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This could be a silly question. In my lecture note on the Lipschitz condition, there is an example, which is

$y(0)=1$ and $f(t,y)=y-t^2+1$, determine whether this IVP is well-posed.

Here we apply the Lipschitz condition, namely $|\frac{\partial f}{\partial y}|=1$. But I think $\frac{\partial f}{\partial y}=\frac{\partial f}{\partial t}\frac{\partial t}{\partial y}+\frac{\partial f}{\partial y}=-2t\frac{\partial t}{\partial y}+1$, which generally contradicts the result above, since $y$ is a function of $t$. How to explain this?

I have considered the condition that $t$ is fixed, which makes $\frac{\partial t}{\partial y}=0$. But then another contradiction pops up, if $t$ is fixed, then $f, y$ should also be constants, why is $|\frac{\partial f}{\partial y}|\neq0$?

Tony Y
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  • The Lipschitz condition is a condition on $f$, which is a two-variable function. It is not a condition on $F(y) = f(t(y), y)$, a one-variable function. To understand this, look at the definition, and look at the theorems involving it. Do you see them forming $F$ and using the condition on it, or do they make use of what the condition says about $f$ itself? Don't talk yourself into making things far more complicated than they have to be. – Paul Sinclair May 03 '23 at 17:54

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