Correct way to show that a Lipschitz condition is satisfied

Question

In one of my homework problems, I have an IVP $$ y'=e^{t-y}, \hspace{5mm}where\hspace{3mm}0≤t≤1,\hspace{3mm}y(0)=1 $$ And I need to show that $$ f(t,y)=e^{t-y} $$

satisfies a Lipshitz condition.

To do this, I used the method shown in class. Same as the one used here: Prove Lipschitz condition

$$ |\dfrac{∂f(t,y)}{∂y}|=e^{t-y} $$

And

$$ \lim_{y \to -∞}e^{t-y}=∞ $$

So since the function isn't bounded from above, I concluded that the function doesn't satisfy a Lipschitz condition.

But I found a solution online that got a contradictory result. That solution substituted the unique solution to the IVP in for y:

$$ y(t)=ln(e^t+e-1) $$ $$ ⇒e^{t-y}=e^{t-ln(e^t+e-1)}=|\dfrac{e^t}{e^t+e-1}| $$ $$ max|\dfrac{e^t}{e^t+e-t}| = |\dfrac{e^1}{e^1+e-1}|≈0.6127 $$ $$ ⇒ L ≈0.6127 $$

So the second method shows that the Lipschitz condition is met. I'm pretty sure that the first method is correct, but I'm not sure. Could I get some clarification?

Thanks

Answer 1

I am guessing, the Lipschitz condition is studied here because it is one of the premises of the Picard-Lindelof Theorem.

While the function $e^{t-y}$ fails, as you showed, to be Lipschitz in the variable $y$ if $y$ is allowed to increase without bound, notice that in your IVP problem, the time variable is restricted to a bounded interval: $0 \leq t \leq 1$.

On this closed and bounded interval, the derivative $y'(t)$ of the solution is continuous, hence is bounded. (This is what is shown in the solution you found online.) Thus, $y(t)$ restricted to the closed and bounded interval $0 \leq t \leq 1$ is Lipschitz.

Answer 2

The question is what exactly you are asked to show. Your interpretation of Lipchitz property does not involve the DE at all. What you find online is a proof of the fact that the solution of the DE is a Lipschitz function of $t$. Certainly $e^{t-y}$ is not a Lipschitz function of $y$ if $y$ is allowed to take all real values.