Introduction
A differential equation is an equation that includes an unknown function and its derivatives A solution to a differential equation is defined as a differentiable function that satisfies the equation over a specific interval For instance, the equation \( x' = 0 \) represents a differential equation involving the unknown function \( x \) and its first derivative concerning an independent variable, such as \( t \) or \( s \) It can be observed that \( e^t \) satisfies the equation \( e^t = 0 \) for all \( t \) in the interval \( I = (-1, 1) \) Consequently, \( x(t) = e^t \) serves as a solution to the differential equation within the specified interval.
A differential equation involving ordinary derivatives is called anordinary dif- ferential equationand one involving partial derivatives is called apartial differential equation For example,x 00 t 2 x 0 C2xD0is an ordinary differential equation, while
@x 2 C @ @y 2 u 2 D0is a partial differential equation In this book, we deal with ordinary differential equations.
By theorderof a differential equation we mean the order of the highest derivative appearing in the equation For example,x 000 C2x 00 3x 0 C2xD 0is a third order differential equation whilex 00 CxD0is second order.
Differential equations are essential in mathematics and are crucial across various fields, including science, engineering, economics, and social sciences They can model diverse phenomena such as electrical current flow, missile motion, mixture behavior, disease spread, and population growth rates Typically, a differential equation expresses a physical quantity x(t) as a function of time t, with its rate of change x'(t) defined by a function f(t, x(t)) that incorporates both time and the physical quantity.
In this chapter, we will focus on first-order linear differential equations, laying the groundwork for understanding more complex topics Subsequent sections will explore differential equations in greater detail, including higher-order equations and systems.
S Ahmad, A Ambrosetti,A Textbook on Ordinary Differential Equations 2nd Ed.,
UNITEXT – La Matematica per il 3+2 88, DOI 10.1007/978-3-319-16408-3_1
2 1 First order linear differential equations
A simple case
Let us begin with the very specific and simple equation x 0 Dkx; k2R: (1.1)
In this article, we will outline a clear method for solving specific equations To begin, we will leverage our understanding of the derivative of the exponential function to tackle the straightforward equation (1.1).
We begin by examining the case where \( k = 1 \) in the differential equation \( x' = x \) A well-known solution to this equation is the exponential function \( x(t) = e^t \) In fact, for any constant \( c \), the function \( x(t) = ce^t \) also satisfies the equation This observation extends to the more general form \( x' = kx \), which has the solution \( x(t) = ce^{kt} \) for any constant \( c \) Furthermore, we will demonstrate that these exponential functions represent the only solutions to this differential equation.
We now illustrate a general procedure that will be used later to solve the most general first order linear differential equations First suppose thatx.t /satisfies the equation x 0 t /Dkx.t /:
Multiplying both sides of the equivalent equationx 0 t /kx.t / D 0bye k t , we have x 0 t /e k t kx.t /e k t D0:
We note that the left-hand side is the derivative of x.t /e k t / Hence we have x.t /e k t / 0 D 0 Integrating, we obtainx.t /e k t D c,8t 2 R, wherecis a con- stant Hencex.t /Dce k t
By substituting a function of the form x(t) = ce^(kt) into equation (1.1), we establish that x(t) is a solution of (1.1) if and only if it can be expressed as x(t) = ce^(kt) for some constant c Thus, x(t) = ce^(kt) is identified as the general solution of (1.1), representing the entire family of solutions to this equation.
Example 1.2.1.Consider the problem of findingx.t /such that x 0 D2x; x.0/D1: (1.2)
An initial value problem seeks a function \( x(t) \) that satisfies a given differential equation along with a specified initial condition, such as \( x(0) = 1 \) The general solution to the equation is expressed as \( x(t) = ce^{2t} \) By substituting \( t = 0 \) into this equation, we find that \( 1 = ce^0 \), leading to the conclusion that \( c = 1 \) Thus, the specific solution to the initial value problem is \( x(t) = e^{2t} \) Since all solutions to the initial value problem take the form \( x(t) = ce^{2t} \) and substituting the initial condition yields a unique constant, we confirm that the solution to the initial value problem exists and is unique.
Some examples arising in applications
Population dynamics
• x.t /denote the number of individuals of a population at timet.
• bdenote the birth rate of the population.
• d the death rate of the population.
The basic model of population growth, based on Malthusian theory, posits that birth and death rates remain constant The change in population from time n to n+1 is determined by the number of new births minus the number of deaths, leading to the equation x(n+1) - x(n) = b*x(n) - d*x(n), which can be simplified to x(n+1) - x(n) = (b - d)*x(n) By defining k as the net growth rate (k = b - d), we derive the recursive equation x(n+1) = k*x(n), applicable for any positive integer n Transitioning to continuous time, the population change over a small time interval t is expressed as x(t + t) - x(t) = k*x(t)*t Dividing by t yields the differential equation dx/dt = k*x(t), which models population growth over continuous time.
The left-hand side is the incremental ratio ofx.t / Lettingt !0, we find x 0 t /Dkx.t /; kDbd; a first order linear differential equation like (1.1), whose solutions arex.t /Dce k t
In population dynamics, if the growth rate \( k \) is positive (i.e., \( k = \frac{f(x_0)}{D x_0} > 0 \)), it indicates that the population will increase over time, represented by the equation \( x(t) = D x_0 e^{kt} \) When the birth rate \( b \) equals the death rate \( d \), the growth rate \( k \) becomes zero, leading to a stable population size over time Conversely, if the birth rate exceeds the death rate (\( b > d \)), the population grows exponentially, approaching a limit as time progresses However, if the birth rate is less than the death rate (\( k < 0 \)), the population will decay exponentially towards extinction This model, while useful, does not account for the potential dependence of birth and death rates on population size, which will be explored in a more comprehensive model in Chapter 3, Section 3.1.1.
4 1 First order linear differential equations t x x
Fig 1.1.Solutions of (1.3), withk > 0(upper curve) andk < 0(lower curve) andk D 0 (dotted curve) which gives rise to the so called “logistic equation” having the formx 0 Dx.˛ˇx/, ˛; ˇpositive constants.
An RC electric circuit
Let us consider an RC circuit with resistanceR and capacityC with no external current or voltage source.
If we denote byx.t /the capacitor voltage (x.t /DV t /in Figure 1.2) and byI.t / the current circulating in the circuit, then, according to the Kirchhoff’s law, we have
Moreover, the constitutive law of capacitor yields
I.t /DC dx.t / dt : SubstitutingI.t /in the first equation, we obtain the first order differential equation
Fig 1.2.An RC circuit namely x 0 t /C x.t /
The equation RC D0 can be expressed in the form (1.1), where k is defined as 1/RC To find a solution x(t) that meets the initial condition x(0) = x0, we consider the initial voltage to be x0 The resulting solution is x(t) = x0e^(t/RC).
The voltage across a capacitor, denoted as x(t) or V(t), exhibits an exponential decay towards zero as time approaches infinity, consistent with observed phenomena The term DRC refers to the RC time constant, which indicates the duration required for the voltage x(t) or V(t) to decrease significantly.
V /D x0e 1 Moreover we can say that the bigger is, the slower the decay As for the intensity of currentI.t /, one finds
Other equations arising in the electric circuit theory will be discussed in Example1.4.3 below, in Example 5.5.5 of Chapter 5 and in Section 11.6 of Chapter 11.
The general case
Now, we solve the general first order linear differential equation x 0 Cp.t /xDq.t / (1.4) wherep.t /; q.t /are continuous functions on an intervalI R.
Ifq.t /0the linear differential equation (1.4) is calledhomogeneous, otherwise it is callednonhomogeneousorinhomogeneous.
Motivated by the above discussion, we try to find a differentiable function.t /, t / > 0fort 2I, such that
An integrating factor is a function that, when multiplied by the differential equation, transforms it into a form that is easier to integrate Specifically, by applying the integrating factor to the equation, the left side becomes equal to a product that can be readily integrated, simplifying the solution process.
Although based on the discussion in the preceding section, one might guess such an integrating factor, we prefer giving a precise method for finding it.
Letx.t /be a solution of (1.4) Multiplying the equation by.t /we have x 0 CpxDq:
Now we wish to findsuch that x 0 CpxD t /x.t // 0 :
6 1 First order linear differential equations
Expanding the right-hand side, we have
Canceling.t /x 0 t /from both sides, we obtain
Assuming thatx.t /6D0and dividing both sides byx.t /, we find
Since.t / > 0we infer d t / Dp.t /dt:
Then, taking the indefinite integrals we obtainR d
t / DR p.t /dtand we find (recall that.t / > 0) that ln t //DR p.t /dt. Thus
In order to obtain the general solution of (1.4), we take the indefinite integral of both sides of
Z t /q.t /dt wherecis a constant Substituting.t /De P.t / , we have x.t /De P.t / cC
If \( x(t) \) satisfies equation (1.4), then there exists a constant \( c \in \mathbb{R} \) such that \( x(t) \) can be expressed in the form of equation (1.6) It is straightforward to verify that for any \( c \in \mathbb{R} \), the function \( x(t) \) defined in (1.6) indeed solves equation (1.4) Therefore, the expression for \( x(t) \) in (1.6) is referred to as the general solution of equation (1.4).
Now, suppose we are interested in solving the initial value problem x 0 Cp.t /xDq.t /; x.t0/Dx0:
To find the constant c in the general solution, we can substitute D t0,x D x0 Alternatively, we can compute the definite integral of (1.5) from t0 to t, rather than using the indefinite integral This approach yields the desired results.
We can also chooseP t /DRt t 0 p.s/dsand then t /De
Remark 1.4.1.We prefer not to have to memorize (1.7) but rather go through this simple procedure each time, starting with integrating factors.
As a special case of (1.6), whenq D 0, the general solution of the homogeneous equation x 0 Cp.t /xD0 (1.8) is x.t /Dc e P.t / ; P t /D
ForcD0we obtain thetrivial solutionx.t /0.
If we are searching for a solution satisfying the initial conditionx.t0/Dx0, then we can solvex0Dx.t0/Dc e P.t 0 / If we takeP t /DR t t 0 p.s/ds, thenP t0/D
0and we findcDx0 Thus x.t /Dx0e
R t t0 p.s/ds is the solution ofx 0 Cp.t /x D 0 such thatx.t0/ D x0, and it is unique As a consequence, ift0is any number inI andx.t0/Dx0, then
If \( x(t) \) is a solution to equation (1.8), it can either be identically zero or consistently positive or negative Notably, if \( x(t) \) equals zero at any point in the interval \( I \), then the only solution is the trivial one, \( x(t) = 0 \) for all \( t \) in \( I \).
The above arguments lead to the following existence and uniqueness result for (1.4), namely forx 0 Cp.t /xDq.t /.
Theorem 1.4.2 Letp.t /; q.t /be continuous inI R Then
1.The general solution of(1.4)is given, for allt 2I, by x.t /De P.t / cC
8 1 First order linear differential equations
2.There is exactly one solutionx.t /satisfying the initial valuex.t0/D x0 for any numberst02Iandx02R Precisely, x.t /De
This theorem can also be deduced from general existence and uniqueness results stated in Chapter 2, Section 2.2.2.
We end this section by demonstrating, through examples, how to solve linear equa- tions.
Example 1.4.3.Find the solution of x 0 t /Ckx.t /Dh; x.0/Dx0; (1.10) whereh; kare constant Equation (1.10) arises in the RC circuit when there is a gen- erator of constant voltagehDV0, see Figure 1.3.
Herep.t / kand hence an integrating factor ise k t Multiplying the equation bye k t yields e k t x 0 Cke k t x.t /Dhe k t ; or d dt x.t /e k t
Integrating, we find x.t /e k t D h ke k t Cc wherecis an arbitrary constant Thus the general solution is x.t /Dce k t C h k:
Fig 1.3.RC circuit with a generator of voltage
To find a solution that satisfies the initial conditionx.0/ D x0 we might simply substitutet D0in the preceding equation, finding x0DcC h k and hence cDx0h k: Hence the unique solution of (1.10) is x.t /D x0h k e k t C h k: (1.11)
Alternatively, we can use (1.9) yielding x.t /De k t x0C
Notice that, ast ! C1,x.t /! k h , from below ifx0< h k (see Figure 1.4a) and from above ifx0> h k (see Figure 1.4b).
The solution (1.11) implies that in this case the capacitor voltagex.t /DV t /does not decay to0but tends, ast ! C1, to the constant voltageh= kDV0=RC.
Example 1.4.4.Find the general solution of x 0 t /C4tx.t /D8t and the solution such thatx.0/D1.
(a) Herep.t /D 4tand hence we can takeP t /D2t 2 We start by multiplying the equation by the integrating factore 2t 2 , which results in the equation e 2t 2 x 0 C4t e 2t 2 x.t /D8t e 2t 2 ; h/k t x
10 1 First order linear differential equations which is the same as d dt x.t /e 2t 2
Integrating both sides, we obtain x.t /e 2t 2 D2e 2t 2 Cc wherecis an arbitrary constant Therefore, the general solution is x.t /D2Cc e 2t 2 :
(b) If we require thatx.0/D1, then the constantcis uniquely determined by the equation1D2Cc e 2 0 D2Cc, that iscD 1and hence x.t /D2e 2t 2 : Alternatively, we can use the general formula (1.9) finding x.t /De 4 R 0 t sds
We make a couple of interesting observations concerning this equation.
In our analysis, we observe that the constant solution xD2 serves as a dividing point for all other solutions, categorizing them into two distinct groups: those that converge towards it from above and those that approach from below as time approaches infinity This behavior is illustrated in Figure 1.5.
2 We could have found the constant solutionx.t / D 2 without even solving the equation, by simply noting that ifx.t /is a constant, thenx 0 t /0and therefore x 0 t /C4tx.t /8timpliesx.t /D2for allt.
Example 1.4.5.Find the general solution of t 2 x 0 C.1Ct /xD 1 t e 1=t ; t > 0: (1.12)
To apply the integrating factor method effectively, we first need to transform the given equation into the correct format This involves dividing both sides of the equation by 26D0, resulting in the expression x' + (1/Ct)t^2 x = (1/t^3)e^(1/t) for t > 0.
We know that an integrating factorcan be determined as
To findP t /we evaluate the indefinite integral
TakingcD0we find that an integrating factor is given by
.t /De 1=t Cln t De 1=t : e ln t Dt e 1=t : Multiplying (1.13) by this integrating factor, we obtain the equation
.t e 1=t x/ 0 Dt e 1=t : 1 t 3 e 1=t D 1 t 2 : Integrating both sides, we get t e 1=t x.t /D 1 t Cc:
The general solution is x.t /D e 1=t t 2 C ce 1=t t D e 1=t t c1 t
All solutions converge to a constant C1, and it can be easily verified that for any constant c, the function x(t) defined by (1.14) satisfies equation (1.13) This indicates that x(t) from (1.14) is a solution to both (1.13) and (1.12) if and only if x(t) takes the form specified in (1.14), establishing it as the general solution of (1.12).
12 1 First order linear differential equations
If we want to solve the equationx 0 C 1 t C 2 t / x D t 1 3 e 1=t fort 6D 0we should distinguish the two casest > 0andt < 0separately As an exercise, the reader might repeat the calculations fort < 0.
Example 1.4.6.Solve the following initial value problem and show that the solution is defined fort > 0and is unique: t 2 x 0 C.1Ct /x D 1 t e 1 t ; x.1/D0: (1.15)
We have shown that the general solution of (1.12) fort > 0isx.t /given by (1.14), wherecis a constant.
Now in order to solve the initial value problem, since all solutions are included in (1.14), we simply substitute the initial values into the equation (1.14), obtaining
Figure 1.6 illustrates the graph of x(t), highlighting the uniqueness of the solution This uniqueness is established by the existence of a single value of c that allows the general solution (1.14) to meet the initial condition x(1) = 0.
The reader can check, as an exercise, that the same result holds if we use the gen- eral formula (1.9). t x
Exercises
1 Find the equation whose general solution isxDc e 5t
4 Find all the solutions to the initial value problemx 0 C 2t 3 t Csin 12 C 5 t C 5 xD0,x.0/D0.
5 Solvex 0 D 2xC3and find the solution satisfyingx.1/D5.
6 Findk such that there exists a solution ofx 0 D kx such thatx.0/ D 1 and x.1/D2.
7 Explain why the solution to the problem x 0 2.cost /x Dcost; x.0/D 1
2 must oscillate, i.e it must have arbitrarily large zeros.
8 In each of the following, find the maximum interval of existence of the solution, guaranteed by the existence theorem
10 Show that there is an infinite family of solutions to the problem t 2 x 0 2txDt 5 ; x.0/D0; all of which exist everywhere on.1;1/ Does this violate the uniqueness property of such equations?
11 Solvex 0 D2txand find the solution satisfyingx.0/D4.
15 Find the solution ofx 0 Cax Dbtsatisfyingx.t0/Dx0.
16 Solve the initial value problems (a)x 0 x D 1 2 t,x.0/D 1, (b)x 0 Cx D 4t, x.1/D0, (c)x 0 2xD2t,x.0/D3.
17 Givenh; k 2 R,k > 0, find the limits ast ! C1of the solutions ofx 0 C kxDh.
(a) For what value ofkwill all solutions tend to2ast ! C1?
(b) Is there any value ofkfor which there exists a non-constant solutionx.t / such thatx.t /! 3ast ! C1? Explain.
14 1 First order linear differential equations
19 Find the limits ast! ˙1of the solution ofx 0 D 1 C 1 t 2 x,x.0/D1.
20 Considerx 0 CkxDh, withk6D 0 Find conditions on the constantsh; ksuch that
(a) all solutions tend to0asttends to + infinity,
(b) it will have only one solution bounded on.0;C1/,
(c) all solutions are asymptotic to the linexD3.
21 Show that for any differentiable functionf t /,t 2R, all solutions ofx 0 CxD f t /Cf 0 t /tend tof t /ast tends toC1.
22 Find a continuous functionq.t /,t2R, such that all solutions ofx 0 Cx Dq.t / (a) approach the linexD7t5ast ! C1,
(b) approach the curvexDt 2 2tC5ast ! C1.
23 Show that ifpis differentiable and such that limt !C1p.t /D C1, then all the solutions ofx 0 Cp 0 t /xD0tend to zero ast ! C1.
24 Ifk 6D 0, show that the constant solutionx.t / D k 1 2 is the only solution of x 0 k 2 xD1such that the limt !C1x.t /is finite.
25 LetR C1 k ¤ 0 and letq.t / be continuous and such that limt !C1q.t / D 0, and
0 e k 2 s q.s/dsD0 Show that the solutionx.t /of the ivp problem x 0 k 2 xDq.t /; x.0/Dx0; tends to0ast ! C1if and only ifx0D0.
26 Show that the solution ofx 0 D k 2 x,x.t0/ D x0, is increasing ifx0 > 0and decreasing ifx0< 0.
27 Show that the solution ofx 0 D kx,x.t0/ D x0is increasing ifkx0 > 0and decreasing ifkx0< 0.
28 Find the locus of minima of the solutions ofx 0 C2xD6t.
29 Find the locus of maxima and minima of the solutions ofx 0 Cx Dat,a6D0.
Theory of first order differential equations
Before exploring methods for solving broader categories of differential equations, it is essential to provide a theoretical foundation on first-order equations and their solutions, establishing a clear framework for the remainder of the book.
Differential equations and their solutions
In Chapter 1, we introduced the concept of differential equations and the significance of their solutions This section will provide a more detailed and general exploration of these topics.
Consider the first order differential equation x 0 Df t; x/ (2.1) wheref t; x/is continuous,.t; x/2, R 2
A solution of (2.1) is a differentiable real valuedfunctionx.t /defined on an in- tervalI Rsuch that x 0 t /f t; x.t //; for.t; x.t //2: (2.2)
An equation in this form is said to be innormal form, to distinguish it from more general differential equations that will be introduced later on.
One of the simplest examples of a first order differential equation isx 0 D h.t /, whereh is continuous on an intervalI R IfH.t / is an antiderivative so that
H 0 t / D h.t /, then all the the solutions are given byx.t / D H.t /Cc,c a real constant.
In Chapter 1, we established that all solutions to the linear equation x' = p(t)x + q(t) constitute a family of functions that depend on a constant This principle extends more broadly, as we will demonstrate that solutions of the form x' = f(t, x) also form a one-parameter family of functions However, it is important to note that in nonlinear cases, there may be isolated solutions that do not fit within this framework.
S Ahmad, A Ambrosetti,A Textbook on Ordinary Differential Equations 2nd Ed.,
UNITEXT – La Matematica per il 3+2 88, DOI 10.1007/978-3-319-16408-3_2
16 2 Theory of first order differential equations
It is essential to remember that the solution of equation (2.2) is a function, unlike algebraic equations that yield real or complex numbers Additionally, it is crucial to highlight that (2.2) represents an identity, valid for all values within the domain of x.t /.
The domain of definition for a solution to equation (2.2) is initially unknown and can be influenced by various factors It is possible that, despite the function f(t, x) being well-defined for all real values of t and x, the solutions may only be valid for a specific subset of R This is illustrated in Example 2.2.2 below.
From a geometric perspective, a solution to the equation (2.1) is represented as a curve \( x(t) \) within a specific set At every point \( (t, x(t)) \) on this curve, the slope of the tangent line matches the function \( f(t, x(t)) \) Consequently, the equation governing this relationship can be expressed as \( x' = f(t, x(t)) \).
The curve defined by the equation \( x = e^t \) in the plane represents a solution to the differential equation \( x' = x \) A generic point \( P \) on this curve is characterized by the coordinates \( P = (t, e^t) \) The equation of the tangent line to the curve at point \( P \) is given by \( x = e^t(t - t_0) + e^t \).
In our discussion, we designate \( t \) as the independent variable and \( x \) as the dependent variable, though alternative designations are equally valid For instance, we could label the dependent variable as \( y \) and the independent variable as \( x \) Under this notation, a first-order differential equation takes the form \( y' = f(x, y) \), where a solution is a differentiable function \( y(x) \) satisfying \( y'(x) = f(x, y(x)) \) for all \( x \) in the domain of \( y(x) \) Regardless of the variable names chosen, the equation clearly indicates which variable is independent and which is dependent.
Dealing with a first order equation, one can distinguish between:
• Linearandnonlinearequations, according to whetherf t; x/is linear with respect toxor not.
• Autonomousandnon-autonomousequations according to whetherf is indepen- dent oft or not.
For example,x 0 D kxCcis linear and autonomous,x 0 Dx 2 CkxCcis nonlin- ear and autonomous; whilex 0 De t xCsint4is linear and non-autonomous, and x 0 Dtx 2 txC3is nonlinear and non-autonomous.
It is important to recognize that even if a function is independent of a variable, its domain must be defined as a suitable subset of R² For instance, in the equation x₀ = p(x), the function f(x) = p(x) is specifically defined for x₀, establishing the domain D as R excluding x₀ Similarly, the equation x₀ = p also adheres to this principle.
1x 2 is defined for1 x 1and henceis the horizontal stripR ạ1x1º.
More generally, letF t; x; p/be a real function of 3 real variables, defined on a setRR 3 Consider the first order differential equation
F t; x; x 0 /D0; whose solution is a differentiable real valued functionx.t / defined on an interval
IfF t; x; p/is of the formF t; x; p/ D pf t; x/, we can write the equation
Even more generally, we may consider systems of n first order equa- tions and n unknowns We may also consider more general scalar equations
In this chapter we deal with first order equations Higher order equations and sys- tems will be discussed starting with Chapter 4.
The Cauchy problem: Existence and uniqueness
Local existence and uniqueness
Theorem 2.2.1 (Local existence and uniqueness) Suppose thatf is continuous in
The Cauchy problem defined by the equation \( x' = f(t, x) \) with initial condition \( x(t_0) = x_0 \) has a unique solution within a closed interval \( I \) that contains the point \( t_0 \) in its interior, given that the function \( f \) has continuous partial derivatives with respect to \( x \).
We will see that this is a particular case of a more general result, see Theorem 2.4.4 below.
We are going to outline the proof of the existence part by using a method intro- duced by Picard, 2 which is based on an approximation scheme.
We define a sequence of approximate solutions by setting x0.t /Dx0; xk C 1.t /Dx0C
One shows that, under the given assumptions, there existsı > 0such thatx k t /con- verges to some functionx.t /, uniformly inŒt0ı; t0Cı Passing to the limit one finds thatxsatisfies x.t /Dx0C
The function x(t) is differentiable, and by applying the Fundamental Theorem of Calculus, we find that x(t) is defined as f(t, x(t)) for all t in the interval [t0, t0 + ε] Additionally, it is evident that x(t0) equals x0, confirming that x(t) is a solution to equation (2.4) within the specified interval For further details, please refer to the Appendix.
Let us show what happens in the particular casef t; x/Dx,t0D0andx0D1. The sequence is constructed as follows: x0.t / D 1; x1.t / D 1C
The sequence xk.t / converges uniformly tox.t / D P 1 kŠt k D e t , which is the solution tox 0 Dx; x.0/D1.
Theorem 2.2.1 is significant because it guarantees the existence and uniqueness of a solution within a specific interval around t0 However, it is essential to understand that the solution may not be defined across the entire set of real numbers, even if the equation itself is valid everywhere.
Example 2.2.2.Consider the ivp ² x 0 Dx 2 x.0/ Da6D0: (2.5)
To solve the equation x' = x², we identify it as a separable equation, which will be explored in detail in Section 3.1 of Chapter 3 Rather than employing the standard method, we will derive the solutions through a direct approach that relies on intuitive reasoning.
We have to find functionsx.t /whose derivatives are equal tox 2 t / One choice could be x.t / D 1 t , becausex 0 D t 1 2 which equalsx 2 D t 1 2 More generally, consider the functions
.tc/ 2 D 2 t; c/; it follows that, for all real constantsc, the functions.t; c/solvex 0 Dx 2
To find the solutionxa.t /of the Cauchy problem (2.5) we impose the requirement that.0; c/Da, that is aD 1
20 2 Theory of first order differential equations
The solution to the initial value problem (IVP) is not defined for all values of \( t \), but only for \( t < \frac{1}{a} \) when \( a > 0 \) and \( t > \frac{1}{a} \) when \( a < 0 \), despite the function \( x^2 \) being defined and continuous everywhere This highlights a unique characteristic of nonlinearity in contrast to linear functions It is important to clarify that stating (2.6) solves (2.5) is somewhat misleading, as we have not specified the interval for consideration.
A solution to a differential equation must be differentiable, which inherently means it is continuous This continuity is not guaranteed when considering the solution defined on the specified domain.
In this section, we expand our discussion to the general case of the Cauchy problem (2.5), which involves finding a function \( x(t) \) defined on an interval \( I \) that contains \( t_0 \) The goal is to satisfy the conditions \( x'(t) = f(t, x(t)) \) for all \( t \in I \) and \( x(t_0) = x_0 \) It is crucial to note that the interval \( I \) may be a subset of the larger set \( S \) where \( x(t) \) is defined This limitation arises because \( x(t) \) could exhibit discontinuities or lack differentiability within \( S \), while a solution to the differential equation must be continuous and differentiable The previously discussed problem (2.5) serves as a specific instance of this broader scenario.
I ăS Actually, in that caseS D ạt 2 RWt Ô 1 a ºwhileI D 1; 1 a /(ifa > 0) orI D 1 a ;C1//(ifa < 0).
Fig 2.2.Solutions ofx 0 Dx 2 ,x.0/Da (a) plot ofxa.t /,a > 0; (b) plot ofxa.t /,a < 0
The following definition is in order.
Definition 2.2.3.We say thatJ Ris themaximal intervalof definition of the so- lutionx.t /of the Cauchy problem (2.4) ifx.t /is defined inJ, andx.t /cannot be extended in an interval greater thanJ.
For example, the maximal interval of definition of the solutionxa.t /ofx 0 D x 2 ; x.t0/ D a ¤ 0, discussed in the preceding Example 2.2.2, is the half line 1; 1 a /(ifa > 0) or the half line a 1 ;C1/(ifa < 0).
Lemma 2.2.4 Letx0.t /be a solution ofx 0 D f t; x/ Suppose, for simplicity, that the setwheref is defined is all ofR 2 IfJ, the maximal interval of definition of the solutionx0.t /, is not all ofR, then it cannot be closed.
Proof By contradiction, letJ DŒ˛; ˇorJ D.1; ˇorJ DŒ˛;C1/ We deal with the first case, the other ones are similar.
The new Cauchy problem is defined as \( x_0 = Df(t, x) \) with the initial condition set at point \( \beta \), where it matches the value of \( x_0(t) \) at that point According to the local existence and uniqueness Theorem 2.2.1, this initial value problem (IVP) has a unique solution \( x_1(t) \) that exists in the interval \( [\beta, \beta + \delta) \) for some \( \delta > 0 \) The overall function \( x(t) \) is created by combining \( x_0 \) and \( x_1 \), such that \( x(t) = x_0(t) \) for \( t < \beta \) and \( x(t) = x_1(t) \) for \( \beta \leq t < \beta + \delta \).
Sincex0.ˇ/Dx1.ˇ/, the functionx.t /is continuous Let us show that it is differ- entiable This is clear for allt 6Dˇ Att Dˇwe consider the left and right limits of t x
22 2 Theory of first order differential equations the difference quotients hlim! 0 x.ˇCh/x.ˇ/ h ; lim h ! 0 C x.ˇCh/x.ˇ/ h :
In the context of calculus, we analyze the limit as \( h \) approaches 0 for the expression involving \( x \) and its differentiable function By evaluating the limits, we establish that the derivative at a point \( x_0 \) is equal to the limit of the difference quotient as \( h \) approaches 0 Similarly, for another point \( x_1 \), we find that the limit as \( h \) approaches 0 from the right gives us the same derivative condition Since \( x_0 \) equals \( x_1 \), it follows that the limit confirms the differentiability of the function \( x(t) \) at the point \( t = \beta \).
We have found a solution ofx 0 Df t; x/defined inŒ˛; ˇCıin contradiction with the fact thatJ D Œ˛; ˇis the maximal interval The argument for the left end point˛is the same.
Proposition 2.2.5 states that if the function f(t, x) meets the criteria of the local existence and uniqueness theorem in R², and if x(t) is a monotone and bounded solution of the differential equation x' = f(t, x), then the maximal interval of definition J for this solution is the entire set of real numbers R.
To prove by contradiction that the interval J is not strictly contained in R, we first assume that J is bounded We then demonstrate that J is closed by considering its right endpoint, denoted as ˇ Since the function x(t) is monotone and bounded, the limit as t approaches ˇ exists and is finite, indicating that x(t) is defined at t = ˇ, which implies that ˇ is included in J A similar argument applies to the left endpoint, ˛, confirming that J is closed This conclusion contradicts the previous lemma, thus establishing that J cannot be strictly contained in R.
To ensure the uniqueness of the solution for the initial value problem (IVP) in equation (2.4), it is essential that the function \( f \) is differentiable with respect to \( x \) An example illustrates that if this differentiability condition is not met, the solution may lack uniqueness.
Example 2.2.6.Consider the Cauchy problem ² x 0 D2p x x.0/ D0: (2.7)
This article discusses a separable equation from Section 3.1 of Chapter 3 One solution is expressed as x(t) = 0, while another solution is x(t) = t^2 for t ≥ 0 It is important to note that for t < 0, the absolute value |t| equals t, which means x(t) = t^2 is not a valid solution in that range.
We have identified two solutions to the initial value problem (IVP) represented by equation (2.7) It can be confirmed that for any value of \( a > 0 \), the functions \( x_a(t) = 0 \) for \( 0 < t < a \) and \( x_a(t) = (t - a)^2 \) for \( t \geq a \) are valid solutions Consequently, equation (2.7) possesses infinitely many solutions, as illustrated in Figure 2.4.
Global existence and uniqueness
As mentioned before, Theorem 2.2.1 is local The nextglobalresult holds, provided the setis a strip andfx is bounded w.r.t.x.
Theorem 2.2.10 (Global Existence and Uniqueness) Let be the strip D Œa; bRand let.t0; x0/be a given point in the interior of Suppose thatf is continuous inand has continuous partial derivative with respect toxand that the
5A statement of the Ascoli Theorem is reported in Chapter 13. partial derivativefx.t; x/is bounded in the strip Then the Cauchy problem ² x 0 Df t; x/ x.t0/ Dx0 has a unique solution defined for allt2Œa; b.
Corollary 2.2.11 IfDR 2 ,DŒa;C1/R, orD.1; bR, andfx.t; x/ is bounded in, then the solution is defined respectively on all ofR, onŒa;C1/, or on.1; b.
Theorem 2.2.10 and Corollary 2.2.11 are particular cases of the more general The- orem 2.4.5 in the next section.
The new feature of the preceding results is that now the solution is defined on the whole intervalŒa; b.
Remark 2.2.12.Example 2.2.2 shows that the condition thatfxis bounded in the strip cannot be removed.
In the context of the linear equation \( x' + p(t)x = q(t) \) discussed in Chapter 1, we define \( f(t, x) = p(t)x + q(t) \) and \( f_x(t, x) = p(t) \), which is bounded on the interval \([a, b]\) This scenario allows us to apply Theorem 2.2.10, offering an alternative proof for the existence and uniqueness of solutions as stated in Theorem 1.4.2 It is important to note that the solutions for the equation \( x' + p(t)x = q(t) \) are valid across the entire interval \([a, b]\) Additionally, Corollary 2.2.11 indicates that if \( p \) and \( q \) are continuous functions on \( \mathbb{R} \), then the solutions are defined for all of \( \mathbb{R} \).
Qualitative properties of solutions
In this section we study some qualitative properties of solutions, using the Global Existence and Uniqueness result stated before.
In the sequel it is understood that the assumptions of this theorem are satisfied. Moreover, for simplicity, we will also assume thatDRR.
A qualitative analysis of a solution can reveal important characteristics without explicitly solving the corresponding equation While only a few equations yield solutions in elementary functions, most require a qualitative study to gain insights into key features such as symmetry, monotonicity, asymptotic behavior, and convexity.
We start with simple symmetry results.
Lemma 2.3.1 Letf x/be odd and letx.t /be a solution ofx 0 Df x/ Thenx.t / is also a solution.
Proof Settingz.t / D x.t /we findz 0 D x 0 D f x/D f z/Sincef is odd thenf z/Df z/andz 0 Df z/.
26 2 Theory of first order differential equations
Lemma 2.3.2 Letf x/be even and letx0.t /be a solution ofx 0 Df x/such that x.0/D0 Thenx0.t /is an odd function.
Proof Settingz.t /D x0.t /we findz 0 t /D x 0 0 t / D f x0.t // D f z/ Sincef is even thenf z/Df z/andz 0 Df z/ Moreover,z.0/Dx0.0/D0. Thus, by uniqueness,z.t /Dx0.t /, namelyx0.t /D x0.t /.
Consider the autonomous equation x 0 Df x/: (2.8)
In the context of the equation (2.8), it is evident that a number \( x_0 \) satisfies \( f(x_0) = 0 \) if and only if \( x_0 \) represents a constant solution Due to the uniqueness of solutions, non-constant solutions cannot intersect with constant ones Consequently, these constant solutions create horizontal divisions in the \( (t, x) \) plane, which may be either bounded or unbounded, and are surrounded by non-constant solutions of the equation Specifically, for any point \( (t_0, x_0) \) in \( \mathbb{R}^2 \), the function \( t \) signifies the solution to the initial value problem defined by \( x' = f(x) \) and \( x(t_0) = x_0 \).
If \( f(x_0) = 0 \), then \( t(x_0) \) is a constant solution If \( f(x_0) > 0 \), then \( t(x_0) \) is increasing, whereas if \( f(x_0) < 0 \), then \( t(x_0) \) is decreasing When \( t(t) \) is defined for all \( t \) and is either increasing or decreasing, it approaches finite or infinite limits \( L \) as \( t \) approaches infinity The finite limits correspond to the zeros of \( f \) For instance, if \( L + C < C_1 \), it can be easily verified that \( t(t) \) approaches 0 as \( t \) approaches \( C_1 \).
Furthermore, iff is differentiable, then.t /is twice differentiable and one has
00 t /D d 0 dt D d dtf t //Df 0 t // 0 t /Df t //f 0 t //: (2.10) This allows us to find the sign of 00 t /and hence the convexity of.
Example 2.3.3.Let˛; ˇ,˛ < ˇ be two consecutive zeroes off and suppose that f x/ > 0for˛ < x < ˇ Takingx0 2.˛; ˇ/, from the discussion above it follows that.t /is increasing,˛ < t / < ˇfor alltand t !1lim t /D˛; lim t !C1.t /Dˇ:
Assuming that the function \( f \) is differentiable, and considering the points \( \alpha \) and \( \beta \) where \( f'(x) = 0 \), we find that \( f'(x) > 0 \) for \( \alpha < x < x \) and \( f'(x) < 0 \) for \( x < x < \beta \) Given that \( t(t) \) is an increasing function with \( \alpha < t(t) < \beta \), the equation \( t(t) = x \) has a unique solution \( t \), where \( \alpha < t(t) < x \) for \( t < t \) and \( x < t(t) < \beta \) for \( t > t \).
Fig 2.5.Qualitative behavior of the solution.t /of (2.9) implies that 00 t /D 0, 00 t / > 0for˛ < t < t and 00 t / < 0fort < t < ˇ. See Figure 2.5 where we have taken˛ < 0 < x < x0< ˇ.
When addressing a general non-autonomous equation \( x_0 Df(t, x) \), the analysis of potential maxima or minima of a solution \( x(t) \) becomes essential If \( 0(t) = 0 \), by setting \( x = x(t) \) and defining \( M_0 = \{(t, x) \in \mathbb{R}^2 \mid f(t, x) = 0\} \), it follows that \( f(t, x) = 0 \) implies \( (t, x) \in M_0 \) Furthermore, if \( f \) is differentiable, the relationship \( 0(t) = Df(t, x(t)) \) indicates a critical connection between the derivative and the function's behavior at that point.
Ift is such that 0 t /D0, then
Hencehas a maximum or a minimum att Dt depending on whetherft.t ; x / 0 In other words, letting
(i) If.t ; x /D.t ; t //2M0\M then.t /has a maximum att Dt (ii) If.t ; x /D.t ; t //2M0\M C then.t /has a minimum att Dt
28 2 Theory of first order differential equations t x
Fig 2.6.The red curve is the parabolax 2 D 4t, where the solutions ofx 0 Dx 2 C4thave minima
Example 2.3.4.(i) Consider the equationx 0 Dx 2 C4t Settingf t; x/Dx 2 C4t we findM0 D ạ.t; x/ 2 R 2 W x 2 D 4tº Furthermoreft.t; x/ D 4 and thus
M D ;andM C DR 2 Therefore the solutions ofx 0 Dx 2 C4thave only minima located on the parabolax 2 D 4t See Figure 2.6.
(ii) Show that the solution.t /ofx 0 Dx 2 t 2 ,x.a/Da¤0has a maximum att Da, ifa > 0, a minimum ifa < 0.
The function has a critical point at \( t = a \), where its second derivative \( 00(a) = 2a \) indicates the nature of this point If \( a > 0 \), the function achieves a maximum at \( t = a \), while if \( a < 0 \), it reaches a minimum at the same point The scenario where \( a = 0 \) is left as an exercise for further exploration.
Fig 2.7.Local behavior neartDaof the solution ofx 0 Dx 2 t 2 ,x.a/Da¤0 (a)a >0; (b)a < 0
The following proposition is a symmetry result for a general non autonomous equation.
Proposition 2.3.5 Suppose thatf t; x/is odd with respect tot, that isf t; x/D f t; x/ Then the solutions ofx 0 Df t; x/are even functions.
Proof Letx.t /be any solution ofx 0 Df t; x/ Settingz.t /Dx.t /one has z 0 t /D x 0 t /D f t; x.t //D f t; z.t //:
Since, by assumption,f t; z/D f t; z/, we deduce z 0 Df t; z/:
The functions x(t) and z(t) satisfy the same equation and share the same initial condition at t = 0, as both have the value x(0) = z(0) Given that f is continuously differentiable, Theorem 2.4.4 confirms their uniqueness, leading to the conclusion that x(t) = z(t) This establishes that x(t) is an even function, as required.
The next result allows us to compare solutions of two differential equations.
Theorem 2.3.6 Letxa.t /; yb.t /be solutions of the Cauchy problems ² x 0 Df t; x/ x.t0/ Da ² y 0 Dg.t; y/ y.t0/ Db defined in a common intervalŒt0; ˇ/ Ifa < bandf t; x/ < g.t; x/, thenxa.t / < yb.t /for allt 2Œt0; ˇ/.
Proof We argue by contradiction Suppose that the set
The set S, defined as S = {xa(t) - yb(t) | t ∈ [t0, t0 + ε]}, is non-empty, and its infimum, or greatest lower bound, is denoted as Given that xa(t0) = a < b = yb(t0), the Sign Permanence Theorem for continuous functions ensures that there exists a δ > 0 such that xa(t) < yb(t) for all t in the interval [t0, t0 + δ], leading to t0 + δ > t0 Since is the infimum of S, there exists a sequence tj > such that tj converges to and xa(tj) approaches yb(tj) Taking the limit reveals that xa(τ) cannot exceed yb(τ); otherwise, the Sign Permanence Theorem would imply xa(t) > yb(t) in a left neighborhood of τ, contradicting the definition of as the infimum of S.
For small values of h less than zero, it follows that xa.Ch/ is less than yb.Ch/ due to DinfS Additionally, since xa./Dyb./ holds true, we can conclude that the incremental ratios demonstrate that xa.Ch/ divided by xa is greater than yb.Ch/ divided by yb for all small h values less than zero.
30 2 Theory of first order differential equations
Passing to the limit ash ! 0,h < 0, we infer that x a 0 / y 0 / But this is impossible, since, by assumption, x a 0 /Df ; xa.// < g.; yb.//Dy 0 /:
We have proved thatSis empty and therefore thatxa.t / < yb.t /for allt 2Œt0; ˇ/. The next examples show how we might apply the comparison theorem.
In Example 2.3.7.(i), we consider a positive solution \( x(t) \) of the differential equation \( x' = f(x) \) with the initial condition \( x(t_0) = a \) Assuming that \( f(x) < kx \) for some constant \( k > 0 \) and that \( x_a(t) \) is defined on the interval \( [t_0, \infty) \), we find that \( x_a(t) \) decays exponentially to 0 as \( t \) approaches infinity To demonstrate this, we define \( y_b(t) \) as the solution to \( y' = ky \) with the initial condition \( y(t_0) = b > \max(a, 0) \), leading to \( y_b(t) = b e^{k(t - t_0)} \) By applying the previous proposition with \( g(y) = ky \), we conclude that \( 0 < x(t) < b e^{k(t - t_0)} \), confirming the exponential decay of \( x(t) \).
The function Letyb.t serves as the solution to the equation ²y 0 Dg.t; y/ with initial condition y.t0/ Db, defined on the interval [t0, C1) Given that g.t; y/ is greater than a positive constant k and b is also positive, it follows that the limit as t approaches infinity for yb.t is infinity By applying the proposition with the function f.t; x/ equal to k and a set to 0, we can deduce that yb.t is greater than or equal to k multiplied by t minus t0, which directly supports the claim.
Improving the existence and uniqueness results
The assumptions in the previous theorems can be relaxed, allowing for an expansion of the results to encompass a broader range of functions Below, we highlight the key extension of this nature.
Definition 2.4.1.The functionf t; x/defined in a setR 2 , islocallylipschitzian 6 (or simply lipschitzian) at a point.t0; x0/ 2 with respect tox, if there exists a neighborhoodU of.t0; x0/and a numberL > 0such that jf t; x/f t; z/j Ljxzj; 8.t; x/; t; z/2U:
We say thatf isgloballylipschitzian onif there existsL > 0, such that jf t; x/f t; z/j Ljxzj; 8.t; x/; t; z/2:
From the definition it immediately follows that any locally lipschitzian function is continuous at.t0; x0/ Moreover, one has:
Lemma 2.4.2 Letf t; x/be continuously differentiable with respect toxin If there exists > 0such thatfx.t; x/is bounded inU D ạjtt0j< ºạjxx0j< º, thenf is lipschitzian onU.
Proof Applying the Mean Value Theorem to the functionf t; x/we infer that f t; x/f t; z/Dfx.t; /.xz/; wherex < < z SinceLDsupạjfx.t; /j W.t; /2Uºis finite by assumption, it follows that jf t; x/f t; z/j Ljxzj; 8.t; x/; t; z/2U; proving the lemma.
Example 2.4.3.(i) The functionf x/ D jxjis globally lipschitzian with constant
The function \( f \) is continuous at \( x = 0 \) but is not differentiable at this point Specifically, the expression \( |f(x) - f(z)| = |x| |z| - |xz| \) holds for all \( x, z \in \mathbb{R} \) Additionally, since \( |x| \) is defined as \( -x \) for \( x < 0 \) and \( x \) for \( x > 0 \), the left derivative of \( f \) at \( x = 0 \) is 1, while the right derivative is +1 Therefore, \( f \) is not differentiable at \( x = 0 \).
The function \( f(x) = x^2 \) is locally Lipschitzian at every point but not globally Lipschitzian on \( \mathbb{R} \) This can be demonstrated by noting that \( f(x) \) is differentiable, with a derivative \( f'(x) = 2x \) that remains bounded on any finite interval of \( \mathbb{R} \) Consequently, \( f \) is locally Lipschitzian at every point However, if \( f \) were globally Lipschitzian on \( \mathbb{R} \), there would exist a constant \( L > 0 \) such that \( |x - z| \leq L |x - y| \) for all \( x, y \in \mathbb{R} \) This leads to the inequality \( |x + z| \leq L \), which is clearly false for all \( x, z \in \mathbb{R} \).
The function \( f(x) = \sqrt{|x|} \) is not Lipschitz continuous at \( x = 0 \) If it were, there would exist constants \( \epsilon > 0 \) and \( L > 0 \) such that \( |\sqrt{|x|} - \sqrt{|z|}| \leq L |x - z| \) for all \( x, z \) in the interval Specifically, by setting \( z = 0 \), we would have \( |\sqrt{|x|}| \leq L |x| \) for all \( x \) in the interval, which leads to a contradiction.
Based on the earlier definition, it is possible to demonstrate a local and global existence result applicable to equations in normal form, thereby extending the existence and uniqueness theorems, specifically Theorems 2.2.1 and 2.2.10, along with Corollary 2.2.11.
Theorem 2.4.4 states that if (t0, x0) is a point within the interior of a domain and the function f is locally Lipschitz continuous with respect to x at (t0, x0), then the Cauchy problem, represented by the equation x' = f(t, x) with the initial condition x(t0) = x0, possesses a unique solution that is defined in a suitable neighborhood around t0.
32 2 Theory of first order differential equations
In Example 2.2.6, we demonstrated that the equation ivpx 0 D2p jxj,x.0/D0 has infinitely many solutions It is important to note that the function f(x) = 2p jxj is not Lipschitzian at x = 0, as illustrated in Example 2.4.3(iii) This indicates that the previous result is sharp, highlighting that we cannot ensure the uniqueness of the Cauchy problem (2.4) if the function f is not Lipschitzian at the point (t0, x0).
Theorem 2.4.5 states that if \( D \subseteq [a, b] \) (or \( D \subseteq \mathbb{R} \)) and the function \( f \) is globally Lipschitz continuous, then for given values \( (t_0, x_0) \), the Cauchy problem defined by \( x' = f(t, x) \) with initial condition \( x(t_0) = x_0 \) has a unique solution that exists for all \( t \) in the interval \( [a, b] \) (or for all \( t \in \mathbb{R} \)).
The proofs are given in the Appendix below.
The Cauchy problem x' = |x| with initial condition x(0) = x0 has a unique solution x(t) for all t in R, due to the global Lipschitz continuity of |x| Specifically, if x0 = 0, then x(t) = 0 for all t For x0 > 0, the solution is x(t) = x0e^t, which remains positive, while for x0 < 0, the solution is x(t) = x0e^t, remaining negative In all cases, x(t) is increasing as long as x0 is not equal to zero.
Appendix: Proof of existence and uniqueness theorems
Proof of Theorem 2.4.5
Let us first prove Theorem 2.4.5 dealing with uniqueness and global existence of the
The Cauchy problem is defined by the equation \( x_0 Df(t, x) = x(t_0) = x_0 \) in the strip \( S = \{(t, x) \in \mathbb{R}^2 | a < t < b, (t_0, y_0) \in S\} \), where the function \( f(t, x) \) is continuous and globally Lipschitz in \( S \) This implies the existence of a constant \( L > 0 \) such that the inequality \( |f(t, x) - f(t, y)| \leq L |x - y| \) holds for all \( (t, x), (t, y) \in S \).
The strategy is to find a sequence of functions that converges to the solution of (2.12). For this, it is convenient to transform the ivp (2.12) into an equivalent integral equa- tion.
Lemma 2.5.1.x.t /is a solution of(2.12)if and only ifx.t /satisfies x.t /Dx0C
Proof Let x.t /be a solution of (2.12) This means that x 0 t / f t; x.t //and hence integrating fromt0tot we find
Sincex.t0/Dx0the first integral is equal tox.t /x.t0/Dx.t /x0and thus,
Conversely, letx.t /satisfy (2.14) If fort 2Œa; bwe set
Z t t 0 f s; x.s//ds by the fundamental theorem of calculusis continuous, differentiable and
Thusx.t /Dx0C.t /is differentiable inŒa; band x 0 t /D 0 t /Df t; x.t //; 8t2Œa; b:
Z t 0 t 0 f s; x.s//dsDx0 and hencex.t /satisfies the ivp (2.12), completing the proof of the lemma.
Define by recurrence a sequence of functions such that for allt 2 Œa; band all integerskD0; 1; 2; : : : x0.t / D x0 x1.t / D x0C
Lemma 2.5.2 The sequencex k t /is uniformly convergent inŒa; b.
34 2 Theory of first order differential equations
Proof Let us start by showing by induction that for allkD1; 2; :: jxk.t /xk 1.t /j M
L jtt0j k L k kŠ ; 8t 2Œa; b (2.15) whereM Dmaxạjf t; x0/j Wt 2Œa; bº.
To simplify notation, we carry out the proof takingt t0 The caset t0requires obvious changes 7 ForkD1we have, using the assumption thatf is lipschitzian, jx2.t /x1.t /j D ˇˇ ˇˇZ t t 0
On the other hand, jx1.s/x0j Dˇˇ ˇˇZ s t 0 f r; x0/d rˇˇ ˇˇˇˇ ˇˇZ s t 0 jf r; x0/jd rˇˇ ˇˇM jst0j and thus jx2.t /x1.t /j ML
2 jtt0j 2 ; 8t 2Œa; b; which proves (2.15) forkD1.
By induction, we assume that (2.15) holds fork1 Repeating the previous ar- guments, we find jxk.t /xk 1.t /j
Z t t 0 jxk 1.s/xk 2.s/jds; 8t 2Œa; b: Using the induction hypothesis we find jx k C 1 t /xk.t /j LM
Therefore (2.15) holds for all natural numbersk.
Since (2.15) holds for allt2Œa; b, then max t 2 Œa;bjxk C 1.t /xk.t /j M
7For example, in the next equation we should writejx2.t /x1.t /j ˇˇR t t 0 jf s; x1.s// f s; x 0 /jdsˇˇ: : :
.ba/ k L k kŠ !0 k! C1/ because the series
XŒL.ba/ k kŠ t k is convergent toe L.b a/t Thus (2.16) implies that the sequencexk.t /is uniformly convergent onŒa; b, as required.
Proof of Theorem 2.4.5.By Lemma 2.5.2, the sequencexk.t / ! x.t /, uniformly inŒa; b Thenf s; xk.s//!f s; x.s//, uniformly inŒa; band we can pass to the limit under the integral in xk C 1.t /Dx0C
According to Lemma 2.5.1 it follows thatx.t /is a solution of the ivp (2.12).
To establish the uniqueness of solutions, we will examine an interval \( jtt_0j \) where \( L_i < 1 \) and \( t \) is within the range \( [a, b] \) In this context, we aim to demonstrate that two solutions, \( x(t) \) and \( y(t) \), of the equation (2.12) are identical within this interval Specifically, we have the relationship \( x(t) - y(t) = 0 \).
.f s; x.s//f s; y.s//ds and hence jx.t /y.t /j ˇˇ ˇˇZ t t 0 jf s; x.s//f s; y.s//jdsˇˇ ˇˇLˇˇ ˇˇZ t t 0 jx.s/y.s/jdsˇˇ ˇˇ
Ljtt0j max j t t 0 j ıjx.t /y.t /j Lı max j t t 0 j ıjx.t /y.t /j: Taking the maximum of the left-hand side onjtt0j ıwe find j t maxt 0 j ıjx.t /y.t /j ıL max j t t 0 j ıjx.t /y.t /j:
By letting \( A D \max_{t} j \xi(x(t), y(t)) j > 0 \), we find a contradiction by dividing by \( A \) and obtaining \( L < 1 \) This leads to the conclusion that \( j \max_{t \leq 0} j \xi(x(t), y(t)) j = 0 \), indicating that \( x(t) = y(t) \) within the interval \( |t - t_0| < \delta \) Specifically, this means \( x(t_0 \pm \delta) = y(t_0 \pm \delta) \) We can then apply the same reasoning in the interval \( [t_0 + \delta, t_0 + 2\delta] \).
The 36 2 Theory of first-order differential equations demonstrates that for all t in the interval [t0 - δ, t0 + δ], the relationship x(t) = y(t) holds true After a finite number of iterations, this relationship is established for all t within the range [a, b] This conclusion finalizes the proof.
Proof of Theorem 2.4.4
In this section, we will demonstrate Theorem 2.4.4, which addresses the local existence and uniqueness of solutions for the initial value problem (IVP) presented in equation (2.12) We assume that the function f(t, x) is defined in R² and exhibits local Lipschitz continuity around the point (t₀, x₀) This implies that there is a neighborhood in which the Lipschitz condition holds true.
U of.t0; x0/and a numberL > 0such that jf t; x/f t; y/j Ljxyj; 8.t; x/; t; y/2U: (2.17)
To derive Theorem 2.4.4 from Theorem 2.4.5, we can assume U DUra represents a closed square in the context of the strip defined by the inequality |x/2W jt t0j r;jxx0j rº| for some r > 0 This approach allows us to build upon the results established in the previous section effectively.
Sr WD ạ.t; x/2R 2 W jtt0j rº and extendf to the functionf WSr7!Rdefined by setting f t; x/D
: f t; x0r / if.t; x/2Sr andxx0r f t; x/ if.t; x/2Ur f t; x0Cr / if.t; x/2Sr andx x0Cr:
It is easy to check thatf is globally lipschitzian onSr For example, ifx; yare such thatx0r < x < x0Cr y, thenf t; x/Df t; x/,f t; y/D f t; x0Cr / and one has jf t; x/f t; y/j D jf t; x/f t; x0Cr /j Ljxx0rj Ljxyj:
Sincef is globally lipschitzian onSr, the global Theorem 2.4.5 yields a solution x.t /, defined onŒt0r; t0Cr , of the ivp ² x 0 D f t; x/ x.t0/D x0:
To address the issue of the function x(t) potentially falling outside the range [x0, r] where f is not defined, we leverage the continuity of x(t) and the condition x(t0) = x0 Consequently, there exists a positive interval ı > 0 such that for t within [t0 - ı, t0 + ı], the difference |x(t) - x0| remains within the bounds of r.
Therefore, fort 2 Œt0ı; t0Cıone has thatf t; x.t // D f t; x.t //and hence x.t /, restricted to such an interval, solves the ivp (2.12).
Exercises
1 Check that the local existence and uniqueness theorem applies to the ivpx 0 D t Cx 2 ,x.0/D0.
2 Show that the functionf x/D jxj p is not lipschitzian atxD0if0 < p < 1.
3 Findasuch that the existence and uniqueness theorem applies to the ivpx 0 D
4 Check that for allt0; athe existence and uniqueness theorem applies to the ivp lnx 0 Dx 2 ; x.t0/Da.
5 Explain whyx 0 C e sin t C t 1x D 0cannot have a solutionx.t /such thatx.1/D 1 andx.2/D 1.
6 Transform the equatione x 0 Dxinto an equation in normal form and show that it has a unique solution such thatx.t0/Da, for allt0and alla > 0.
7 Find the equation whose solution is the catenaryx.t /Dcosht D 1 2 e t Ce t /.
8 Check that the functionsx.t /1and
1 ift > 2 are solutions of the ivpx 0 Dp
9 Finda 0such that the Cauchy problemx 0 D jxj 1=4 ,x.0/Dahas a unique solution.
10 Show that ifp > 1the solution ofx 0 Dx p ,x.0/Da > 0, is not defined for all t 0.
11 Show that if0 < p1, the solution ofx 0 D jxj p ,x.0/Da > 0, is defined for allt 0.
12 Show that the solutions ofx 0 Dsinxare defined on allt 2R.
13 Show that the solutions ofx 0 Darctanxare defined on allt 2R.
14 Show that the solutions ofx 0 Dln.1Cx 2 /are defined on allt 2R.
15 Show that the ivpx 0 Dmaxạ1; xº,x.0/D1, has a unique solution defined for allt and find it.
16 Show that the ivpx 0 Dmaxạ1;xº,x.0/D 1, has a unique solution defined for allt and find it.
17 Show that the solution ofx 0 Dt 2 x 4 C1,x.0/D0is odd.
18 Show that, iff x/ > 0, resp.f x/ < 0, the solutions ofx 0 D f x/cannot be even.
38 2 Theory of first order differential equations
19 Show that the solutionx.t /of the Cauchy problemx 0 D 2Csinx,x.0/ D0, cannot vanish fort > 0.
20 Letf x/be continuously differentiable and such thatf 0/D 0 Show that the solutions ofx 0 Df x/h.t /cannot change sign.
21 Show that the solutions ofx 0 Dsin.tx/are even.
22 Find the limits, ast ! ˙1, of the solution.t /of the ivpx 0 D.xC2/.1x 4 /, x.0/D0.
23 Show that for everya,1 < a < 1the solution.t /of the ivpx 0 D x 3 x, x.0/Dais such that limt !C1.t /D0.
24 Show that the solutions ofx 0 DarctanxCtcannot have maxima.
25 Show that the solutions ofx 0 De x t cannot have minima.
26 Let.t /be the solution of the ivpx 0 Dtxt 3 ,x.a/Da 2 witha¤0 Show thathas a maximum att Da.
27 Let.t /be the solution of the ivpx 0 Dtxt 3 ,x.0/Da 2 witha¤0 Show thathas a minimum att D0.
28 Show that the solution.t /ofx 0 D x 2 t 2 ,x.0/D0, has an inflection point att D0.
29 Suppose thatg.x/is continuously differentiable and let.t /be the solution of x 0 Dtg.x/,x.0/Da Ifg.a/ > 0, show that the function.t /has a minimum att D0, for alla2R.
30 Show that the solutionxa.t /ofx 0 D 2t Cg.x/,xa.0/ D a > 0 satisfies xa.t / tCt 2 fort > 0, providedg.x/ 1.
31 Letxa.t /be the solution of ofx 0 D t Cg.x/,xa.0/ D a,with0 < a < 2.
Ifxa.t /is defined for allt > 0andg.x/ x, show using the comparisonTheorem 2.3.6 that the equationxa.t /D 0has at least one positive solution in.0; 2/.
First order nonlinear differential equations
The main focus of this chapter is on learning how to solve certain classes of nonlinear differential equations of first order.
Separable equations
The logistic equation
A notable example of a separable equation is the population model proposed by P.F Verhulst, which posits that the change in population size (x') is proportional to the current population (x) multiplied by a factor of αβx This leads to the differential equation x'(t) = x(t)(αβx(t)), where α and β are both positive constants.
Contrary to the Malthusian model discussed in the first chapter, here the constant factorkis substituted by the function˛ˇx The fact that this function is decreas-
The logistic equation describes how the difficulty of finding essential resources, such as food and space, increases with the size of a population (x), making survival more challenging.
In this model, \( x(t) \) represents the population of a species, and we focus on solutions where \( x(t) > 0 \) By solving the equation \( x(\alpha - \beta x) = 0 \), we identify the equilibrium solutions \( x = 0 \) and \( x = \frac{\beta}{\alpha} \) These equilibrium points are crucial for analyzing the trajectories of solutions over time We will now examine the solutions \( x(t) \) that remain positive for all \( t \geq 0 \.
The uniqueness of the solutions ensures that no other trajectories can intersect the lines defined by the two equilibrium solutions, specifically x = 0 and x = α/β As a result, any non-constant solution x(t) will not equal zero or α/β for any time t This creates a clear division of trajectories into two categories: those that exist above the line x = β/α and those that fall between the lines x = 0 and x = β/α.
In the previous chapter's Section 2.3, we explored techniques for understanding the qualitative behavior of solutions without directly solving the equations Now, we will examine methods for obtaining solutions, whether explicitly or implicitly.
To solve the logistic equation, we begin by separating the variables, leading to the expression dx / (x(α - x)) = dt, under the conditions that x is not equal to 0 and x is not equal to α The left side of the equation can be integrated using the method of partial fractions, where we seek constants A and B to facilitate the integration process.
1 x.˛ˇx/ D A x C B ˛ˇx and findAD 1 ˛ andBDˇAD ˇ ˛ Therefore, we have
1 ˛lnjxj 1 ˛lnj˛ˇxj D 1 ˛lnj x ˛ˇxj DtCc from which we obtain ˇˇˇˇ x.t / ˛ˇx.t / ˇˇˇˇDe ˛.t C c/ Dke ˛t ; k De ˛c /: (3.5)
This is the general solution in implicit form To solve forxwe distinguish the cases
0 < x.t / < ˇ ˛ andx.t / > ˛ ˇ In the first case one has that ˛ x.t / ˇ x.t / > 0 Then (3.5) becomes x.t / ˛ˇx.t / Dke ˛t and, solving forx, x.t /D ˛ke ˛t
Ifx.t / > ˇ ˛ one has that ˛ x.t / ˇ x.t / < 0 Then (3.5) becomes x.t / ˛ˇx.t / Dke ˛t and, solving forx, x.t /D k˛e ˛t
In any case, one finds that t !C1lim x.t /D ˛ ˇ:
All non-constant solutions converge towards the equilibrium solution \( x(t) = \frac{\alpha}{\beta} \) as time \( t \) approaches infinity, with some solutions approaching from above the line \( x = \frac{\beta}{\alpha} \) and others from below This behavior significantly differs from that observed in the Malthusian model.
Assuming that the initial size of the poulation isa, we can determine the size of the population at any timetby solving the Cauchy problem ´x 0 Dx.˛ˇx/; x.0/ Da:
Fig 3.2.Solutions of the logistic equation forx.0/ > ˛=ˇandx.0/ < ˛=ˇ
44 3 First order nonlinear differential equations
Assuminga > 0anda ¤ ˇ ˛ , we use (3.5) and the initial condition to findk D ˇˇˇ˛ a aˇˇˇˇ If0 < a < ˛ ˇ thenk D ˛ a aˇ , while ifa > ˛ ˇ thenk D ˛ a aˇ , namely kD ˛ a aˇ Therefore both (3.6) and (3.7) imply x.t /D a˛ ˛aˇe ˛t
Example 3.1.2.Solve (i)x 0 Dx.2x/,x.0/D 1and (ii)x 0 Dx.2x/,x.0/D
3 This is a logistic equation with˛D2,ˇD1, where the initial population size is given byaD1in case (i) andaD3in case (ii) Then we find
Exact equations
Notice that in (3.8) we useyas dependent variable andxas the independent variable, while in (3.9) the roles ofxandyare exchanged:xis now the dependent variable whileyis the independent variable.
The preceding equations can be stated, in a differential form, as
Let us associate with (3.10) the differential form
We say that (3.10) is anexact equationif!is the exact differential of a function; that is, there exists an antiderivativeF x; y/such thatdF D!, which means that ² Fx.x; y/ DM.x; y/
M.x; y/dxCN.x; y/dy DdF x; y/DFx.x; y/dxCFy.x; y/dy:
Proposition 3.2.1 states that for continuous functions M and N in R², if the equation M(x, y)dx + N(x, y)dy = 0 is exact, then an antiderivative F(x, y) exists If y(x) is a solution to the equation, it follows that F(x, y(x)) = c for some constant c in R Conversely, if y(x) is continuously differentiable and satisfies F(x, y(x)) = c, then y(x) also solves the original equation.
Proof Lety.x/be a solution of (3.8) The functionF x; y.x//is differentiable and d dx F x; y.x//DFx.x; y.x//CFy.x; y.x//dy dx: SincedF D!, thenFx DM; Fy DN and we find d dxF x; y.x//DM.x; y.x//CN.x; y.x// dy dx:
Assuming that y(x) is a solution to the equation (3.8), it follows that the derivative of F with respect to x, evaluated at (x, y(x)), is zero, indicating that F(x, y(x)) is constant, specifically F(x, y(x)) = c, where c is a real number Conversely, if y(x) is continuously differentiable and satisfies F(x, y(x)) = c for some real number c, differentiating this equation leads to the result that the derivative of F with respect to x, combined with the partial derivatives of F and the derivative of y with respect to x, equals zero.
SinceFx DM andFy DN we deduce that
M.x; y.x//CN.x; y.x//dy dx D0 and this means thaty.x/is a solution of (3.8).
Proposition 3.2.2 states that for continuous functions M and N in R², if the equation M(x, y)dx + N(x, y)dy = 0 is exact, then there exists an antiderivative F(x, y) such that F(x(y), y) = c for some constant c in R, indicating that x(y) is a solution to the equation (3.9) Conversely, if x(y) is continuously differentiable and satisfies F(x(y), y) = c for some constant c in R, then x(y) also satisfies equation (3.9).
We have seen that the solutions of the exact equation (3.10) are those defined by
F x; y/Dc, whereF is an antiderivative of! We will say thatF x; y/Dcis the general solution of (3.10).
The constantcdepends on the initial conditions IfF x; y/ D cis the general solution of (3.10), the solution curve passing throughP0 D x0; y0/is given by
46 3 First order nonlinear differential equations
Notice that at any point.x; y/2such thatN.x; y/¤0, equation (3.8) can be put in the normal form dy dx D M.x; y/
Similarly, at any point.x; y/2such thatM.x; y/¤0, equation (3.9) can be put in the normal form dx dy D N.x; y/
To these equations we can apply the existence and uniqueness Theorems discussed in the preceding Chapter.
Remark 3.2.3.At points.x ; y /such thatM.x ; y /D N.x ; y /D 0(3.10) is neither equivalent to (3.11) nor to (3.12) In Examples 3.2.4-3.2.5 below we illustrate some typical behavior of solutions near such points.
The simplest case of exact equations is whenM DM.x/andN DN.y/ In this case, one hasMy DNx D0 Notice that the corresponding equations dy dx D M.x/
M.x/ (M 6D0) are also separable equations An antiderivative of!DMdxCNdyis given by
Z N.x/dx; sinceFx DM.x/; FyDN.y/by the Fundamental Theorem of Calculus.
We now discuss some examples of exact equations, starting with the simple case
Example 3.2.4.(i) The equationxdxCydy D0is exact and an antiderivative of
!DxdxCydyisF x; y/D 1 2 x 2 C 1 2 y 2 Then the general solution is given by x 2 Cy 2 Dc:
Ifc > 0this is a family of circles centered at.0; 0/ Ifc D 0thenx 2 Cy 2 D c reduces to the point.0; 0/ Notice that hereM Dx,N Dyand they both vanish at 0; 0/.
(ii) The equationxdxydy D0is also exact and the general solution is given by x 2 y 2 Dc:
The family of hyperbolas can be represented by the equation \(x^2 - y^2 = 0\), which simplifies to the pair of straight lines \(y = \pm x\) It is important to note that the variables \(M\) and \(N\) correspond to \(x\) and \(y\) respectively, and both lines intersect at the origin point \((0, 0)\).
Example 3.2.5.Find the solutions of2xdxC3.1y 2 /dyD0passing through the points.0; 1/and.0;1/and discuss their behavior.
HereM D2xandN D3.1y 2 /and hence the equation is exact An antideriva- tive of!DMdxCNdyis
Thus the general solution is x 2 C3yy 3 Dc:
IfxD0; yD1we findcD2 Solvingx 2 Dy 3 3yC2with respect tox, we find xD ˙ y/D ˙p y 3 3yC2:
Sincey 3 3yC2D.yC2/.y1/ 2 then ˙ is defined fory 2and one has ˙ y/D ˙p
This makes it clear that ˙ y/have a cusp point atyD1, while lim y ! 2 C ˙ 0 y/D ˙1.
IfxD0; yD 1we findcD 2 Solvingx 2 Dy 3 3y2with respect tox, we find xD ˙ y/D ˙p y 3 3y2:
Sincey 3 3y2D.y2/.yC1/ 2 then ˙ is defined on the setạ1º [ ạy 2º and one has ˙.y/D ˙p
The point (0, ±1) is an isolated point on the graph defined by the equation x² + y³ + 3y = c, where c = ±2 Additionally, as y approaches 2 from the right, the limit approaches ±1 In the graph, both M and N equal zero at the coordinates (0, ±1).
Understanding exact equations is crucial in solving them effectively First, we need to learn how to identify an exact equation Once we confirm that an equation is exact, we can proceed to solve it using the appropriate methods The following theorem and its proof will guide us through this process.
Theorem 3.2.6 Assume thatM.x; y/andN.x; y/are continuous, with continuous partial derivatives with respect toxandy, onD.˛1; ˛2/.ˇ1; ˇ2/.
(i) If!DM.x; y/dxCN.x; y/dyis exact, thenMy.x; y/DNx.x; y/.
(ii) IfMy.x; y/DNx.x; y/, then!DM.x; y/dxCN.x; y/dyis exact.
Remark 3.2.7.The reader should be aware that in the previous theorem we assume thatM; N are defined in a rectangular region, only for simplicity In general, one
48 3 First order nonlinear differential equations
The plot of x²y³C₃yD² (in black) and x²y³C₃yD² (in red) illustrates that these functions can be applied to any domain R², provided that any closed continuous curve within this domain ensures that the enclosed set is entirely contained within it This condition is satisfied by any convex domain; however, it does not hold true for R² excluding the origin.
Proof of Theorem3.2.6 i /First let us assume that!is exact Then there exists a differentiable functionF x; y/such thatdF D! This means that ² Fx.x; y/ DM.x; y/;
Since the mixed second derivatives ofF are equal, that isFxy.x; y/D Fyx.x; y/, we deduce thatMy.x; y/DNx.x; y/.
We provide two methods for proving part (ii), which can also be used for solving exact equations in general.
(ii-1) Now, we assume thatMy.x; y/ D Nx.x; y/and seek a functionF x; y/ such thatFx.x; y/DM.x; y/andFy.x; y/DN.x; y/ Let
To establish the existence of a differentiable function \( h(y) \), we start with the equation \( F(x, y) = M(x, y) \), which already satisfies part of our requirements due to the Fundamental Theorem of Calculus, as \( F_x(x, y) = M(x, y) \) Our goal is to demonstrate that a function \( h(y) \) exists such that \( F_y(x, y) = N(x, y) \) This relationship holds true if and only if certain conditions are met.
This means that if we choose h.y/ to be any antiderivative of N.x; y/
R M.x; y/dx, thenF x; y/ will be the desired antiderivative of! and we are done But we can chooseh.y/in this manner only if we can show thatN.x; y/
The function R M.x; y/dx is dependent solely on y, as indicated by the absence of any terms involving x on the left side If it were otherwise, we would encounter h 0 y/, a function of y, alongside a two-variable function of x and y on the right side, which is illogical To demonstrate that the right side is indeed a function of y only, we will prove that its derivative with respect to x equals zero.
RM.x; y/dxhas continuous mixed partial derivatives, we have
In the above proof, we could have just as easily chosen
Z N.x; y/dyCh.x/ and determinedh.x/as we obtainedh.y/above.
(ii-2) Let.x0; y0/; x; y/be two arbitrary points in the rectangle Consider the path D.Œx0; x ạy0º/[.ạxº Œy0; y/, which is contained in, see Figure 3.4, and defineF x; y/by
The differential form can be integrated along a specified path, denoted as N.x; s/ds; (3.13) To demonstrate that F serves as an antiderivative of this form, we establish that Fx equals DM and Fy equals DN By applying the fundamental theorem of Calculus, we can derive the necessary results.
We may recall from Calculus, or show independently, by using the definition of the derivative and the Mean Value Theorem, that
50 3 First order nonlinear differential equations
Since, by assumption,Nx DMywe infer that
To prove thatFy DN, we notice that the first integral is a function ofxonly So,
In the above discussion, we could have also taken the path1 D.ạx0º Œy0; y/[ Œx0; x ạyº/yielding
Example 3.2.8.Solve.2xCy/dxC.xC2y/dy D0.
The equation is exact because
Using (3.13), withx0 Dy0 D0, we have
Therefore the general solution is given by x 2 CxyCy 2 Dc:
The equation \( x^2 + Cy^2 = 1 \) implies that \( x^2 + CxyCy^2 = 0 \) for all \( x \) and \( y \), leading to the conclusion that \( C = 0 \) When \( C > 0 \), the result represents a family of ellipses centered at the origin To better understand this, we can simplify the equation by applying a change of coordinates, setting \( x = Du + Cv \) and \( y = u v \).
In the u-v plane, the equation \( u^2 + v^2 + 2uv = c \) simplifies to \( 3u^2 + v^2 = c \) This indicates that when \( c > 0 \), the equation \( x^2 + xy + y^2 = c \) represents a family of ellipses centered at the origin, as illustrated in Figure 3.5.
Ifc D 0the ellipse degenerates to the point.0; 0/, the only point whereM D 2xCyandN DxC2yvanish Ifc < 0the equationx 2 CxyCy 2 D chas no real solution.
HereM.x; y/D2xyandN.x; y/Dx 2 Cy 2 SinceMyD2xDNx, the equation is exact We give four solutions, using the four methods discussed above. v u x y
52 3 First order nonlinear differential equations
To solve the differential equation Fx D 2xy D M(x, y), we need to determine h(y) such that Fy D x² + y² = N(x, y) Given that F(x, y) = x²y + h(y), we find that x² + h'(y) = x² + y², leading to h'(y) = y² Integrating, we obtain h(y) = (1/3)y³ + k Consequently, the solution to the differential equation is F(x, y) = x²y + (1/3)y³.
In the equation \( h(y) = \frac{1}{3} y^3 + k \), we set \( k = 0 \) for simplicity If \( k \) were not zero, the equation would be expressed as \( F(x, y) = x^2 y + \frac{1}{3} y^3 + k = c \), where \( c \) remains an arbitrary constant This means we could rename \( c \) as \( l \), leading to the equation \( x^2 y + \frac{1}{3} y^3 = l \), which essentially modifies the constant's label without altering the overall relationship.
To determine \( h(x) \) such that \( F(x, y) = 2xy = M(x, y) \), we start with \( F(x, y) = x^2 y + \frac{1}{3} y^3 + h(x) \) This leads to the equation \( 2xy + h'(x) = 2xy \), resulting in \( h(x) = k \) For simplicity, we can choose \( k = 0 \), which gives us \( F(x, y) = x^2 y + \frac{1}{3} y^3 \) Consequently, the general solution is \( x^2 y + 1 \).
Method 3.We now use the method whereF x; y/is given by
The integrating factor
In this section we learn how to deal with equation
When dealing with differential equations, an equation may initially appear non-exact, such as M(x, y)dx + N(x, y)dy = 0 However, it can become exact by multiplying it with an appropriate function For instance, the equation ydx + xdy = 0, defined for x > 0 and y > 0, is not exact at first Yet, by multiplying it by the function y^(1/2), the modified equation y^(1/2)dx + xy^(2)dy = 0 becomes exact.
An integrating factor for a nonzero function \( f(x, y) \) is defined as a function that, when multiplied by \( f(x, y) \), transforms it into an exact equation While integrating factors generally exist, identifying them can often be challenging However, in certain special cases, finding an integrating factor may be straightforward and worth pursuing It is important to note that, as demonstrated in the following example, an integrating factor is not necessarily unique.
Example 3.3.1.The reader should verify that forx; y > 0, all of the three functions
1 xy; 1 x 2 ; 1 y 2 ; are integrating factors ofy dxx dyD0:
One of the cases where finding an integrating factor can be quite simple is when the equation
The equation M.x; y/ dx + N.x; y/ dy = 0 possesses an integrating factor that depends solely on x or y Assuming it has an integrating factor that is a function of x only, we can multiply the entire equation by this function to facilitate the integration process.
In order for this equation to be exact, we must have (notice that @.x/@yD0)
If‰is a function ofxonly, then integrating 0 x/D‰.x/.x/, we obtain
If‰ is not a function ofxonly, then we may try to find an integrating factor.y/ which is a function ofyonly.
Multiplying the differential equation by.y/and following the same procedure, we obtain the equation
If‰is a function ofyonly, then integrating 0 y/D‰.y/, we obtain
Example 3.3.2.Find an integrating factor for the equation xsiny dxC.xC1/cosy dyD0:
To determine the existence of an integrating factor, we can refer to the formula (3.15) However, it is advisable for students to understand the derivation process rather than memorizing the formula By applying this method and multiplying by the integrating factor, we can proceed with our calculations effectively.
.x/xsiny dxC.x/.xC1/cosy dyD0;
Dividing by cosy(assuming cosy ¤0), we have x.x/D.xC1/ 0 x/C.x/
56 3 First order nonlinear differential equations and hence
Multiplying the given equationxsiny dxC.xC1/cosy dyD0by this.x/we have e x x
The equation can be applied to both the half-space where x > 1 and x < 1 Given that M(x, y) = (x e + x^2) sin(y), setting x0 = y0 = 0 results in M(x, y0) = 0 for all x Consequently, this leads to the conclusion that the integral from x0 to x of M(s, y0) ds equals zero.
Thus the general solution is e x
Notice that for c D 0 the solutions are straight lines given byy D k,k D 0;˙1;˙2; : : :.
We could have found these constant solutions without solving the equation, sim- ply by observing that dy dx D xsiny
.xC1/cosy: For example, it is easy to see thaty.x/is a solution sincey 0 D0and also xsin xC1/cos D0:
The equation yCxyCy 2 /dxC.x/.xC2y/dyD0 is not exact due to the condition MyD1CxC2y while Nx D1 To address this, we will search for an integrating factor, denoted as x(.), and equate the partial derivatives to simplify the problem.
.x/.1CxC2y/D 0 x/.xC2y/C.x/ and hence.xC2y/.x/D.xC2y/ 0 x/ Therefore 0 x/D.x/and so we can take.x/De x Now!De x yCxyCy 2 /dxCe x xC2y/dyis exact HereD
R 2 and we can use Method 3 to find an antiderivative Sincee x yCxyCy 2 /D0 foryD0, one has
Z y 0 e x xC2s/dsDe x xyCy 2 / and hence the general solution ise x xyCy 2 /Dc.
Example 3.3.4.Consider the equationydxC.2xC3y/dy D0 SinceMy D16D
The equation Nx D 2 is not exact, prompting the search for an integrating factor of the form y The equation yydx + y(2x + 3y)dy = 0 becomes exact when y is applied as the integrating factor This leads to the expression y dy/dy + y = D2(y) H, which simplifies to y = Dy Consequently, an antiderivative of y^2 dx + y(2x + 3y) dy = 0 can be derived.
Z y 0 s.2xC3s/dsDxy 2 Cy 3 and hencexy 2 Cy 3 D cis the general solution of our equation Ifc D 0we find y D 0andy D x Ifc > 0, theny 2 xCy/D c > 0impliesy >x, while if c < 0, theny 2 xCy/Dc < 0impliesy 0 c>0 c