Let $\phi(t, x_0)$ be a solution of the one-dimensional differential equation $$\dot{x}= f(x),$$ with $\phi(0, x_0) = x_0$. Show that its derivative $\frac{\partial}{\partial x_0}\phi(t, x_0)$ is given by $$\frac{\partial}{\partial x_0}\phi(t, x_0) = exp \Big(\int_{0}^{t}f'(\phi(s, x_0))ds\Big)$$

Here in my proposed solution.

By the fundamental theorem of calculus, we have

$$\phi(t, x_0)=x_0 + \int_{0}^{t}f(\phi(s, x_0))ds$$

because $\frac{\partial{\phi}}{\partial t}(t, x_0)=f(\phi(t, x_0))$ and $\phi(0, x_0)=x_0$.

If we differentiate this solution with respect to $x_0$, we obtain via the chain rule that

$$\frac{\partial{\phi}}{\partial x_0}(t, x_0)=1 + \int_{0}^{t}\frac{\partial{f}}{\partial{x_0}}(\phi(s, x_0)) \cdot \frac{\partial{\phi}}{\partial x_0}(s, x_0)ds$$

Let $$z(t)= \frac{\partial{\phi}}{\partial x_0}(t, x_0)$$


$$z(0)= \frac{\partial{\phi}}{\partial x_0}(0, x_0)=1$$

by the analysis above. Therefore, if we differentiate $z$ with respect to $t$, we find that

\begin{equation} \begin{split} z'(t) & = \frac{\partial{f}}{\partial{x_0}}(\phi(t, x_0)) \cdot \frac{\partial{\phi}}{\partial x_0}(t, x_0)\\ & = \frac{\partial{f}}{\partial{x_0}}(\phi(t, x_0)) \cdot z(t) \end{split} \end{equation}

We do not have an explicit solution to $\phi(t,x_0)$, but the above equation tells us that $z(t)$ solves the following differential equation,

$$z'(t)= \frac{\partial{f}}{\partial{x_0}}(\phi(t, x_0)) \cdot z(t)$$


$$\frac{dz}{dt}= \frac{\partial{f}}{\partial{x_0}}(\phi(t, x_0)) \cdot z(t)$$

Rearranging terms produces

$$\frac{1}{z(t)}{dz}= \frac{\partial{f}}{\partial{x_0}}(\phi(t, x_0))dt$$


$$ln(z(t))= \int_{0}^{t}\frac{\partial{f}}{\partial{x_0}}(\phi(s, x_0))ds$$

And taking the exponential of both sides produces

$$z(t)= exp \Big(\int_{0}^{t}\frac{\partial{f}}{\partial{x_0}}(\phi(s, x_0))ds\Big)$$

As $z(t)=\frac{\partial{\phi}}{\partial x_0}(t, x_0)=exp \Big(\int_{0}^{t}f'(\phi(s, x_0))ds\Big)$, we are done.

I'm not certain if there is a more direct approach. Please let me know if the solution can be improved.

  • $\begingroup$ Could you correct the missing partial derivative in the chain rule application and regard that $f$ has the variable $x$, thus its derivative is for $x$, not $x_0$? Btw., exponentiation is different from multiplying with $e$. $\endgroup$ – LutzL Dec 20 '18 at 19:25
  • $\begingroup$ I'm not sure what is wrong with the partial derivative. Does $\phi(t, x_0)=1 + \int_{0}^{t}\frac{\partial{f}}{\partial{x_0}}(\phi(s, x_0)) \cdot \frac{\partial{\phi}}{\partial x}(s, x_0)ds$ need to be changed to $\phi(t, x_0)=1 + \int_{0}^{t}\frac{\partial{f}}{\partial{x}}(\phi(s, x_0)) \cdot \frac{\partial{\phi}}{\partial x_0}(s, x_0)ds$? $\endgroup$ – Axion004 Dec 20 '18 at 19:40
  • $\begingroup$ Yes, that too, but on the left side the derivative is missing. $\endgroup$ – LutzL Dec 20 '18 at 19:52
  • $\begingroup$ Yes, I had that correct in my original solution and forgot to write it down. I don't see why I would need to write $\frac{\partial {f}}{\partial x}$ instead of $\frac{\partial {f}}{\partial x_0}$, as $f(\phi(s, x_0))$ is being differentiated with respect to $x_0$. $\endgroup$ – Axion004 Dec 20 '18 at 19:59
  • $\begingroup$ Why not to differentiate your equation directly? $\dot x=f(x)\implies \frac{d}{dt}\frac{\partial x}{\partial x_0}=f'(x)\frac{\partial x }{\partial x_0}$, which immediately yields the required conclusion? $\endgroup$ – Artem Dec 20 '18 at 20:18

From the comments above, here is a shorter answer:

As $\dot{x}=f(x) \implies \frac{dx}{dt}=f(x)$, we can differentiate both sides with respect to $x_0$ to form


Therefore, as $\phi(t,x_0)$ is a solution to the ODE, we know that $x(t)=\phi(t,x_0)$. Hence,


Now, let $z(t)=\frac{\partial\phi}{\partial{x_0}}(t,x_0)$. Then,




Hence, by the fundamental theorem of calculus,


Therefore, if we take the exponential of both sides,


So, $\frac{\partial\phi}{\partial{x_0}}(t,x_0)=exp\Big(\int_0^t{f'(\phi(s,x_0))ds}\Big)$.


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