I'm studying the Finite Element Method and I came across the following form [1], where: u is a vector field representing velocity; v is a given test function; $\Omega$ is a given spatial domain.
$$a(u; v) = \int_{\Omega} (grad(u) \cdot u)\cdot v \; dx $$
Given that u is a vector field, as far as I understood here the gradient represents the Jacobian of u in respect to space. In my reference book [2] this form is then linearized for u considering $ u = \hat{u} + \delta u$, so we have: $$a(u; v) = a(\hat{u}; v) + a'(\hat{u}; v)\delta u + O(\delta u^2)$$ What I don't understand is the definition of $a'(u; v)$. My reference books gives this definition: $$a'(\hat{u}; v) \cdot \delta u = \int_{\Omega}(grad(\delta u) \cdot \hat{u}) \cdot v + (grad(\hat{u}) \cdot \delta u) \cdot v \; dx$$ But, applying the basic definition for derivative, what I get is: $$a'(u; v) = \frac{\partial}{\partial u}a(u;v) = \int_{\Omega}(\frac{\partial}{\partial u}grad(u)\cdot u + grad(u)\cdot \frac{\partial}{\partial u} u) \cdot v \; dx = \\ = \int_{\Omega}(grad(\frac{\partial}{\partial u} u)\cdot u + grad(u) \frac{\partial}{\partial u} u)\cdot v \; dx = \\ = \int_{\Omega}(grad(1)\cdot u + grad(u)\cdot 1)\cdot v \; dx = \\ = \int_{\Omega}(0 \cdot u + grad(u))\cdot v \; dx = \int_{\Omega}grad(u)\cdot v \; dx$$ Thus $a(\hat{u}; v) \cdot \delta u$ should be simply: $$a'(\hat{u}; v) = \int_{\Omega}grad(\hat{u})\cdot \delta u \cdot v \; dx$$ Does someone knows what I am missing?
Notes:
- [1] FYI, the form comes from the Navier-Stokes equations for incompressible fluids.
- [2] This is the text: link
Let's look at the function $a(u;v):=(\nabla u\cdot u)\cdot v $. If you let $u = \hat u +\delta u$, then
$$a(u;v) = \left(\nabla (\hat u + \delta u)\cdot (\hat u + \delta u)\right)\cdot v$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\left(\nabla\hat u\cdot\hat u+\nabla\hat u\cdot\delta u+\nabla(\delta u)\cdot \hat u+\nabla(\delta u)\cdot\delta u\right)\cdot v$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=a(\hat u;v) +\left(\nabla\hat u\cdot\delta u+\nabla(\delta u)\cdot \hat u\right)\cdot v+\mathcal{O}(\delta u^2)\cdot v,$$
So the function they are calling $a'(\hat u;v)\cdot\delta u$ is the function
$$a'(\hat u;v)\cdot\delta u=(\nabla\hat u\cdot\delta u+\nabla(\delta u)\cdot \hat u)\cdot v$$ which essentially is the derivative of $a$ at $(\hat u,v)$ in the direction $\delta u$. I think the confusion comes from the odd notation they have used, as what they are calling $a'(\hat u;v)\cdot\delta u$ is not actually in the form "something$\cdot\delta u$".
If you want to formally calculate the derivative, let's define the functions $a_1(u):=\nabla u$ and $a_2(u):=u$, so that $a(u;v) = (a_1(u)\cdot a_2(u))\cdot v$. Then, by the product rule we have that
$$a'(\hat u;v)\cdot\delta u = \lim_{\varepsilon\to 0}\frac{a(\hat u+\varepsilon\delta u;v)-a(\hat u;v)}{\varepsilon}$$ $$= \lim_{\varepsilon\to 0}\frac{(a_1(\hat u+\varepsilon\delta u)\cdot a_2(\hat u+\varepsilon\delta u))\cdot v-(a_1(\hat u)\cdot a_2(\hat u))\cdot v}{\varepsilon}$$ $$= \left(a_2(\hat u)\cdot\lim_{\varepsilon\to 0}\frac{a_1(\hat u+\varepsilon\delta u)-a_1(\hat u)}{\varepsilon}+a_1(\hat u)\cdot\lim_{\varepsilon\to 0}\frac{a_2(\hat u+\varepsilon\delta u)-a_2(\hat u)}{\varepsilon}\right)\cdot v$$ $$= \left(\hat u\cdot\lim_{\varepsilon\to 0}\frac{\nabla\hat u+\varepsilon\nabla \delta u-\nabla\hat u}{\varepsilon}+\nabla\hat u\cdot\lim_{\varepsilon\to 0}\frac{\hat u+\varepsilon\delta u-\hat u}{\varepsilon}\right)\cdot v$$ $$=(\hat u\cdot \nabla \delta u+\nabla\hat u\cdot\delta u)\cdot v$$