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 1st].

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


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

Let signify F differentiated with respect to its nth variable.

(integrating by parts)

Due to the constraints of and , .

As this must be equal to 0 for all h satisfying the endpoint constraints, we must have that , or in other terms, .

Special cases (without derivations):

In the case that F does not depend on y', we have

In the case that F does not depend on y, we have

In the case that F does not depend on x, we have