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Consider the following initial value problem

$y'(t) = f(y(t)), $ $0 < t$

$y(0) = y_0$, where $y_0$ is a fixed constant.

Here, $y'(t)$ is given only for $t > 0$, not including $t = 0$. That is, $y'(0)$ is not prescribed.

Yet, if one tries to solve this numerically, say by Forward Euler, almost all textbooks that I saw starts by computing

$y(h) \approx y(0) + hy'(0) = y_0 + f(y(0)) = y_0 + f(y_0)$, as if $y'(t) = f(y(t)) $ holds at $t = 0$.

How can this be justified?

Is there a theorem that says that the above initial value problem is equivalent to the following initial value problem

$y'(t) = f(y(t)), $ $0 \leq t$

$y(0) = y_0$, where $y_0$

where we actually require a solution $y(t)$ to be differentiable and satisfy the ode at $t = 0$?

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It is curious that you wrote down an autonomous differential equation where the domain of $f$ does not contain the time. Thus the condition $t>0$ and your question regarding it are vacuous.

If $f$ is defined on some domain $D\subset \Bbb R^n$, then the only condition on the initial value is $y(0)\in D$. Under reasonable assumptions, the Picard integral equation $$ y(t)=y_0+\int_0^t f(y(s))\,ds $$ is at least solvable for $t\in(-\delta,\delta)$ with some small $\delta$, so that $y$ is also differentiable in $t=0$ and $y'(0)=f(y(0))$ holds true.

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  • $\begingroup$ What if $y(0)$ is not in $D$? $\endgroup$ – mononono Apr 3 '15 at 3:51
  • $\begingroup$ Then you do not have a problem. An initial value problem that is. $\endgroup$ – LutzL Apr 3 '15 at 7:13
  • $\begingroup$ I was thinking something like $y'(t) = 1/y$ with the initial condition $y(0) = 0$. You are saying that this is not well-posed? $\endgroup$ – mononono Apr 3 '15 at 17:34
  • $\begingroup$ Of course not. Even if there is an obvious solution $y(t)=\sqrt{2t}$, it can't be a solution to this IVP since it is not differentiable in $t=0$. $\endgroup$ – LutzL Apr 3 '15 at 18:06
  • $\begingroup$ But, in my original formulation of IVP, I only require $y'(t)$ to be differentiable for $t>0$, so $y(t)= \sqrt(2t)$ is a solution to my IVP $\endgroup$ – mononono Apr 3 '15 at 19:10

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