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How do we solve $\frac{d}{dx_t}[\int_0^T y(x_s) ds]$ where T>t and $y(x_t)$ is a function of $x_t$ and $x_t$ is a function of $t$

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I don't understand your notation. What does $\frac{d}{dx_t}$ mean? –  copper.hat Jul 25 '12 at 7:37

1 Answer 1

You can't take the derivative with respect to a single function value of the function $x_t$; the value of the integral wouldn't change if you changed just that one function value.

The usual approach to varying functions would go like this: Consider the functional

$$I[f]=\int_0^Ty(f(s))\,\mathrm ds\;,$$

where $f(s)$ corresponds to your $x_s$. If $y$ is differentiable, the (first) variation of $I$ is

$$\delta I=\frac{\partial I[f+\epsilon\delta f]}{\partial\epsilon}=\frac{\partial}{\partial\epsilon}\int_0^Ty(f(s)+\epsilon\delta f(s))\,\mathrm ds=\int_0^Ty'(f(s))\delta f(s)\,\mathrm ds\;.$$

Now you can ask things like when is $I$ stationary, i.e. for which $f$ does the first variation vanish for all $\delta f$. You can also take the limit $\delta f(s)=\delta(s-t)$ to get something similar to what you asked for, a measure of the change in $I$ if you change $f$ only in an arbitrarily small neighbourhood of $t$.

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