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I am not sure about the boundary conditions for a variational problem when $~f~$ is a function of higher order derivatives of $~y~.$

i.e. if we have: $~f(x,y,y',y'')~$ then what is the natural boundary condition in this case?

Any help is highly appreciated.

nmasanta
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Kashif
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1 Answers1

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  • When the functional is of the form $$I[y(x)] = \int_{x_1}^{x_2} F \left(x,y(x),y'(x)\right) \,\rm{d}x~,$$then the natural boundary conditions are given in my answer of the following question: What are natural boundary conditions in the calculus of variations?

  • If the functional involves higher derivatives of $~y(x)~,$ such as, $$I[y(x)] = \int_{x_1}^{x_2} F \left(x,y(x),y'(x),y''(x)\right) \,\rm{d}x$$then the minimum (or maximum) of $~I[y(x)]~$ satisfies the following Euler-Lagrange equation, $$\dfrac{\partial F}{\partial y}-\dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y'}\right)+\dfrac{d^2}{dx^2}\left(\dfrac{\partial F}{\partial y''}\right)=0$$provided that $~y(x)~$ is subjected to the following essential or natural boundary conditions,

\begin{array}{|c|c|} \hline \text{essential}& \text{natural}\\ \hline y(x_1)=y_1 &\frac{\partial F}{\partial y'}-\frac{d}{dx}\left(\frac{\partial F}{\partial y''}\right)=0~~\text{at}~~x=x_1 \\ \hline y'(x_1)=y_1' & \frac{\partial F}{\partial y''}=0 ~~\text{at}~~x=x_1 \\ \hline \hline y(x_2)=y_2 & \frac{\partial F}{\partial y'}-\frac{d}{dx}\left(\frac{\partial F}{\partial y''}\right)=0~~\text{at}~~x=x_2 \\ \hline y'(x_2)=y_2'& \frac{\partial F}{\partial y''}=0 ~~\text{at}~~x=x_2 \\ \hline \end{array}

nmasanta
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  • can you please tell me form where it is given in Simple way about natural boundary conditions in calculus of variations. – neelkanth Mar 20 '23 at 14:55
  • Most probably it's from a research paper. But I am unable to remember the exact source as it is an old post... – nmasanta Mar 22 '23 at 01:19