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I'm not too sure what a 'sharper' version of the existence and uniqueness theorem refers to here.

Suppose I'm given:

$$f(x,t) = |t|sin(x)$$

I know that for the theorem to apply, I show that $f(x,t)$ and $\frac{\partial f}{\partial x}$ are continuous in $(x,t)$. But I'm not sure how to necessarily apply the 'sharper' version of this theorem.

For this example, I know that $f(x,t)$ is continuous on all $(x,t)$ and that $\frac{\partial f}{\partial x} = |t|cos(x)$ is also continuous on all $(x,t)$.

How would I go about finding an $x_0$ and $t_0$ such that a sharper version of this theorem applies?

Any help is appreciated!

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  • $\begingroup$ It depends on what sharper means in this context. Choosing $x_0=0$ will result in an exact answer which is fairly sharp :-). $\endgroup$ – copper.hat Feb 19 at 21:45
  • $\begingroup$ The exact words of the prompt are: "find the values of $t_0$ and $x_0$ for which the sharper version of the existence and uniqueness theorem implies that the differential equation $\frac{dx}{dt} = f(x, t)$ with the initial condition $x(t_0) = x_0$ has a solution and state whether it is guaranteed to be unique." Utterly confused! $\endgroup$ – Ced Feb 19 at 21:49
  • $\begingroup$ I don't know. The solution exists and is unique everywhere. $\endgroup$ – copper.hat Feb 19 at 21:51
  • $\begingroup$ That's what I thought. Might have to ask the professor directly. Thank you! $\endgroup$ – Ced Feb 19 at 21:52

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