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November 8th, 2018, 07:39 AM   #11
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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.
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November 8th, 2018, 09:49 AM   #12
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Zylo, the derivative of a functional needs to be a linear functional. Your answer is not a linear functional. It is not linear.
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November 12th, 2018, 11:04 AM   #13
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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 11:16 AM.
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