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On page $46$, there is

($1.87$) $E[L]=\int \int \{y(x)-t\}^2p(x,t)dxdt$

Calculus of variations is used to give

($1.88$) $\dfrac{\partial E[L]}{\partial{y(x)}} = $2$ \int \{y(x)-t\}p(x,t)dt = 0$

The reader is referred to appendix $D$ on calculus of variations, but I am still confused. How does one get from ($1.87$) to ($1.88$), step by step?

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1 Answer 1

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Rename $\hat x$ as $x$, then interchange the order of integration, so that we integrate with respect to $x$ last. Then Equation (1.87) is $$ \int\int[y(x)-t]^2p(x,t)\,dt\,dx $$which is of the form $$ \int G(y(x),y'(x),x)\,dx\tag{D.5} $$ where $$ G(y,y',x)=\int[y-t]^2p(x,t)\,dt.\tag{*}$$ By the Euler-Lagrange equations we require $$ \frac{\partial G}{\partial y} -\frac d{dx}\left(\frac{\partial G}{\partial y'}\right)=0.\tag{D.8} $$ In this case the function $G$ doesn't depend on $y'$ so the LHS of the Euler-Lagrange equations simplifies to $$\frac{\partial G}{\partial y}=\int 2[y-t]p(x,t)\,dt,$$ obtained by differentiating (*) under the integral sign.

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  • $\begingroup$ Why G(y,y',t) in * and not G(y,y',x)? $\endgroup$ Feb 5, 2017 at 15:55
  • $\begingroup$ @user99889 That was a typo. Fixed. $\endgroup$
    – grand_chat
    Feb 5, 2017 at 17:39

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