# Calculating the derivative of a one-dimensional ODE

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

Then,

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

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

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

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,

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

Therefore,

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

Rearranging terms produces

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

Hence,

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

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.

• 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 $e$. – LutzL Dec 20 '18 at 19:25
• 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 $\phi(t, x_0)=1 + \int_{0}^{t}\frac{\partial{f}}{\partial{x}}(\phi(s, x_0)) \cdot \frac{\partial{\phi}}{\partial x_0}(s, x_0)ds$? – Axion004 Dec 20 '18 at 19:40
• Yes, that too, but on the left side the derivative is missing. – LutzL Dec 20 '18 at 19:52
• 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 $x_0$. – Axion004 Dec 20 '18 at 19:59
• 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? – Artem Dec 20 '18 at 20:18

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

$$\frac{d}{dt}\frac{\partial{x}}{\partial{x_0}}=\frac{df(x)}{dx}\frac{\partial{x}}{\partial{x_0}}=f'(x)\frac{\partial{x}}{\partial{x_0}}$$

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

$$\frac{d}{dt}\frac{\partial\phi}{\partial{x_0}}(t,x_0)=f'(\phi(t,x_0))\frac{\partial\phi}{\partial{x_0}}(t,x_0)$$

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

$$\frac{d}{dt}z(t)=f'(\phi(t,x_0))z(t)$$

So,

$$\frac{1}{z(t)}d{z(t)}=f'(\phi(t,x_0))dt$$

Hence, by the fundamental theorem of calculus,

$$ln(z(t))=\int_0^t{f'(\phi(s,x_0))ds}$$

Therefore, if we take the exponential of both sides,

$$z(t)=exp\Big(\int_0^t{f'(\phi(s,x_0))ds}\Big)$$

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