(Uniqueness of) solution to $f'(t)=-\int_0^{f(t)}\!\mathrm dg(x)$ Let $g$ be a measure over the unit interval. Can we show that $f'(t)=-\int_0^{f(t)}\!\mathrm d g(x)$ has a unique nonzero solution $f$ ? (The boundary condition is that $f(0)>0$ and there is some $s>0$ with $f(t)=0$ for $t> s$.)
 A: In general, no.
Note that $\int_0^s dg(x) = g(s)$, hence the equation can be simplified to $$
f'(t) = -g(f(t))
$$
If $g(0)=0$ and $g$ is Lipschitz in a neighborhood of $0$, then the Picard–Lindelöf theorem implies that $f(t)\equiv 0$ is the unique solution satisfying $f(t)=0$ for some $t$. (Proof: If $f(t_0) = 0$ then the IVP $f'(t)=g(f(t))$ and $f(t_0)=0$ has $0$ as its unique solution. Since $t_0$ was arbitrary, this means that $0$ is the only solution which ever equals $0$.)
To solve in full generality, we try solving the equation like this: For $s\in[0,1]$, suppose that there exists a solution such that $f_s(t)>0$ for all $t\in [0,s)$ and $f_s(t)=0$ for $t\ge s$ (continuity of $f$, which is implied by differentiability, means we can use a non-strict inequality here). Then, for $t<s$, \begin{eqnarray}
-\frac{f_s'(t)}{g(f_s(t))} &=& 1\\
\int_t^s \frac{f_s'(\tau)}{g(f_s(\tau))}d\tau &=& t-s\\
\int_{f_s(t)}^0 \frac{1}{g(u)}du &=& t-s\end{eqnarray}
which makes sense only if $1/g(t)$ is integrable. If $1/g(t)$ is not integrable, then we have arrived at a contradiction as the LHS is infinite and the RHS is finite, so our assumption of a nonzero solution is false. Note also that $g(0)=0$ and $g(t)$ being Lipschitz in a neighborhood of $0$ implies that $1/g(t)$ is not integrable, so this covers our previous case as well. On the other hand, if $1/g(t)$ is integrable, we can define $H(t) = \int_0^t \frac1{g(u)}du$ and obtain that $H(f_s(t)) = s-t$, which uniquely defines $f_s$ on $[0,s]$ because $H$ is strictly increasing and its range contains the interval $[0,s]$. Thus, for each $s\in[0,1]$ there exists a unique solution with $f_s(s)=0$ and $f_s(t)>0$ for $t<s$ if and only if $1/g(t)$ is integrable. Note that if $g(0)\ne 0$ this solution does not satisfy the equation for $t\ge s$.
Thus we get the following result:
Theorem: If $1/g(t)$ is not integrable, then $f(t)\equiv 0$ is the only solution to $f'(t)=-g(f(t))$ such that $f(t)=0$ for some $t$. If $1/g(t)$ is integrable, then for each $s\in[0,1]$, there exists a unique $f_s$ such that for all $t<s$ we have $f_s'(t)=-g(f_s(t))$ and $f_s(t)>0$, and $f_s(t)=0$ for $t= s$. This solution is given implicitly by $$
\int_0^{f_s(t)} \frac1{g(y)} dy = s - t
$$
In general, this is not defined for $t>s$. However, if $g(0)=0$, then you can define $f_s(t)=0$ for $t>s$ and obtain a solution with larger domain.

Examples:

*

*$g(t) = t$ : $1/g(t)$ is not integrable, so $f(t)=0$ is the unique solution.

*$g(t) = \sqrt{t}$ : $1/g(t)$ is integrable, so for each $s\in [0,1]$, we have a solution given implicitly (for $t\le s$) by $$
\int_0^{f_s(t)} \frac1{\sqrt{\tau}}d\tau = s-t
$$
which we can solve \begin{eqnarray}
&&s-t=\int_0^{f_s(t)} \frac1{\sqrt{\tau}}d\tau = 2\sqrt{f_s(t)}\\
&\implies& f_s(t) = \begin{cases}\frac{(s-t)^2}{4} & t \le s\\
0 & t>s\end{cases}
\end{eqnarray}

*$g(t) = 1$ : $1/g(t)$ is integrable, so for each $s\in [0,1]$, we have (for $t\le s$) $$
\int_0^{f_s(t)} 1 d\tau = s-t
$$
which implies $f_s(t) = s-t$ for $t\le s$. Note that $$
f_s(t) = \begin{cases}s-t& t \le s\\
0 & t>s\end{cases}
$$
does not solve $f' = -g(f)$ for $t>s$, so the only solution here that works for all $t\le 1$ is $f_1(t) = 1-t$.

