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FIRST ORDER DIFFERENTIAL EQUATIONS
MIKE WILLS
Contents
- Introduction 1
- Alternative Texts 2
- Calculus 3 Notation and Facts 2
- Picard’s Existence and Uniqueness Theorem 4
- Separable Differential Equations 6
- Autonomous Equations 7
- Qualitative Analysis- Isoclines 8
- Linear Homogeneous Equations 9
- The Superposition Principle and Inhomogeneous Equations 9
- Method of Undetermined Coefficients 10
- Integrating Factor 11
- Variation of Parameters 12
- Other First Order Equations 13
- Exact Equations 14
- Introduction As your correspondent was writing up lecture notes and selecting exercises for chapter 2, he realized that the presentation given in the text was not to his liking. The purpose of this handout is to give a summary of the key ideas in what he hopes is a somewhat more orderly manner. Before we proceed, we should make something clear: for most math classes below the 3000 level at this university, the textbooks are selected by the department. As a professor, your correspondent can choose any text that he pleases provided the said text covers the material that is needed for the course; however, it is usually best to go with the department recommended text. As such, our textbook was selected passively. Having gone through the first two chapters of the text, your correspondent can safely say that he is not impressed. The book uses a lot to say a little and the order of topics in chapter 2 is, in this writer’s view, confusing. More importantly, it is difficult to identify where the book placed several of the key algorithms. This is a problem in a introductory analytically oriented differential equation course such as ours, where cookbook recipes are de rigeur. Hence, the purpose of the handout is to present the core material from chapter 2 in an orderly, logically consistent manner. The examples in the text are reasonable, so the presentation below will have relatively few examples. Also, the goal is to present the material in 1
2 MIKE WILLS
an elementary manner. As such, mathematical rigor is given relatively short shrift. Your instructor hopes that you find the exposition helpful, orderly, and elegant. Since not every one has taken vector calculus, some concepts from vector calculus which we will need from time to time are also discussed.
- Alternative Texts Here are some alternative texts that the readers may find useful. The last time your professor taught DEs it was out of Blanchard/Devaney/Hall (BDH). The isbn is 0495012653. Your professor likes the book a lot more than Boyce/Diprima and now wishes he’d used this book, despite its hefty price tag, and even though supplemental material would be needed. It should be pointed out that Amazon reviews for both our book and BDH are similar. Another book that was considered is by Zill whose isbn is 0534418783. Some people swear by the book, but it too is expensive and has decidedly mixed reviews. As a graduate student teacher at UCSB, your correspondent used Differential Equations and Linear Algebra by Farlow, Hall, McDill, and West. The isbn is
- Unfortunately, it’s not suitable for our course which is a pure d.e. course. Also, it seems to be especially unpopular with reviewers. For the reader with a background in linear algebra, this would be a good reference. For a more advanced perspective, Hirsch and Smale is a very good text. The isbn is 0123495504. The text assumes that the reader has had linear algebra and calculus 3, and perhaps even some real analysis. The text by Perko (isbn 0387974431) is also pretty good but it is arguably more difficult than Hirsch and Smale. Ultimately, the best supplemental text for most readers would probably be Schaum’s outline whose isbn is 0071456872. Notice that it is cheap. Regardless of which differential equations texts one uses, one would be well- advised to have one’s calculus book on-hand for the entire course.
- Calculus 3 Notation and Facts Differential equations involve, to a certain, extent, the calculus of functions of more than one variable. For the most part, this is not a big deal since the fine details of vector calculus are not required for our course. This section summarizes the key ideas that we will need.
Notation 3.1. The set R will for this handout and for our course be a rectangle in the plane. Specifically,
(3.1) R = {(t, y) ∈ R^2 | a ≤ t ≤ b, c ≤ y ≤ d} = [a, b] × [c, d].
To avoid silly irritations, we will assume that a < t < b and that c < y < d. The interior of R is
(3.2) R◦^ = {(t, y) ∈ R^2 | a < t < b, c < y < d} = (a, b) × (c, d).
We will also assume that the point (t 0 , y 0 ) is in R◦.
We will often have cause to deal with real-valued functions of more than one independent variable. Specifically, functions whose domain is the rectangle R and whose codomain is R.
4 MIKE WILLS
Theorem 3.8 (Chain Rule). Suppose t = φ(τ ) and y = ψ(τ ). Suppose that the partial derivatives of f : R → R exist and that φ and ψ are differentiable with respect to τ. Then f is differentiable with respect to τ and
(3.8) df dτ
= ft φ˙ + fy ψ˙
where in this case the dot derivative is with respect to τ. In particular, if t = τ , we have
(3.9)
df dτ
= ft + fy ψ.˙
If f is continuous, we can talk about its partial integrals. One can define partial integrals in terms of Riemann sums, much as we did in calculus. However, for our purposes, all we need to do is integrate f with respect to one of the variables, pretending that the other variable is constant. There is a slight twist which we illustrate in the following example.
Example 3.9. Let f (t, y) = t^3 y^2 + y. We compute the partial integrals
∫ f dt^ and f dy.
∫ f dt =
(t^3 y^2 + y)dt =
t^4 y^2 4
(t^3 y^2 + y)dy = t^3 y^3 3
y^2 2
where φ is an arbitrary function of y and ψ is an arbitrary function of t.
The point is that when (partially) integrating functions with more than one independent variable, the arbitrary constant so beloved from calculus becomes an arbitrary function of the variables that are not involved in the integration.
- Picard’s Existence and Uniqueness Theorem The general first order ordinary differential equation (ODE) is an equation of the form
(4.1) F (t, y, y˙) = 0.
Notice that F can be thought of as a function of three variables. The primary objective is to find a function, y = φ(t) that solves this equation. Ideally, φ will be differentiable, and typically we hope that φ˙ is continuous as well. Sometimes this is possible and sometimes it is not. Equation (4.1) is a bit too general for our purposes. The generic equation that we will work with is
(4.2) y˙ = f (t, y),
where we will typically assume that f is continuous on some rectangle, R = [a, b] × [c, d]. The corresponding initial value problem is y˙ = f (t, y) y(t 0 ) = y 0 ,
where the point (t 0 , y 0 ) is in the interior of R.
FIRST ORDER DIFFERENTIAL EQUATIONS 5
A primary reason to study differential equations is because they model real world phenomena. The only justification for such models is that they are expected to work^1. This means that the models are soluble in some sense, and make accurate predictions. In our context, this translates to requiring (at the very least) that the initial value problem (4.3) have a unique solution. There is a general theorem which we now state that gives sufficient conditions to guarantee this happy ending.
Theorem 4.1 (Picard’s Existence and Uniqueness Theorem). Consider the initial value problem (4.3). If f and fy are continuous^2 in R then there exists some interval I = [t 0 − h, t 0 + h] ⊂ [a, b] and a function y = φ(t) such that φ solves (4.3) uniquely on the interval I.
Proof. The proof of the result is sketched in the extra credit assignment from section 2.8. The basic idea is to define
(4.4) y 1 (t) = y 0 +
∫ (^) t
t 0
f (s, y 0 )ds
and inductively
(4.5) yn+1(t) = yn(t) +
∫ (^) t
t 0
f (s, yn(s))ds.
Then one shows that the sequence of functions {yn(t)} converges near t 0 to a solution of (4.3). Finally one shows that the difference between any two solutions of (4.3) is zero, thus guaranteeing that there is really only one solution. There are several potential pitfalls which have to be dealt with when proving the result. The full details require a good deal of real analysis. Some of the pitfalls are addressed in the text.
Remark 4.2. To guarantee existence to a solution of the initial value problem (4.3), Peano showed that one only needs f to be continuous. To guarantee uniqueness one needs a condition called Lipschitz continuity which is defined in the exercises.
Remark 4.3. Picard’s theorem can be extended to first order systems of ODEs. Since any ODE can be converted to a system of ODEs (as illustrated in the following example) it then follows that we have a general existence and uniqueness theorem for ODEs. Similar results for partial differential equations (PDEs) require significant strengthening of the hypotheses, and thus it is reasonable to state that there is no general existence theorem for PDEs.
Example 4.4. Convert the second order ode
(4.6) ¨y + y = 0
into a first order system.
Solution 4.5. Let v = ˙y. Then
(4.7) v˙ = ¨y = −y.
(^1) This is a paraphrase of something John von Neumann (a demigod to mathematicians) said. (^2) It is of passing interest that the continuity of fy does not guarantee the continuity of f.
FIRST ORDER DIFFERENTIAL EQUATIONS 7
Since g(y) is continuous, h(y) := (^) g(^1 y) is continuous away from the zeros of g.
Since f (t) and h(y) are continuous, there exist functions F (t) and H(y) such that F˙ = f and dH dy =^ h. We compute: y˙ = f (t)g(y) dy g(y)
= f (t)dt
h(y)dy = f (t)dt ∫ h(y)dy =
f (t)dt
H(y) = F (t) + C,
where C is an arbitrary constant. Given an initial condition y(t 0 ) = y 0 , the initial value problem y˙ = f (t)g(y) y(t 0 ) = y 0
has, by Picard’s theorem, a unique solution provided dgdy is continuous. In this situation, if y 0 is a zero of g the solution is y(t) = y 0. If y 0 is not a zero of g, then H is invertible near y 0 , C = H(y 0 ) − F (t 0 ), and
(5.8) y(t) = H−^1
F (t) + H(y 0 ) − F (t 0 )
This algorithm (called separation of variables) always works in principal but there are a few pitfalls.
- It is easy to forget to check for the constant solutions. Further, it may be impossible to find the roots of g explicitly.
- It may be difficult or impossible to compute H and/or F explicitly.
- The last equation in (5.6) is usually valid only for the non-constant solutions. Hence, the general solution to ˙y = f (t)g(y) is the last equation in (5.6) together with the equilibria.
- Even if one has computed H and F explicitly, finding H−^1 may not be possible.
- Separation of variables may work even if gy is not continuous, as long as g is continuous. However, one will likely lose uniqueness of solutions. A canonical example is given in example 3 on page 71 in the text.
- Autonomous Equations A special case of a separable first order ODE is an autonomous first order ODE.
Definition 6.1. A first order autonomous ODE is an ODE that can be written in the form
(6.1) y˙ = g(y).
(Hence, f (t) = 1, and, using the notation of the previous section, F (t) = t.)
When dgdy is continuous, the solution to the initial value problem y˙ = g(y) y(t 0 ) = y 0
8 MIKE WILLS
is y(t) = y 0 if g(y 0 ) = 0. If g(y 0 ) 6 = 0, the solution is
(6.3) y(t) = H−^1
t + H(y 0 ) − t 0
where the notation from the previous section is in use.
- Qualitative Analysis- Isoclines In principal, separable first order ODEs are about the easiest class of DEs to solve analytically. However, because of the difficulties inherent in integration, the practice is often rather different. Even when analytic methods work, it is often helpful to draw pretty pictures to help get a feel for how solutions behave. The simplest qualitative tool is the direction field. Drawing direction fields by hand can be tiresome. However, there are some techniques which cut down on the tedium. Consider the first order ODE
(7.1) y˙ = f (t, y).
Definition 7.1. Let c ∈ R. The set Lc = {(t, y) | f (t, y) = c} is called a level set (or level curve) of f. In the context of qualitative analysis of ODEs, Lc is also known as an isocline.
Typically, Lc will be a smooth curve or union of curves in the ty plane. The key point is that all the slope lines in the direction field that lie on Lc will have the same slope, namely c. Hence, sketching a direction field reduces to sketching level curves and then on each level curve, Lc draw a series of slope lines each with slope c. This is much easier than working out the slope at a bunch of points in a grid. The only downside is that it adds an extra layer of information which may confuse matters.
Example 7.2. If f (t, y) = y−t^2 then the isoclines are parabolas given by y = c+t^2. Each parabola is centered on the y-axis and has vertex (0, c). Hence, the slope lines on a given parabola have slope c.
Example 7.3. If f (t, y) is a constant function with f (t, y) = a, then there is only one level set and it is the entire plane. All the slope lines have the same slope, namely a.
Example 7.4. If f (t, y) = g(y) (the ODE is autonomous) the isoclines are hori- zontal lines, provided g is not constant.
Example 7.5. If f (t, y) = f (t) (the ODE is directly integrable) the isoclines are vertical lines, provided f is not constant.
Example 7.6. If f (t, y) = ty, then the t and y axes are the isoclines for c = 0. The other isoclines are hyperboli of the form y = ct where c 6 = 0.
We sketch these examples at the end of the handout. Notice in particular, that sketching direction fields for autonomous equations is relatively easy.
10 MIKE WILLS
Theorem 9.1. Suppose yp is a particular solution of (9.1). Then y = yh + yp is the general solution of (9.1).
Proof. We are given that L(yp) = b(t) and we know that L(yh) = 0. Hence,
(9.3) L(yh + yp) = L(yh) + L(yp) = 0 + b(t).
Thus, yh + yp solves (9.1). Conversely, if L(y 1 ) = b(t), (that is y 1 is any particular solution) then
(9.4) L(y 1 − yp) = L(y 1 ) − L(yp) = b(t) − b(t) = 0
and so y 1 − yp solves the homogeneous equation L(y) = 0. Consequently, there exists a solution yk of the homogeneous equation such that y 1 = yk + yp.
The proof shows first that yh + yp is a solution of (9.1) and then shows that any solution of (9.1) can be written in the form yk + yp for some yk a specific solution to L(y) = 0. This is exactly what we needed to show. This result is sometimes referred to as the extended superposition principle (or extended linearity principle.) Since computing yh is straightforward, we now search for ways of finding partic- ular solutions to (9.1).
- Method of Undetermined Coefficients
Example 10.1. If ˙y + ay = b where a 6 = 0 and b are constants, then yp = ba is a particular solution. Hence, the general solution is
(10.1) y = Ce−at^ +
b a
Notice that C = y(0) − ba.
This was essentially a lucky guess. If a is constant, and b is a nice function, lucky guesses like this can often work. The fancy name for the lucky guesses the ‘method of undetermined coefficients’. Here is the basic idea. Consider the equation
(10.2) y˙ + ay = b(t).
Here a is constant and b is continuous. We guess that yp looks similar to b(t), perhaps differing by some constants which we can determine by plugging our guess into the ODE. The process is best illustrated with examples.
Example 10.2. Suppose ˙y + y = e^2 t. Here b(t) = e^2 t. We guess that yp = me^2 t. Now y˙p = 2me^2 t. Plugging yp into the left hand side of the ODE, we get
(10.3) y˙p + yp = 2me^2 t^ + me^2 t^ = 3me^2 t.
Hence, yp solves the ODE provided 3m = 1; that is, m = 13.
Hence, the general solution is y = Ce−t^ + e
2 t
Example 10.3. Suppose that ˙y − 2 y = t^2. The right hand side is a polynomial of degree 2. We therefore guess that yp = mt^2 + nt + k where m, n, and k are to be determined. Now y˙p = 2mt + n. Plugging yp into the left hand side of the ODE, we get
(10.4) y˙p − 2 yp = 2mt + n − 2 mt^2 − nt − k = − 2 mt^2 + (2m − n)t − k.
FIRST ORDER DIFFERENTIAL EQUATIONS 11
Hence, if yp is to solve the ODE, we find by equating coefficients that − 2 m = 1,
2 m − n = 0, and −k = 0. Hence, m = − 12 , n = −1, and k = 0. Thus, yp = − t
2 2 −^ t solves the ODE. The general solution is
(10.5) y = Ce^2 t^ −
t^2 2
− t.
Here is a summary of what will often work. Throughout we assume that a is constant. (i) If b(t) is a polynomial of degree n, Take yp to be a polynomial of degree n with coefficients to be determined. (ii) If b(t) = lekt, try yp = mekt^ where m is to be determined. This will not work if a = −k. (iii) If b(t) = k 1 sin(lt) + k 2 cos(lt), try yp = m 1 sin(lt) + m 2 cos(lt) where m 1 and m 2 are to be determined. (iv) If b(t) = lekt^ + t, try yp = m 1 ekt^ + m 2 t + m 3 where m 1 , m 2 , and m 3 are to be determined. The interested reader will no doubt find further generalizations. The method breaks down if a is not constant or if one’s guess for yp turns out to be a solution to the homogeneous equation. There are various ways of dealing with this, but the bottom line is that the method is simply not that systematic. This is probably why the text barely gives the method a mention until chapter 3 in the context of second order equations. However, this method has the advantage of being fairly simple for sufficiently nice b. The more systematic methods are not as simple.
- Integrating Factor Once again, we look at the ODE
(11.1) y˙ + a(t)y = b(t)
where a and b are continuous. Let μ(t) be a non-zero continuous function. Multi- plying the ODE by μ, we obtain
(11.2) μ y˙ + μay = μb,
where we have suppressed the arguments of the functions a, b, and μ since they are starting to irritate us. Suppose that we can choose μ in such a way that
(11.3) (μy)′^ = μ y˙ + μay.
By the product rule, this currently hypothetical equation becomes
(11.4) μy˙ + μ y˙ = μ y˙ + μay.
Then
(11.5) μy˙ = μay
and so
(11.6) μ˙ = μa
away from y = 0. We now have a linear homogenous ODE in μ. If A is an antiderivative of a, then μ(t) = eA(t)^ solves our ODE. In other words, we have found our candidate μ. Having done so, we now note that
(11.7) (μy)′^ = μ y˙ + μay = μb.
FIRST ORDER DIFFERENTIAL EQUATIONS 13
However, let’s assume that it does and solve for v. We have:
y˙ + a(t)y = b(t) d dt
v(t)e−
a(t)dt
a(t)dt (^) = b(t)
ve˙ −
a(t)dt (^) + v d dt
e−
a(t)dt
a(t)dt (^) = b(t)
ve˙−
a(t)dt (^) − ve−
a(t)dt d dt
a(t)dt + ave−
a(t)dt (^) = b(t)
ve˙−
a(t)dt (^) − ve−
a(t)dta + ave−
a(t)dt (^) = b(t)
ve˙ −
a(t)dt (^) = b(t)
v˙ = b(t)e
a(t)dt
v =
b(t)e
a(t)dtdt.
Since we are searching for a particular solution, we take the constants of in-
tegration to be zero. We have just shown that when v =
b(t)e
a(t)dtdt, then
yp = ve−
a(t)dt (^) is a particular solution of ˙y + a(t)y = b(t). The general solution is
(12.3) y = (C + v)e−
a(t)dt.
It is the fifth line in the above calculation where the miracle takes place. We get a nice calculation which allows us to compute v explicitly, albeit in terms of integrals which as usual may be impossible to compute. When employing variation of parameters for a specific ODE, we usually repeat the above process rather than memorize the formula for v at the end. Some of the exercises in 2.1 deal with this method. Variation of parameters is probably the hardest of the three methods to im- plement, but it is systematic (unlike undetermined coefficients) and generalizable (unlike integrating factor). As such, variation of parameters is arguably the best method for linear equations. The text discusses variation of parameters for second order equations in chapter 3.
- Other First Order Equations Most first order ODEs are neither separable nor linear. Hence, none of the above methods are applicable to the general situation. There are a variety of techniques which allow certain classes of ODEs to be converted to linear or separable ODEs. Some of these are discussed in the exercises in section 2.4. Most of these techniques are highly specialized, do not generalize, and are quite frankly ad hoc. One class of equations that we consider in detail in this class is the collection of exact equations. However, for most first order ODES, the canonical way to arrive at solutions is via numerical techniques. Euler’s method is a good start to this method. Your correspondent is reasonably happy with the presentation of Euler’s method in the text, so our last section in this handout will be on exact equations.
14 MIKE WILLS
- Exact Equations
Definition 14.1. Let R be a rectangle in R^2 and suppose that M, N : R → R^2 are continuous. Suppose that there exists a function ψ : R → R^2 such that (i): ψ(t, y) = c implicitly defines y as a function of t for some (arbitrary) constant c (ii) ∂ψ∂t = M (iii) ∂ψ∂y = N Then the first order ODE
(14.1) M + N y˙ = 0
is said to be exact.
Let τ = t so we can think of y, and hence ψ as functions of τ as well. Let us differentiate the equation ψ = c with respect to τ. By the chain rule (in the case where t = τ )
ψ(t, y(t)) = c ∂ ∂τ
ψ(t, y(t)) = 0 ∂ψ ∂t
∂ψ ∂y y˙ = 0
M + N y˙ = 0,
where the dot derivative is with respect to τ. Thus, the equation ψ = c implicitly defines a solution to the exact equation. If My and Ny are both continuous, then by Picard’s theorem, the solution will be unique for a given initial condition. It is not immediately clear how to check whether such a ψ exists. Luckily, there is a theorem (2.6.1 in the text) which will help. We restate it here for completeness but the text does a good job of stating and proving the result.
Theorem 14.2. If M, N, My , and Nt are all continuous on R◦^ then M + N y˙ = 0 is exact if and only if My = Nt. In this situation,
(14.3) ψ =
M dt =
N dy.
Thus, a straightforward way to check if our candidate ODE is exact if simply to compute
M dt and
N dy. If one can make these two partial integrals equal, the ODE is exact, and we have found an implicit solution. The proof given in the text can also be mimicked for specific situations.
