most recent 30 from http://mathoverflow.net 2013-06-19T16:47:23Z http://mathoverflow.net/feeds/question/42820 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42820/expressions-for-the-square-of-an-integral blackj4ck 2010-10-19T19:22:08Z 2010-10-20T05:47:19Z

<p>Is there a way to simplify the following expression: </p> <p>$\lgroup{\int^A_0 x(s)ds}\rgroup ^2$</p> <p>I'm looking for an expression that can possibly get rid of the squared term, so that I can have just an integral of the first order.</p>

http://mathoverflow.net/questions/42820/expressions-for-the-square-of-an-integral/42826#42826 Pietro Majer 2010-10-19T19:44:32Z 2010-10-19T19:44:32Z

<p>For instance, the derivatives wrto $A$ of the two expressions coincide choosing $s(x):=2r(x)\int_0^xr(\xi)u(\xi)d\xi$. So the two expressions coincide for all $A$ since they both vanish at $A=0$.</p>

http://mathoverflow.net/questions/42820/expressions-for-the-square-of-an-integral/42869#42869 Gilead 2010-10-20T03:27:02Z 2010-10-20T03:32:17Z

<p>I'm not sure about simplifying, but you can easily write your objective functional in Bolza form like this:</p> <p><code>$$ \begin{align} &amp;\min_{u(t) \in \Omega(t)} \, J = z(T)^2 + \int_{0}^{T} s(t)u(t)dt \\ s.t. &amp;\frac{dz(t)}{dt} = r(t)u(t),\quad z(0) = 0 \end{align} $$</code></p>