12-267/Derivation of Euler-Lagrange: Difference between revisions

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Disclamer: This is a student prepared note based on [http://drorbn.net/dbnvp/12-267-121002-2.php the lecure of Tuesday October 2nd].
Disclamer: This is a student prepared note based on [http://drorbn.net/dbnvp/12-267-121002-2.php the lecure of Tuesday October 2nd].


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<math> = \int_a^b (F_2(x, y, y') \cdot h + F_3(x, y, y') \cdot h') dx </math>
<math> = \int_a^b (F_2(x, y, y') \cdot h + F_3(x, y, y') \cdot h') dx </math>


<math> = \int_a^b (F_2 \cdot h + [\frac{d}{dx} F_3] \cdot h) dx + F_3 \cdot h |_a^b</math> (integrating by parts)
<math> = \int_a^b (F_2 \cdot h - [\frac{d}{dx} F_3] \cdot h) dx + F_3 \cdot h |_a^b</math> (integrating by parts)


Due to the constraints of <math>h(a) = 0</math> and <math>h(b) = 0</math>, <math>F_3 \cdot h |_a^b = 0</math>.
Due to the constraints of <math>h(a) = 0</math> and <math>h(b) = 0</math>, <math>F_3 \cdot h |_a^b = 0</math>.
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As this must be equal to 0 for all h satisfying the endpoint constraints, we must have that <math>F_2 = \frac{d}{dx} F_3</math>, or in other terms, <math>F_y - \frac{d}{dx} F_y' = 0</math>.
As this must be equal to 0 for all h satisfying the endpoint constraints, we must have that <math>F_2 = \frac{d}{dx} F_3</math>, or in other terms, <math>F_y - \frac{d}{dx} F_y' = 0</math>.

Special cases (without derivations):

In the case that F does not depend on y', we have <math>F_y = 0</math>

In the case that F does not depend on y, we have <math>F_{y'} = c</math>

In the case that F does not depend on x, we have <math>F - y'F_{y'} = c</math>

Latest revision as of 17:18, 24 October 2012

Disclamer: This is a student prepared note based on the lecure of Tuesday October 2nd.

For a function [math]\displaystyle{ y(x) }[/math] defined on [math]\displaystyle{ [a, b] }[/math] to be an extremum of [math]\displaystyle{ J(y) = \int_a^b F(x, y, y') dx }[/math], it must be that for any function [math]\displaystyle{ h(x) }[/math] defined on [math]\displaystyle{ [a, b] }[/math] that preserves the endpoints of [math]\displaystyle{ y }[/math] (that is, [math]\displaystyle{ h(a) = 0 }[/math] and [math]\displaystyle{ h(b) = 0 }[/math]), we have [math]\displaystyle{ \frac{d}{d \epsilon } J(y + \epsilon h) }[/math][math]\displaystyle{ |_{\epsilon = 0} = 0 }[/math].

[math]\displaystyle{ \frac{d}{d \epsilon } J(y + \epsilon h) }[/math][math]\displaystyle{ |_{\epsilon = 0} }[/math]

[math]\displaystyle{ = \frac{d}{d \epsilon } \int_a^b F(x, y + \epsilon h, y' + \epsilon h') dx |_{\epsilon = 0} }[/math]

Let [math]\displaystyle{ F_n }[/math] signify F differentiated with respect to its nth variable.

[math]\displaystyle{ = \int_a^b (F_1 \cdot 0 + F_2 \cdot h + F_3 \cdot h') dx |_{\epsilon = 0} }[/math]

[math]\displaystyle{ = \int_a^b (F_2(x, y, y') \cdot h + F_3(x, y, y') \cdot h') dx }[/math]

[math]\displaystyle{ = \int_a^b (F_2 \cdot h - [\frac{d}{dx} F_3] \cdot h) dx + F_3 \cdot h |_a^b }[/math] (integrating by parts)

Due to the constraints of [math]\displaystyle{ h(a) = 0 }[/math] and [math]\displaystyle{ h(b) = 0 }[/math], [math]\displaystyle{ F_3 \cdot h |_a^b = 0 }[/math].

[math]\displaystyle{ = \int_a^b (F_2 - \frac{d}{dx} F_3) \cdot h dx }[/math]

As this must be equal to 0 for all h satisfying the endpoint constraints, we must have that [math]\displaystyle{ F_2 = \frac{d}{dx} F_3 }[/math], or in other terms, [math]\displaystyle{ F_y - \frac{d}{dx} F_y' = 0 }[/math].

Special cases (without derivations):

In the case that F does not depend on y', we have [math]\displaystyle{ F_y = 0 }[/math]

In the case that F does not depend on y, we have [math]\displaystyle{ F_{y'} = c }[/math]

In the case that F does not depend on x, we have [math]\displaystyle{ F - y'F_{y'} = c }[/math]