ, t − τ))f(ξ, τ) dξ|dτ

+ | ∫_{0}^{t}∫_{0}^{1} γ(x − ξ, t − τ)f(ξ, τ) dξdτ − ∫_{0}^{t} f(x, τ) dτ|

=: I_{1} + I_{2}.

First we consider the inner integral of I_{1} to obtain the estimation

| ∫_{0}^{1} (Γ(x, t; ξ, τ) − γ(x − ξ, t − τ))f(ξ, τ) dξ| ≤ A + B

with the integral

A = ∫_{0}^{1} ∑_{0≠n∈ℤ} exp ( − \frac{n^{2}}{t − τ} + n\frac{x − ξ}{t − τ} )γ(x − ξ, t − τ)|f(ξ, τ)| dξ

and the other integral

B = ∫_{0}^{1} ∑_{n=−∞}^{∞} exp ( − \frac{n^{2}}{t − τ} + n\frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )γ(x − ξ, t − τ)|f(ξ, τ)| dξ.

As usual we write A = A_{1} + A_{2} + A_{3} and, obviously, we gain the inequality

\begin{aligned}A_{1} & = ∫_{0}^{1} \underbrace{∑_{|n|≥2} exp ( − \frac{n^{2}}{t − τ} + n\frac{x − ξ}{t − τ} )\frac{(t − τ)^{− \frac{α}{2}}}{\sqrt{4π(t − τ)}}}_{≤c(T)} (t − τ)^{\frac{α}{2}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ \\ & ≤ c(T) (t − τ)^{\frac{α}{2}} ||f||_{C_{0}^{α,0}(\overline{Q})}. \end{aligned}

On a Class of Integro-Differential Equations 175

For A_{2} (n = +1) and A_{3} (n = −1) there are no difficulties to show the same inequality, so we omit it. Considering B = B_{1} + B_{2} + B_{3} + B_{4} we get for the integral B_{1}

B_{1} = ∫_{0}^{1} \underbrace{∑_{|n|≥2} exp ( − \frac{n^{2}}{t − τ} + n\frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )\frac{(t − τ)^{− \frac{α}{2}}}{\sqrt{4π(t − τ)}}}_{≤c(T)}

× (t − τ)^{\frac{α}{2}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ

≤ c(T)(t − τ)^{\frac{α}{2}} ||f||_{C_{0}^{α,0}(\overline{Q})}.

In the integral B_{2} we apply

|f(ξ, τ)| = |f(ξ, τ) − f(0, τ)| ≤ ξ^{α}||f||_{C_{0}^{α,0}(\overline{Q})}

and derive the inequality

B_{2} ≤ ∫_{0}^{1} \frac{ξ^{α}}{\sqrt{4π(t − τ)}} exp −\frac{xξ}{t − τ} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

= ∫_{0}^{1} \frac{ξ^{α}}{\sqrt{4π(t − τ)}} exp −\frac{(x + ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}.

Then we estimate further

B_{2} ≤ ∫_{0}^{1} \frac{ξ^{α}}{\sqrt{4π(t − τ)}} exp −\frac{ξ^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

and substitute by φ(ξ) = ξ\sqrt{4(t − τ)} to obtain

\begin{aligned}B_{2} & ≤ ∫_{−∞}^{+∞}ξ^{α} exp −ξ^{2}dξ (4(t − τ))^{\frac{α}{2}} 4π^{− \frac{1}{2}} ||f||_{C_{0}^{α,0}(\overline{Q})} \\ & ≤ c(α, T)(t − τ)^{\frac{α}{2}} ||f||_{C_{0}^{α,0}(\overline{Q})}. \end{aligned}

In the integral B_{3} the inequality

|f(ξ, τ)| = |f(1, τ) − f(ξ, τ)| ≤ (1 − ξ)^{α}||f||_{C_{0}^{α,0}(\overline{Q})}

leads us to

\begin{aligned}B_{3} & ≤ ∫_{0}^{1}(1 − ξ)^{α} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(1 − x + 1 − ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})} \\ & ≤ ∫_{0}^{1}(1 − ξ)^{α} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(1 − ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})} \end{aligned}

176 W. Kohl

and substitution via φ(ξ) = ξ\sqrt{4(t − τ)} + 1 yields the inequality

B_{3} ≤ c(α, T)(t − τ)^{\frac{α}{2}} ||f||_{C_{0}^{α,0}(\overline{Q})}. Concerning the integral B_{4} no difficulties occur in proofing the same kind of esti-

mation. We summarize our results so far

| ∫_{0}^{1} (Γ(x, t; ξ, τ) − γ(x − ξ, t − τ))f(ξ, τ) dξ| ≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{\frac{α}{2}} . (23)

For the investigation of the integral I_{2} we use the property ∫_{−∞}^{+∞} γ(x−ξ, t−τ) dξ = 1 and extend the function f ∈ C_{0}^{α,0}(\overline{Q}) by 0 in the set ℝ×[0, T]\\overline{Q} to obtain the extension f̂ ∈ C_{0}^{α,0}(ℝ × [0, T]) with ||f||_{C_{0}^{α,0}(\overline{Q})} = ||f̂||_{C_{0}^{α,0}(ℝ×[0,1])},

| ∫_{0}^{1} γ(x − ξ, t − τ)f(ξ, τ) dξ − ∫_{−∞}^{+∞}γ(x − ξ, t − τ)f(x, τ) dξ|

= | ∫_{−∞}^{+∞}γ(x − ξ, t − τ)(f̂(ξ, τ) − f̂(x, τ)) dξ|

≤ ∫_{−∞}^{+∞}\frac{1}{\sqrt{4πt − τ}} |x − ξ|^{α} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ ||f̂||_{C_{0}^{α,0}(ℝ×[0,1])}.

Finally, substitution with φ(ξ) = x + ξ\sqrt{4(t − τ)} yields the inequality

| ∫_{0}^{1} γ(x − ξ, t − τ)f(ξ, τ) dξ − f(x, τ)| ≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{\frac{α}{2}} .  (24)

Therefore we deduce

I_{1} + I_{2} ≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} t^{1+ \frac{α}{2}}

and our assertion is proved.

Part (c): Obviously, we may apply Lemma 1 to get u_{x} ∈ C^{0}(\overline{Q}) and u_{x}(x, 0) = 0 for all x ∈ [0, 1]. In order to proof the inequality

|u_{x}(x, t)| ≤ c_{2}(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} t^{\frac{1+α}{2}}

it suffices to convince ourselves that both inequalities

\begin{aligned}I_{1} & := ∫_{0}^{1} |θ_{x}(x, t; ξ, τ)γ(x − ξ, t − τ)f(ξ, τ)| dξ \\ & ≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{− \frac{1}{2} + \frac{α}{2}} \end{aligned}  (25)

On a Class of Integro-Differential Equations 177

and

\begin{aligned}I_{2} & := ∫_{0}^{1} |θ(x, t; ξ, τ)γ_{x}(x − ξ, t − τ)f(ξ, τ)| dξ \\ & ≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{− \frac{1}{2} + \frac{α}{2}} \end{aligned}  (26)

hold. As usual we employ the integral

A = ∫_{0}^{1} ∑_{n=−∞}^{n=+∞} \frac{|n|}{t − τ} exp ( − \frac{n^{2}}{t − τ} + n\frac{x − ξ}{t − τ} )γ(x − ξ, t − τ) dξ

and the other integral

B = ∫_{0}^{1} ∑_{n=−∞}^{n=+∞} \frac{|n − ξ|}{t − τ} exp ( − \frac{n^{2}}{t − τ} + n\frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )γ(x − ξ, t − τ) dξ

to estimate I_{1} by I_{1} ≤ A + B. Writing A as sum of integrals A = A_{1} + A_{2} + A_{3} we get

A_{1} = ∫_{0}^{1} \underbrace{∑_{|n|≥2} \frac{|n|}{t − τ} exp ( − \frac{n^{2}}{t − τ} + n\frac{x − ξ}{t − τ} )\frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} \frac{1}{(t − τ)^{− \frac{1}{2} + \frac{α}{2}}}}_{≤c(α,T)}

× (t − τ)^{− \frac{1}{2} + \frac{α}{2}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ

≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{− \frac{1}{2} + \frac{α}{2}}

and remark that we can reach the same estimation for the integrals A_{2} and A_{3}.

In view of B = B_{1} + B_{2} + B_{3} + B_{4} we obtain for B_{1} the estimation

B_{1} ≤ ∫_{0}^{1} ∑_{|n|≥2} \frac{|n − ξ|}{t − τ} exp ( − \frac{n^{2}}{t − τ} + n\frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )\frac{(4π)^{− \frac{1}{2}}}{(t − τ)^{\frac{α}{2}}}

× (t − τ)^{− \frac{1}{2} + \frac{α}{2}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ

≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{− \frac{1}{2} + \frac{α}{2}} .

The integral B_{2} is treated by

∫_{0}^{1} \frac{ξ}{t − τ} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{xξ}{t − τ} exp −\frac{(x − ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ

≤ ∫_{0}^{1} \frac{ξ^{1+α}}{t − τ} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{(x + ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

≤ ∫_{0}^{1} \frac{ξ^{1+α}}{t − τ} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{ξ^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

and substitution with φ(ξ) = ξ\sqrt{4(t − τ)} leads to

B_{2} ≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{− \frac{1}{2} + \frac{α}{2}} .

The integral B_{3} will be estimated in the following way:

\begin{aligned}B_{3} & ≤ ∫_{0}^{1} \frac{1 − ξ}{t − τ} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} (1 − ξ)^{α} exp −\frac{((1 − x) + (1 − ξ))^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})} \\ & ≤ ∫_{0}^{1} \frac{(1 − ξ)^{1+α}}{t − τ} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{(1 − ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}. \end{aligned}

Now substitution with φ(ξ) = ξ\sqrt{4(t − τ)} + 1 yields the desired estimation

B_{3} ≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{− \frac{1}{2} + \frac{α}{2}} .

For the integral B_{4} (n = −1) we obtain without difficulties the same kind of estimation. In order to reach the desired inequality for the integral I_{2} we estimate it with the help of two integrals

I_{2} ≤ ∫_{0}^{1} |θ(x, t; ξ, τ) − 1| |γ_{x}(x − ξ, t − τ)| |f(ξ, τ)| dξ

+ | ∫_{0}^{1} γ_{x}(x − ξ, t − τ)f(ξ, τ) dξ|

=: J_{1} + J_{2}.

We estimate J_{1} ≤ C + D with the integrals

C = ∫_{0}^{1} ∑_{n=−∞}^{n=+∞} exp ( − \frac{n^{2}}{t − τ} + n\frac{x − ξ}{t − τ} )|γ_{x}(x − ξ, t − τ)| dξ

and

D = ∫_{0}^{1} ∑_{n=−∞}^{n=+∞} exp ( − \frac{n^{2}}{t − τ} + n\frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )|γ_{x}(x − ξ, t − τ)| dξ.

The integral C can be treated in the usual way, so we turn at once to the integral D = D_{1} + D_{2} + D_{3} + D_{4}. Here we restrict ourselves to the investigation of the integrals D_{2} and D_{3}, because the way to estimate the other two integrals is clear. For the integral

On a Class of Integro-Differential Equations 179

D_{2} we obtain

∫_{0}^{1} \frac{|x − ξ|}{2(t − τ)} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{xξ}{t − τ} exp −\frac{(x − ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ

≤ ∫_{0}^{1} ξ^{α} \frac{|x + ξ|}{2(t − τ)} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{(x + ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

≤ ∫_{0}^{1} \frac{(x + ξ)^{1+α}}{2(t − τ)} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{(x + ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

and substitution via φ(ξ) = ξ\sqrt{4(t − τ)} − x leads to the inequality

D_{2} ≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{− \frac{1}{2} + \frac{α}{2}} .

At last we estimate the integral D_{3} in the following way:

∫_{0}^{1} \frac{|x − ξ|}{2(t − τ)} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp ( − \frac{1}{t − τ} + \frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} ) exp −\frac{(x − ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ

≤ ∫_{0}^{1}(1 − ξ)^{α} \frac{2 − (x + ξ)}{2(t − τ)} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{(2 − (x + ξ))^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

≤ ∫_{0}^{1} \frac{(2 − (x + ξ))^{1+α}}{2(t − τ)} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{(2 − (x + ξ))^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}.

Then we substitute by φ(ξ) = ξ\sqrt{4(t − τ)} − x + 2 to derive the desired inequality.

With regard to the integral J_{2} we apply the identity ∫_{−∞}^{+∞} γ_{x}(x−ξ, t−τ) dξ = 0 and employ the extended function f̂ ∈ C_{0}^{α,0}(ℝ × [0, T]) of the function f (see p. 176) to get

J_{2} = | ∫_{0}^{1} γ_{x}(x − ξ, t − τ)f(ξ, τ) dξ − ∫_{−∞}^{+∞}γ_{x}(x − ξ, t − τ) dξ f(x, τ)|

≤ ∫_{−∞}^{+∞}γ_{x}(x − ξ, t − τ)|f̂(ξ, τ) − f̂(x, τ)| dξ

≤ ∫_{−∞}^{+∞}\frac{|x − ξ|^{1+α}}{2(t − τ)} \frac{(4π)^{− \frac{1}{2}}}{\sqrt{t − τ}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}.

Finally the substitution φ(ξ) = ξ\sqrt{4(t − τ)} + x yield inequality (26).

Part (d): To derive the existence of the second derivative u_{xx} of the heat potential we show that the inequality

| ∫_{0}^{1} Γ_{xx}(x, t; ξ, τ)f(ξ, τ) dξ| ≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{\frac{α}{2} −1}  (27)

holds. After applying the product rule we estimate this integral by four integrals:

| ∫_{0}^{1} Γ_{xx}(x, t; ξ, τ)f(ξ, τ) dξ|

≤ ∫_{0}^{1} |θ_{xx}(x, t; ξ, τ)γ(x − ξ, t − τ)f(ξ, τ)| dξ

+ 2 ∫_{0}^{1} |θ_{x}(x, t; ξ, τ)γ_{x}(x − ξ, t − τ)f(ξ, τ)| dξ

+ ∫_{0}^{1} |(θ(x, t; ξ, τ) − 1)γ_{xx}(x − ξ, t − τ)f(ξ, τ)| dξ

+ | ∫_{0}^{1} γ_{xx}(x − ξ, t − τ)f(ξ, τ) dξ|

=: I_{1} + 2I_{2} + I_{3} + I_{4}.

These integrals will be investigated separately. First we estimate the integral I_{1} by the sum of the two integrals

A = ∫_{0}^{1} ∑_{n=−∞}^{n=+∞} \frac{n^{2}}{(t − τ)^{2}} exp ( − \frac{n^{2}}{t − τ} + n\frac{x − ξ}{t − τ} )\frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ

and

\begin{aligned}B & = ∫_{0}^{1} ∑_{n=−∞}^{n=+∞} (\frac{|n − ξ|}{t − τ})^{2} exp ( − \frac{n^{2}}{t − τ} + n\frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} ) \\ & × \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ. \end{aligned}

We turn at once to the two interesting parts of the integral B. For the one part we obtain

∫_{0}^{1} \frac{ξ^{2}}{(t − τ)^{2}} exp ( − \frac{xξ}{t − τ} )\frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ

≤ ∫_{0}^{1} \frac{ξ^{2+α}}{(t − τ)^{2}} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{ξ^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

≤ c(α, T) ∥f∥_{C_{0}^{α,0}(\overline{Q})}(t − τ)^{\frac{α}{2} −1}

where we employed the substitution φ(ξ) = ξ\sqrt{4(t − τ)}. Similiar calculations lead to

∫_{0}^{1} \frac{(1 − ξ)^{2}}{(t − τ)^{2}} exp ( − \frac{1}{t − τ} + \frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )\frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ

≤ ∫_{0}^{1} \frac{(1 − ξ)^{2+α}}{(t − τ)^{2}} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(1 − ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

≤ ∫_{0}^{1} \frac{(1 − ξ)^{2+α}}{(t − τ)^{2}} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(1 − ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{\frac{α}{2} −1}

for the other part. In the treatment of the integral I_{2} we proceed in the same way. We have

∫_{0}^{1} \frac{ξ}{t − τ} exp ( − \frac{xξ}{t − τ} )\frac{|x − ξ|}{2(t − τ)} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x − ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ

≤ ∫_{0}^{1} \frac{(x + ξ)^{2+α}}{2(t − τ)^{2}} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(x + ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{\frac{α}{2} −1}

where c(α, t) is a positive constant obtained via the transformation φ(ξ) = ξ\sqrt{4(t − T)}− x. Then we estimate

∫_{0}^{1} \frac{1 − ξ}{t − τ} exp(− \frac{1}{t − τ} + \frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} )\frac{1}{\sqrt{4π(t − τ)}} \frac{|x − ξ|}{2(t − τ)} exp −\frac{(x − ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ

≤ ∫_{0}^{1} \frac{(1 − ξ)^{1+α}}{t − τ} \frac{1}{\sqrt{4π(t − τ)}} \frac{2 − x − ξ}{2(t − τ)} exp −\frac{(2 − x − ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

≤ ∫_{0}^{1} \frac{(2 − x − ξ)^{2+α}}{t − τ} \frac{1}{\sqrt{4π(t − τ)}} exp −\frac{(2 − x − ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{\frac{α}{2} −1}

182 W. Kohl

using the substitution φ(ξ) = ξ\sqrt{4(t − T)} + x − 2. Also, by the integral I_{3} we restrict ourselves to the following two cases. First we calculate

∫_{0}^{1} exp ( − \frac{xξ}{t − τ} )\frac{1}{\sqrt{4π(t − τ)}}(\frac{1}{2(t − τ)} + \frac{(x − ξ)^{2}}{4(t − τ)^{2}} ) exp −\frac{(x − ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ

≤ ∫_{0}^{1} \frac{1}{\sqrt{4π(t − τ)}}ξ^{α}(\frac{1}{2(t − τ)} + \frac{(x − ξ)^{2}}{4(t − τ)^{2}} ) exp −\frac{(x + ξ)^{2}}{4(t − τ)} |f(ξ, τ)| dξ

≤ ∫_{0}^{1} \frac{1}{\sqrt{4π(t − τ)}}(\frac{(x + ξ)^{α}}{2(t − τ)} + \frac{(x + ξ)^{2+α}}{4(t − τ)^{2}} ) exp −\frac{(x + ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{\frac{α}{2} −1}

where we used the substitution φ(ξ) = ξ\sqrt{4(t − τ)} − x. Next we estimate

∫_{0}^{1} exp ( − \frac{1}{t − τ} + \frac{x + ξ}{t − τ} − \frac{xξ}{t − τ} − \frac{(x − ξ)^{2}}{4(t − τ)} )

× \frac{(4π)^{− \frac{1}{2}}}{\sqrt{(t − τ)}}(\frac{1}{2(t − τ)} + \frac{(x − ξ)^{2}}{4(t − τ)^{2}} )|f(ξ, τ)| dξ

≤ ∫_{0}^{1} \frac{1}{\sqrt{4π(t − τ)}}(\frac{(x − ξ)^{α}}{2(t − τ)} + \frac{(x − ξ)^{2+α}}{4(t − τ)^{2}} ) exp −\frac{(2 − x − ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

≤ ∫_{0}^{1} \frac{1}{\sqrt{4π(t − τ)}}(\frac{(2 − x − ξ)^{α}}{2(t − τ)} + \frac{(2 − x − ξ)^{2+α}}{4(t − τ)^{2}} ) exp −\frac{(2 − x − ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}

≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{\frac{α}{2} −1} employing the substitution φ(ξ) = ξ\sqrt{4(t − τ)} + 2 − x.

At last, it remains to look at the integral I_{4}. Here we apply the identity ∫_{−∞}^{+∞} γ_{xx}(x− ξ, t − τ) dξ = 0 and use the extension f̂ of the function f (see p. 176) to get

I_{4} = | ∫_{0}^{1} γ_{xx}(x − ξ, t − τ)f(ξ, τ) dξ|

= | ∫_{−∞}^{+∞}γ_{xx}(x − ξ, t − τ)(f̂(ξ, τ) − f̂(x, τ)) dξ|

≤ ∫_{−∞}^{+∞}|γ_{xx}(x − ξ, t − τ)| |f̂(ξ, τ) − f̂(x, τ)| dξ

≤ ∫_{−∞}^{+∞}\frac{1}{\sqrt{4π(t − τ)}}(\frac{|x − ξ|^{α}}{2(t − τ)} + \frac{(x − ξ)^{α+2}}{4(t − τ)^{2}} ) exp −\frac{(x − ξ)^{2}}{4(t − τ)} dξ ||f||_{C_{0}^{α,0}(\overline{Q})}.

Substituting with φ(ξ) = x + ξ\sqrt{4(t − τ)} yields the asserted inequality. Now we may conclude that the function

p(x, t) = ∫_{0}^{t}∫_{0}^{1} Γ_{xx}(x, t, ξ, τ)f(ξ, τ) dξdτ

is well-defined and satisfies the inequality

|p(x, t)| ≤ ∫_{0}^{t} | ∫_{0}^{1} Γ_{xx}(x, t, ξ, τ)f(ξ, τ) dξ|dτ

≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} ∫_{0}^{t} \frac{1}{(t − τ)^{\frac{α}{2} −1}} dτ

= c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})}t^{\frac{α}{2}} .

Hence the function p(·, t) is uniformly bounded on [0, 1] for each t and p(x, t) → 0 uniformly on [0, 1] as t ↘ 0. Considering the difference p − \frac{∂^{2}u_{h}}{∂x^{2}} , we obtain

|p(x, t) − \frac{∂^{2}u_{h}}{∂x^{2}} (x, t)| ≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} h^{\frac{α}{2}} ((x, t) ∈ [0, 1] × [ε, T]).

We conclude similiarly as in Lemma 1/Part (a), p ∈ C^{0}(\overline{Q}) with p(x, 0) = 0 for all x ∈ [0, 1]. Moreover, we have p|R = 0. Since the functions \frac{∂u_{h}}{∂x} (·, t) are continuously differentiable on [0, 1] for each t, we may apply the fundamental theorem of calculus to get

\frac{∂u_{h}}{∂x} (x, t) − \frac{∂u_{h}}{∂x} (0, t) = ∫_{0}^{x} \frac{∂^{2}u_{h}}{∂x^{2}} (ξ, t) dξ ((x, t) ∈ [0, 1] × [ε, T]).

Obviously, we gain further

\frac{∂u}{∂x}(x, t) − \frac{∂u}{∂x}(0, t) = ∫_{0}^{x} p(ξ, t) dξ ((x, t) ∈ [0, 1] × (0, T])

as h ↘ 0, and this equation is also true for t = 0. Differentiating with respect to x yields u_{xx}(x, t) = p(x, t) for all (x, t) ∈ \overline{Q}.

Part (e): Assuming t ∈ [0, δ^{2}] we calculate by virtue of Part (d)

|u_{xx}(x + δ, t) − u_{xx}(x, t)| ≤ 2 sup_{x∈[0,1]} |u_{xx}(x, t)|
≤ 2c_{3}(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} t^{\frac{α}{2}}
≤ 2c_{3}(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})}δ^{α}

and the claimed inequality is valid.

In the case t ∈ (δ^{2}, T] we estimate with the help of three integrals

|u_{xx}(x + δ, t) − u_{xx}(x, t)| ≤ | ∫_{0}^{t−δ^{2}} ∫_{0}^{1} (Γ_{xx}(x + δ, t; ξ, τ) − Γ_{xx}(x, t; ξ, τ))f(ξ, τ) dξdτ|

+ | ∫_{t−δ^{2}}^{t} ∫_{0}^{1} Γ_{xx}(x + δ, t; ξ, τ)f(ξ, τ) dξdτ|

+ | ∫_{t−δ^{2}}^{t} ∫_{0}^{1} Γ_{xx}(x, t; ξ, τ)f(ξ, τ) dξdτ|

=: I_{1} + I_{2} + I_{3}.

Of course, we deduce a suitable inequality for I_{2} + I_{3}

I_{2} + I_{3} ≤ 2c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (δ^{2})^{\frac{α}{2}} .

So it remains to consider the integral I_{1}. Applying the mean value theorem we obtain for the inner integral of I_{1}

∫_{0}^{1} Γ_{xxx}(y, t; ξ, τ)f(ξ, τ) dξ δ

where y lies between x and x + δ. The product rule and further estimations lead to the investigation of integrals which have the form

A_{kl} = ∫_{0}^{1} |\frac{∂^{k}θ}{∂x^{k}} (y, t; ξ, τ)\frac{∂^{l}γ}{∂x^{l}} (y − ξ, t − τ)f(ξ, τ|dξ (k + l = 3, k, l ∈ N_{0}).

We remark that each integral may be estimated by

A_{kl} ≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} (t − τ)^{− \frac{3}{2} + \frac{α}{2}}

using simliar calculations as in Part (d). Hence we know

I_{1} ≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} ∫_{0}^{t−δ^{2}}(t − τ)^{− \frac{3}{2} + \frac{α}{2}} dτ δ

= c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})}(\frac{2}{1 − α}(t − τ)^{\frac{−1+α}{2}} |_{0}^{t−δ^{2}})δ

= c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} \frac{2}{1 − α}(δ^{−1+α} − t^{\frac{−1+α}{2}} )δ

≤ c(α, T)\frac{2}{1 − α} ||f||_{C_{0}^{α,0}(\overline{Q})}δ^{α}

and our assertion is proved.

Part (f): We conclude with the help of the results of Part (a) and the inequalities

|\frac{u(x, t) − u(x, 0)}{t} − f(x, 0)|

≤ |\frac{u(x, t)}{t} − \frac{1}{t} ∫_{0}^{t} f(x, τ) dτ| + |\frac{1}{t} ∫_{0}^{t} (f(x, τ) − f(x, 0)) dτ|

≤ c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} t^{\frac{α}{2}} + \frac{1}{t} ∫_{0}^{t} |f(x, τ) − f(x, 0)| dτ

→ 0

as t ↘ 0, uniformly for all x ∈ [0, 1], and this yields the property u_{t}(x, 0) = f(x, 0). Next we consider the estimation

| ∫_{0}^{1} Γ(x, t; ξ, t − h)f(ξ, t − h) dξ − f(x, t)|

≤ | ∫_{0}^{1} (Γ(x, t; ξ, t − h) − γ(x − ξ, h))f(ξ, t − h) dξ|

+ | ∫_{0}^{1} γ(x − ξ, h)f(ξ, t − h) dξ − f(x, t − h)|

+ |f(x, t − h) − f(x, t)|

=: I_{1} + I_{2} + I_{3}.

Applying inequalities (23) and (24) for τ = t − h we get I_{1} + I_{2} ≤ 2c(α, T) ||f||_{C_{0}^{α,0}(\overline{Q})} h^{\frac{α}{2}} ,

and from the uniform continuity of the function f on [0, 1] × [ε, T] we deduce the relationship

∫_{0}^{1} Γ(x, t; ξ, t − h)f(ξ, t − h) dξ → f(x, t) as h ↘ 0 uniformly on [0, 1] × [ε, T].

With regard to equality (18) we notice that

\frac{∂u_{h}}{∂t} (x, t) → \frac{∂^{2}u}{∂x^{2}}(x, t) + f(x, t) as h ↘ 0 uniformly on [0, 1] × [ε, T].

Hence u_{t}(x, t) exists for all (x, t) ∈ [0, 1] × (0, T]. We include the case t = 0 to yield

u_{t}(x, t) = u_{xx}(x, t) + f(x, t) ((x, t) ∈ \overline{Q}).

The properties u_{t} ∈ C_{0}^{α,0}(\overline{Q}) and sup_{x∈[0,1]} |u_{t}(x, t)| ≤ ||f||_{C_{0}^{α,0}(\overline{Q})}(1+c_{3}(α, T) t^{\frac{α}{2}} ) follow now from the identity above in connection with Part (d)

3. The Barbashin operator

In this section we state sufficient conditions under which both (4) and (5) are continuous operator functions mapping C_{0}^{α,0}(\overline{Q}) into itself and estimate their norm.

Lemma 3. Suppose that c ∈ C^{α,0}(\overline{Q}). Then the corresponding multiplication operator (4) is bounded in C_{0}^{α,0}(\overline{Q}) and ||C|| ≤ ||c||_{C^{α,0}(\overline{Q})}.

Proof. From the definition of C_{0}^{α,0}(\overline{Q}) and the hypothesis on the function c we conclude directly that Cu ∈ C^{0}(\overline{Q}) for u ∈ C_{0}^{α,0}(\overline{Q}) and that the function Cu satisfies the boundary condition (22). From the estimates

|c(x, t)u(x, t) − c(y, t)u(y, t)|
≤ |c(x, t)u(x, t) − c(x, t)u(y, t)| + |c(x, t)u(y, t) − c(y, t)u(y, t)|
≤ |c(x, t)| |u(x, t) − u(y, t)| + |c(x, t) − c(y, t)| |u(y, t)|
≤ ||c||_{∞}höl_{α}(u(·, t), [0, 1]) |x − y|^{α} + höl_{α}(c(·, t), [0, 1]) |x − y|^{α}||u||_{∞}
≤ höl_{α}(u(·, t), [0, 1]) |x − y|^{α}||c||_{C^{α,0}(\overline{Q})}

it follows that Cu ∈ C_{0}^{α,0}(\overline{Q}) and ||C|| ≤ ||c||_{C^{α,0}(\overline{Q})}

Lemma 4. Suppose that the function k : [0, 1] × [0, 1] → ℝ has the following properties:

(a) k(·, x) is measurable for each x ∈ [0, 1].

(b) k(s, ·) ∈ C^{α}([0, 1]) uniformly for all s ∈ [0, 1], i.e. there exists a constant q̃ ∈ ℝ

with

|k(s, x) − k(s, y)| ≤ q̃|x − y|^{α}  (s ∈ [0, 1]).  (28)

(c) k(s, 0) = k(s, 1) = 0 for all s ∈ [0, 1].

Then the corresponding partial integral operator (5) is bounded in C_{0}^{α,0}(\overline{Q}) with ||K|| ≤ \frac{q}{α+1}, where

q = sup_{s∈[0,1]} höl_{α}(k(s, ·), [0, 1]).  (29)

Proof. The function F (·, x, t) = k(·, x)u(·, t) is measurable and bounded on the interval [0, 1] for fixed (x, t) ∈ [0, 1] × [0, T], while the function F (s, ·, ·) = k(s, ·)u(s, ·) is continuous on \overline{Q} for fixed s ∈ [0, 1]. Since

|F (s, x, t)| ≤ q ||u||_{∞} ((s, x, t) ∈ [0, 1] × [0, 1] × [0, T]),

we conclude that the integral over F (·, x, t) depends continuously on the parameters x ∈ [0, 1] and t ∈ [0, T]; this means that Ku ∈ C^{0}(\overline{Q}). It is clear that the function Ku fulfills the boundary condition (22). Finally, from

|Ku(x, t) − Ku(y, t)| = |∫_{0}^{1}[k(s, x) − k(s, y)]u(s, t) ds|

≤ ∫_{0}^{1} |k(s, x) − k(s, y)| |u(s, t) − u(0, t)| ds

≤ sup_{s∈[0,1]} höl_{α}(k(s, ·), [0, 1]) |x − y|^{α} ∫_{0}^{1} |u(s, t) − u(0, t)| ds

≤ q |x − y|^{α} ∫_{0}^{1} s^{α} höl_{α}(u(·, t), [0, 1]) ds

we get

\frac{|Ku(x, t) − Ku(y, t)|}{|x − y|^{α}} ≤ q höl_{α}(u(·, t), [0, 1]) ∫_{0}^{1} s^{α} ds,

hence

höl_{α}(Ku(·, t), [0, 1]) ≤ \frac{q}{α + 1} höl_{α}(u(·, t), [0, 1]).

Passing to the supremum in the interval [0, T] leads to ||Ku||_{C_{0}^{α,0}(\overline{Q})} ≤ \frac{q}{α+1}||u||_{C_{0}^{α,0}(\overline{Q})} as claimed

4. The linear problem

Now we turn from the parabolic differential equation (6) to the equivalent operator equation (8). We calculate the spectral radius of the operator L^{−1}(C + K) and give existence and uniqueness results for equation (6).

First of all, we need the following

Lemma 5. For f ∈ C_{0}^{α,0}(\overline{Q}), the following two statements are equivalent:

(A) u ∈ C^{0}(\overline{Q}) has the properties u_{x} ∈ C^{0}(\overline{Q}), u_{t}, u_{xx} ∈ C^{0}(Q) and solves the boundary value problem

\begin{matrix}Lu & = (C + K)u + f in Q \\ u & = 0  on R. \end{matrix} \}  (30)

(B) u ∈ C_{0}^{α,0}(\overline{Q}) satisfies the linear operator equation (8).

Proof. Let u be as in statement (A). We fix (x, t) ∈ Q and observe that for 0 < t_{0} < t the vector field F : [0, 1] × [0, t_{0}] → ℝ^{2} defined by

F (ξ, τ) = (Γ(x, t; ξ, τ)u_{ξ}(ξ, τ) − Γ_{ξ}(x, t; ξ, τ)u(ξ, τ), −Γ(x, t; ξ, τ)u(ξ, τ))

is continuous on [0, 1] × [0, t_{0}] and continuously differentiable on (0, 1) × (0, t_{0}) with

\begin{aligned}div F (ξ, τ) & = −Γ(x, t; ξ, τ)Lu(ξ, τ) − u(ξ, τ)[Γ_{ξξ}(x, t; ξ, τ) + Γ_{τ} (x, t; ξ, τ)] \\ & = −Γ(x, t; ξ, τ)[(C + K)u(ξ, τ) + f(ξ, τ)]. \end{aligned}

So the divergence of the vector field F is continuous and bounded on (0, 1) × (0, t_{0}) and we may apply the Gauss theorem to obtain

∫_{0}^{t_{0}} ∫_{0}^{1} −Γ(x, t; ξ, τ)[(C + K)u(ξ, τ) + f(ξ, τ)]dξdτ = ∫_{0}^{1} −Γ(x, t; ξ, t_{0})u(ξ, t_{0}) dξ.

Letting t_{0} → t we get the identity

∫_{0}^{t}∫_{0}^{1} Γ(x, t; ξ, τ)[(C + K)u(ξ, τ) + f(ξ, τ)] dξdτ = u(x, t)  ((x, t) ∈ Q).  (31)

The function on the left-hand side of (31) is continuous on \overline{Q} by Lemma 1 and we have u ∈ C^{0}(\overline{Q}), so the above equation holds for all (x, t) ∈ \overline{Q}. Of course, u ∈ C_{0}^{α,0}(\overline{Q}) and [I − L^{−1}(C + K)]u = L^{−1}f.

Conversely, let u be as in (B). Since f ∈ C_{0}^{α,0}(\overline{Q}), the same is true for the function (C + K)u + f. Moreover, from the identity L^{−1}[(C + K)u + f] = u and from Lemmas 1 and 2 it follows that the function u has the regularity properties stated in (A) and satisfies (30)

Lemma 6. The spectral radius r(A) of the operator A = L^{−1}(C + K) : C_{0}^{α,0}(\overline{Q}) → C_{0}^{α,0}(\overline{Q}) is zero.

Proof. We use the classical Gel’fand formula

r(A) = lim_{n→∞} ^{n} \sqrt{||A^{n}||}.

First of all, the inequalities ||Cv||_{∞} ≤ ||c||_{∞} ||v||_{∞} and ||Kv||_{∞} ≤ q ||v||_{∞}, with q as in (29), combined with property (a) in Lemma 1, lead to the estimate

|Av(x, t)| ≤ t c_{1} (||c||_{∞} + q) ||v||_{∞}.

By induction, we get then

|A^{n}v(x, t)| ≤ \frac{t^{n}}{n!} [c_{1} (||c||_{∞} + q)]^{n} ||v||_{∞}  (n ∈ ℕ).  (32)

Furthermore, for arbitrary x, z ∈ [0, 1] we have, by the mean value theorem,

\frac{|A^{n}v(x, t) − A^{n}v(z, t)|}{|x − z|^{α}}

≤ \frac{|A^{n}v(x, t) − A^{n}v(z, t)|}{|x − z|}

= |∫_{0}^{t}∫_{0}^{1} Γ_{x}(y, t; ξ, τ)(C + K)A^{n−1}v(ξ, τ) dξdτ|

=: I(y) for some y between x and z. Applying inequalities (20) and (32) we obtain

I(y) ≤ ∫_{0}^{t}∫_{0}^{1} |Γ_{x}(y, t; ξ, τ)|[|c(ξ, τ)| |A^{n−1}v(ξ, τ)| + q ∫_{0}^{1} |A^{n−1}v(s, τ)| ds]dξ dτ

≤ ∫_{0}^{t} \frac{c_{2}}{\sqrt{t − τ}} (||c||_{∞} + q)\frac{τ^{n−1}}{(n − 1)!}[c_{1}(||c||_{∞} + q)]^{n−1}||v||_{∞} dτ

≤ (∫_{0}^{t} \frac{τ^{n−1}}{\sqrt{t − τ}} \frac{1}{(n − 1)!} dτ) [a(||c||_{∞} + q)]^{n}||v||_{∞}

with a = max {c_{1}, c_{2}}. The identity

∫_{0}^{t} \frac{τ^{n−1}}{\sqrt{t − τ}} \frac{1}{(n − 1)!} dτ = \frac{2^{n}}{1 · 3 · 5 ⋯ (2n − 1)} t^{n− \frac{1}{2}}  (33)

leads to

(∫_{0}^{t} \frac{τ^{n−1}}{\sqrt{t − τ}} \frac{1}{(n − 1)!} dτ) [a(||c||_{∞} + q)]^{n}||v||_{∞}

≤ \frac{2^{n}}{1 · 3 · 5 ⋯ (2n − 1)} t^{n− \frac{1}{2}} [a(||c||_{∞} + q)]^{n}||v||_{∞}

≤ \frac{[2aT (||c||_{∞} + q)]^{n}}{n!} T^{− \frac{1}{2}} ||v||_{∞}.

Consequently, we obtain the estimate

||A^{n}v||_{C_{0}^{α,0}(\overline{Q})} ≤ \frac{[2aT (||c||_{∞} + q)]^{n}}{n!} T^{− \frac{1}{2}} ||v||_{C_{0}^{α,0}(\overline{Q})} (n ∈ ℕ).

From this estimate we deduce

n \sqrt{||A^{n}||} ≤ 2aT (||c||_{∞} + q) ^{n} \sqrt{T^{− \frac{1}{2}}} ^{n} \sqrt{\frac{1}{n!}} → 0 (n → ∞)

as claimed Building on the results of the previous sections we are now able to prove the following Theorem 1. The inhomogeneous linear equation (8) has for each f ∈ C_{0}^{α,0}(\overline{Q}) a

unique solution u ∈ C_{0}^{α,0}(\overline{Q}). This solution can be represented as infinite series

u = ∑_{n=0}^{∞}[L^{−1}(C + K)]^{n}(L^{−1}f)  (34)

and depends continuously on the data f ∈ C_{0}^{α,0}(\overline{Q}).

Proof. The operator A = L^{−1}(C+K) is a continuous endomorphism of the Banach space C_{0}^{α,0}(\overline{Q}). From Lemma 6 we know that the Neumann series ∑_{n=0}^{∞} A^{n} converges to the inverse of the operator I − A. Consequently, for f ∈ C_{0}^{α,0}(\overline{Q}) the inhomogeneous linear equation (8) has a unique solution u = (I −A)^{−1}(L^{−1}f) ∈ C_{0}^{α,0}(\overline{Q}) which depends continuously on f and has the representation (34)

From the proof of Lemma 6 we see that the norm of (I − A)^{−1} may be estimated by

||(I − A)^{−1}|| ≤ 1 + ∑_{n=1}^{∞} ||A^{n}|| ≤ 1 + \frac{exp [2aT (||c||_{∞} + q)] − 1}{\sqrt{T}}.

Next we consider the Dirichlet problem for the linear equation (8) with prescribed boundary function φ, which belongs to the set

C^{1}(R) = {φ ∈ C^{0}(R)| φ(·, 0) ∈ C^{1}([0, 1]) and φ(0, ·), φ(1, ·) ∈ C^{1}([0, T])}.

Theorem 2. Let f ∈ C_{0}^{α,0}(\overline{Q}) and φ ∈ C^{1}(R). Then the problem

\begin{aligned}Lu & = (C + K)u + f \\ u(x, 0) & = φ(x, 0) \\ u(0, t) & = φ(0, t) \\ u(1, t) & = φ(1, t) \end{aligned}  \begin{matrix}in Q \\ (x ∈ [0, 1]) \\ (t ∈ (0, T)) \\ (t ∈ (0, T]) \end{matrix} \}  (35)

has a unique solution u ∈ C^{0}(\overline{Q}) with u_{x} ∈ C^{0}(\overline{Q}) and u_{t}, u_{xx} ∈ C^{0}(Q).

Proof. If u_{1} and u_{2} are two solutions of problem (35), we see that the function u = u_{1} − u_{2} solves problem (30), and hence u ≡ 0

As usual, we obtain a representation of a solution u of problem (35) if we add the solution of problem (30) to the solution of the homogeneous heat equation Lu = 0 with u|R = φ, which we denote by Sφ, with

\begin{aligned}Sφ(x, t) & = ∫_{0}^{1} Γ(x, t; ξ, 0)φ(ξ, 0) dξ \\ & + ∫_{0}^{t} Γ_{ξ}(x, t; 0, τ)φ(0, τ) dτ − ∫_{0}^{t} Γ_{ξ}(x, t; 1, τ)φ(1, τ) dτ. \end{aligned}  (36)

So we have explicitly

u(x, t) = ∑_{n=0}^{∞}[L^{−1}(C + K)]^{n}(L^{−1}f)(x, t) + Sφ(x, t).

5. The nonlinear problem

In the nonlinear case we first give sufficient conditions under which the nonlinear operator KH, with K given by (5) and H given by (10), acts on C_{0}^{α,0}(\overline{Q}) and satisfies a Lipschitz condition in order to apply a classical fixed point principle.

Lemma 7. Suppose that the function k : [0, 1] × [0, 1] → ℝ satisfies the three conditions (a) - (c) stated in Lemma 4. Moreover, let h : \overline{Q} × ℝ → ℝ be a continuous function satisfying a Lipschitz condition

|h(x, t, u) − h(x, t, v)| ≤ L |u − v|.  (37)

Then the nonlinear operator KH acts on C_{0}^{α,0}(\overline{Q}) with

\begin{matrix}||KHu − KHv||_{∞} & ≤ qL ||u − v||_{∞} \\ ||KHu − KHv||_{C_{0}^{α,0}(\overline{Q})} & ≤ qL ||u − v||_{C_{0}^{α,0}(\overline{Q})} \end{matrix} \}

where q is given by (29).

Proof. It is easy to see that the function KHu is continuous if u is continuous. Furthermore, the function KHu satisfies the boundary condition KHu(0, t) = KHu(1, t) = 0 for all t ∈ [0, T]. From the estimate

\begin{aligned}|KHu(x, t) − KHu(y, t)| & ≤ ∫_{0}^{1} |k(s, x) − k(s, y)| |h(s, t, u(s, t))| ds \\ & ≤ q |x − y|^{α} max_{(s,t)∈\overline{Q}} |h(s, t, u(s, t))| \end{aligned}

we see that the operator KH maps C^{0}(\overline{Q}) into C_{0}^{α,0}(\overline{Q}), and hence KH : C_{0}^{α,0}(\overline{Q}) → C_{0}^{α,0}(\overline{Q}). For functions u, v ∈ C_{0}^{α,0}(\overline{Q}) we have

\frac{|(KHu − KHv)(x, t) − (KHu − KHv)(y, t)|}{|x − y|^{α}}

≤ \frac{1}{|x − y|^{α}} ∫_{0}^{1} |k(s, x) − k(s, y)| |h(s, t, u(s, t)) − h(s, t, v(s, t))| ds

≤ qL ||u − v||_{∞}

≤ qL ||u − v||_{C_{0}^{α,0}(\overline{Q})}.

From this the assertion follows

In view of the nonlinear operator equation (12) with imposed boundary conditions we define the function spaces

C_{0}^{1}([0, 1]) = {g ∈ C^{1}([0, 1])| g(0) = g(1) = 0}

and

C_{0}^{1,0}(\overline{Q}) = \{u| u, u_{x} ∈ C^{0}(\overline{Q}) and u(0, t) = u(1, t) = 0 for all t ∈ [0, T]\}.

Equipped with the norms

||g||_{C_{0}^{1}([0,1])} = sup_{x∈[0,1]} |g′(x)|  and  ||u||_{C_{0}^{1,0}(\overline{Q})} = sup_{(x,t)∈\overline{Q}} |u_{x}(x, t)|,

respectively, both function spaces are Banach spaces, and we can state the following

Lemma 8. For g ∈ C_{0}^{1}([0, 1]), the boundary operator S with

(Sg)(x, t) = ∫_{0}^{1} Γ(x, t; ξ, 0)g(ξ) dξ

is a continuous operator from C_{0}^{1}([0, 1]) into C_{0}^{1,0}(\overline{Q}) and ||S|| = 1.

Proof. The function r = Sg satisfies the homogenuous heat equation Lr(x, t) = 0 for all (x, t) ∈ Q with the boundary conditions r(x, 0) = g(x) for all x ∈ [0, 1] and r(0, t) = r(1, t) = 0 for all t ∈ [0, T]. If we extend g to the odd function g̃ on the interval [−1, 1] and continue g̃ to the periodic function ĝ with period 2, we remark that the function ∫_{−∞}^{+∞} γ(x − ξ, t)ĝ(ξ) dξ is also a solution of the Dirichlet problem above. Hence a unicity argument yields

r(x, t) = ∫_{−∞}^{+∞}γ(x − ξ, t)ĝ(ξ) dξ ((x, t) ∈ \overline{Q}).

Obviously, ĝ and ĝ′ are bounded and continuous functions on the whole real line. First we have r ∈ C^{0}(\overline{Q}). Considering the difference quotients

I_{h} = \frac{r(x + h, t) − r(x, t)}{h}

= \frac{1}{h} ( ∫_{−∞}^{+∞}γ(x + h − ξ, t)ĝ(ξ) dξ − ∫_{−∞}^{+∞}γ(x − ξ, t)ĝ(ξ) dξ)

= ∫_{−∞}^{+∞}γ(ξ, t)\frac{ĝ(ξ + x + h) − ĝ(ξ + x)}{h} dξ

we notice that the integrand is dominated by

|γ(ξ, t)\frac{ĝ(ξ + x + h) − ĝ(ξ + x)}{h}| ≤ \underbrace{γ(ξ, t) · sup_{ℝ} |ĝ′|}_{∈L^{1}(ℝ)} (ξ ∈ ℝ).

Now Lebesgue’s Dominated Convergence Theorem insures that

I_{h} → ∫_{−∞}^{+∞}γ(ξ, t)ĝ′(ξ + x) dξ = ∫_{−∞}^{+∞}γ(x − ξ, t)ĝ′(ξ) dξ

as h → 0. Thus

r_{x}(x, t) = ∫_{−∞}^{+∞}γ(x − ξ, t)ĝ′(ξ) dξ  and  r_{x} ∈ C^{0}(\overline{Q})

hold. Moreover, from the inequality

sup_{(x,t)∈\overline{Q}} |r_{x}(x, t)| ≤ ∫_{−∞}^{+∞}γ(x − ξ, t) dξ sup_{ξ∈ℝ} |ĝ′(ξ)| = sup_{ξ∈[0,1]} |g′(ξ)|

we deduce ||S|| ≤ 1. For the function g(x) = sin πx ∈ C_{0}^{1}([0, 1]) we have explicitly

Sg(x, t) = exp −π^{2}t sin πx ∈ C_{0}^{1}(\overline{Q}) and ||Sg||_{C_{0}^{1}(\overline{Q})} = ||g||_{C_{0}^{1}([0,1])} = π,

so ||S|| cannot be less than 1

Before we turn to the nonlinear operator equation (12), we remark that equations (9) and (12) are equivalent. Even more is true, namely (9)/(2) is equivalent to a nonlinear operator equation with an imposed boundary operator in the sense of the following

Lemma 9. For f ∈ C_{0}^{α,0}(\overline{Q}) and g ∈ C_{0}^{1}([0, 1]), the following two statements are equivalent:

(A) u ∈ C^{0}(\overline{Q}) has the properties u_{x} ∈ C^{0}(\overline{Q}), u_{t}, u_{xx} ∈ C^{0}(Q) and solves the boundary value problem

\begin{aligned}Lu & = (C + KH)u + f \\ u(x, 0) & = g(x) \\ u(0, t) & = u(1, t) = 0 \end{aligned}  \begin{matrix}in Q \\ (x ∈ [0, 1]) \\ (t ∈ [0, T]). \end{matrix} \}  (38)

(B) u ∈ C_{0}^{α,0}(\overline{Q}) satisfies the nonlinear operator equation

u − L^{−1}(C + KH)u = L^{−1}f + Sg.

Proof. It follows the pattern of the proof of Lemma 5 with only minor modifications. Hence it is omitted

Theorem 3. The nonlinear operator B : C_{0}^{α,0}(\overline{Q}) → C_{0}^{α,0}(\overline{Q}) defined by Bu = L^{−1}(C +KH)u+L^{−1}f +Sg has precisely one fixed point w ∈ C_{0}^{α,0}(\overline{Q}). This fixed point may be obtained as limit of the successive approximations v_{n} = B^{n}v_{0} with arbitrary v_{0} ∈ C_{0}^{α,0}(\overline{Q}).

Proof. First of all, the inequalities

\begin{matrix}||C(u − v)|| & ≤ ||c||_{∞}||u − v||_{∞} \\ ||KHu − KHv||_{∞} & ≤ q L||u − v||_{∞} \end{matrix} \}

with q given by (29) and L by (37), lead to

|Bu(x, t) − Bv(x, t)| = |L^{−1}C(u − v)(x, t) + L^{−1}K(Hu − Hv)(x, t)|
≤ |L^{−1}C(u − v)(x, t)| + |L^{−1}K(Hu − Hv)(x, t)|
≤ tc_{1} (||c||_{∞} + qL) ||u − v||_{∞}.

By induction, the inequality

|B^{n}u(x, t) − B^{n}v(x, t)| ≤ \frac{t^{n}}{n!} [c_{1}(||c||_{∞} + qL)]^{n}||u − v||_{∞}

may be proved for arbitrary n ∈ ℕ. In order to estimate the norm ||B^{n}u − B^{n}v||_{C_{0}^{α,0}(\overline{Q})} we fix x, z ∈ [0, 1] and get, by the mean value theorem,

|(B^{n}u − B^{n}v)(x, t) − (B^{n}u − B^{n}v)(z, t)|

= |∫_{0}^{t}∫_{0}^{1} [Γ(x, t; ξ, τ) − Γ(z, t; ξ, τ)] [(C + KH)B^{n−1}u − (C + KH)B^{n−1}v](ξ, τ) dξdτ|

= |x − z| |∫_{0}^{t}∫_{0}^{1} Γ_{x}(y, t; ξ, τ)[(C + KH)B^{n−1}u − (C + KH)B^{n−1}v](ξ, τ) dξdτ|

=: J(y)

for some y between x and z. Furthermore,

J(y) ≤ |x − z|^{α} ∫_{0}^{t}∫_{0}^{1} |Γ_{x}(y, t; ξ, τ)| |(C + KH)B^{n−1}u − (C + KH)B^{n−1}v)(ξ, τ)| dξdτ

≤ |x − z|^{α} ∫_{0}^{t}∫_{0}^{1} |Γ_{x}(y, t; ξ, τ)| (||c||_{∞} + qL) |(B^{n−1}u − B^{n−1}v)(ξ, τ)| dξdτ

≤ |x − z|^{α} ∫_{0}^{t} \frac{c_{2}(||c||_{∞} + qL)}{\sqrt{t − τ}} \frac{τ^{n−1}[c_{1}(||c||_{∞} + qL)]^{n−1}||u − v||_{∞}}{(n − 1)!} dτ

≤ |x − z|^{α} (∫_{0}^{t} \frac{1}{\sqrt{t − τ}} \frac{τ^{n−1}}{(n − 1)!} dτ) [a(||c||_{∞} + qL)]^{n}||u − v||_{∞}

with a = max {c_{1}, c_{2}}. Using again identity (33) we obtain

höl_{α}((B^{n}u − B^{n}v)(·, t)) ≤ \frac{2^{n}t^{n− \frac{1}{2}} [a(||c||_{∞} + qL)]^{n}}{1 · 3 · 5 ⋯ (2n − 1)} ||u − v||_{∞}.

This implies that ||B^{n}u − B^{n}v||_{C_{0}^{α,0}(\overline{Q})} ≤ d_{n}||u − v||_{C_{0}^{α,0}(\overline{Q})} where

d_{n} = \frac{2^{n}T^{n− \frac{1}{2}} [a(||c||_{∞} + qL)]^{n}}{1 · 3 · 5 ⋯ (2n − 1)} .

Obviously, we can choose n_{0} ∈ ℕ such that

\frac{d_{n+1}}{d_{n}} = \frac{2T a(||c||_{∞} + qL)}{2n + 1} < \frac{3}{4},

say, for n ≥ n_{0}. Consequently, the series ∑_{n=1}^{∞} d_{n} converges. By Weissinger’s fixed point theorem of [7], the operator B has a unique fixed point w ∈ C_{0}^{α,0}(\overline{Q}), which can be obtained by the successive approximation v_{n+1} = Bv_{n}, with v_{0} ∈ C_{0}^{α,0}(\overline{Q}) arbitrary. Moreover, the error estimate ||w − v_{n}||_{C_{0}^{α,0}(\overline{Q})} ≤ ||Bv_{0} − v_{0}||_{C_{0}^{α,0}(\overline{Q})} ∑_{k=n}^{∞} d_{k} is true

As a corollary of Theorem 3, we get the following

Theorem 4. Let f ∈ C_{0}^{α,0}(\overline{Q}) and φ ∈ C^{0}(R). Then the problem

\begin{aligned}Lu & = (C + KH)u + f \\ u & = φ \end{aligned}  \begin{matrix}in Q \\ on R \end{matrix} \}  (39)

has a at least one solution u ∈ C^{0}(\overline{Q}) such that u_{x}, u_{t}, u_{xx} ∈ C^{0}(Q). One solution can be represented in the form

\begin{aligned}u(x, t) & = w(x, t) + ∫_{0}^{1} Γ(x, t; ξ, 0)φ(ξ, 0) dξ \\ & + ∫_{0}^{t} Γ_{ξ}(x, t; 0, τ)φ(0, τ) dτ − ∫_{0}^{t} Γ_{ξ}(x, t; 1, τ)φ(1, τ) dτ \end{aligned}

where w ∈ C_{0}^{α,0}(\overline{Q}) is the unique fixed point of the nonlinear operator B̃u = L^{−1}(C + KH)u + L^{−1}f.

Theorem 5. The solution of the boundary value problem (38) depends continuously on the functions f ∈ C_{0}^{α,0}(\overline{Q}) and g ∈ C_{0}^{1}([0, 1]).

Proof. Given f, h ∈ C_{0}^{α,0}(\overline{Q}) and g, j ∈ C_{0}^{1}([0, 1]), denote by v, w ∈ C_{0}^{α,0}(\overline{Q}) the unique solutions of the operator equations

\begin{aligned}v & = L^{−1}(C + KH)v + L^{−1}f + Sg \\ w & = L^{−1}(C + KH)w + L^{−1}h + Sj, \end{aligned}

196 W. Kohl

respectively. Differencing the derivatives with respect to x and estimating yields

|v_{x}(x, t) − w_{x}(x, t)|

≤ ||Sg − Sj||_{C_{0}^{1,0}(\overline{Q})}

+ ∫_{0}^{t}∫_{0}^{1} |Γ_{x}(x, t; ξ, τ)|[|Cv − Cw| + |KHv − KHw| + |f − h|](ξ, τ) dξdτ

≤ c_{2}(T)\sqrt{t} ||f − h||_{C_{0}^{α,0}(\overline{Q})} + ||g − j||_{C_{0}^{1}([0,1])}

+ ∫_{0}^{t} \frac{c(T)}{\sqrt{t − τ}} (∥c∥_{∞} + qL) sup_{ξ∈[0,1]} |(v − w)(ξ, τ)| dτ.

We apply the mean value theorem,

|(v − w)(ξ, t)| = |(v − w)(ξ, t) − (v − w)(0, t)| = |(v_{x} − w_{x})(z, t)| |ξ|

for z ∈ (0, ξ), to obtain the relationship

sup_{ξ∈[0,1]} |(v − w)(ξ, t)| ≤ sup_{x∈[0,1]} |v_{x}(x, t) − w_{x}(x, t)| := φ(t).

Employing this inequality and passing to the supremum over the interval [0, 1] yields

φ(t) ≤ c_{2}(T)\sqrt{t} ∥f − h∥_{C_{0}^{α,0}(\overline{Q})} + ||g − j||_{C_{0}^{1}([0,1])} + ∫_{0}^{t} \frac{c(T)}{\sqrt{t − τ}} (∥c∥_{∞} + qL)φ(τ) dτ.

By virtue of the generalized Gronwall’s inequality (see, e.g., [2: p. 304/ Lemma 17.7.1]) the estimate

φ(t) ≤ c̃_{2}(T)(c_{2}(T)\sqrt{t}∥f − h∥_{C_{0}^{α,0}(\overline{Q})} + ||g − j||_{C_{0}^{1}([0,1])}))

holds. Hence we get the inequality

∥v − w∥_{C_{0}^{1,0}(\overline{Q})} ≤ c̃(T)(∥f − h∥_{C_{0}^{α,0}(\overline{Q})} + ||g − j||_{C_{0}^{1}([0,1])})  (40)

with a certain constant c̃(T). This concludes the proof

To illustrate the existence and uniqueness results of the previous section, let us consider a very simple example. Let ω : [0, 1] → ℝ be defined by

ω(x) = \{ \begin{matrix}x^{α} & for 0 ≤ x ≤ \frac{1}{2} \\ (1 − x)^{α} & for \frac{1}{2} ≤ x ≤ 1. \end{matrix}

Put

k(s, x) = k̂(s)ω(x)
h(x, t, u) = ĥ(x, t) arctan u
f(x, t) = ω(x)f̂(t)
g(x) = sin πx

where k̂ : (0, 1) → ℝ is measureable and bounded, while ĥ : Q → ℝ and f̂: [0, T] → ℝ are continuous. Obviously, k satisfies the hypotheses of Lemma 4 with q = ||k̂||_{L^{∞}((0,1))}. Moreover, h satisfies (37) with L = max\{|ĥ(x, t)| : (x, t) ∈ \overline{Q}\}, f ∈ C_{0}^{α,0}(\overline{Q}) with ||f||_{C_{0}^{α,0}(\overline{Q})} = ||f̂||_{C^{0}([0,T])} and g ∈ C_{0}^{1}([0, 1]). As multiplicator we may choose, for example, c(x, t) = x^{α}p(t) with p ∈ C^{0}([0, T]; from Lemma 3 we know then that ||C|| ≤ 2||p||_{C^{0}([0,T])}. For this choice of data, the operator C + KH has the form

(C + KH)u(x, t) = x^{α}u(x, t) + ω(x) ∫_{0}^{1} k̂(s)ĥ(s, t) arctan u(s, t) ds.  (41)

From Theorem 3 we conclude that the sequence of successive approximations

v_{0}(x, t) ≡ 0 ; ⋮ ; v_{n+1}(x, t) = L^{−1}(C + KH)v_{n}(x, t) + L^{−1}f(x, t) + Sg(x, t) }

has a well-defined limit w ∈ C_{0}^{α,0}(\overline{Q}). If f̂(t) ≡ 0 and g ≡ 0, we have of course u(x, t) ≡ 0, and this is the only solution of problem (38), by Lemma 9 and Theorem 3. On the other hand, if f̂(t) ≢ 0 and g ≢ 0, from Theorem 5 we may conclude not only that u(x, t) = w(x, t) is the unique solution of problem (38), but also that this solution depends continuously on f̂. In particular, u(x, t) → 0 uniformly on \overline{Q} if ||f̂||_{∞} → 0 and ||g||_{C_{0}^{1}([0,1])} → 0.

6. The extension of the operator L^{−1}

This last section is concerned with some generalizations of the preceding results. In order to solve the inhomogenuous heat equation with zero boundary values, we chose for technical reasons the heat source f from the Hölder space C_{0}^{α,0}(\overline{Q}). On this space the operator L^{−1} has particularly nice properties. Actually, one can take the larger Hölder space C^{α,0}(\overline{Q}) as underlying Banach space of the boundary value problem (7).

Together with the solution operator S of the homogenuous heat equation with C^{1}- boundary values, where Sφ is given by (36), with the projection operator P,

P f(x, t) = f(0, t) + x(f(1, t) − f(0, t)),  (42)

and with the Volterra operator V,

V f(x, t) = ∫_{0}^{t} f(x, τ) dτ,  (43)

we can represent the unique solution of the boundary value problem (7) with the help of the extended operator L_{e}^{−1}

L_{e}^{−1} = L^{−1}(I − P) + (I − S)V P  (44)

in the form

u = L_{e}^{−1}f + Sφ.

In fact, we have u, u_{x} ∈ C^{0}(\overline{Q}), and direct calculations yield u|R = φ and Lu = f.

According to the plan in the introduction we formulate now sufficient conditions that the operators C and KH act continuously on C^{α,0}(\overline{Q}).

Lemma 10. Suppose that c ∈ C^{α,0}(\overline{Q}). Then the corresponding multiplication operator (4) is bounded in C^{α,0}(\overline{Q}) and ||C|| ≤ ||c||_{C^{α,0}(\overline{Q})}.

Lemma 11. Suppose that the function k : [0, 1] × [0, 1] → ℝ has the following properties:

(a) k(x, ·) ∈ L^{1}([0, 1]) for each x ∈ [0, 1].

(b) k(·, s) ∈ C^{α}([0, 1]) for almost every s ∈ [0, 1], such that there exists a function q ∈ L^{1}([0, 1]) with the property

|k(x, s) − k(y, s)| ≤ q(s) |x − y|^{α}  for a.e. s ∈ [0, 1].  (45)

Then the corresponding partial integral operator (5) is bounded in C^{α,0}(\overline{Q}) with

||K|| ≤ ||q||_{L^{1}([0,1])} + sup_{x∈[0,1]} ||k(x, ·)||_{L^{1}([0,1])}.

Lemma 12. Suppose that the function k : [0, 1] × [0, 1] → ℝ satisfies conditions (a) and (b) stated in Lemma 11. Moreover, let h : \overline{Q} × ℝ → ℝ be a continuous function satisfying the Lipschitz condition (37). Then the nonlinear operator KH acts on C^{α,0}(\overline{Q}) with

\begin{matrix}||KHu − KHv||_{∞} & ≤ L sup_{x∈[0,1]} ||k(x, ·)||_{L^{1}([0,1])}||u − v||_{∞} \\ ||KHu − KHv||_{C^{α,0}(\overline{Q})} & ≤ L(||q||_{L^{1}([0,1])} + sup_{x∈[0,1]} ||k(x, ·)||_{L^{1}([0,1])})||u − v||_{∞}. \end{matrix} \}

We omit the proofs of these three lemmata, because their proofs follow the pattern of that given in Lemma 3, 4 and 7 with only minor modifications.

After modifying the proof of Lemma 5 we are able to establish the next result.

Lemma 13. For f ∈ C^{α,0}(\overline{Q}) and φ ∈ C^{1}(R), the following two statements are equivalent:

(A) u ∈ C^{0}(\overline{Q}) has the properties u_{x} ∈ C^{0}(\overline{Q}) and u_{t}, u_{xx} ∈ C^{0}(Q) and solves the boundary value problem

\begin{aligned}Lu & = (C + KH)u + f \\ u|R & = φ \end{aligned}  \begin{matrix}in Q \\ on R \end{matrix} \}  (46)

(B) u ∈ C^{α,0}(\overline{Q}) satisfies the nonlinear operator equation

u − L_{e}^{−1}(C + KH)u = L_{e}^{−1}f + Sg.

Our main existence and uniqueness result reads as follows.

Theorem 6. For f ∈ C^{α,0}(\overline{Q}) and initial data φ ∈ C^{1}(R) the boundary value problem

\begin{matrix}Lu & = (C + KH)u + f \\ u|R & = φ \end{matrix} \}

possesses a unique solution.

Proof. By Lemma 13 it suffices to show that the integral equation admits a unique solution. For a < b the set

B([a, b]) = {v : [0, 1] × [a, b] → ℝ| v, v_{x} ∈ C^{0}([0, 1] × [a, b])}

becomes together with the norm

||v||_{[a,b]} = sup_{(x,t)∈[0,1]×[a,b]} |v(x, t)| + sup_{(x,t)∈[0,1]×[a,b]} |v_{x}(x, t)|

a Banach space. The nonlinear operator Au = L_{e}^{−1}(C + KH)u + L_{e}^{−1}f + Sφ maps B([0, η]) into B([0, η]) for 0 < η ≤ T. Employing Lemma 1 we know that for v ∈ B([0, η]) the estimation

||L^{−1}v||_{[0,η]} ≤ c(T) (η + \sqrt{η})||v||_{[0,η]}

holds. Actually, the same kind of estimation is valid for the operator L_{e}^{−1}. Applying this and the assumptions on the continuity of the operator C +KH we obtain the inequality

||Au − Av||_{[0,η]} ≤ c̃(η + \sqrt{η})||u − v||_{[0,η]}

with a constant c̃(T, ||C||, ||K||, L). Choosing n ∈ ℕ such that the estimation

c̃(η + \sqrt{η}) < 1  (46)

holds for η = \frac{T}{n}, the operator A is a contraction of B([0, η]) into B([0, η]), and according to the Banach fixed point theorem the operator A possesses a unique fixed point w_{1} ∈ B([0, η]). Assuming that the integral equation possesses a unique solution w_{k} ∈ B([0, kη]) for k ≤ n − 1 we introduce the function r : [0, 1] × [kη, (k + 1)η] → ℝ,

r(x, t) = ∫_{0}^{1} Γ(x, t; ξ, 0)φ(ξ, 0) dξ

+ ∫_{0}^{kη} ∫_{0}^{1} Γ(x, t; ξ, τ)(I − P)[(C + KH)w_{k}(ξ, τ) + f(ξ, τ)] dξdτ

+ ∫_{0}^{kη}P [(C + KH)w_{k}(x, τ) + f(x, τ)] dτ

− ∫_{0}^{kη}Γ_{ξ}(x, t; 0, τ) (∫_{0}^{τ} (C + KH)w_{k}(0, s) + f(0, s) ds − φ(0, τ)) dτ

+ ∫_{0}^{kη}Γ_{ξ}(x, t; 1, τ) (∫_{0}^{τ} (C + KH)w_{k}(1, s) + f(1, s) ds − φ(1, τ)) dτ.

Next we consider the operator A_{1},

A_{1}v(x, t) = r(x, t) + ∫_{kη}^{t}∫_{0}^{1} Γ(x, t; ξ, τ)(I − P)[(C + KH)v(ξ, τ) + f(ξ, τ)] dξdτ

+ ∫_{kη}^{t} P [(C + KH)v(x, τ) + f(x, τ)] dτ

− ∫_{kη}^{t} Γ_{ξ}(x, t; 0, τ) (∫_{kη}^{τ} (C + KH)v(0, s) + f(0, s) ds − φ(0, τ)) dτ

+ ∫_{kη}^{t} Γ_{ξ}(x, t; 1, τ) (∫_{kη}^{τ} (C + KH)v(1, s) + f(1, s) ds − φ(1, τ)) dτ.

Of course, A_{1} maps B([kη, (k + 1)η]) into B([kη, (k + 1)η]) and the estimate

||A_{1}u − A_{1}v||_{[kη,(k+1)η]} ≤ c̃(η + \sqrt{η})||u − v||_{[kη,(k+1)η]}

holds. From (46), we conclude that A_{1} is a contraction and hence possesses a unique fixed point u ∈ B([kη, (k + 1)η]). Since the fixed point of A_{1} matches continuously with w_{k}(x, kη) and \frac{∂}{∂x}w_{k}(x, kη), we see that the function w_{k+1},

w_{k+1}(x, t) = \{ \begin{matrix}w_{k}(x, t) & if (x, t) ∈ [0, 1] × [0, kη] \\ u(x, t) & if (x, t) ∈ [0, 1] × [kη, (k + 1)η] \end{matrix}

is the unique solution of the integral equation in B([0, (k+1)η]). Applying this argument we can construct inductively the unique solution of the integral equation in B([0, T])

On a Class of Integro-Differential Equations 201

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Received 13.04.1999