Show that each of the following initial-value problems has a unique solution ($0 ≤ t ≤ 1 , y(0) = 1$).
$$y' = \exp(t-y)$$
Theorem 1: Suppose that $D=\{(t,y)|a≤t≤b, −∞< y<∞\}$ and that $f(t,y)$ is continuous on $D$. If $f$ satisfies a Lipschitz condition on $D$ in the variable $y$, then the initial-value problem $y′(t)=f(t,y)$, $a≤t≤b$, $y(a)=α$, has a unique solution $y(t) for a ≤ t$ ≤ b.
Can theorem 1 be applied?
Best Answer
The theorem cannot be applied directly since the function $e^{t-y}$ is not Lipschitz. However, we can work around this by noting that $y$ must always be positive, as $y(0)>0$ and $y'(t)>0$ for all $t$. Thus we can replace $y'=e^{t-y}$ with $$y'=\begin{cases} e^{t-y} &\text{if } y > 0\\ e^t &\text{otherwise} \end{cases}$$ without changing the problem, and this function is indeed Lipschitz, so theorem 1 can be applied.