## Abstract

The differential equation with *f*(*t*) positive, periodic and continuous is studied. After describing some physical applications of this equation, we construct a variety of invariant sets for it, thereby partitioning the phase plane into regions in which solutions grow without bound and also those in which bounded periodic solutions exist.

## 1. Introduction

We construct invariant sets for the differential equation(1.1)where *γ*>0, *μ*>1 and *f*(*t*) is a continuous, bounded, positive, non-constant, periodic function with finite period *τ* and mean . In order that solutions remain real, when *μ* is not an integer we assume that equation (1.1) only applies for *x*≥0. It will be convenient to write the bounds on *f*(*t*) as and with *F*>*f*>0. Since *f*(*t*) is non-constant and continuous, 〈*f*〉<*F*^{μ}. We re-write the differential equation as(1.2)This equation with constant *f*(*t*) is relatively trivial. The case of non-constant *f*(*t*) arises in at least three different contexts.

A simple electronic circuit, shown in figure 1 and known as the resistor–inductor–varactor circuit, is described, after linear rescaling (Deane & Marsh 2004; Marsh in preparation), by equation (1.1) provided that

*x*>0, ∀*t*. The varactor is a particular type of diode, which is a nonlinear electronic device analogous to a nonlinear spring—one for which Hooke's law is modified to read ‘applied force is proportional to*x*^{μ}’, where*x*is the extension and typically*μ*∈[1.5, 2.5]. We present results for the representative value*μ*=2 and also for the more general case*μ*>1. The mechanical analogies of the resistor and the inductor are, respectively, a source of linear damping and a constant mass. The full model for this circuit, i.e. one in which the restriction*x*>0 is removed, possesses a nonlinearity of a different form,*c*_{1}exp(*c*_{2}|*x*|),*c*_{1},*c*_{2}constants, for*x*≤0, and has been extensively studied—see for instance Azzouz*et al*. (1983) and Matsumoto*et al*. (1984).Studies of ship roll and capsize have led to investigations of the behaviour of the ODE (Thompson 1997). Substituting gives , which is equation (1.1) with and

*μ*=2.Stationary wave solutions of a perturbed Korteweg-de Vries (KdV) equation are described by a special case of equation (1.1) with

*γ*=0 and*μ*=2. Following (Blyuss 2002), we start with a perturbed KdV equation , where*f*(*u*,*ξ*−*Vτ*) is taken to be , and subscripts refer to partial differentiation. The standard transformation to a moving frame, , results in in the steady state , with*v*=*V*±*c*and*C*a constant of integration. Finally, letting*u*=2*βx*+*v*and re-naming*ξ*′ as*t*, we again obtain equation (1.1) with ,*γ*=0 and*μ*=2.

## 2. Invariant sets, *μ*=2

We define an invariant set, , as a subset of the phase plane such that solutions starting from an initial condition in *S* remain in *S* for all time. We use the term ‘absorbing set’ for an invariant set of finite area, with the intention that any bounded limit cycle solutions of equation (1.1) can be shown to lie within such a set, and two such sets, and , are constructed.

In order to construct invariant sets for equation (1.1) we need to prove certain inequalities. To this end, we first assume that *γ*=0, *μ*=2 and *f*(*t*)=*A*, a constant. Then, with , equation (1.1) becomes(2.1)which can be integrated to give(2.2)where *x*_{0}, *y*_{0} are the initial conditions. This elliptic curve plays an important role in the construction of an invariant set, , of initial conditions for solutions all of which eventually grow without bound.

Depending on the parameters appearing in equation (2.2), the expression for *y*^{2} can have one, two or three real roots; when there are two, one of these must be a pair of repeated roots, and the condition for this is easily seen from equation (2.2) to be , *y*_{0}=0. When this is satisfied, the curve *y*(*x*) is known as the separatrix since it separates the two qualitatively different types of behaviour (solutions that grow without bound, and those that are bounded and periodic) displayed by equation (2.1).

### (a) Construction of set

We now construct , shown in figure 2, whose boundary consists of three curves, AB, BC and CD, where points A and D are at infinity. The technique used for all boundaries is essentially as follows. We define the two-dimensional vector fieldsThe importance of the second and third fields is that, for any initial condition (*x*,*y*)=(*x*(*t*_{0}), *y*(*t*_{0})), the direction of ** ϕ**(

*t*) is such that

**(**

*ϕ**t*)=

*μ*

_{1}

*ϕ*_{F}+

*μ*

_{2}

*ϕ*_{f}, where

*μ*

_{1},

*μ*

_{2}are non-negative scalars which sum to one. (Put loosely,

**(**

*ϕ**t*) ‘lies between’

*ϕ*_{f}and

*ϕ*_{F}). This simple observation allows us to project the three-dimensional system equation (1.2) onto the

*x*,

*y*phase plane.

If a given curve in the plane is defined by *G*(*x*,*y*)=0, then two normals to it are * n*=±(∂

*G*/∂

*x*, ∂

*G*/∂

*y*). The choice of sign determines whether the normal is inward or outward pointing. To prove that the flow is always in a particular direction across a curve defined by

*G*, we then only have to show that

*.*

**n****, which is proportional to the cosine of the angle between the normal to the curve and the vector field,**

*ϕ***, is of a given sign at all points on the curve. Hence it will be unnecessary to normalize either**

*ϕ**or*

**n****, as only the sign of the dot product is important.**

*ϕ*#### (i) Boundary *AB*

*Let curve* AB *be defined by G*_{AB}(*x, y*)*=y*^{2}*−λ*^{2}(2*F−x*)(*x+F*)^{2}*=*0 *for x∈*(*−∞, −F*] *with y≥*0. *Then ϕ*(

*t*)

*is into*AB

*in the direction of decreasing y, along its entire length and for all time, provided that λ*

^{2}≤2/3.

Since *y*≥0, curve AB is for *x*∈(−∞,−*F*]. The required normal to AB, pointing in the direction of negative *y*, is * n*=(−3

*λ*

^{2}(

*x*

^{2}−

*F*

^{2}),−2

*y*). At any point (

*x*,

*y*), the

*y*-component of

*ϕ*_{F}is greater than that of

*ϕ*_{f}, so to show that for all time the flow is through AB in the direction of negative

*y*, we should prove that

*.*

**n**

*ϕ*_{F}≥0 for

*x*∈(−∞,−

*F*]. The fact that

*.*

**n**

*ϕ*_{f}≥0 then automatically follows, and hence positivity of the dot product for all time. Now,since

*y*≥0, provided that (2−3

*λ*

^{2})≥0. ▪

In order to make as large as possible we should maximize the area between AB and the *x*-axis and hence take .

#### (ii) Boundary *BC*

*Let curve* BC *be defined by* *for x∈*[*−F*,*x*_{C}] *with y≤*0. *Then ϕ*(

*t*)

*is into*BC

*in the direction of decreasing y, along its entire length and for all time, provided that*

The required normal must be directed towards negative *y* and is therefore . For the same reason used in the construction of AB, we need only prove that * n*.

*ϕ*_{F}≥0 for

*x*∈[−

*F*,

*x*

_{C}]. This dot product is and since

*y*≤0, it is non-negative provided that for

*x*∈[−

*F*,

*x*

_{C}]. Furthermore, since

*γ*>0, the inequality can only make sense if

*x*

_{C}>2

*F*. Squaring both sides of the inequality and solving for

*x*

_{C}we find thatSince

*F*>0, the square root is greater than unity, so the second solution has

*x*

_{C}<2

*F*and can therefore be rejected. ▪

Since our objective is to make as large as possible, we should choose the *x*_{C} that minimizesand as this is monotonically increasing with *x*_{C}>−*F*, we take the minimal value(2.3)

#### (iii) Boundary *CD*

*Let curve* CD *be defined by G*_{CD}(*x, y*)*=y*^{2}*−*(*x*_{C}*−x*)[(*x*_{C}*−x*)^{2}*+b*^{2}]*=*0 *for x∈*(*−∞*,*x*_{C}) *with y≥*0. *Then ϕ*(

*t*)

*is into*CD

*in the direction of increasing y, along its entire length and for all time, provided that*

*where x*

_{C}

*is defined in*

*lemma 2.2*.

*There always exist real values of b such that this inequality holds*.

For this curve, the normal we require is in the direction of increasing *y* and so * n*=(3(

*x*

_{C}−

*x*)

^{2}+

*b*

^{2}, 2

*y*). We need to consider

*ϕ*_{f}this time since

*.*

**n**

*ϕ*_{f}≥0 implies

*.*

**n**

*ϕ*_{F}≥0. The dot product is where

*u*=(

*x*

_{C}−

*x*)∈[0,∞). Since

*y*≥0, for the dot product to be non-negative we need(2.4)and since

*γ*is also positive, this inequality is only feasible over the required range of

*u*if (i) . With this assumed, we can square both sides of inequality (2.4), substitute for

*y*in terms of

*x*, and simplify to obtain(2.5)Now, as well as (i), we need all the coefficients of the powers of

*u*in this inequality to be non-negative; the coefficients of

*u*

^{4},

*u*

^{2}

*u*

^{0}are obviously so, and those of

*u*

^{3}and

*u*are so too, provided (ii) 2

*x*

_{C}−γ

^{2}≥0 and (iii) . We now show that the definition of

*x*

_{C}in equation (2.3) implies (ii), and that (i) is implied by (iii).

From the definition of *x*_{C} in equation (2.3), we have with *q*=1+(1+8*F*/γ^{2})^{1/2}>2; hence, 2*x*_{C}>3*γ*^{2}, so (ii) is automatically satisfied.

Dividing (iii) by 2*x*_{C}>0 gives and since the right-hand side is positive, (iii) implies (i).

In the light of (ii), it now becomes clear that there always exists *b* large enough that inequality (2.5) is true; and so the lemma is proved. ▪

In order to maximize the area of , we should choose the minimal value of *b*, which, from (iii), is given by(2.6)We have therefore proved the following.

*Define* *with**where x*_{C} *is given by equation* *(2.3)* *and b by equation* *(2.6)*. *Then* *is an invariant set such that all solutions to equation* *(2.1)* *starting within* *remain in* *for all time*.

### (b) Dynamics within

In fact, we can say more about solutions to the differential equation which are confined to : all solutions in eventually enter a subset, , which is itself invariant, and once here, grow without limit. The proof, which is somewhat lengthy, is contained in lemmas 2.5–2.7, 2.9–2.13.

It is first necessary to split into seven subsets and these are illustrated in figure 3. Also shown there are the two parabolas P_{f}: *y*=(*f*^{2}−*x*^{2})/*γ* and P_{F}: *y*=(*F*^{2}−*x*^{2})γ, the importance of these being that for *y*>P_{F}, ; for *y*<P_{f}, ; with being either positive, negative or zero only in the region between P_{F} and P_{f}.

We first define and show that it is invariant.

*Define**where ξ<−F is the least real root of* . *Then* *is invariant*.

First, note that and that ; hence, is a local minimum of *h*(*x*) and so *h*(*x*) decreases monotonically from +∞, for increasing . Furthermore, *h*(−*F*)=γ^{2}(*f*^{2}−*F*^{2})<0 and so there is exactly one real root of *h*(*x*), *ξ*∈(−∞,−*F*).

We now consider the three boundaries of separately. The boundary P_{F} for *x*≤*ξ* can be treated by the same technique used for constructing . The required normal to P_{F} is * n*=(2

*x*/

*γ*, 1) and the appropriate choice for

**is**

*ϕ*

*ϕ*_{f}, giving

*.*

**n**

*ϕ*_{f}=

*h*(

*x*)≥0 for

*x*∈(−∞,

*ξ*]. Hence, the flow is through this part of P

_{F}and is towards negative

*x*, for all time.

Along the boundary *x*=*ξ*, 0≥*y*≥(*F*^{2}−*ξ*^{2})/*γ*, and so ; thus, the flow is through this boundary towards negative *x* for all time here also.

The remaining boundary is *y*=0, *x*≤*ξ*, along which since *ξ*<−*F*, and so the flow is through this boundary in the negative *y* direction for all time. Therefore, is an invariant set. ▪

We now prove several lemmas that, in combination, enable us to show that all solutions starting in end up in .

*All solutions x*(

*t*)

*=*(

*x*(

*t*)

*, y*(

*t*))

*initially in*

*or*

*enter*

*in finite time*.

The points O–W are shown in figure 4, which is referred to throughout this proof. The curve OPQRS is a typical orbit starting in and entering in finite time and we now prove that all solutions starting in or behave in this way.

Consider first any initial point (*x*(*t*_{0}),*y*(*t*_{0})) in the subset of with *x*≤3*F*/2. The line *x*=3*F*/2 lies to the left of point C, whose *x*-co-ordinate *x*_{C}>2*F*. Hence, for all *x* in this region, *y*≥*y*_{T}, the (strictly positive) *y*-co-ordinate of point T, since solutions cannot cross CD, and so *x*(*t*)≥*x*(*t*_{0})+(*t*−*t*_{0})*y*_{T}. Thus, any solution crosses the line *x*=3*F*/2 in finite time.

Consider now a solution starting at point P on the line *x*=3*F*/2; since, while the solution remains in , *y*≥0 and so *x*≥3*F*/2. Solving this linear differential inequality, we obtain *y*(*t*)≤e^{−γt}[*y*_{P}+5*F*^{2}/4*γ*]−5*F*^{2}/(4*γ*) and so the solution reaches point Q on the *x*-axis in finite time.

The *x*-co-ordinate of Q>*x*_{C}>2*F* and so starting from Q, we have , and so *y*(*t*)≤3*F*^{2}(e^{−γt}−1)/*γ*. Hence, the solution must cross the line *y*=*Y*_{1}<0 in finite time, provided that *Y*_{1} satisfies the condition (a) that *Y*_{1}>−3*F*^{2}/*γ*, so that lim_{t→∞}*y*(*t*)<*Y*_{1}. Three additional conditions on *Y*_{1} are required, these being that (b) *Y*_{1} is sufficiently small that the *x*-co-ordinate of *R*≥2*F*, so that remains true while the solution moves from Q to R; (c) the line *y*=*Y*_{1} does not intersect the parabola P_{F}: *y*=(*F*^{2}−*x*^{2})/*γ* for *x*>0 in , so ensuring that everywhere in the subset of in which *y*≥*Y*_{1}; and (d) *Y*_{1}>*Y*_{U}, the *y*-co-ordinate of U, the point at which curve BC and the parabola P_{f} intersect for *x*<0. Such a *Y*_{1} can always be chosen because Q and C are both to the right of the point *x*=*F* where P_{F} intersects the *x*-axis. With these conditions on *Y*_{1}, point R is reached in finite time. By the definition of *Y*_{1}, once R has been reached, then line VW must also be crossed in finite time, this being the vertical line from the point of intersection of BC with *y*=*Y*_{1}. This is because so *x*(*t*) decreases at least linearly with time; and, by the definition of *Y*_{1}, all along VR—so VR cannot be re-crossed.

From the crossing of VW, still applies and so *x* decreases at least linearly with time until S is reached. The part of in the vicinity of S is below P_{f} and so and ; hence the solution must cross P_{f} at some point S and hence enter . Since each of the steps described takes finite time, we conclude that point S is reached from any point in or in finite time. ▪

We now consider the fate of solutions starting in the rest of .

*A solution initially at point* B*,* (*−F*, 0) *eventually moves further into* .

Consider the case that * x*(

*t*

_{0}) is point at some time

*t*

_{0}; then either (i)

*f*(

*t*

_{0})<

*F*

^{2}or (ii)

*f*(

*t*

_{0})≡

*F*

^{2}. From the differential equation, we have in case (i) that but and so the solution moves vertically away from B in the direction of decreasing

*y*, and so further into . In case (ii), we have and for

*t*∈[

*t*

_{0},

*t*

_{1}) where

*t*

_{1}is the time at which

*f*(

*t*

_{1}) is first less than

*F*

^{2};

*t*

_{1}is necessarily finite by the conditions imposed on

*f*(

*t*). Hence, the solution remains at point B only until

*t*=

*t*

_{1}, at which time and the subsequent behaviour is the same as in case (i). ▪

To proceed further, we need the following definition:

*A double crossing is said to occur when the solution x*(

*t*)

*crosses both the lines y=Y*

_{0}

*and y=*2

*Y*

_{0}

*, in either order. Here, Y*

_{0}

*<*0

*and is sufficiently small that (1) the intersection of y=*2

*Y*

_{0}

*with*P

_{f}for x<0 takes place within*, and that (2) X*

_{0}

*>ξ, with X*

_{0}

*the negative root of*

*(see*

*figure*5

*)*.

*The existence of the k-th double crossing implies that ∃ a _{k},*

*with b*(

_{k}>a_{k}such that either y*a*)

_{k}*=Y*

_{0}

*and y*(

*b*)

_{k}*=*2

*Y*

_{0}

*(falling) or y*(

*a*)

_{k}*=*2

*Y*

_{0}

*and y*(

*b*)

_{k}*=Y*

_{0}

*(rising).*

Now, by continuity, any pair of successive double crossings must consist either of a falling double crossing followed by a rising one or vice versa. Then we have the following.

*An infinite sequence of double crossings confined to* *cannot occur, in either (i) finite or (ii) infinite time.*

First, we consider the sets that solutions leaving can enter: by continuity, these must be , , or . The first two are impossible, by definition and because and only meet at point B, and passing through this in the direction of increasing *y* is impossible by lemma 2.7.

The other two transitions *can* take place; to because , along their common boundary, and to because can have any sign along the common boundary.

To prove the lemma, first observe that we have so . Now, so(2.7)By hypothesis, for all *t*≥0, so, by equation (2.7), with *v*>0.

Consider first case (i), in which infinitely many double crossings occur within during the time interval [0,*T*] with *T*>0 finite. Define the sequence of crossing times where *I*_{k}=[*a*_{k},*b*_{k}] with *a*_{k}≥*b*_{k−1}, so that ; then *y*(*t*)∈[2*Y*_{0},*Y*_{0}] if . Hence,and since *v* and *T* are both finite, the latter by hypothesis, this leads to a contradiction and so an infinite sequence of double crossings within cannot occur in finite time.

Now consider case (ii), in which infinitely many double crossings occur during infinite time. We introduce the sequence as before but now . In order that * x*(

*t*) remains in , we must have (see figure 5). From the differential equation we have(2.8)Now, , ∃

*N*=

*N*(

*k*) such that and, by hypothesis, . Furthermore, given positive real constants

*c*and

*c*′, we have, by the continuity of curve BC, that and so(2.9)Using this in equation (2.8) results inAlso, from we have that is bounded, as is , since

*x*(

*b*

_{k})≥

*X*

_{0}, and sowhere

*K*is a finite constant. Remembering that

*f*(

*t*) is continuous and periodic and that

*f*(

*t*)≤

*F*

^{2}, ∀

*t*, the coefficient of

*Nτ*in the above can be made negative by choosing |

*Y*

_{0}| sufficiently small. Hence, finally, we have proved that for a constant

*K*′, which is in contradiction with our initial hypothesis that

*y*(

*b*

_{k})=

*Y*

_{0}or 2

*Y*

_{0}. Hence, an infinite number of double crossings, confined to , cannot occur either in finite or infinite time. ▪

*No solution can remain in* *indefinitely.*

From lemma 2.9 we now have the result that an infinite number of double crossings cannot occur, and so ∃ *t*_{0}<∞ such that ∀*t*>*t*_{0}, while , either (i) *y*(*t*)<*Y*_{0} or (ii) *y*(*t*)>2*Y*_{0}—there are no further double crossings after *t*=*t*_{0}. We can now prove the lemma as follows.

We consider case (i) first. Here, for all *t*>*t*_{0}, and since *Y*_{0}<0, . Hence, in case (i), no solution can remain in indefinitely.

Turning now to case (ii), we have and so, with ,where the last term on the RHS is obtained from equation (2.9). Since *y*(*t*)<0 for all *t*>*t*_{0}, we havewhere *X*_{1} is shown in figure 5. Thus,and since |*Y*_{0}| can be made as small as desired, we have . Hence, in case (ii) too, no solution can remain in indefinitely. ▪

*All solutions initially in* , *with x>ξ, or* *, cross the line x=ξ in the direction of decreasing x, in finite time.*

Consider first solutions initially in . By lemma 2.10, solutions here must eventually leave . As shown in the proof of lemma 2.9, solutions leaving can only enter or .

To show that all solutions cross the line *x*=*ξ* in the direction of decreasing *x*, consider first a solution that leaves and enters at a point (*x*_{0}, *y*_{0}) at a time *t*_{0}. Then we need to prove two facts: (i) *y*<*y*_{0} for all *t*>*t*_{0}, and (ii) *y*_{0}<0.

We prove (i) as follows. While the solution remains in , , so , and so the lemma is plainly true for a solution that remains in . The solution may, however, re-cross P_{F} and enter or , in both of which can have either sign. If the point at which the solution re-crossed P_{F} is (*x*_{1}, *y*_{1}) then clearly *y*_{1}<*y*_{0}. Suppose now a further crossing occurs, from back into , this time at a point (*x*_{2}, *y*_{2}); then *y*_{2}<*y*_{1}<*y*_{0} despite the ambiguity of the sign of in . This is because *y*<0 always, so *x* decreases at least linearly in time, so *x*_{2}<*x*_{1}<*x*_{0}; but the crossing points are all on P_{F} and so *y*_{2}<*y*_{1}<*y*_{0}. Hence, we conclude that *y*<*y*_{0} for all *t*>*t*_{0}.

The proof of (ii) relies on lemma 2.7. A solution at point B can initially only move vertically downwards away from B, and hence cannot enter from point B. Hence, any solution that enters from does so at *y*=*y*_{0}<0. Thereafter, the argument used in case (i) applies.

Consider now a solution that leaves and enters . According to lemma 2.10, although solutions may oscillate between and , they cannot do so indefinitely and so there must be a last crossing into , let us say at time *t*_{0}. Thereafter, as before, *y*<*y*_{0}=2*Y*_{0}, whether or not the solution subsequently remains in or moves into .

Hence, we conclude that in all cases, for all *t*>*t*_{0}, and so all solutions eventually cross the line *x*=*ξ*. ▪

The previous lemma shows that solutions with *x*>*ξ* eventually cross the line *x*=*ξ*. Hence, solutions eventually enter (invariant), or with *x*<*ξ*, and we now deal with the fate of solutions in the latter case.

*All solutions initially in* *with x≤ξ enter* *in finite time and remain there*.

We prove this lemma by contradiction: we assume that with *x*≤*ξ* for all *t* and find an inconsistency. Set *t*=0 when *x*=*ξ* to simplify notation, and write *y*(0)=*y*_{0}. Then by the assumption that * x*(

*t*) is between P

_{F}and P

_{f}. Integrating gives(2.10)with

*a*=(

*F*

^{2}−

*ξ*

^{2})/

*γ*a negative constant. Also, by equation (2.7) we have and so integrating from 0 to

*t*gives(2.11)Now, as long as

*(*

**x***t*) remains in , we must have , but, by equation (2.10), . Hence, in order to remain in for all

*t*>0,

*y*(

*t*) has to be bounded above by a function that decreases as −

*t*

^{2}, and, from equation (2.11), bounded below by one that goes as −

*t*. These two requirements are mutually incompatible, and so no solutions can remain in indefinitely. ▪

*All solutions initially in* *enter* *in finite time*.

For all , *x*<−*F* and *y*≥0. Solving the differential equation from initial condition *x*(0)<−*F*, *y*(0)≥0, we haveand, while the solution is in , −*x*^{2}<−*F*^{2}, sowhere we have used *x*(*nτ*)<−*F*. Since, as before, 〈*f*〉−*F*^{2}<0, the second term above ensures that *y* decreases with time and so the solution eventually crosses the *x*-axis at *x*<−*F*, and so enters or ; if , then lemmas 2.11 and 2.12 apply, and in either case is entered in finite time. ▪

*All solutions in* *grow without limit as t→∞.*

For all , and . Furthermore, by lemma 2.5, is invariant, and so both *x* and *y* tend to −∞ in . ▪

*All solutions initially in* *remain in* , *eventually entering* *, where they remain and grow without limit.*

The invariance of is proved in theorem 2.4. The rest of the theorem is a direct consequence of lemmas 2.5–2.7, 2.9–2.14. ▪

### (c) Set

As an easy corollary to theorems 2.4 and 2.15, we can now construct a set which is such that any finite area absorbing set .

*Any finite area absorbing set* , *where*

The boundaries of set consist of curve BC, the part of curve CD with *x*≥−*F* and the line *x*=−*F* between *y*=0 and , the latter being the intersection of this line with CD. Since this line has along its entire length, flow must be into along this boundary. Furthermore, if is any absorbing set, then since, by theorem 2.15, any initial conditions in lead to solutions that grow without bound. ▪

We can now visualize how good an approximation set is to the set of initial conditions which lead to solutions that do not blow up. We make the following definition:

*Let X*(

*x*

_{0},

*y*

_{0},

*t*

_{0};

*t*)

*be the solution of equation*

*(1.1)*

*for given t*

_{0}

*, with initial conditions x*(

*t*

_{0})

*=x*

_{0}

*, y*(

*t*

_{0})

*=y*

_{0}

*, and define*

*Then*

Clearly, . For an illustration of a numerical approximation to this set, see figure 6.

### (d) Absorbing set

We now construct a polygonal absorbing set. The underlying method for this construction is the same as for , but by contrast to that case, there are now constraints on the parameters *F*, *f* and *γ* additional to *F*>*f*>0 and *γ*>0.

Three preliminary observations are in order. First, any absorbing set must lie partly above and partly below the *x*-axis: a set entirely above the *x*-axis would always have and no such set whose boundary is a closed curve could be absorbing over its whole boundary.

Second, where possible, we choose the sides of the absorbing sets to be parallel to either the *x* or the *y*-axis, since this simplifies the proof that the flow is into that side. On the other hand, a rectangular set is not possible: for instance, the right-hand vertical boundary above the *x*-axis would necessarily entail flow out of the set, since . The same applies to the left-hand vertical boundary below the *x*-axis. Hence, there must be boundaries not parallel to either axis.

Third, as pointed out previously, can only be zero if * x*=(

*x*,

*y*)∈

*I*, where

*I*is the region between P

_{f}and P

_{F}. This suggests the possibility of basing the non-vertical/horizontal boundaries on scaled versions of these parabolas, and this is the case for .

*Let γ*^{2}*≥*8*F and define the set* *as the closed hexagon* GHIJKL*, whose vertices are**and whose edges are straight lines except that**Here**and λ*_{2} *is such that**Then at least one λ*_{2}*∈*(1*,* 2) *exists and the set* *as defined is an absorbing set for all t.*

To show that , illustrated in figure 7, is an absorbing set, we need to prove that * n*.

**(**

*ϕ**t*)≥0 over the entire boundary of and for all

*t*, with the appropriate choice for

**, and where**

*ϕ**is the inward normal to the boundary.*

**n**This is trivial for the horizontal and vertical portions of the boundary, i.e. GH, IJ, JK and LG. Taking GH as a horizontal example, we have *x*≥*f* and * n*=(0,−1) is an inward-pointing normal. By the same reasoning as used in the construction of , we need to prove that

*.*

**n**

*ϕ*_{F}=

*x*

^{2}−

*f*

^{2}≥0 for

*x*≥

*f*, which is clearly true. Taking IJ as a vertical example, we have

*y*≤0 and an inward-pointing normal is

*=(−1, 0). Then*

**n***.*

**n**

*ϕ*_{f}=

*.*

**n**

*ϕ*_{F}=−

*y*≥0, also obviously true.

We define HI as the curve , from which the inward normal * n*=(−2

*λ*

_{1}

*x*/γ, −1). Here, real

*λ*

_{1}>1 is to be found. The appropriate choice for

**is**

*ϕ*

*ϕ*_{F}and so we need(2.12)where

*x*

_{H}is the

*x*-co-ordinate of H. Now,

*F*

^{2}−

*x*

^{2}≥0 so we require . This is guaranteed for all

*x*∈[

*x*

_{H},

*F*] if , which is a quadratic in

*λ*

_{1}in which has a negative coefficient; hence inequality (2.12) is satisfied for , where . Letting

*r*=8

*F*/

*γ*

^{2}>0, so for

*λ*

_{1}to be real,

*r*∈(0,1], we have , which is clearly monotonically increasing from 1, at

*r*=0, to 2, at

*r*=1. Hence, for the smallest possible we choose , which is guaranteed to be positive, as, by construction, it must be. Similarly, we define KL by and so

*=(2*

**n***λ*

_{2}

*x*/γ, 1) is the required normal, where

*λ*

_{2}>1 is to be found. The inequality to be satisfied this time is for

*x*∈[

*f*,

*x*

_{K}], where

*x*

_{K}is the

*x*-coordinate of K. The first bracket is clearly non-positive, so(2.13)guarantees flow through KL in the direction of increasing

*y*. Additionally, we have

*y*

_{K}=

*y*

_{J}=(

*f*

^{2}−

*F*

^{2})/

*γ*and so . Eliminating

*x*

_{K}between this and equation (2.13) results in(2.14)Now,

*h*(1)=4

*F*

^{2}>0 and

*h*(2)=32(

*F*

^{2}+

*f*

^{2})−

*γ*

^{4}; but

*γ*

^{2}≥8

*F*so

*h*(2)≤32 (

*f*

^{2}−

*F*

^{2})<0, and so, by the Intermediate Value Theorem, there is at least one real root of

*h*(

*λ*

_{2}) such that

*λ*

_{2}∈(1,2). ▪

## 3. Absorbing set

We now construct an absorbing set, for all *μ*>1. The condition *μ*>1 is required for the nonlinear function in the differential equation, *x ^{μ}*, to be Lipschitz, thereby guaranteeing uniqueness of solutions (Hirsch & Smale 1974). We redefine the vector field and its bounds as

By analogy with the construction of , we observe that the set *I* in which has indefinite sign is now . We can then construct a hexagonal absorbing set whose vertices are MNOPQR—see figure 8. In the course of the construction, we make several assumptions whose purpose is to make the problem tractable, these being (i) MN, QP are horizontal, MR and OP are vertical; (ii) the *x*-co-ordinates of N and Q are *x*_{N}=*F* and *x*_{Q}=*f* respectively; (iii) the gradients of lines NO and RQ are both −*γ*; and (iv) M lies on P_{F}: *y*=(*F*^{μ}−*x*^{μ})/γ and P lies on P_{f}: *y*=(*f*^{μ}−*x*^{μ})/*γ*. Constraint (iv) arises in an attempt to make as small as possible: for instance, M cannot be below P_{F} because could then be positive, which would allow flow out of MN. With these assumptions in place, the co-ordinates of all the vertices of can be expressed in terms of *x*_{M}, as will be seen from the following.

*Let μ>*1*, f ^{μ}+fγ*

^{2}

*≥*(

*F+F*

^{μ}/γ^{2})

^{μ}and define the set*as the closed hexagon*MNOPQR

*whose edges are straight lines and whose vertices are*

*where*

*and x*

_{M}

*is the smallest real root of*(3.1)

*such that*0

*≤x*

_{M}

*<f. Then at least one such x*

_{M}

*exists and the set*

*is an absorbing set for all t and all μ>*1.

To show that , see figure 8, is an absorbing set, as before, we need to prove that * n*.

**(**

*ϕ**t*)≥0 over the entire boundary of for all

*t*, with the appropriate choice for

**and where**

*ϕ**is the inward normal to the boundary. This is trivial for the horizontal and vertical portions of the boundary, i.e. OP, PQ and RM (see the proof of theorem 2.18).*

**n**We define NO as the straight line *G*_{NO}(*x*, *y*)=*y*−*λ*_{3}(*x*_{O}−*x*) where *x*_{O} is the *x*-coordinate of point O and *λ*_{3}>0, from which the inward normal * n*=(−

*λ*

_{3},−1). The appropriate choice for

**is**

*ϕ*

*ϕ*_{F}here, so we need(3.2)where

*x*

_{N}has been set equal to

*F*: with hindsight, we shall see that this choice greatly simplifies the construction. Here

*λ*

_{3}(

*x*

_{O}−

*x*)≥0 and (

*x*

^{μ}−

*F*

^{μ})≥0 if and only if

*μ*≥0, so we require (−

*λ*

_{3}+

*γ*)≥0, and therefore choose

*λ*

_{3}=

*γ*, which guarantees that the boundary NO has

*.*

**n****(**

*ϕ**t*)≥0. This choice for

*λ*

_{3}also simplifies the construction.

Similarly, we define QR by *G*_{QR}(*x*, *y*)=*y*−*λ*_{4}(*x*_{R}−*x*), where *x*_{R} is the *x*-coordinate of R and *λ*_{4}>0, and so * n*=(

*λ*

_{4}, 1) is the required normal. The inequality to be obeyed is now for

*x*∈[

*x*

_{R},

*f*], where

*x*

_{Q}has been set equal to

*f*, again for simplicity. Here (

*f*

^{μ}−

*x*

^{μ})≥0 if

*μ*≥0 but

*λ*

_{4}(

*x*

_{R}−

*x*) is non-positive, so we need (

*λ*

_{4}−

*γ*)≤0. In choosing

*λ*

_{4}=

*γ*we guarantee that the boundary QR has

*.*

**n****(**

*ϕ**t*)≥0.

Having determined the constants *λ*_{3} and *λ*_{4} which guarantee that * n*.

**(**

*ϕ**t*)≥0 over the entire boundary of , we now need to find its vertices. In addition to

*x*

_{N}=

*F*,

*x*

_{Q}=

*f*and

*y*

_{O}=

*y*

_{R}=0, we know the following:which results in the simultaneous equations(3.3)between which

*x*

_{O}can be eliminated to give equation (3.1). When this can be solved, it gives all the vertices of MNOPQR in terms of

*γ*,

*μ*,

*F*,

*f*and

*x*

_{M}. The conditions under which a solution

*x*

_{M}of equation (3.1) exists satisfying 0≤

*x*

_{M}<

*f*and

*x*

_{O}>

*F*are easily derived. Consider equation (3.3), solved for

*x*

_{O}in terms of

*x*

_{M}, givingrespectively. At

*x*

_{M}=

*f*, we have for curve (a) and

*x*

_{O}=

*f*for curve (b), and so (a) is above (b) at

*x*

_{M}=

*f*, provided

*μ*>1. Therefore, to guarantee at least one solution of equation (3.3), we need (b) to be above (a) at

*x*

_{M}=0, and so we must havewhich is the condition given in the theorem. ▪

### (a) Comparison of _{1} and _{2}

We can now compare the restrictions on the parameters imposed in theorems 2.18 and 3.1 when *μ*=2. These are that *γ*^{2}≥8*F* (set ) and that (set ). Let *γ*^{2}=8*F* so that the second inequality becomes 64*F*^{2}*f*(8*F*+*f*)=81*F*^{4}. Solving for *f* in terms of *F* gives . Hence we conclude that, subject to *F*>*f*>0, if , the constraint on *γ* is weaker for set ; otherwise, the constraint on *γ* is weaker for set .

## 4. Conclusions

We have constructed a variety of invariant sets for the differential equation (1.1), the sets being one for which solutions grow without bound; one which must contain any bounded limit cycles, both of these sets requiring no constraints other than those given in the first paragraph of the paper. We have additionally found two absorbing sets, both of which require additional parameter constraints (effectively, large dissipation), one of which is valid for all *μ*>1. We have also described three areas of physical interest in which the differential equation arises.

We have had little to say here about the dynamics of this equation, since our aim was to construct some important invariant sets. Numerical results presented in Deane & Marsh (2004), Marsh (in preparation) indicate the presence of co-existing periodic attractors but not chaos. There are also some results on dynamics for large *γ* and *μ*=2 in Gentile *et al*. (2005), in particular a study of the analyticity properties of the orbit with the same period as *f*(*t*) that, numerically at least, appears to be the only bounded periodic solution to equation (1.1) in this case. Additionally, there are in principle ways to understand which periodic orbits should occur (Bartuccelli *et al*. 2004), in particular when the dissipation and the oscillatory part of *f*(*t*) are both small.

At least two interesting open questions remain. The first concerns the basin of attraction of periodic solutions, which clearly must lie in , but the construction of this set does not exclude the possibility that this basin has an infinite ‘tail’ lying between curves AB and CD as *x*→−∞. The second concerns the rate at which solutions that grow without bound approach infinity: do they do so in finite or infinite time? In the unperturbed version of the differential equation (*γ*=0, *μ*=2, *f*(*t*) constant), exact solutions exist and these can be expressed in terms of Weierstrass elliptic functions, which do indeed blow up in finite time, but it is not clear whether this property is inherited by solutions to equation (1.1).

## Footnotes

- Received March 3, 2005.
- Accepted August 16, 2005.

- © 2005 The Royal Society