I'm trying to evaluate Euler Lagrange equation from the following relation: $$ F[f(\vec{r})]=\int_{\vec{r_1}}^{\vec{r_2}} d^n r J[f(\vec{r}),\nabla f (\vec{r}),H f(\vec{r}) ] $$

where $H$ is the hessian and $\vec{r}$ represent $n$ independent variables. Our boundary condition is that $\eta $ and its derivatives shout be zero at end points.

what I tried is:

$$ \left [ \frac{\partial F}{\partial \alpha}\right ]_{\alpha=0} =0 $$ and

$$ f(\vec{r},\alpha)=f(\vec{r},0)+\alpha \eta (\vec{r}) \\ \nabla f(\vec{r},\alpha)=\nabla f(\vec{r},0)+\alpha \nabla \eta (\vec{r}) \\ H f(\vec{r},\alpha)=H f(\vec{r},0)+\alpha H \eta (\vec{r}) $$ so we can write $$ \int_{\vec{r_1}}^{\vec{r_2}} \left [ \frac{\partial J}{\partial f}\frac{\partial f}{\partial \alpha}+\frac{\partial J}{\partial \nabla f} \frac{\partial \nabla f}{\partial \alpha}+\frac{\partial J}{\partial H f}\frac{\partial H f}{\partial \alpha} \right ] d^n r \\ \int \left [ \frac{\partial J}{\partial f}\eta(\vec{r})+\frac{\partial J}{\partial \nabla f} \nabla \eta(\vec{r})+\frac{\partial J}{\partial H f} H \eta(\vec{r}) \right ] d^n r $$ for the second and third terms in bracket I use the integrating by parts. For the second term I wrote: $$ = \eta(\vec{r}) \frac{\partial J}{\partial \nabla f}-\int \eta(\vec{r}) \nabla .\left ( \frac{\partial J}{\partial \nabla f} \right ) d^n r $$ and for the second term including the hessian: $$ =\nabla \eta(\vec{r}) \frac{\partial J}{\partial H f}-\int \nabla \eta(\vec{r}) \nabla .\left ( \frac{\partial J}{\partial \nabla f} \right ) d^n r\\ = \eta(\vec{r})\left (\nabla . \frac{\partial J}{\partial H f} \right )+\int \eta(\vec{r}) \nabla \left ( \nabla .\frac{\partial J}{\partial Hf} \right )d^nr $$ The terms including $\eta$ and its derivative are zero with regard to the boundary conditions. Finally we can write the Euler Lagrange equation as:

$$ \frac{\partial J}{\partial f(\vec{r})}-\nabla .\left(\frac{\partial J}{\partial \nabla f(\vec{r})}\right )+ \nabla\left ( \nabla .\frac{\partial J}{\partial Hf(\vec{r})} \right )=0 $$ I'm wondering is it true? Edit: Is it true to integrate from hessian and get gradient?


1 Answer 1


Yes it is. You can write it explicitly in terms of derivatives, i.e. by seeing the hessian as $\frac{n(n+1)}2$ parameters :

$$ J(Hf)=J(\{\partial _i \partial _j f\}) $$ hence (with summation over repeated indices) $$\delta J =\frac{\partial^2 J}{\partial _i\partial _j f } \delta(\partial _i\partial _j f ) = \frac{\partial^2 J}{\partial _i\partial _j f } \delta\alpha\partial _i\partial _j \eta$$ Integrating by parts twice yields your correct result.

  • $\begingroup$ note enough reputation to vote up. Thanks. $\endgroup$
    – Abolfazl
    Oct 23, 2014 at 19:55

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