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Let $\frac{dy}{dt}=Ay+g(y)$ and consider $\lVert y(t)\rVert^2=\langle y,y\rangle$. I would like to prove that $$ \frac{d}{dt}\langle y,y\rangle= 2\langle\frac{dy}{dt},y\rangle. $$

To do so, I made the following start:

$$ \langle y+\Delta y,y+\Delta y\rangle=\langle y,y\rangle+2\langle\Delta y,y\rangle+o(\lVert \Delta y\rVert)\text{ as }\Delta y\to 0. $$

So the derivative seems to arise from the summand $$ 2\langle\Delta y,y\rangle. $$ But how? It seems to be $$ 2\langle\Delta y,y\rangle=2\langle\dot{y},y\rangle\Delta y? $$

mike
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Salamo
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  • Well, to keep your notation, actually $$\Delta y=\dot{y}\Delta t+o(\Delta t)$$ which is what you want, since you are deriving in $t$. The existence of $\dot{y}$ implies that $o(\Vert \Delta y\Vert)=o(\Delta t)$ as well. –  Dec 07 '15 at 16:38
  • @G.Sassatelli Did not get it, sorry. What exactly do you mean? – Salamo Dec 07 '15 at 16:49
  • You are deriving in $t$. So, at some point, you'll have to expand it as a function of $t$, not of $y$. –  Dec 07 '15 at 16:51

2 Answers2

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It is straight forward: \begin{align} \frac{d}{dt} (y \cdot y) &= \frac{d}{dt}\sum_i y_i^2 \\ &= \sum_i 2 y_i \dot{y_i} \\ &= 2 \sum_i y_i \dot{y_i} \\ &= 2(y \cdot \dot{y}) \end{align}

mvw
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The derivative $\frac{d}{dt}\left<y(t),y(t)\right>$ is defined as $$\frac{d}{dt}\left<y(t),y(t)\right> \equiv \lim_{\Delta t\to 0}\frac{1}{\Delta t}\left[\left<y(t+\Delta t),\,y(t+\Delta t)\right> - \left<y(t),y(t)\right>\right]$$

To prove the result in the question we can try to rewrite the right hand side above (forget about the limit until the end) using the laws of an inner-product

  • Bi-linearity: $\left<x+y,z\right>=\left<x,z\right> + \left<y,z\right>$
  • Multiplication by scalar: $c\left<x,z\right>=\left<cx,z\right>$ for $c\in\mathbb{C}$.
  • Symmetry: $\left<x,y\right>=\left<y,x\right>$

to try to put it on a form resembling $2\left<\frac{dy(t)}{dt},y(t)\right>$. Doing this we find

$$\frac{1}{\Delta t}\left[\left<y(t+\Delta t),\,y(t+\Delta t)\right> - \left<y(t),y(t)\right>\right] = \left<\frac{y(t+\Delta t) - y(t)}{\Delta t},y(t)\right> + \left<y(t),\frac{y(t+\Delta t)-y(t)}{\Delta t}\right> + \Delta t\left<\frac{y(t+\Delta t) - y(t)}{\Delta t},\frac{y(t+\Delta t) - y(t)}{\Delta t}\right>$$

And from this form it's just a matter of taking the limit $\Delta t\to 0$ to arrive at the result (you will need to use that "the limit of a product equals the product of the limits" for the last term). Also note that the inner-product is a continuous function in both arguments so the limit can be moved inside the brackets.

Community
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Winther
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  • Thank you, that's a nice proof. But my original problem is, that the professor just wrote $\langle y+\Delta y,y+\Delta y\rangle=\langle y,y\rangle+2\langle\Delta y,y\rangle+o(\lVert \Delta y\rVert)\text{ as }\Delta y\to 0$ on the blackboard and only from this, he concluded that $2\langle \dot{y},y\rangle$ is the derivative. I could not understand it that quickly and therefore I now try to understand it how it can be seen so fast. – Salamo Dec 07 '15 at 17:03
  • @Salamo Your professor does basically the same as me here, but with a more casual notation. I tried to make it a bit more basic so that all steps that go into it would be more visible. The full equation you have above (using the laws I mentioned above) is: $\left<y+\Delta y,y+\Delta y\right> = \left<y,y\right> +2\left<\Delta y,y\right> + \left<\Delta y,\Delta y\right>$. The last term equals $||\Delta y||^2$ making it $o(|\Delta y||)$. Now subtracting $\left<y,y\right>$ from both sides, dividing by $\Delta x$ and taking the limit gives the result. – Winther Dec 07 '15 at 17:17
  • Am a bit confused since he does not use $\Delta t$ (what makes more sense to me), but $\Delta y$. – Salamo Dec 07 '15 at 17:25
  • I guess, with $\Delta y$, he means $y(t+\Delta t)-y(t)$? – Salamo Dec 07 '15 at 17:32
  • Your professor is doing a quick-and-dirty derivation. That is fine if one understands what is going on, but otherwise it's easy to get confused. There is indeed a $\Delta t$ there, but it's 'under the cover'. The term $\Delta y$ is the change in $y$ when we change $t$ by $\Delta t$. – Winther Dec 07 '15 at 17:32
  • @Salamo Exactly! – Winther Dec 07 '15 at 17:32