Note: Ode for Exponential

Steve Trettel

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$$\newcommand{\RR}{\mathbb{R}}$$

Here’s a self-contained proof that the solution to $y^\prime = y$ 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 $E\colon\mathbb{R}\to\mathbb{R}$ satisfying the law of exponents $$E(x+y)=E(x)E(y)$$ for all $x,y\in\mathbb{R}$.

First a uniqueness result we’ll need:

Let $f,g$ be two solutions to the differential equation $y^\prime =y$. Then they are constant multiples of one another.

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

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

Thus $h^\prime(x)=0$ 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 $g^\prime = g$ and has $g(0)=1$. Then $g$ is an exponential.

Let $g\colon\RR\to\RR$ solve $Y^\prime = Y$ and satisfy $g(0)=1$. We wish to show that $g(x+y)=g(x)g(y)$ for all $x,y\in\RR$.

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,

$$ \begin{align*} L^\prime(x)&=\left(g(x+y)\right)^\prime\
&=g(x+y)(x+y)^\prime\
&=g(x+y)\
&=L(x) \end{align*} $$

$$ \begin{align*} R^\prime(x)&=\left(g(x)g(y)\right)^\prime\
&=(g(x))^\prime g(y)\
&=g(x)g(y)\
&=R(x) \end{align*} $$

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

$$\frac{L(x)}{R(x)}=k\hspace{1cm} \forall x\in\RR$$

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 $1$:

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

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
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