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This article says:

It also turns out that these two equations [those given in question title], together with the conditions $g(0)=1$ and $h(0)=0$ [...] uniquely determine the functions $g$ and $h$.

Why is that so?

user97890
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    Do you mean $h'=g$ and $g'=-h$? – Fly by Night May 16 '17 at 17:50
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    Think about Taylor series. Note that $g(0)=1$ and $h(0)=0$ are necessary for uniqueness. $g=\sin$, $h=-\cos$ also satisfy $h'=g$ and $g'=-h$, but they do not satisfy $g(0)=1$ and $h(0)=0$. – kccu May 16 '17 at 17:57

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By derivation we see that the functions $g$ and $h$ satisfy the differential equation $y''=-y$. And with $g(0)=1$ and $g'(0)=-h(0)=0$ we get that $g=\cos$ and it's the unique solution due to the Cauchy Lipschitz theorem. The same reason for $h=\sin$.

Matthew Leingang
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user296113
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  • Can you show how this problem gets mapped to that theorem? More specifically, what should $f$ be in $y'(t) = f(t, y(t))$ so as to obtain $y=\sin$ as the solution? Please explain in a bit more simpler terms. – user97890 May 16 '17 at 18:25
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This is an alternative to the existing answer that only uses elementary calculus.

You can define $G(x) = g(x) - \sin x$ and $H(x) = h(x) - \cos x$.

Consider $F(x) = G(x)^2 + H(x)^2$. Then $$F'(x) = 2(g(x) - \sin x)(g'(x) - \cos x) + 2 (h(x) - \cos x)(h'(x) + \sin x).$$ You can use the relationships $g' = h$ and $h' = -g$ to write this as $$F'(x) = 2(g(x) - \sin x)(h(x) - \cos x) + 2 (h(x) - \cos x)(-g(x) + \sin x)$$ which conveniently reduces to $0$. This means $F$ is constant. Moreover $g(0) = 0$ and $h(0) = 1$ implies that $F(0) = 0$, so $F$ is identically zero.

Since $G(x)^2 + H(x)^2 = 0$ for all $x$, it follows that $G(x)$ and $H(x)$ are both identically zero. Consequently $g(x) = \sin x$ and $h(x) = \cos x$.

Umberto P.
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  • This is very interesting. Can this be considered formally a "proof of uniqueness"? I'm very curious about about the thought-process involved in the construction of this "proof". – user97890 May 16 '17 at 18:37
  • @user97890 it is a rock-solid proof of uniqueness. As far as the thought-process goes, like a lot of mathematics I was reminded of a trick I saw once upon a time. See the accepted answer to this question: https://math.stackexchange.com/questions/349143/how-do-we-find-specific-values-of-sin-and-cos-given-the-series-definition – Umberto P. May 16 '17 at 18:43