Deriving the Euler-Lagrange Equations for the Chan-Vese Model I am reading the Chan-Vese paper "An Active Contour Model without Edges" and IPOL tutorial "Chan-Vese Segmentation" by Pascal Getreuer.  The energy for this model is given by:
$$
E(\varphi) = {\mu \int_{\Omega} \delta(\varphi(x)) | \nabla \varphi(x)|dx} + \nu\int_{\Omega} H(\varphi(x)) dx \;+$$
$$\lambda_1\int_{\Omega}|f - c_1| H(\varphi(x))+\lambda_2\int_{\Omega}|f - c_2|(1 - H(\varphi(x))dx
$$
Above, $\mu, \nu, \lambda_1, \lambda_2$ are real parameters, $c_1, c_2$ are constants determined for segmentation, $-1 \leq \varphi(x) \leq 1$ is the level-set function in which $\varphi(x) = 0$ specifies the interface,  $f$ is the original image, $H$ is the heavyside function in 1 dimension centered at $0$ and $\delta = H'$.  The Euler-Lagrange equations can be derived by inspecting the following derivative:
$$
\left.\frac{d}{d\epsilon} E(\varphi(x) + \epsilon \eta(x))\right|_{t = 0}
$$
for some test function $\eta(x)$.  Writing the Euler-Lagrange Equations as a gradient descent both papers claim:
$$
\begin{cases}
\frac{d\varphi}{dt} = \delta(\varphi(x))\left[\mathrm{div}\left( \frac{\nabla \varphi}{\mu |\nabla \varphi|}\right) - \nu - \lambda_1(f- c_1)^2 + \lambda_2(f-c_2)^2\right]\\
\frac{\delta(\phi(x))}{|\nabla \varphi|} \nabla \varphi \cdot \mathbf{n} = 0 &amp; x \in \partial \Omega.
\end{cases}
$$
My problem is coming from the differentiation of
$$
{\int_{\Omega} \delta(\varphi(x)) | \nabla \varphi(x)|dx}.
$$
in the variation of the energy.  It should look like:
$$
\int_{\Omega} \underbrace{\left(\frac{d}{d\epsilon} \delta( \varphi (x) + \epsilon \eta(x) )\right)|\nabla \varphi(x)|}_{(*)} + \underbrace{\delta(\varphi(x)) \left(\frac{d}{d\epsilon}|\nabla\varphi(x) + \epsilon \nabla\eta(x)|\right)}_{(**)}.
$$
Now, $(**)$ gives us the curvature/divergence term in the Euler-Lagrange equations as well as the boundary condition after applying Green's first identity.  However, I don't understand how one might argue in this case why $(*)$ goes to 0 or why, for practical considerations, we can just ignore it.
Thanks to Luminita Vese, who responded to this question via email. I will post the answer here.  Let $\varphi_\epsilon(x) = \varphi(x) + \epsilon\eta(x)$ for some test function $\eta(x)$.

\begin{eqnarray}
\int_{\Omega} \frac{d}{d\epsilon} \left[ \delta(\varphi_\epsilon(x)) |\nabla (\varphi_\epsilon(x))|\right]dx &amp;=&amp; \int_{\Omega} \left[\frac d{d\epsilon} \delta(\varphi_\epsilon(x))\right] |\nabla (\varphi(x))|+  \delta(\varphi(x)) \left[\frac{d}{d\epsilon}|\nabla (\varphi_\epsilon(x))|\right]dx\\
&amp;=&amp;\int_\Omega \delta'(\varphi(x))\eta(x) |\nabla \varphi(x)| + \delta(\varphi(x))\left[\frac{d}{d\epsilon} \sqrt{\left&lt;\nabla\varphi_\epsilon(x), \nabla\varphi_\epsilon(x)\right&gt;}\right]dx\\
&amp;=&amp;\int_\Omega \delta'(\varphi(x))\eta(x) |\nabla \varphi(x)| + \delta(\varphi(x))\left[ \frac{\frac{d}{d\epsilon}{\left&lt;\nabla\varphi_\epsilon(x), \nabla \varphi_\epsilon(x)\right&gt;}}{2|\nabla(\varphi(x)|}\right]dx\\
&amp;=&amp;\int_\Omega \delta'(\varphi(x))\eta(x) |\nabla \varphi(x)| \\
&amp;\quad&amp;+ \delta(\varphi(x))\left[ \frac{{\left&lt;\nabla\varphi (x), \nabla \eta(x)\right&gt;}}{|\nabla(\varphi(x)|}\right]\\
&amp;=&amp;\int_\Omega \delta'(\varphi(x))\eta(x) |\nabla \varphi(x)| \\
&amp;\quad&amp;+ \frac{{\left&lt;\delta(\varphi(x))\nabla\varphi (x), \nabla \eta(x)\right&gt;}}{|\nabla(\varphi(x)|}dx \quad (*)\\
&amp;=&amp;\int_{\Omega}\delta'(\varphi(x))\eta(x)|\nabla \varphi(x)|\\
&amp;\quad&amp;-\delta(\varphi(x))\mathrm{div}\left(\frac{\nabla \varphi}{|\nabla \varphi|}\right) \eta(x) \\
&amp;\quad&amp;- \delta'(\varphi(x))\frac{|\nabla(\varphi(x))|^2}{|\nabla \varphi(x)|}\eta(x)dx\quad(**)\\
&amp;=&amp;-\int_{\Omega}\delta(\varphi(x))\mathrm{div}\left(\frac{\nabla \varphi}{|\nabla \varphi|}\right) \eta(x) dx
\end{eqnarray}
Above, I am writing $\left&lt; \cdot , \cdot \right&gt;$ to denote the vector inner product, From (*) to (**), Green's first identity was applied as well as the product rule for divergence.