# Calculus of Variations and Natural BCs

## Main Question or Discussion Point

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)?

Last edited:

jambaugh
Gold Member
Your first variation is not well formed. $y$ is your (function valued) variable and so your variation of $J$ should manifest as a linear functional on the variation $\delta y$ of $y$. This term is critical to formulating your solutions.

You should see something like:
$$J[y] = \int_D L(y,y',...)dV;\quad \delta J[y] = \int_D A(y,y',...) \delta y dV + \int_{\partial D} B(y, y',...)\delta y dS$$

I'm also a bit confused by your apparent single variable prime notation for derivatives vs your use of multivariable integrals over volumes and surfaces. Now it's perfectly fine to speak of the generalized derivative of a function to and/or from vector spaces but in so doing one should be explicit as to the vector on which such an operator valued derivative acts. It would thus be most helpful to me if you could clarify your asserted variation by giving the original functional and explaining the independent variables on which the function $y$ depends and whether it is a scalar or vector valued function. All of this does relate to your specific question.

What is typically argued, in these variational problems is that the functional is stationary for arbitrary variations including those which are constrained to be zero on the boundary of the region of integration. This allows you to independently set the volume integral in the variation to zero rather than the sum of the volume and boundary integrals. Your variation $\delta y$ which is absent here should depend on the integration variables so its occurrence within the integral and its (interior) arbitrariness is what allows you to argue that to get zero for the whole integral for all cases the actual integrand must be zero... your Euler-Lagrange equation manifests.

You can then apply this result with the more general variations which do not disappear at the boundary whereby the sum of boundary and interior integrals must also be zero, thus having already zeroed out the interior, you get an independent boundary condition. Again the occurrence of the arbitrary $\delta y$ factor in the surface integral will require that the integrand must be zero.

Your first variation is not well formed. $y$ is your (function valued) variable and so your variation of $J$ should manifest as a linear functional on the variation $\delta y$ of $y$. This term is critical to formulating your solutions.

You should see something like:
$$J[y] = \int_D L(y,y',...)dV;\quad \delta J[y] = \int_D A(y,y',...) \delta y dV + \int_{\partial D} B(y, y',...)\delta y dS$$

I'm also a bit confused by your apparent single variable prime notation for derivatives vs your use of multivariable integrals over volumes and surfaces. Now it's perfectly fine to speak of the generalized derivative of a function to and/or from vector spaces but in so doing one should be explicit as to the vector on which such an operator valued derivative acts. It would thus be most helpful to me if you could clarify your asserted variation by giving the original functional and explaining the independent variables on which the function $y$ depends and whether it is a scalar or vector valued function. All of this does relate to your specific question.
I copied this from a text I'm going through, though not verbatim. Here I'll copy it verbatim and do my best to interpret each component as you requested. The first variation actually looks like this

$$\delta J[\vec x] = \int_\Gamma(aN-\Delta N)N\,d\Gamma + \int_\gamma (\chi N + N_e)N\, d\gamma$$

where $\gamma$ is the boundary of $\Gamma$, $\Delta$ is the Laplace-Beltrami operator, $\chi$ and $a$ are geometric parameters, $e$ is a direction tangent to $\gamma$ so $N_e$ is a partial derivative of $N$, and lastly, $\vec{e_1}\cdot \vec{\delta x} \sin\alpha = N$ where $\vec{x}$ is a position vector on the surface $\Gamma$ and $\vec{\delta x}$ is a small displacement from $\Gamma$.

What is typically argued, in these variational problems is that the functional is stationary for arbitrary variations including those which are constrained to be zero on the boundary of the region of integration. This allows you to independently set the volume integral in the variation to zero rather than the sum of the volume and boundary integrals. Your variation $\delta y$ which is absent here should depend on the integration variables so its occurrence within the integral and its (interior) arbitrariness is what allows you to argue that to get zero for the whole integral for all cases the actual integrand must be zero... your Euler-Lagrange equation manifests.
Since $N$ is proportional to $\delta y$, are things looking better?

Last edited:
Have I left out something that requires further elaboration? Please let me know, as I'm very curious about the natural boundary condition for this problem.