# Finding $x_0$ and $t_0$ such that a sharper version of the existence and uniqueness theorem applies?

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!

• It depends on what sharper means in this context. Choosing $x_0=0$ will result in an exact answer which is fairly sharp :-). – copper.hat Feb 19 at 21:45
• 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! – Ced Feb 19 at 21:49
• I don't know. The solution exists and is unique everywhere. – copper.hat Feb 19 at 21:51
• That's what I thought. Might have to ask the professor directly. Thank you! – Ced Feb 19 at 21:52