16th Conference on Applied Mathematics, Univ. of Central Oklahoma, Electronic Journal of Differential Equations, Conf. 07, 2001, pp. 89–97. ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu (login: ftp)

Properties of the solution map for a first order linear problem ^{∗}

James L. Moseley

Abstract

We are interested in establishing properties of the general mathematical model

\frac{du⃗}{dt} = T (t, u⃗) + b⃗ + g⃗(t), u⃗(t_{0}) = u⃗_{0}

for the dynamical system defined by the (possibly nonlinear) operator T (t, ·) : V → V with state space V. For one state variable where V = ℝ this may be written as dy/dx = f(x, y), y(x_{0}) = y_{0}. This paper establishes some mapping properties for the operator L[y] = dy/dx + p(x)y with y(x_{0}) = y_{0} where f(x, y) = −p(x)y + g(x) and T (x, y) = −p(x)y is linear. The conditions for the one-to-one property of the solution map as a function of p(x) appear to be new or at least undocumented. This property is needed in the development of a solution technique for a nonlinear model for the agglomeration of point particles in a confined space (reactor).

1 Introduction

We begin with a family of initial value problems (IVP’s) each consisting of a (possibly nonlinear) first order ordinary differential equation (ODE)

\frac{dy}{dx} = f(x, y),  (1)

and an arbitrary initial condition (IC) at an arbitrary point:

y(x_{0}) = y_{0} .  (2)

To specify a problem in this family, we must give the point (x_{0}, y_{0}) in the plane ℝ^{2} = {(x, y) : x, y ∈ ℝ} where ℝ is the set of real numbers and the function f : Ω → ℝ where Ω is an open connected region in ℝ^{2} containing (x_{0}, y_{0}). Constraints on f are needed to make the problem reasonable and more likely to have a physical application. We assume immediately that Ω ⊇ R. where R is a closed rectangle containing (x_{0}, y_{0}) in its interior and that f is continuous on R. Thus to specify a problem, we choose, in order, (x_{0}, y_{0}) ∈ ℝ^{2}, R ⊆ Ω (the exact definition of Ω really does not matter) as a closed rectangle in ℝ^{2} containing (x_{0}, y_{0}) in its interior, and f ∈ C(R), the set of functions f : R → ℝ that are continuous. We let Ω_{CR}(x_{0}, y_{0}) be the set of all closed rectangles in ℝ^{2} that contain (x_{0}, y_{0}) in their interior, and C(Ω_{CR}(x_{0}, y_{0})) = {f ∈ C(R) : R ∈ Ω_{CR}(x_{0}, y_{0})}. Then ℝ^{2} × Ω_{CR}(x_{0}, y_{0}) × C(Ω_{CR}(x_{0}, y_{0})) is in a one-to-one correspondence with the set of problems of interest.

∗Mathematics Subject Classifications: 34A99.

Key words: First order linear ordinary differential equation.

\begin{matrix}c \\ ◯ \end{matrix}2001 Southwest Texas State University and University of North Texas.

Published July 20, 2001.

If (1) is nonlinear, the interval of validity (i.e, the open interval I containing x_{0} where (1) and (2) are satisfied) is part of the problem which is therefore impredicative. However, we state the problem predicatively by assuming that I is given and look for solutions to (1) on I. That is, we look for solutions to (1) in a set Σ(I) of functions whose common domain is I (i.e., a subset of F (I) = {f : I → ℝ}). Let I_{IV} (x_{0}, y_{0}, R) be the set of intervals I containing x_{0} where I × {y_{0}} ⊆ R ⊆ ℝ^{2} and Prob((x_{0}, y_{0}), R, f, I) denote the initial value problem (1) and (2) associated with (x_{0}, y_{0}) ∈ ℝ^{2}, R ∈ Ω_{CR}(x_{0}, y_{0}), f ∈ C(R), and I ∈ I_{IV} (x_{0}, y_{0}, R). A minimum requirement for this IVP to be well-posed is that there is exactly one solution in Σ(I) that satisfies both (1) and (2). Since we elect to specify the interval I, we denote by Prob(ℝ^{2} × Ω_{CR}(x_{0}, y_{0}) × C(Ω_{CR}(x_{0}, y_{0})) × I_{IV} (x_{0}, y_{0}, R)) the set of all initial value problems of interest.

There are at least four problem solving contexts for (1)–(2).

Traditional: If f(x, y) is given as an elementary function and has one of several specific forms, the solution process starts with the ODE and uses calculus to obtain the “general” solution (i.e., a formula for all or at least almost all solutions) to the ODE as a parameterized family of functions (or curves). The IC is then applied to obtain the parameter and hence the (name of the) unique solution function. The interval of validity is then obtained as the largest (open) interval where the solution is valid. The solution algorithm (usually) establishes uniqueness and, if all steps are reversible, existence. If all steps are not reversible, existence can be established by substituting the proposed solution back into the ODE and the IC. (Or this can be used simply as a check.) Often a formula can be found for the (name of the) solution function for a whole class of problems by allowing parameters such as x_{0} and y_{0} to be arbitrary.

Classical: For a class of problems, existence and uniqueness of a solution in Σ(I) is established using properties of f(x, y), without necessarily obtaining a solution algorithm to obtain the (name of the) solution function. Classical I: Σ(I) = A(I) = {y : I → ℝ : y is analytic on I} (e.g., if f is analytic). Classical II: Σ(I) = C^{1}(I) = {y : I → ℝ : y′ exists and is continuous on I}.

Modern: A weak form of the problem is developed which allows weak solutions; that is, things that need not be functions (e.g., equivalence classes of functions and distributions).

In the Classical II context, the standard condition that f, ∂f/∂y ∈ C(R) assures local uniqueness (i.e., that for any I ∈ I_{IV} (x_{0}, y_{0}, R), Σ(I) = C^{1}(I) contains at most one solution), but only local (and not global) existence (i.e., there exists an I ∈ I_{IV} (x_{0}, y_{0}, R) such that Σ(I) = C^{1}(I) contains a solution). Thus, in this context, the problem then focuses on finding the extent of the interval of validity for a class of problems rather than on finding the (name of the) solution function for a specific problem.

2 The linear solution map

Even though different contexts may define the problem differently, Traditional, Classical, and Modern all come together with the assumption of linearity, that is, when f(x, y) = −p(x)y + g(x). In this case we have the ordinary differential equation (ODE)

\frac{dy}{dx} + p(x)y = g(x)  (3)

with the initial condition (IC)

y(x_{0}) = y_{0} .  (4)

We switch from viewing (1) as an equation to viewing (3) and (4) as a mapping problem. Thus we keep x_{0}, y_{0}, I, and p fixed and only vary g. To keep solutions as functions, we continue with the Classical II context, let Σ(I) = C^{1}(I), and assume p, g ∈ C(I) = {f : I → ℝ that are continuous} where x_{0} ∈ I so that f, ∂f/y ∈ C(R) where R is the strip R = I×ℝ = {(x, y) : x ∈ I}. Now let L_{p}[y] : C^{1}(I) → C(I) be defined by L_{p}[y] = dy/dx + p(x)y and N_{p,y_{0}}; D_{y_{0}}(I) → C(I) be the restriction of L_{p} to the hyper-plane D_{y_{0}}(I) = {y ∈ C^{1}(I) : y(x_{0}) = y_{0}}. Not only are we assured that for any g ∈ C(I), a unique (global) solution to the IVP (3) and (4) exists in Σ(I) = C^{1}(I) so that inverse mapping exists, but, using the integrating factor μ_{p}(x) = exp{∫_{t=x_{0}}^{t=x} p(t)dt}, we have a calculus formula for y(x) = N_{p,y_{0}}^{−1}[g](x):

y(x) = (y_{0} + ∫_{t=x_{0}}^{t=x} g(t) exp{∫_{s=x_{0}}^{s=t} p(s)ds}dt) exp{− ∫_{t=x_{0}}^{t=x} p(t)dt}
= y_{0} exp{− ∫_{t=x_{0}}^{t=x} p(t)dt} + ∫_{t=x_{0}}^{t=x} g(t) exp{∫_{s=x}^{s=t} p(s)ds}dt  (5)

If y_{0} ≠ 0, D_{y_{0}}(I) will not pass through the origin and not be a subspace of C^{1}(I) so that N_{p,y_{0}} and N_{p,y_{0}}^{−1} will not be linear operators. However,

y(x) = N_{p,y_{0}}^{−1}[g](x) = y_{0}μ_{−p}(x) + L_{p,0}^{−1}[g](x)  (6)

where the (linear) Voltera operator L_{p,0}^{−1}[g](x) = ∫_{t=x_{0}}^{t=x} G(x, t)g(t)dt with kernel (Green’s function) G(x, t) = exp{∫_{s=x}^{s=t} p(s)ds} is the inverse of N_{p,0} which we might also call L_{p,0} since when y_{0} = 0, N_{p,y_{0}}is linear.

In a traditional context, p and g are given elementary functions in C(I). If possible, the Riemann integrals in (5) are computed explicitly. In a classical context, the interval of validity I = (a, b) can be extended to closed (and halfopen) intervals by requiring p and g to be analytic (in a neighborhood of) or continuous at the end points. In one modern context, the Riemann integrals become Lebesgue integrals and act on equivalence classes of functions, for example, piecewise continuous functions which need not be defined at the points of discontinuity (as these points form a set of measure zero). We continue to keep x_{0} and I fixed, but now allow y_{0} and p as well as g to vary. To simplify our notation, we write y(x) = y(x; y_{0}, p, g) instead of y(x) = N_{p,y_{0}}^{−1}[g](x) = y_{0}μ_{−p}(x) + L_{p,0}^{−1}[g](x).

3 One-to-one properties

To understand how y(x; y_{0}, p, g) given by (5) depends on each of the parameters y_{0}, p, and g, we need several relations. For i=1,2, let y_{i} be the solution to the IVP (3) and (4) (on I where I may be open or closed) when p = p_{i}, g = g_{i}, and y_{0} = y_{i}^{0}. If p_{1} = p_{2} = p, g_{1} = g_{2} = g and p, g ∈ C(I), then for all x in I using (5) we obtain

\begin{matrix}|y_{1}(x) − y_{2}(x)| = & |y0_{1} − y0_{2}| exp\{− ∫_{t=x_{0}}^{t=x} p(t)dt\} \\ ≤ & |y0_{1} − y0_{2}| exp\{∫_{t=x_{0}}^{t=x} |p(t)| dt\} \end{matrix}  (7)

If y_{1}^{0} = y_{2}^{0} = y_{0}, p_{1} = p_{2} = p and p, g_{1}, g_{2} ∈ C(I), then for all x in I using (3) we obtain

d(y_{1} − y_{2})/dx + p(x)(y_{1}(x) − y_{2}(x)) = g_{1}(x) − g_{2}(x)  (8)

and using (5)

|y_{1}(x) − y_{2}(x)| = |∫_{t=x_{0}}^{t=x} [g_{1}(t) − g_{2}(t)] exp{∫_{s=x}^{s=t} p(s)ds}dt|  (9)
≤ ∫_{t=x_{0}}^{t−x} |g_{1}(t) − g_{2}(t)| exp{∫_{s=x}^{s=t} |p(s)| ds}dt  (10)

If y_{1}^{0} = y_{2}^{0} = y_{0}, g_{2} = g_{1} = g and p_{1}, p_{2}, g ∈ C(I), then for all x in I from (3) we obtain

d(y_{1} − y_{2})/dx + p_{1}(x)y_{1}(x) − p_{2}(x)y_{2}(x) = 0  (11)

and using (5)

|y_{1}(x) − y_{2}(x)|
=|y_{0} exp{− ∫_{t=x_{0}}^{t=x} p_{1}(t)dt} − y_{0} exp{− ∫_{t=x_{0}}^{t=x} p_{2}(t)dt}
+ ∫_{t=x_{0}}^{t=x} g(t) exp{∫_{s=x}^{s=t} p_{1}(s)ds}dt − ∫_{t=x_{0}}^{t=x} g(t) exp{∫_{s=x}^{s=t} p_{2}(s)ds}dt| (12)
=|y_{0} exp{− ∫_{t=x_{0}}^{t=x} p_{1}(t)dt}[1 − exp{∫_{t=x_{0}}^{t=x}[−p_{2}(t) + p_{1}(t)]dt}]
+ ∫_{t=x_{0}}^{t=x} g(t)[ exp{∫_{s=x}^{s=t} p_{1}(s)ds}][1 − exp{∫_{s=x}^{s=t}[p_{2}(s) − p_{1}(s)]ds}]dt|
≤|y_{0}| exp{∫_{t=x_{0}}^{t=x} |p_{1}(t)| dt}[ exp{∫_{t=x_{0}}^{t=x} |p_{2}(t) − p_{1}(t)|dt} − 1]

+ ∫_{t=x_{0}}^{t=x} |g(t)| [ exp{∫_{s=x}^{s=t} |p_{1}(s)| ds}][ exp{∫_{s=x}^{s=t} |p_{2}(s) − p_{1}(s)|ds} − 1]dt (13)

where we have used the inequality |1 − e^{b}| = |e^{b} − 1| ≤ |e^{|b|} − 1| = e^{|b|} − 1.

Standard theory [1] implies that for each y_{0} ∈ ℝ and p ∈ C(I), N_{p,y_{0}}^{−1}[g](x) provides a one-to-one correspondence between g ∈ C(I) and y ∈ D_{y_{0}}(I) as well as establishing that for fixed p and g, the solutions to (3) (parameterized by y_{0}) do not cross each other. Interestingly, with some restrictions, the solution map from (y_{0}, p, g) ∈ ℝ×C(I) × C(I) to y ∈ C^{1}(I) is one-to-one if any two of these three variables are held constant.

Theorem 1 If p, g ∈ C(I), then the solution map defined by (5) from y_{0} ∈ ℝ to y ∈ C^{1}(I) is one-to-one. Also, the solutions to (3) in C^{1}(I) do not cross each other.

Proof. If y_{1}^{0} and y_{2}^{0} are different initial conditions and p, g ∈ C(I), then for all x in I we have from (7) that |y_{1}(x) − y_{2}(x)| = |y_{1}^{0} − y_{2}^{0}|{exp − ∫_{t=x_{0}}^{t=x} p(t)dt}. Hence if y_{1}(x) = y_{2}(x) for any x ∈ I, then we must have y_{1}^{0} = y_{2}^{0}. That is, if we change the initial condition, the solution changes everywhere. If p and g are fixed, not only is the the mapping from y_{0} ∈ ℝ to y ∈ C^{1}(I) one-to-one, but the family of solutions to (3) parameterized by y_{0} do not cross each other.

Theorem 2 If y_{0} ∈ I, p ∈ C(I), then the solution map y(x) = N_{p,y_{0}}^{−1}[g](x) defined by (5) from g ∈ C(I) to y ∈ C^{1}(I) is one-to-one. Also N_{p,y_{0}}^{−1}[g](x) provides a one-to-one correspondence between C(I) and D_{y_{0}}(I).

Clearly L_{p} maps C^{1}(I) into C(I). N_{p,y_{0}}^{−1} given by (6) shows N_{p,y_{0}} is oneto-one and that the domain of N_{p,y_{0}}^{−1} is all of C(I). Hence N_{p,y_{0}}^{−1} is one-to-one. Alternately, this follows directly from (8) and is just the statement that N_{p,y_{0}}=(N_{p,y_{0}}^{−1})^{−1}is a well defined operator. Hence N_{p,y_{0}}^{−1} provides a one-to-one correspondence between C(I) and D_{y_{0}}(I). (And N_{p,y_{0}} provides a one-to-one correspondence between D_{y_{0}}(I) and C(I).)

Restrictions on y_{0} and g are needed for the mapping from p ∈ C(I) to y ∈ C^{1}(I) to be one-to-one. To see why, note from (5) that if y_{0} = 0 and g is the zero function, then for any p in C(I), y is identically zero.

Definition f ∈ C(I) is said to have only dispersed zeros on an interval I (either open, closed, or half-opened), if there exists no open interval J ⊆ I where f is identically zero; that is, if for any open interval J ⊆ I there exists x ∈ J such that f(x) ≠ 0 (i.e., Z_{f} = {x ∈ I : f(x) = 0} has no interior points).

We show that for any y_{0} and p, if g has only dispersed zeros, then the solution y also has only dispersed zeros. On the other hand, if g is identically zero, then y is either never zero or always zero, depending on y_{0}.

Theorem 3 If g ∈ C(I) and has only dispersed zeros on an interval I, then the solution y has only dispersed zeros on I.

Proof. We prove the contrapositive. Assume y does not have only dispersed zeros on I. By definition there exists an open interval J ⊆ I such that y(x) = 0 for all x in J. Then dy/dx = 0 on J and by (3), g(x) = 0 on J. Hence g does not have only dispersed zeros on I.

Theorem 4 Let I be an interval. If g(x) = 0 on an interval J ⊆ I, then either y is identically zero on J or is never zero on J. In particular, if g(x) = 0 on I, then either y_{0} = 0 and y is identically zero or y_{0} ≠ 0 and y is never zero on I.

Proof. Suppose g(x) = 0 on an interval J ⊆ I. Choose x_{1} ∈ J. Applying (5) at x_{1} with g(x) = 0 on J, we have that y(x) = y(x_{1})exp{− ∫_{t=x_{0}}^{t=x} p(t)dt} on J. If y(x_{1}) = 0, then y is identically zero on J. If y(x_{1}) ≠ 0, then y is never zero on J. Similarly for I.

Corollary 1 If y has only dispersed zeros on an interval I, then on any interval J ⊆ I where g(x) = 0, y is never zero.

Proof. Assume y has only dispersed zeros and that g(x) = 0 on an interval J ⊆ I. But if y has only dispersed zeros, there exists x_{1} ∈ J such that y(x_{1}) ≠ 0. Then by Theorem 4, we have that y(x) ≠ 0 for all x in J.

Theorem 5 If g is identically zero on an interval I and y_{0} ≠ 0, then the solution map defined by (5) from p ∈ C(I) to y ∈ C^{1}(I) is one-to-one.

Proof. Suppose y_{0} ≠ 0, g = 0 and p_{1}, p_{2} ∈ C(I). If y_{1} = y_{2} = y, then for all x ∈ I we have from (11) that (p_{1}(x) − p_{2}(x))y(x) = 0. From Theorem 4, y(x) is never zero so that for all x ∈ I, p_{1}(x) = p_{2}(x). Hence p_{1} = p_{2} so that the solution map is one-to-one.

Theorem 6 If g has only dispersed zeros on I, then the solution map defined by (5) from p ∈ C(I) to y ∈ C^{1}(I) is one-to-one.

Proof. Suppose g has only dispersed zeros on I, and p_{1}, p_{2}, g ∈ C(I). If y_{1} = y_{2} = y, then for all x in I we have from (11) that (p_{1}(x)−p_{2}(x))y(x) = 0. Hence if y(x) ≠ 0, then p_{1}(x) = p_{2}(x). Given x ∈ I, if there exists a sequence {x_{n}}_{n=1}^{∞} such that y(x_{n}) ≠ 0 and lim_{n→∞} x_{n} = x; then by continuity, p_{1}(x) = p_{2}(x). Thus p_{1} = p_{2} everywhere if the zeros of y are dispersed. It remains to show that if g has only dispersed zeros on I, then y has only dispersed zeros on I. But this is just Theorem 3.

4 Continuity properties

Finally, for the IVP (3) and (4) to be a well-posed problem, the solution map should be continuous with respect to y_{0}, p,andg. From (7), we see that y(x; y_{1}^{0}, p(x), g(x)) will be pointwise close to y(x; y_{2}^{0}, p(x), g(x)) if y_{0}^{1} is close to y_{0}^{2}. From (10), we see that y(x; y_{0}, p(x), g_{1}(x)) will be pointwise close to y(x; y_{0}, p(x), g_{2}(x)) if g_{1}(x) is everywhere pointwise close to g_{2}(x). Finally, from (13), we see that y(x; y_{0}, p_{1}(x), g(x)) will be pointwise close to y(x; y_{0}, p_{2}(x), g(x)) if p_{1}(x) is everywhere pointwise close to p_{2}(x).

To obtain a global notion of closeness, we require the domain of y to be the closed interval \overline{I} = [a, b] where I = (a, b) and redefine C(I) and C^{1}(I) as C(I) = \{f : \overline{I} → ℝ such that f is continuous on I = (a, b)\} and C^{1}(I) = \{f : \overline{I} → ℝ such that f′ exists and is continuous on I = (a, b)\} ⊆ C(I). Then C(\overline{I}) = \{f : I → ℝ such that f is continuous on I = [a, b]\} ⊆ C(I), and C^{1}(\overline{I}) = \{f : I → ℝ such that, using one-sided limits, f′ exists and is continuous on \overline{I} = [a, b]\} ⊆ C^{1}(I). As we have said, if p, g ∈ C(\overline{I}), then, using one-sided limits, y given by (5) can be considered to solve (3) on \overline{I} so that Σ(\overline{I}) = C^{1}(\overline{I}). Now let D_{y_{0}}(\overline{I}) = \{y ∈ C^{1}(\overline{I}) : y(x_{0}) = y_{0}\}. Then N_{p,y_{0}} maps D_{y_{0}}(\overline{I}) to C(\overline{I}), the solution maps (y_{0}, p, g) ∈ ℝ × C(\overline{I}) × C(\overline{I}) to y ∈ C^{1}(\overline{I}) and Theorems 1, 2, 6, and 7 remain valid with C(I) replaced by C(\overline{I}), C^{1}(I) replaced by C^{1}(\overline{I}) and D_{y_{0}}(I) replaced by D_{y_{0}}(\overline{I}). We have D_{y_{0}}(\overline{I}) ⊆ C^{1}(\overline{I}) ⊆ C(\overline{I}) ∩ C^{1}(I) ⊆ C(I).

Now recall that L^{∞}(\overline{I}) is the set of equivalent classes of functions that are equal except on a set of Lebesgue measure zero such that ess sup_{x∈\overline{I}} |f(x)| < ∞. Since C(\overline{I}) can be considered as a subset of L^{∞}(\overline{I}) we can use the L^{∞}(\overline{I}) norm, ∥f∥_{∞} = ess sup_{x∈\overline{I}} |f(x)| for functions in C(\overline{I}) where ∥f∥_{∞} = max_{x∈\overline{I}} |f(x)| as well as for those in all of its subsets: D_{y_{0}}(\overline{I}) ⊆ C^{1}(\overline{I}) ⊆ C(\overline{I}) ∩ C^{1}(I) ⊆ C(\overline{I}) ⊆ L^{∞}(\overline{I}). In this restricted context, we can obtain the following global inequalities. If p_{1} = p_{2} = p, g_{1} = g_{2} = g and p, g ∈ C(\overline{I}), then for all x in \overline{I} = [a, b] we have using (7) that

|y_{1}(x) − y_{2}(x)| ≤ |y_{1}^{0} − y_{2}^{0}| exp{∥p∥_{∞} (b − a)}

so that

∥y_{1}(x) − y_{2}(x)∥_{∞} ≤ |y_{1}^{0} − y_{2}^{0}| exp{∥p∥_{∞} (b − a)}  (14)

If y_{1}^{0} = y_{2}^{0} = y_{0}, p_{1} = p_{2} = p, and p, g_{1}, g_{2} ∈ C(\overline{I}), then for all x ∈ \overline{I} = [a, b] we have from (9) that

|y_{1}(x) − y_{2}(x)| ≤ ∥g_{1} − g_{2}∥_{∞} (b − a) exp{∥p∥_{∞} (b − a)}

so that

∥y_{1} − y_{2}∥_{∞} ≤ ∥g_{1} − g_{2}∥_{∞} (b − a) exp{∥p∥_{∞} (b − a)}  (15)

If y_{1}^{0} = y_{2}^{0} = y_{0}, g_{1} = g_{2} = g, and p_{1}, p_{2}, g, p ∈ C(\overline{I}), then for all x ∈ \overline{I} = [a, b] we have from (13) that

|y_{1}(x) − y_{2}(x)| ≤ |y_{0}| exp{∥p_{1}∥_{∞} (b − a)}[exp{∥p_{1} − p_{2}∥ (b − a)} − 1]
+ ∥g∥_{∞} exp{∥p_{1}∥_{∞} (b − a)}[exp{∥p∥_{∞} (b − a)} − 1](b − a)

so that

∥y_{1} − y_{2}∥ ≤|y_{0}| exp{∥p_{1}∥_{∞} (b − a)}[exp{∥p_{1} − p_{2}∥ (b − a)} − 1]  (16)
+ ∥g∥_{∞} exp{∥p_{1}∥_{∞}(b − a)}[exp{∥p_{1} − p_{2}∥_{∞}(b − a)} − 1](b − a)

To show that y(x; y_{0}, p, g) depends continuously on y_{0}, p,and g when two are fixed, we use the norm (and hence metric) topologies for ℝ and that induced on C(\overline{I}) from L^{∞}(\overline{I}).

Theorem 7 If p, g ∈ C(\overline{I}), then the solution map defined by (5) from y_{0} ∈ ℝ to y ∈ C^{1}(\overline{I}) is continuous.

Proof. Let p_{1} = p_{2} = p, g_{1} = g_{2} = g, p, g ∈ C(\overline{I}), and ϵ > 0. Then from (14), there exists δ > 0 such that |y_{1}^{0} − y_{2}^{0}| < δ implies ∥y_{1} − y_{2}∥ < ϵ. Hence the mapping from y_{0} ∈ ℝ to y ∈ C^{1}(\overline{I}) is continuous.

Theorem 8 If y_{0} ∈ I, p ∈ C(\overline{I}), then the solution map defined by (5) from g ∈ C(\overline{I}) to y∈ C^{1}(\overline{I}) is continuous.

Proof. Let y_{1}^{0} = y_{2}^{0} = y_{0}, p_{1} = p_{2} = p, p, g_{1}, g_{2} ∈ C(\overline{I}), and ϵ > 0. Then from (15), there exists δ > 0 such that ∥g_{1} − g_{2}∥ < δ implies ∥y_{1} − y_{2}∥ < ϵ. Hence the mapping is continuous.

Theorem 9 If y_{0} ∈ I, p ∈ C(\overline{I}), then the solution map defined by (5) from p ∈ C(\overline{I}) to y ∈ C^{1}(\overline{I}) is continuous.

Proof. Let y_{1}^{0} = y_{2}^{0} = y_{0}, g_{1} = g_{2} = g, p_{1}, p_{2}, g ∈ C(\overline{I}), and ϵ > 0. Then from (16) there exists δ > 0 such that ∥p_{1} − p_{2}∥ < δ implies ∥y_{1} − y_{2}∥ < ϵ . Hence the mapping is continuous.

Summary. Let y_{0} ∈ R and p, g ∈ C(I) where I is an interval and consider the solution map for the IVP (3) and (4) given by (5) that maps (y_{0}, p, g) ∈ ℝ × C(I) × C(I) to y ∈ C^{1}(I).

1. If p and g are fixed, the mapping is one-to-one and solutions do not cross.

2. If y_{0} and p are fixed, the mapping is one-to-one and has an inverse mapping that is linear if and only if y_{0} = 0.

3. If y_{0} and g are fixed and either y_{0} ≠ 0 and g is identically zero or g has only dispersed zeros, then the mapping is one-to-one.

4. If I is a closed interval, and two of y_{0}, p and g are fixed, then the solution map is continuous using the usual topology for ℝ and the norm (metric) topologies for C(\overline{I}) and C^{1}(\overline{I}) as subspaces of L^{∞}(\overline{I}).

References

[1] Boyce, E. and R. C. DiPrima, Elementary Differential Equations and Boundary Value Problems (sixth edition), John Wiley & Sons, Inc., New York, 1997.

[2] Douglass, S. A., Introduction to Mathematical Analysis, Addition-Wesley Publishing Company, New York, 1996.

[3] Moseley, J. L., Properties of the Solution Map for a First Order Linear Problem with One State Variable. Applied Math Report #16, AMR#16, West Virginia University, Morgatown, West Virginia, March, 2000.

James L. Moseley West Virginia University Morgantown, West Virginia 26506-6310 USA e-mail: moseley@math.wvu.edu Telephone: 304-293-2011