1. Introduction to Differential Equations
Def: Equations in which the unknown function or the vector function appears under the sign of
the derivative or the differential are called differential equations.
The following are some examples of differential equations:
𝑑𝑥
(1.1)
= −𝑘𝑥
𝑑𝑡
is the equation of radioactive disintegration (𝑘 is the disintegration constant, 𝑥 is the quantity of
𝑑𝑥
undisintegrated substance at time 𝑡, and
is the rate of decay proportional to the quantity of
𝑑𝑡
disintegrating substance).
(1.2)
𝑑2𝑟
𝑑𝑟
𝑚 2 = 𝐹(𝑡, 𝑟, )
𝑑𝑡
𝑑𝑡
is the equation of motion of a particle of mass m under the influence of a force F dependent on
the time; the position of the particle (which is determined by the radius vector r), and its velocity
𝑑𝑟
: The force is·equal to the product of the mass by the acceleration.
𝑑𝑡
(1.3)
𝜕2𝑢 𝜕2𝑢 𝜕2𝑢
+
+
=
4𝜋𝜌(𝑥,
𝑦,
𝑧)
𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2
is Poisson's equation, which for example is satisfied by the potential u(x, y, z) of an electrostatic
field, 𝜌(𝑥, 𝑦, 𝑧) is the charge density.
Def: If an equation involves the derivative of one variable with respect to another, then the former
is called a dependent variable and the latter an independent variable.
Thus, in the equation (1.4)
(1.4)
𝑑2𝑥
𝑑𝑥
+
𝑎
+
𝑘𝑥
=
0,
𝑑𝑡 2
𝑑𝑡
t is the independent variable and x is the dependent variable. We refer to a and k as coefficients
in equation (1.4).
Def: If in a differential equation the unknown functions or the vector functions (dependent
variables) are functions of one independent variable, then the differential equation is called
ordinary (for example, (1.1) or (1.2)). But if the unknown function appearing in the differential
equation is a function of two or more independent variables, the differential equation is called a
partial differential equation ((1.3) is an instance).
Def: The order of a differential equation is the highest order of the derivative (or differential) of
the unknown function.
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Def: A solution of a differential equation is a function which, when substituted into the differential
equation, reduces it to an identity.
To illustrate, the equation of radioactive disintegration (1.1)
𝑑𝑥
= −𝑘𝑥
𝑑𝑡
has the solution
𝑥 = 𝐶𝑒 −𝑘𝑡 ,
where 𝐶 is an arbitrary constant.
It is obvious that the solution of the differential equation (1.1) does not yet fully determine the
law of disintegration x = x(t). For a full determination, one must know the quantity of
disintegrating substance 𝑥0 at some initial instant of time 𝑡0 . If 𝑥0 is known, then, taking into
account the condition 𝑥(𝑡0 ) = 𝑥0 from the solution of the differential equation (1.1), we find the
law of radioactive disintegration:
𝑥 = 𝑥0 𝑒 −𝑘(𝑡−𝑡0) .
In the example above the condition 𝑥(𝑡0 ) = 𝑥0 is called initial condition. The number of initial
conditions that are required for a given differential equation will depend upon the order of the
differential equation.
Def: Initial Value Problem (or IVP) is a differential equation along with an appropriate number
of initial conditions.
Def: The procedure of finding the solutions of a differential equation is called integration of the
differential equation.
The notation for differentiation varies depending upon the author and upon which notation is most
useful for the task at hand. For example,
 The Leibniz's notation
𝑑𝑦 𝑑 2 𝑦
𝑑𝑛 𝑦
,
,…, 𝑛
𝑑𝑥 𝑑𝑥 2
𝑑𝑥
is more useful for differentiation and integration;
 Lagrange's notation
𝑦 ′ , 𝑦 ′′ , … , 𝑦 (𝑛)
is more useful for representing derivatives of any order compactly;
 Newton's notation
𝑦̇ , 𝑦̈ , 𝑦⃛
is often used in physics for representing derivatives of low order with respect to time.
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2. Classification of Differential Equations
There are many types of differential equations, and we classify them into different categories
based on their properties. Let us quickly go over the most basic classification. We already saw
the distinction between ordinary and partial differential equations.
Def: If there are several equations working together, we have a so-called system of differential
equations.
For example,
𝑦 ′ = 𝑥, 𝑥 ′ = 𝑦
is a simple system of ordinary differential equations.
Def: An equation is linear if the dependent variable (or variables) and their derivatives appear
linearly. Otherwise, the equation is called nonlinear. For example, an ordinary differential
equation is linear if it can be put into the form:
𝑑𝑛 𝑦
𝑑 𝑛−1 𝑦
𝑑𝑦
𝑎𝑛 (𝑥) 𝑛 + 𝑎𝑛−1 (𝑥) 𝑛−1 + ⋯ + 𝑎1 (𝑥)
+ 𝑎0 (𝑥)𝑦 = 𝑏(𝑥)
𝑑𝑥
𝑑𝑥
𝑑𝑥
The linear equation is allowed to depend arbitrarily on the independent variable. So
𝑑2𝑦
𝑑𝑦
1
𝑥
2
𝑒
+
𝑠𝑖𝑛𝑥
+
𝑥
𝑦
=
𝑑𝑥 2
𝑑𝑥
𝑥
is still a linear equation as y and its derivatives only appear linearly.
(2.1)
Let us see some nonlinear equations. For example, Burger's equation,
𝜕𝑦
𝜕𝑦
𝜕2𝑦
+ 𝑦 𝜕 = 𝜈 2,
𝜕𝑡
𝜕𝑥
𝜕𝑥
𝜕𝑦
is a nonlinear second order partial differential equation. It is nonlinear because y and
are
𝜕𝑥
multiplied together.
The equation
𝑑𝑥
= 𝑥2
𝑑𝑡
is a nonlinear first order differential equation as there is a second power of the dependent variable.
Def: A linear equation may further be called homogeneous if all terms depend on the dependent
variable. Otherwise, the equation is called nonhomogeneous or inhomogeneous. A homogeneous
linear ODE can be put into the form:
𝑑𝑛 𝑦
𝑑 𝑛−1 𝑦
𝑑𝑦
𝑎𝑛 (𝑥) 𝑛 + 𝑎𝑛−1 (𝑥) 𝑛−1 + ⋯ + 𝑎1 (𝑥)
+ 𝑎0 (𝑥)𝑦 = 0
𝑑𝑥
𝑑𝑥
𝑑𝑥
(2.2)
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Def: If the coefficients of a linear equation are actually constant functions, then the equation is
said to have constant coefficients. A constant coefficient nonhomogeneous ODE is an equation
of the form
(2.3)
𝑑𝑛 𝑦
𝑑 𝑛−1 𝑦
𝑑𝑦
𝑎𝑛 𝑛 + 𝑎𝑛−1 𝑛−1 + ⋯ + 𝑎1
+ 𝑎0 𝑦 = 𝑏(𝑥)
𝑑𝑥
𝑑𝑥
𝑑𝑥
where 𝑎0 , 𝑎1 , 𝑎2 , … , 𝑎𝑛 are all constants, but b may depend on the independent variable 𝑥.
Def: Finally, an equation (or system) is called autonomous if the equation does not depend on the
independent variable.
For example, the simplest autonomous equation
𝑥̇ = 𝑥.
The examples in this section have mostly focused on the classification of ODEs. The
classification of PDEs is similar but more involved. PDEs can also be classified by linearity or
nonlinearity, order and constant or variable coefficients. More important is the classification that
identifies a PDE as hyperbolic, parabolic, or elliptic. PDEs will be discussed in upcoming
lectures.
3. Classification of methods for solving IVP (the Cauchy problem) for ordinary
differential equations
All methods for solving the initial value problem (the Cauchy problem) for ODEs are divided
into exact, approximate and numerical.
Def: Exact methods are studied in courses of differential equations and allow one to express the
solution of an equation in terms of elementary functions or using quadratures of elementary
functions. The class of problems that can be solved obtained by exact methods is relatively
narrow.
Def: When using approximate methods, the solution of the Cauchy problem for an ODE is defined
as the limit of some sequence of functions. Moreover, each member of the sequence is expressed
in terms of elementary functions or quadratures from elementary functions.
Def: Numerical methods for ordinary differential equation are methods used to find numerical
approximations to the solutions of ODEs. The desired solution is obtained in the form of a table.
Without loss of generality to higher-order systems, we restrict ourselves to first-order differential
equations, because a higher-order ODEs can be converted into a larger system of first-order
equations by introducing extra variables.
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4. First-order Differential Equation: main definitions, existence of the solution and its
uniqueness
Def: An ordinary first-order differential equation of the first degree may be represented as
follows:
(4.1)
𝐹(𝑥, 𝑦, 𝑦 ′ ) = 0
or, solving for the derivative:
𝑑𝑦
(4.2)
= 𝑓(𝑥, 𝑦)
𝑑𝑥
Def: The general solution of a first-order ODE is the function
𝑦 = 𝜑(𝑥, 𝐶)
or the relation
Φ(𝑥, 𝑦, 𝐶) = 0
which depends on a single arbitrary constant 𝐶
(4.3)
(4.4)
In the beginning of the lecture it was shown that for the equation (1.1)
𝑑𝑥
= −𝑘𝑥
𝑑𝑡
the general solution is
𝑥 = 𝐶𝑒 −𝑘𝑡
From the geometric viewpoint, the general solution is a family of curves in a coordinate plane,
which depends on a single arbitrary constant 𝐶. These curves are called integral curves of the
given differential equation. Thus, in the example above, the general solution is geometrically
depicted by a family of the exponential functions. Fig. 1 shows the curves of a family that
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associated with certain values of the constant: 𝐶 = 1, 2, 3, , ….
2
5
Fig. 1
Def: A particular solution is any functions
(4.5)
𝑦 = 𝜑(𝑥, 𝐶0 )
or relations
(4.6)
Φ(𝑥, 𝑦, 𝐶0 ) = 0
which is obtained from the general solution by assigning to the arbitrary constant 𝐶 a definite
value 𝐶 = 𝐶0 .
From geometric viewpoint, a particular solution is associated with one curve of the family that
passes through a certain given point of the plane (Fig. 2).
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Fig. 2
Def: In the case of a first-order equation, the initial value problem is of the form
𝑦 ′ = 𝑓(𝑥, 𝑦),
𝑦(𝑥0 ) = 𝑦0
(4.7)
Example 1: For the first-order differential equation
𝑑𝑦
𝑦
=−
𝑑𝑥
𝑥
𝐶
the general solution is 𝑦 = (Fig. 3a).
𝑥
2
A particular solution that will satisfy the following initial condition 𝑦(2) = 1 is 𝑦 = (Fig. 3b)
𝑥
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Fig. 3
In order to apply to (4.7) methods of approximate integration of a differential equation, it is first
necessary to be sure of the existence of the desired solution and also of the uniqueness of the
solution, because in the absence of uniqueness it will not be clear what solution is to be
determined.
4.1. A Lipschitz condition
At the beginning we are going to discuss the idea of a Lipschitz condition. Such a condition takes
the form of a fundamental inequality (4.8) for gaining information about the solution to (4.7),
such as existence, uniqueness and approximation.
We say 𝑓(𝑥, 𝑦) satisfies a Lipschitz condition on a set D if there is a constant 𝐿 ≥ 0 such that
|𝑓(𝑥, 𝑦1 ) − 𝑓(𝑥, 𝑦2 )| ≤ 𝐿|𝑦1 − 𝑦2 |
for all (𝑥, 𝑦1 ), (𝑥, 𝑦2 ) ∈ 𝐷.
(4.8)
Example 2: Let 𝐷 = 𝑅2 and let 𝑓(𝑥, 𝑦) = 𝑥 2 − 𝑦. For each (𝑥, 𝑦1 ), (𝑥, 𝑦2 ) ∈ 𝐷 consider
|𝑓(𝑥, 𝑦1 ) − 𝑓(𝑥, 𝑦2 )| = |(𝑥 2 + 𝑦1 ) − (𝑥 2 + 𝑦2 )| = 2|𝑦1 − 𝑦2 |
and so the given function 𝑓(𝑥, 𝑦) does satisfy a Lipschitz condition on D = R2 with L = 2.
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In opposite of the example 2, it is not usually that easy to verify inequality (4.8). Thus, the
following lemma is essentially what this chapter is centered around as the researchers would like
to have some practical conditions under which it is easy to tell when 𝑓(𝑥, 𝑦) satisfies a Lipschitz
condition. So, the following lemma provides some sufficient conditions that guarantee the
Lipschitz condition will be satisfy.
Lemma. Let D be the rectangle
𝑥0 − 𝑎 ≤ 𝑥 ≤ 𝑥0 + 𝑎,
If 𝑓(𝑥, 𝑦) and
𝜕𝑓
𝜕𝑦
𝑦0 − 𝑏 ≤ 𝑦 ≤ 𝑦0 + 𝑏.
are continuous functions in the rectangle D and there is some constant 𝐾 ≥ 0
such that
|
𝜕𝑓
| ≤ 𝐾 𝑓𝑜𝑟 𝑎𝑙𝑙 (𝑥, 𝑦) ∈ 𝐷
𝜕𝑦
(4.9)
then (4.8) holds with 𝐿 = 𝐾.
Proof. By the fundamental theorem of calculus, for all (𝑥, 𝑦1 ), (𝑥, 𝑦2 ) ∈ 𝐷 we have
𝑦2
𝑓(𝑥, 𝑦1 ) − 𝑓(𝑥, 𝑦2 ) = ∫
𝑦1
and thus
𝑦2
|𝑓(𝑥, 𝑦1 ) − 𝑓(𝑥, 𝑦2 )| = | ∫
𝑦1
■
𝜕𝑓
𝑑𝑦
𝜕𝑦
𝑦2
𝑦2
𝜕𝑓
𝜕𝑓
𝑑𝑦| ≤ | ∫ | | 𝑑𝑦| ≤ | ∫ 𝐾𝑑𝑦| = 𝐾|𝑦1 − 𝑦2 |.
𝜕𝑦
𝜕𝑦
𝑦1
𝑦1
Example 3: Let
𝑓(𝑥, 𝑦) = cos(𝑥) + 𝑦 3
and let D be the rectangle
0 ≤ 𝑥 ≤ 1,
−1 ≤ 𝑦 ≤ 1
Then for all (𝑥, 𝑦) ∈ 𝐷 we have
𝜕𝑓
| = |3𝑦 2 | ≤ 3
𝜕𝑦
are continuous functions in the rectangle D. Thus (4.9) holds with K = 3. By
|
and 𝑓(𝑥, 𝑦) and
𝜕𝑓
𝜕𝑦
lemma the given function 𝑓(𝑥, 𝑦) = cos(𝑥) + 𝑦 3 is a Lipschitz continuous with 𝐿 = 𝐾 = 3.
That is,
|(cos(𝑥) + 𝑦13 ) − (cos(𝑥) + 𝑦23 )| ≤ 3|𝑦1 − 𝑦2 |.
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The next part of the lecture will represent the Picard–Lindelöf theorem, Picard's existence
theorem, Cauchy–Lipschitz theorem, or existence and uniqueness theorem which gives a set
of conditions under which IVP (4.7) has a unique solution.
4.2. Picard–Lindelöf theorem
Picard–Lindelöf theorem
If in the IVP
𝑦 ′ = 𝑓(𝑥, 𝑦), 𝑦(𝑥0 ) = 𝑦0
the function f(x, y) is continuous in the rectangle D:
𝑥0 − 𝑎 ≤ 𝑥 ≤ 𝑥0 + 𝑎, 𝑦0 − 𝑏 ≤ 𝑦 ≤ 𝑦0 + 𝑏
and satisfies, in D, the Lipschitz condition
|𝑓(𝑥, 𝑦1 ) − 𝑓(𝑥, 𝑦2 )| ≤ 𝐿|𝑦1 − 𝑦2 |
then there exists a unique solution 𝑦 = 𝜑(𝑥) on some interval containing 𝑥0 .
The preceding theorem tells us two things. First, when an equation satisfies the hypotheses of
Picard–Lindelöf theorem, we are assured that a solution to the initial value problem exists.
Second, when the hypotheses are satisfied, there is a unique solution to the initial value problem.
Notice that the existence and uniqueness of the solution holds only in some neighborhood 𝑥0 −
𝐻 ≤ 𝑥 ≤ 𝑥0 + 𝐻 (Fig. 4).
Fig. 4
In view of the proven above lemma we can state the following special case of Picard–Lindelöf
theorem.
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Theorem 1.
If in the IVP
𝑦 ′ = 𝑓(𝑥, 𝑦),
the function f(x, y) and its partial derivative
𝜕𝑓
𝜕𝑦
𝑦(𝑥0 ) = 𝑦0
are continuous in the rectangle D:
𝑥0 − 𝑎 ≤ 𝑥 ≤ 𝑥0 + 𝑎, 𝑦0 − 𝑏 ≤ 𝑦 ≤ 𝑦0 + 𝑏
and there is some constant 𝐾 ≥ 0 such that
𝜕𝑓
| | ≤ 𝐾 𝑓𝑜𝑟 𝑎𝑙𝑙 (𝑥, 𝑦) ∈ 𝐷
𝜕𝑦
then the IVP has a unique solution 𝑦 = 𝜑(𝑥) on some interval containing 𝑥0 .
Note: In fact, the Theorem1 is seen in many textbooks on ordinary differential equation rather
than the Picard–Lindelöf theorem. The theorem 1 is a simpler thing to check.
Example 4: For the initial value problem
𝑑𝑦
= 𝑥 √𝑦
𝑦(1) = 4
𝑑𝑥
does Theorem 1 imply the existence of a unique solution?
Solution:
1) We identify 𝑓(𝑥, 𝑦) as 𝑥√𝑦:
𝑓(𝑥, 𝑦) = 𝑥√𝑦
′
and partial derivative 𝑓′𝑦 as (𝑥√𝑦) :
𝑦
𝜕𝑦
𝑥
=
𝜕𝑥 2√𝑦
Both of these functions are continuous in any rectangle containing the point (1, 4), so the
hypotheses of Theorem 1 are satisfied. It then follows from the Theorem 1 that the initial value
problem of this example has a unique solution in an interval about x = 1.
Example 5: In Example 4 suppose the initial condition is changed to (1, 0). Here. again, 𝑓(𝑥, 𝑦) =
𝜕𝑦
𝑥
𝜕𝑦
𝑥 √𝑦 and =
. Unfortunately is not continuous or even defined when y = 0. Consequently,
𝜕𝑥
2√𝑦
𝜕𝑥
𝜕𝑦
there is no rectangle containing (1, 0) in which both 𝑓(𝑥, 𝑦) and are continuous. Because the
𝜕𝑥
hypotheses of Theorem 1 do not hold, we cannot use Theorem 1 to determine whether the initial
value problem does or does not have a unique solution. It turns out that this initial value problem
has more than one solution.
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5. Well-posed problems
In situations where a physical process is described (modelled) by an initial value problem for
ODEs, then it is desirable that any errors made in the measurement of either initial data or the
vector field, do not influence the solution very much. In mathematical terms, this is known as
continuous dependence of solution of an IVP, on the data present in the problem. In fact, the
following Theorem 2 asserts that solution to an IVP has not only continuous dependence on initial
data but also on the vector field f (parameters).
Theorem 2 (on the continuous dependence of a solution on a parameter and on the initial
values)
If the right side of the differential equation
𝑑𝑦
(5.1)
= 𝑓(𝑥, 𝑦, 𝜇)
𝑑𝑥
is continuous with respect to 𝜇 for 𝜇0 ≤ 𝜇 ≤ 𝜇1 and satisfies the conditions of the existence and
uniqueness theorem, and the Lipschitz constant L does not depend on 𝜇, then the solution 𝜑(x, 𝜇)
of this equation that satisfies the condition 𝑦(𝑥0 ) = 𝑦0 depends continuously on 𝜇.
This theorem shows continuous dependence on parameters if, in addition to the hypotheses of the
Picard–Lindelöf theorem, the right-hand side of the equation (5.1) is assumed to be Lipschitz
continuous with respect to the parameter (on some interval 𝑥0 ≤ 𝑥 ≤ 𝑏). The connection between
(4.7) and (5.1) shows that the hypotheses of the Picard–Lindelöf theorem are sufficient to
guarantee continuous dependence on initial conditions. (Fig. 5).
Fig. 5
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Def: A solution that changes but slightly for an arbitrary but sufficiently small variation of the
initial values, given arbitrarily large values of the argument, is called stable. Stable solutions will
be examined in more detail in the upcoming lectures.
Def: A mathematical problem is said to be well-posed if it has the EUC property (Hadamard
criteria):
(1) Existence: The problem should have at least one solution.
(2) Uniqueness: The problem has at most one solution.
(3) Continuous dependence: The solution depends continuously on the data that are present in the
problem.
Def: An ill-posed problem is one which doesn’t meet the three Hadamard criteria for being well
posed.
Example 6: Consider the Cauchy problem (the initial value problem):
𝑦′ = 𝑦 − 𝑥,
𝑦(0) = 1, 𝑥 ∈ [0; 100].
The general solution of the equation is of the form:
𝑦(𝑥, 𝐶) = 1 + 𝑥 + 𝐶𝑒 𝑥
and contains one arbitrary constant.
From initial conditions we find that
𝐶 = 0 𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 𝑦(100) = 101.
Let’s slightly change the initial condition. For example, let it be:
𝑦′ = 𝑦 − 𝑥,
𝑦(0) = 1.000001, 𝑥 ∈ [0; 100].
The arbitrary constant:
𝐶 = 10−6 (change slightly)
However
𝑦(100) = 101 + 10−6 𝑒 100 ≈ 2.7 ∙ 1037
and the solution has changed a lot.
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