What is an Ordinary Differential Equation (ODE)?

Summary:
In this class, we explore Ordinary Differential Equations (ODEs) of order k, starting with their definition and their representation in both normal and general form. Through concepts such as the Jacobian matrix and the Implicit Function Theorem, we build the foundation for understanding the solutions of these equations and the associated properties, such as the domain of definition and explicit and implicit solutions.

LEARNING OBJECTIVES

By the end of this class, the student will be able to:

1. Recall the definition and basic characteristics of an Ordinary Differential Equation (ODE).
2. Explain the relationship between an ODE and its possible solutions.

INDEX
The Ordinary Differential Equation (ODE) of Order k
Implicit Function Theorem
The Solution of an Ordinary Differential Equation
Beware of the domain of definition of the solutions
Extended solution and maximal solution
Explicit solution and implicit solution

With what we have seen so far, we have a fairly clear idea of what a differential equation is and the many applications it can have. We will now pause to study some definitions and properties with the aim of establishing a solid common foundation to continue this study.

The ODE of Order k

An Ordinary Differential Equation (ODE) is an equation involving an independent variable x, a function y(x), and some of its ordinary derivatives. The first-order ordinary derivatives of y(x) are denoted by symbols such as \frac{dy(x)}{dx} or y'(x), the second-order ones as \frac{d^2y(x)}{dx^2} or y''(x), and in general, of order n, as \frac{d^ny(x)}{dx^n} or y^{(n)}(x). The highest value k for which y^{(k)}(x) appears in the equation is what we call the Order of the Equation. Thus, the General Form of an ODE of order k is:

F\left(x,y(x),y'(x), \cdots, y^{(k)}(x)\right)=0.

An ODE of order k is said to be in normal form if it is expressed by solving for y^{(k)}(x) from the above equation, that is:

y^{(k)}(x) = f\left(x,y(x),y'(x), \cdots, y^{(k-1)}(x)\right).

In general, the function y is a function \mathbb{R} \longrightarrow \mathbb{R}^n, so that it and all its derivatives evaluated at some point x\in\mathbb{R} are vectors in \mathbb{R}^n. With this in mind, we see that since the function F that describes the ODE of order k has 1+(k+1) variables, it holds that \text{Dom}(F)\subset \mathbb{R}^{1+n(k+1)} and \text{Range}(F)\subset \mathbb{R}; and similarly, \text{Dom}(f) = \mathbb{R}^{1+nk} and \text{Range}(f)\subset \mathbb{R}^n.

The transition from the General Expression of an ODE of order k to its Normal Form is made possible by the Implicit Function Theorem.

Implicit Function Theorem

Let F be a class \mathcal{C}^1 function over an open set U \subset \mathbb{R}^n with real values. And let (a_1,\cdots, a_n) \in U such that F(a_1,\cdots, a_n) = 0 and

\displaystyle \frac{\partial F(a_1,\cdots, a_n)}{\partial x_n} \neq 0

Then there exists a neighborhood V of (a_1, \cdots, a_{n-1}) \in \mathbb{R}^{n-1} and a function \varphi:V \longrightarrow \mathbb{R} such that:

1. V \times \varphi(V) \subset U
2. F(x_1,\cdots,x_{n-1},x_n) = 0 \leftrightarrow x_n = \varphi(x_1,\cdots, x_{n-1})
3. \varphi is differentiable and

   \displaystyle\dfrac{\partial \varphi (a_1,\cdots, a_{n-1})}{\partial x_i} = - \dfrac{ \dfrac{\partial F (a_1,\cdots, a_n)}{\partial x_i} }{ \dfrac{\partial F (a_1,\cdots, a_n)}{\partial x_n} }

Proof of the Implicit Function Theorem

Development from the Jacobian Matrix

Let \psi(x_1,\cdots,x_{n-1}, x_n) = (x_1,\cdots,x_{n-1}, F(x_1,\cdots, x_n)). If we compute its Jacobian matrix, shown below:

\displaystyle \left( \dfrac{\partial \psi(x_1,\cdots, x_n)}{\partial(x_1,\cdots, x_n)} \right) = \left( \begin{array}{cccc} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & \vdots \\ \vdots &\vdots & \ddots & \vdots \\ \displaystyle \dfrac{\partial F(x_1, \cdots, x_n)}{\partial x_1} & \dfrac{\partial F(x_1, \cdots, x_n)}{\partial x_2} & \cdots & \dfrac{\partial F(x_1, \cdots, x_n)}{\partial x_n} \end{array}\right),

we will see that its determinant is non-zero at (a_1,\cdots, a_n), precisely because, as established at the beginning, \partial F(a_1,\cdots, a_n)/\partial x_n \neq 0. From this, we can say that \psi has an inverse over an open set W containing (a_1,\cdots, a_n).

Solution Development

Now, let us consider a set

\tilde{V}=\psi(W)\ni \psi(a_1,\cdots,a_{n}) = (a_1,\cdots,a_{n-1},F(a_1,\cdots,a_{n}))=(a_1,\cdots,a_{n-1},0).

From this, we can define another set

V=\{(x_1,\cdots,x_{n-1}) \;|\; (x_1,\cdots,x_{n-1},0)\in \tilde{V}\}\ni (a_1,\cdots,a_{n-1})

The set V is, consequently, an open set containing (a_1,\cdots,a_{n-1})\in\mathbb{R}^{n-1}.

Moreover, since \psi has an inverse (on W), there exists a unique (y_1,\cdots,y_n)\in W such that \psi(y_1,\cdots,y_n) = (x_1,\cdots,x_{n-1},0). This means that:

\begin{array}{rl} y_1 &= x_1 \\ \\ \vdots & \vdots \\ \\ y_{n-1} &= x_{n-1} \\ \\ F(x_1,\cdots,x_{n-1},y_n) &= 0 \end{array}

Thus, we can define \varphi(x_1,\cdots,x_{n-1}) = y_n, so that:

\psi^{-1}(x_1,\cdots,x_{n-1},0) = (x_1,\cdots,x_{n-1},\varphi(x_1,\cdots,x_{n-1}))

and

F(x_1,\cdots,x_{n-1},\varphi(x_1,\cdots,x_{n-1})) = 0

From this, we have that \varphi(V)\ni a_n, and consequently V\times\varphi(V) \subset U, and also:

F(x_1,\cdots,x_{n-1},x_n) = 0 \leftrightarrow x_n = \varphi(x_1,\cdots,x_{n-1})

Differentiability

And finally, the differentiability of \psi leads to the differentiability of \psi^{-1}, which in turn leads to the differentiability of \varphi over V. Taking this into account, we can define a function g through the relation:

g(x_1, \cdots,x_{n-1}) = F(x_1,\cdots,x_{n-1},\varphi(x_1,\cdots,x_{n-1})) = 0

And then, using the chain rule, we have:

\displaystyle \frac{\partial g}{\partial x_i} = \frac{\partial F}{\partial x_i} + \frac{\partial F}{\partial x_n}\frac{\partial \varphi }{\partial x_i} = 0,

where i=1,\cdots, n-1. It is from this last equation that we obtain:

\displaystyle \dfrac{\partial \varphi(a_1,\cdots,a_{n-1})}{\partial x_i} = - \dfrac{\dfrac{\partial F(a_1,\cdots,a_{n})}{\partial x_i}}{\dfrac{\partial F(a_1,\cdots,a_{n})}{\partial x_n}}

And with this, everything we wanted to prove is concluded ■

The solution of an ordinary differential equation

Let us consider an ODE expressed in normal form

y^{(n)} = f(x,y(x),y^\prime(x),\cdots,y^{(n-1)(x)})

Then, a function \varphi : I_\phi \longmapsto \mathbb{R}^n, where I_\phi is an interval in \mathbb{R}, is said to be a solution of the ODE if:

\left(\forall x \in I_\phi \right) \left(\varphi^{(n)}(x) = f(x,\varphi(x),\varphi^\prime(x),\cdots,\varphi^{(n-1)(x)}\right)

Beware of the domain of definition of the solutions

At this point, it is necessary to emphasize the importance of explicitly stating the domain of the solution of the differential equation. For example, the domain of the function \phi mentioned in the previous paragraph is the interval I_\phi. This is important because a common mistake when working with differential equations is to consider two solutions \phi_1 and \phi_2 as equal just because \left(\forall x \in I_{\phi_1}\cap I_{\phi_2}\right)\left(\phi_1(x) = \phi_2(x)\right), even though I_{\phi_1}\neq I_{\phi_2}. To explain this point, let’s examine the differential equation:

y^\prime = -y^2.

A possible solution for this ODE is the function \psi_1 : ]0,+\infty[ \longrightarrow \mathbb{R}^+\setminus\{0\} defined by \psi_1(x)=1/x, because \psi_1^{\prime} = -1/x^2 = -\psi_1^2 for any x\in]0,+\infty[. But with a little algebraic manipulation, we can go from this to a completely different solution if we’re not careful with the details. For example, it is clear that:

\displaystyle \frac{1}{x} = \frac{1}{1 - (1-x)},

and the right-hand side of this equality is the result of the geometric series:

\displaystyle \sum_{n=0}^{+\infty} (1-x)^n = \frac{1}{1 - (1-x)}

So an untrained eye in these arcane arts might venture to think that the functions \psi_1
and \psi_2 = \sum_{n=0}^{+\infty} (1-x)^n offer the same solution to the differential equation posed at the beginning, because they indeed agree in their results; however, they will have overlooked the fact that this geometric series is only valid when |1-x| \lt 1, that is, when x\in]0,2[). But there’s more, since ]0,2[\subset]0,+\infty[, we also have that \psi_1 extends \psi_2 because wherever \psi_2 is valid, \psi_1 is also valid and even beyond.

Extended solution and maximal solution

Let us consider two functions \phi_1 and \phi_2 defined on the intervals I_{\phi_1} and I_{\phi_2}, respectively, which are solutions of a differential equation. If I_{\phi_1}\subset I_{\phi_2}, then we say that the solution \phi_2 extends the solution \phi_1, or that the solution \phi_2 is more general than the solution \phi_1. A solution \phi is called “maximal” if there is no other solution that extends it in a non-trivial way.

Explicit solution and implicit solution

A function \phi is considered a solution of the ODE of order n (written in normal form)

y^{(n)}(x)=f(x,y(x),y^\prime(x),\cdots,y^{(n-1)}(x)),

within an interval I if

(\forall x\in I)\left(\phi^{n}(x) = f(x,\phi(x),\phi^\prime(x),\cdots,\phi^{(n-1)}(x))\right)

What we reviewed several paragraphs ago is what is known as an Explicit Solution of the Differential Equation on the interval I. As the name suggests, there is also an implicit way to define solutions. A relation \Phi(x,y)=0 is said to be an Implicit Solution of the Differential Equation on I if it defines two or more implicit solutions on I.

Conclusion

In this class, we have broken down the notion of an ordinary differential equation with a rigorous yet accessible approach, establishing the formal foundations that allow us not only to recognize an ODE but also to understand the logic behind its solutions. Thanks to the Implicit Function Theorem, it was possible to clearly justify the transition from its general form to its normal form, which translates into a crucial technical ability to tackle concrete problems.

Moreover, we precisely distinguished the different ways a solution can be understood: as an explicit or implicit solution, extended or maximal, and we emphasized the —often underestimated— importance of properly declaring its domain. These distinctions are not merely formal: they are operative. Ignoring them can lead us, as we saw, to serious conceptual errors when interpreting the results obtained.

With this conclusion, we now have a well-honed first tool. The understanding of an ODE cannot be limited to solving a formula: it requires critical thinking, attention to detail, and a solid conceptual foundation that allows progress without losing the thread. This is just the beginning.