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joshmccraney

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Hi PF!

Given a functional ##J[y]##, if the first variation is $$\delta J[y] = \int_D(ay+y'')y \, dV + \int_{\partial D} (y'+by)y\,dS$$

am I correct to think that when finding stationary points of ##J[y]##, I would solve ##ay+y''=0## on ##D## subject to boundary conditions, which would either be ##y|_{\partial D}=0## or ##(y'+by)|_{\partial D}=0##?

Is it correct to say ##(y'+by)|_{\partial D}=0## is the natural boundary condition? Isn't it true that if I solve ##ay+y''=0## and no value is specified for ##y## at ##\partial D##, then I must enforce ##(y'+by)|_{\partial D}=0## for the solution to be valid (the solution will not automatically satisfy ##(y'+by)|_{\partial D}=0## unless I enforce this, right)?

Given a functional ##J[y]##, if the first variation is $$\delta J[y] = \int_D(ay+y'')y \, dV + \int_{\partial D} (y'+by)y\,dS$$

am I correct to think that when finding stationary points of ##J[y]##, I would solve ##ay+y''=0## on ##D## subject to boundary conditions, which would either be ##y|_{\partial D}=0## or ##(y'+by)|_{\partial D}=0##?

Is it correct to say ##(y'+by)|_{\partial D}=0## is the natural boundary condition? Isn't it true that if I solve ##ay+y''=0## and no value is specified for ##y## at ##\partial D##, then I must enforce ##(y'+by)|_{\partial D}=0## for the solution to be valid (the solution will not automatically satisfy ##(y'+by)|_{\partial D}=0## unless I enforce this, right)?

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