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 November 8th, 2018, 06:39 AM #11 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Everything I've done so far is correct, though I admit I stumbled briefly on $\displaystyle \Delta y$. This should make it clearer: $\displaystyle J(y)= \int_{0}^{1}g(y)dx$ where $\displaystyle y=y(x)$ Let a be a small arbitrary constant. Then $\displaystyle J(y+a)= \int_{0}^{1}g(y+a)dx \approx \int_{0}^{1}g(y)dx+\int_{0}^{1}ag'(y)dx$ $\displaystyle J'(y) = \lim_{a\rightarrow0}\frac{J(y+a)-J(y)}{a} = \int_{0}^{1}g'(y)dx$ The connection with calculus of variations (minimize integral of a function by varying the function) is: J(y) is a function of the function y which makes it a functional, and y+a is a specific variation of y. November 8th, 2018, 08:49 AM #12 Senior Member   Joined: Oct 2009 Posts: 850 Thanks: 327 Zylo, the derivative of a functional needs to be a linear functional. Your answer is not a linear functional. It is not linear. November 12th, 2018, 10:04 AM #13 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Functional Derivative Consider the following variation of the Functional F(x,y,y') $\displaystyle F(x,y+\epsilon\eta ,y+\epsilon\eta')$, where $\displaystyle \eta$ is an arbitrary function of x. Then, to first order variation (MacLaurens Series): $\displaystyle \delta F= F_{y}\epsilon\eta+ F_{y}\epsilon\eta'$\\ If $\displaystyle J(y)=\int_{a}^{b}F(x,y,y')dx$ $\displaystyle \delta J(y) =\int_{a}^{b}(F_{y}-\frac{\mathrm{d} }{\mathrm{d} x}F_{y}')\epsilon\eta dx$ after integration by parts. At an extremum, $\displaystyle \delta J(y)$ = 0: $\displaystyle \int_{a}^{b}(F_{y}-\frac{\mathrm{d} }{\mathrm{d} x}F_{y}')\epsilon\eta dx=0 \rightarrow (F_{y}-\frac{\mathrm{d} }{\mathrm{d} x}F_{y}')=0$ because $\displaystyle \eta$ arbitary requires coefficient of $\displaystyle \eta$ to be zero because otherwise you could make integral positive by suitable choice of $\displaystyle \eta$. Finally $\displaystyle (F_{y}-\frac{\mathrm{d} }{\mathrm{d} x}F_{y}')\equiv \frac{\delta J}{\delta y}$ is called (not defined as) the functional derivative of J(y), presumably as a matter of semantics so you can say you have an extremum of J when the fractional derivative of J is zero. Other than this semantic device, there is no such thing as the functional "derivative" of J. You can logically speak of the derivative of the functional J(y), which was the subject of this thread previous to the jargon attack. Further jargon attacks will be ignored- hunting them down is too much work. Last edited by zylo; November 12th, 2018 at 10:16 AM. Tags calculate, derivative, functional Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post el keke Calculus 0 October 31st, 2014 12:07 PM Imaxium Calculus 2 April 13th, 2014 09:42 PM hultr Applied Math 1 April 29th, 2013 01:14 PM Dougy Real Analysis 5 October 3rd, 2012 02:43 PM cris(c) Applied Math 0 July 6th, 2012 04:58 PM

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