While we learn to work with equations relatively early on in our mathematical education, we are almost never encouraged to stop and ask what a solution to such an equation actually means. Such a philosophical question may initially evoke reactions of disgust from many quantitative thinkers, but they too will eventually be convinced that the time required to answer it is well spent.
Let me clarify these ideas with the example of equation
(1.2.6). This was the equation
\begin{equation*}
f(\theta) = (\cos \theta , \sin \theta), \text{ where } 0 \leq \theta \lt 2\pi.
\end{equation*}
giving the parametrization of the unit circle. However, any teenager will tell you that the equation for the unit circle is
\begin{equation}
x^2 + y^2 = 1.\tag{1.3.1}
\end{equation}
Of course, both equations describe the circle, but they do it in very different ways. The first equation expresses a set of points in the plane as the image of the function \(f(\theta)\text{.}\) In doing so, it places a number at each point on the unit circle (which is the angle made with the positive \(x\)-axis). This number, or parameter, is extra information and so the first equation is not just describing the unit circle, it is parametrizing it! Indeed, we could have considered
\begin{equation*}
g(\theta ) = (-\sin 2\theta , \cos 2 \theta ) \text{ where } 0 \leq \theta \lt \pi
\end{equation*}
and we would have a new parametrization of the same circle. There are in fact infinitely many parametrizations which are all correct descriptions of the same geometric object.
On the other hand, equation
(1.3.1) is nearly unique (you can do some silly things like multiply both sides by
\(2\text{,}\) but this doesn’t really change the equation). It does not tell you, for example, that the point on the positive
\(y\)-axis is
\(\pi / 2\) radians away from the point on the positive
\(x\)-axis. In fact, there is no clearly defined parameter on the points solving it, just the points themselves. This type of equation is often called
implicit or a constraint equation because it is not identifying the solutions, but characterizing them implicitly.
Now, one possible answer to our initial philosophical question could be :
| Convert an implicit equation to an explicit one. |
Of course, our first experience with this is solving a simple equation such as
\begin{equation*}
2x - 1 = 5
\end{equation*}
with a solution
\begin{equation*}
x = 3.
\end{equation*}
So how is this solution a parametrization? Well, I contend it is a parametrization of a point by a point and could (but maybe not should) be written
\begin{equation*}
x(\bullet ) = 3
\end{equation*}
for a function \(x : \{\bullet\} \to \mathbb{R}\text{.}\) Perhaps a more convincing example would be solving the implicit equation
\begin{equation*}
ax^2 + bx + c = 0.
\end{equation*}
As we all know, this was solved one million years ago with a formula
\begin{equation*}
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
\end{equation*}
In this case, it is easier to see the parametrization. It is a function of two points, which we could call \(-\) and \(+\) and could be written
\begin{equation*}
x (\square ) = \frac{-b \square \sqrt{b^2 - 4ac}}{2a}.
\end{equation*}
In practice, solving more general equations can be more difficult or simply impossible. For example, the distance formula between two points \(\mathbf{x} = (x_1, \ldots, x_n)\) and \(\mathbf{y} = (y_1, \ldots, y_n)\) in \(n\)-dimensions is
\begin{equation*}
\text{distance} (\mathbf{x} , \mathbf{y}) = \sqrt{(x_1 - y_1)^2 + \ldots + (x_n - y_n)^2}
\end{equation*}
Thus the implicit equation for the unit sphere in \(\mathbb{R}^3\) with coordinates \((x,y,z)\) is given by
\begin{equation*}
x^2 + y^2 + z^2 = 1.
\end{equation*}
However, parametrizing this sphere is a little tricky. A bit of help from spherical coordinates will show that
\begin{equation*}
f (\theta , \phi ) = ( \cos \theta \sin \phi, \sin \theta \sin \phi , \cos \phi) \text{ where } 0 \leq \theta \lt 2\pi \text{ and } 0 \leq \phi \leq \pi
\end{equation*}
is a solution.
To summarize the difference in the two types of equations, the implicit equation is an equation of the form
\begin{equation}
\mathbf{F} (\text{possible solution} ) = \text{ something special}\tag{1.3.2}
\end{equation}
where \(\mathbf{F} : P \to V\) is a function from a set of possible solutions to a set of values and ‘something special’ is one of those values. The actual solutions are possible solutions that satisfy the equation.
On the other hand, an explicit solution is of the form
\begin{equation}
\mathbf{f} (\text{something known} ) = \text{ actual solutions}.\tag{1.3.3}
\end{equation}
where \(\mathbf{f} : K \to S\) is often an onto function from ‘something known’ (usually a point, real numbers or \(\mathbb{R}^k\)) to the set of actual solutions of an interesting problem or points of a space.
We have seen several examples above where the sets \(P, V, K, S\) are subspaces of real space, but this template persists in an even broader context.
Subsection 1.3.1 General First Order ODE
To illustrate these ideas, we consider the differential equation:
\begin{equation}
\mathbf{F} (y^\prime , y, x ) = 0.\tag{1.3.4}
\end{equation}
Let us understand the general form of a first order ordinary differential equation (ODE) given in equation
(1.3.4). The set
\(P\) of ‘possible solutions’ here is the set of differentiable functions from
\(\mathbb{R}\) to
\(\mathbb{R}\) (or from a possible smaller domain in
\(\mathbb{R}\)). The set of values
\(V\) in this case is also a set of functions, albeit maybe only continuous functions. The ‘something special’ is the constant function equal to zero (so in truth, the right hand side is a function, not a number). To solve this equation we would need to find a function
\(\mathbf{f}\) on a known set of parameters
\(K\) to
\(P\) giving all the actual solutions (the ‘all’ part is saying that
\(\mathbf{f}\) is onto).
Perhaps it will bother the uninitiated student to hear that equation
(1.3.4) in its general form is absolutely impossible to solve and, in some cases, extremely difficult to understand qualitatively. However, if we take the example of a first order
linear ODE, we can indeed solve it completely. This is the case when there are functions
\(p(x)\) and
\(q(x)\) for which the equation becomes
\begin{equation}
\mathbf{F} (y^\prime , y, x) = y^\prime + p(x) y - q (x) = 0.\tag{1.3.5}
\end{equation}
To solve equation
(1.3.5) we take the practical step of first considering
\(q(x) = 0\) which is known as the homogeneous case. The equation then reduces to
\begin{equation*}
\frac{y^\prime}{y} = - p(x).
\end{equation*}
We may take anti-derivatives of both sides with respect to \(x\) to see that
\begin{equation*}
\ln (y) = - \int p(x) \diff x + \tilde{C}
\end{equation*}
or (after exponentiating),
\begin{equation*}
y = C e^{- \int p(x) \diff x}
\end{equation*}
This solution may be used to help find the solution when \(q(x) \ne 0\text{.}\) To see that consider the expression
\begin{equation*}
y = C v e^{-\int p(x) \diff x}
\end{equation*}
where
\(v\) is a function of
\(x\text{.}\) Then using the product and chain rule we have that equation
(1.3.5) becomes
\begin{equation*}
C v^\prime e^{-\int p(x) \diff x} = q(x)
\end{equation*}
or
\begin{equation*}
v = C^{-1} \left[\int e^{\int p(x) \diff x} q(x) \diff x + D \right]
\end{equation*}
Putting this back into the expression for \(y\) gives our general solution
\begin{equation*}
y = e^{-\int p(x) \diff x}\left[ \int e^{\int p(x) \diff x} q (x) \diff x + D \right].
\end{equation*}
Note that the solution expressed here is in fact a function
\begin{equation*}
\mathbf{f} : \mathbb{R} \to \text{actual solutions}
\end{equation*}
where \(y = \mathbf{f} (D)\) for any real number \(D\text{.}\)
This exposition on the solution to a first order linear ODE has been from the computational and abstract viewpoint. Later in the course, we will see a geometric viewpoint as well.
