Note: Ode for Exponential

Steve Trettel

May 4, 2024 |

Here’s a self-contained proof that the solution to $y^{'} = y$ with $y (0) = 1$ is an exponential function. This is one possible means of proving the existence of exponentials, when defined functionally:

(Exponential Functions) An exponential function is a continuous nonconstant function

E : R \to R

satisfying the law of exponents

E (x + y) = E (x) E (y)

for all

x, y \in R

First a uniqueness result we’ll need:

Let

f, g

be two solutions to the differential equation

y^{'} = y

. Then they are constant multiples of one another.

Consider the function $h (x) = \frac{f (x)}{g (x)}$ . Differentiating with the quotient rule,

$\begin{aligned} h^{'} (x) & = \frac{f^{'} (x) g (x) - f (x) g^{'} (x)}{g (x)^{2}} & = \frac{f (x) g (x) - f (x) g (x)}{g (x)^{2}} & = \frac{0}{g (x)^{2}} & = 0 \end{aligned}$

Thus $h^{'} (x) = 0$ for all $x$ , which implies $h = f / g$ is a constant function, and $g$ is a constant multiple of $f$ as claimed.

Now we’re ready for the main theorem:

Let

g

be any differentiable function which solves

g^{'} = g

and has

g (0) = 1

. Then

g

is an exponential.

Let $g : R \to R$ solve $Y^{'} = Y$ and satisfy $g (0) = 1$ . We wish to show that $g (x + y) = g (x) g (y)$ for all $x, y \in R$ .

So, fix an arbitrary $y$ , and consider each of these separately, defining functions $L (x) = g (x + y)$ and $R (x) = g (x) g (y)$ .
Differentiating,

$\begin{aligned} L^{'} (x) & = {(g (x + y))}^{'} & = g (x + y) (x + y)^{'} & = g (x + y) & = L (x) \end{aligned}$

$\begin{aligned} R^{'} (x) & = {(g (x) g (y))}^{'} & = (g (x))^{'} g (y) & = g (x) g (y) & = R (x) \end{aligned}$

Thus, both $L$ and $R$ satisfy the differential equation $Y^{'} = Y$ . Our previous proposition implies they are constant multiples of one another,

$\frac{L (x)}{R (x)} = k \forall x \in R$

To find this constant we evaluate at $x = 0$ where (using $g (0) = 1$ ) we have $L (0) = g (0 + y) = g (y)$ $R (0) = g (0) g (y) = g (y)$

They are equal at $0$ so the constant is $1$ :

$\frac{L (x)}{R (x)} = \frac{L (0)}{R (0)} = \frac{g (y)}{g (y)} = 1$ $⟹ L = R$

But these two functions are precisely the left and right side of the law of exponents for $g$ . Thus their equality is equivalent to $g$ sayisfying the law of exponents for this fixed value of $y$ :

$\forall x,,, L (x) = g (x + y) = g (x) g (y) = R (x)$

As $y$ was arbitrary, this holds for all $y$ , and $g$ is an exponential.

Note this proof does not establish the existence of a solution to this ODE, it only says if you have a solution then its an exponential. But this provides a much easier means of rigorously constructing an exponential: if we can develop techniques to build such a function (eg power series, and term by term differentiation) we will immediately know the result satisfies the law of exponents.

◀ Note: Ode Existence Uniqueness

Note: Ftc From Axioms ▶