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There are many questions on this site about this topic, but as far as I could find, all omitted the step I am not certain about. $I \subseteq \mathbb{R},u_0 \in \mathbb{R}$

The IVP is given as $$\begin{cases} u'(t)=2tu(t)=f(t,u(t)) & \text{for}\ t\in I \\ u(0)=u_0 \end{cases} $$

I want to first check for existence and uniqueness, so I want to know if it satisfies the Lipschitz Condition in the second variable. I know that if $$\frac{\partial f}{u(t)} $$ is bounded, then by the MVT the Lipschitz Condition is satisfied.

But how do I differentiate $\frac{\partial f}{u(t)} $?

I looked at questions like these: Derivative of a function with respect to another function., but I can't make that work for my example and in examples often times only the solution is given, which I can not follow.

Jonathan
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    It should read $\frac{\partial f}{\partial u(t)}$, not $\frac{\partial f}{u(t)}$. You are better off replacing $u(t) $ by some other variable $x$, say, so $f(t,x) = 2tx$. Now $\frac{\partial f}{\partial x}$ is just $2t$. – Thomas Jun 04 '17 at 13:35
  • My mistake. But are they not dependent on one another? Can I simply treat $u(t)$ as a separate variable and treat $t$ as constant? – Jonathan Jun 04 '17 at 13:37

2 Answers2

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Focus on $f:[a,b] \times \Bbb R \to \Bbb R $ and forget for a moment about the IVP itself.

One standard condition for uniqueness of ODE solutions is that if $f$ is Lipschitz in its second argument you have uniqueness of solutions wherever they overlap. That is to say there is a positive constant $C$ such that $|f(t,y) - f(t,x)| \le C |y-x|$. Notice that we don't have to consider potential solutions to check this condition.

In this case a direct argument shows that the condition is satisfied. $$|f(t,x) - f(t,y)| = |2t(x-y)| \le |2t||x-y| \le |2b||x-y|$$ Setting $C = |2b|$ shows that $f$ is Lipschitz in its second argument.

Bernard W
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A first order IVP is of the form, $$y'=f(x,y(x))\;\;\;,y(x_0)=y_0$$

Here $x$ denotes the independent variable whereas, $y$ is the dependent one.

Computationally, $\displaystyle \frac{\partial f(x,y(x))}{\partial y(x)}$ and $\displaystyle \frac{\partial f}{\partial y}$ are one and the same as, $y$ in itself is a variable.

So for computational ease, we replace $y(x)$ by $y$ and proceed with usual partial differentiation.

Naive
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