From Example Problems
Jump to: navigation, search

Derive the Euler-Lagrange equation from the attempt to minimize the functional

T(y)=\int _{a}^{b}L(y,y',x)\,dx\,

Let h\, be a function such that h(a)=0,h(b)=0\,.

Consider a small change in the function y\,.

T(y+\epsilon h)=\int _{a}^{b}L(y+\epsilon h,y'+\epsilon h',x)\,dx\,

In general, the Taylor expansion for a function g\, of three variables is

g(x+\epsilon _{1},y+\epsilon _{2},z+\epsilon _{3})=g(x,y,z)+\left(\epsilon _{1}{\frac  {\partial g}{\partial x}}+\epsilon _{2}{\frac  {\partial g}{\partial y}}+\epsilon _{3}{\frac  {\partial g}{\partial z}}\right){\bigg |}_{{(x,y,z)}}+...

T(y+\epsilon h)=\int _{a}^{b}\left[L(y,y',x)+{\frac  {\partial L}{\partial y}}\epsilon h(x)+{\frac  {\partial L}{\partial y'}}\epsilon h'(x)+...\right]\,dx\,

=T(y)+\epsilon \int _{a}^{b}\left(h(x){\frac  {\partial L}{\partial y}}+h'(x){\frac  {\partial L}{\partial y'}}\right)\,dx+...\,

\int _{a}^{b}h(x){\frac  {\partial L}{\partial y}}\,dx+\int _{a}^{b}h'(x){\frac  {\partial L}{\partial y'}}\,dx=0\,

0=\int _{a}^{b}h(x){\frac  {\partial L}{\partial y}}\,dx+h(x){\frac  {\partial L}{\partial y'}}{\bigg |}_{a}^{b}-\int _{a}^{b}h(x){\frac  {d}{dx}}{\frac  {\partial L}{\partial y'}}\,dx\,

The evaluated term is identically zero because of the boundary conditions of h\,.

0=\int _{a}^{b}h(x)\left[{\frac  {\partial L}{\partial y}}-{\frac  {d}{dx}}{\frac  {\partial L}{\partial y'}}\right]\,dx\,\forall h\implies {\frac  {\partial L}{\partial y}}-{\frac  {d}{dx}}{\frac  {\partial L}{\partial y'}}=0\,.

Calculus of Variations

Main Page