# Differentiable dependency of solution on initial data

Let $$I\subset\mathbb R$$ be an ope interval and $$D\subset\mathbb R^n$$ an open set. For any $$f\in C^{0,1}(I\times D;\mathbb R^n)$$ and for any choice of $$t_0\in I, x_0\in D$$ let$$t\mapsto\Phi(t,t_0,x_0)$$ be the unique maximal solution of Cauchy Problem $$\begin{cases}x'=f(t,x)\\x(t_0)=x_0\end{cases}$$

Let $$\displaystyle y(t):=\frac{\partial\Phi}{\partial t_0}(t,t_0,x_0)$$. Derive initial value problem satisfied by $$y(t)$$ if $$f(t,x)=e^{-t}\arctan\left(e^{x^2}\right)$$

If $$\displaystyle y=\frac{\partial\Phi}{\partial t_0}$$, we have to differentiate $$f$$ with repect to $$t_0$$ and it exists since $$f$$ is $$C^1$$ in $$x$$. Now;

$$\begin{cases}y'=f_{\Phi(t,t_0,x_0)}(t,\Phi(t,t_0,x_0))y+f_t(t,\Phi(t,t_0,x_0))\underbrace{\frac{\partial t}{\partial t_0}}_{0}=\large\frac{2e^{-t}\Phi(t,t_0,x_0)e^{\Phi^2(t,t_0,x_0)}}{1+e^{2\Phi^2(t,t_0,x_0)}}y\\y(t_0)=?\end{cases}$$

I have difficulties to express $$y(t_0)$$, I thought it is $$y(t_0)=\frac{\partial x(t_0)}{\partial t_0}=f(t_0,x_0)$$ but it must be $$\color{red}-f(t_0,x_0)$$, where does the sign come from ?

• Differentiating $f$ with respect to $t_0$ does not actually make sense; $f$ does not depend on $t_0$, only $\Phi$ does. It may help to rewrite the differential equation as an integral equation.
– Ian
Dec 26 '14 at 16:43
• @Ian Can I not write $y(t_0)=y(t\to t_0)=\frac{\partial\Phi}{t_0}(t\to t_0,t_0,x_0)=\lim\limits_{t\to t_0}\frac{\partial\Phi}{t_0}(t,t_0,x_0)=x'(t_0)=f(t_0,x_0)$ or is the minus sign really correct ?
– OBDA
Dec 26 '14 at 19:11
• @Ian Actually, it does. See my answer below. Dec 30 '14 at 23:01
• @OBDA Where is this problem from? Dec 31 '14 at 13:17
• – OBDA
Dec 31 '14 at 20:02

It took me some time to figure it out, here is my attempt is to explain to you what is happening. First, there is one piece missing in your statement of the problem: When one is talking about differentiation with respect to the parameter, it means that to actually use it we need the point of the parameter at which we know the solution of the problem. Hence, instead of variables $t,x,t_0,x_0$ you need one extra variable, which I will call $\tau$, such that you have the problem $$\dot x=f(t,x),\quad x(\tau)=a,\tag{1}$$ please note that I use $\tau$ instead on your $t_0$ and $a$ instead of your $x_0$. The general theory tells us that there exists solution $$x=x(t,\tau,a)$$ depending differentiably on both parameters $\tau$ and $a$. Since $a$ is not a parameter, I will use the notation $$x(t,\tau)$$ to simplify things a little.

Now assume that we know the solution for the specific values of $\tau=t_0$. $t_0$ is fixed, and the solution is $$x_0(t,t_0).$$ The fact that one can differentiate with respect to the parameter $\tau$ means that at least close to $\tau=t_0$ the unknown solution $x(t,\tau)$ can be represented as $$x(t,\tau)=x_0(t,t_0)+(\tau-t_0)x_1(t,t_0)+\ldots,\tag{2}$$ where, obviously, $$x_1(t,t_0)=\frac{\partial x(t,\tau)}{\partial\tau}|_{\tau=t_0}$$ Your goal is to set up the problem for $x_1(t,t_0)$.

Now plug $(2)$ into $(1)$ and get that $$\dot x_0(t,t_0)+(\tau-t_0)\dot x_1(t,t_0)+\ldots\\=f(t,x(t,t_0))+\frac{\partial f}{\partial x}(t,x(t,t_0))\frac{\partial x(x,\tau)}{\partial \tau}|_{\tau=t_0}(\tau-t_0)+\ldots,$$ where I used the properties of $f$ to expand it into Taylor series around $\tau=t_0$. Now I find the coefficients at the same degrees of $(\tau-t_0)$ and get for the zero's degree $$\dot x_0(t,t_0)=f(t,x_0(t,t_0)),\quad x_0(t_0,t_0)=a,$$ the initial condition is taken directly from the initial condition in $(1)$ and representation $(2)$. Note that I used the fact that $x_0(t,t_0)=x(t,t_0)$.

For the first degree, i.e., for $(\tau-t_0)$, I have $$\dot x_1(t,t_0)=\frac{\partial f}{\partial x}(t,x_0(t,x_0))x_1(t,t_0),$$ which is (up to the notations) the problem you found yourself for the unknown $x_1(t,t_0)$. So what is the initial condition?

From $(2)$ and the initial condition for $(1)$ we have $$x_0(\tau,t_0)+(\tau-t_0)x_1(\tau,t_0)+\ldots=a,$$ by differentiating with respect to $\tau$ and plugging $\tau=t_0$ I find $$\frac{\partial x_0}{\partial \tau}(\tau,t_0)|_{\tau=t_0}+x_1(t_0,t_0)=0,$$ or $$x_1(t_0,t_0)=-\frac{\partial x_0}{\partial \tau}(\tau,t_0)|_{\tau=t_0}.$$ But notice that $x_0(\tau,t_0)$ solves $$\frac{d x_0}{d\tau}(\tau,t_0)=f(\tau,x_0(\tau,t_0)),$$ therefore finally I get, after plugging $\tau=t_0$ in the last equation, the required initial condition $$x_1(t_0,t_0)=-f(t_0,a).$$ Hope someone will read it :)

• sorry I don't get it, what was the advantage of introducing $g$, do you mean $y(\tau=0)=0$, and the last line is Taylor ?
– OBDA
Dec 26 '14 at 19:03
• The advantage is to make the initial conditions into parameters. Dec 26 '14 at 19:32
• @OBDA I wrote a different solution. Dec 30 '14 at 22:55