6
$\begingroup$

I trying to wrap my head of derivation of the analytic FEM Jacobian for the Newton method. Say we have a nonlinear Poisson problem of the (weak) form

$$ \int a(u)\nabla\ u\cdot \nabla v = \int f v $$

where $a(u)$ is a coefficient. From the derivation of the Newton method the Jacobian matrix (in the Newton update $\Delta u = -J(u_k)R(u_k) $ will be

$$ J(u) = \int a(u)\nabla\ \phi_i\cdot \nabla \phi_j + \int \frac{\partial a(u)}{\partial u}\phi_i\nabla\ u\cdot \nabla \phi_j $$

which essentially is the original stiffness matrix plus a contribution due to the derivative of $a(u)$. Now if for example $a(u)=u^2$ then the 2nd Jacobian term will be $J_2=\int 2u\ \phi_i\nabla\ u\cdot \nabla \phi_j$ which is ok. But if now $a(u)=(\frac{\partial u}{\partial x})^2$ or even $a(u)=\frac{\partial u}{\partial x}\frac{\partial u}{\partial y}$, how does this work as the problem is still nonlinear but $\frac{\partial a(u)}{\partial u}$ is essentially zero?

EDIT: Could someone verify that these Jacobian forms are indeed correct so that I have understood it correctly.

$$ a(u) = u\frac{\partial u}{\partial x} \Rightarrow J_2 = \int (u\frac{\partial \phi_i}{\partial x} + \frac{\partial u}{\partial x}\phi_i)(\nabla\ u\cdot \nabla \phi_j) $$

$$ a(u) = (\frac{\partial u}{\partial x})^2 \Rightarrow J_2 = \int 2\frac{\partial u}{\partial x}\frac{\partial \phi_i}{\partial x}(\nabla\ u\cdot \nabla \phi_j) $$

$$ a(u) = \frac{\partial u}{\partial x}\frac{\partial u}{\partial y} \Rightarrow J_2 = \int (\frac{\partial u}{\partial x}\frac{\partial \phi_i}{\partial y} + \frac{\partial u}{\partial y}\frac{\partial \phi_i}{\partial x})(\nabla\ u\cdot \nabla \phi_j) $$

where $u$ and $\nabla u$ are evaluated explicitly with the current solution in the assembly.

$\endgroup$
11
$\begingroup$

Your example is a pretty good indication that the two derivatives (with respect to $x$ and with respect to $u$) do not commute :) (In fact, they're very different beasts -- one is a Fréchet derivative, the other a weak derivative.)

Rather, you should consider $a(u) = (\partial_x u)^2$ as the composition $a = f\circ g$, of two functions $$ f: v(x)\mapsto v(x)^2$$ and $$ g: v(x) \mapsto (\partial_x v)(x)$$ and apply the chain rule $$ \partial_u a(u) = \partial_u f(g(u)) \circ \partial_u g(u).$$ Here, it is important to note that derivatives of such a mapping $a$ between functions are actually linear operators, not functions (and the function spaces are important, but I'll ignore that here) -- the derivative $\partial_u a(u)$ acts on a given function $h$ to show how the output $a(u)$ changes if the input $u$ is perturbed in the direction of $h$. When computing such derivatives, it is therefore very helpful to keep these directions around until the end!

For $f$ (which is called a superposition or Nemytskii operator), one can under some conditions (again, ignored here) show that for any function $h$ $$\partial_u f(v) h = 2 v * h,$$ where $*$ denotes the pointwise multiplication. (Basically, the rule is if you have a mapping $v(x)\mapsto f(v(x))$ for some differentiable $f:\mathbb{R}\to\mathbb{R}$, then its derivative in $u$ -- if it exists -- is given by the pointwise multiplication with $f'(u(x))$.)

For $g$, you can use that the (weak) derivative is linear and so the derivative is the linear operator itself, i.e., $$\partial_u g(v)h = \partial_x h.$$

Together, you obtain $$\partial_u a(u)h = 2\partial_x u * \partial_x h,$$ where $*$ denotes the pointwise multiplication. In your equation for $J(u)$, $\phi_i$ is taking the place of $h$, so the corresponding Jacobian term would be $$ \int_\Omega 2\partial_x u(x) \partial_x \phi_i(x) (\nabla u(x) \cdot \nabla \phi_j(x))\,dx.$$

In your second example, you proceed similar using the product rule instead of the chain rule to get $$\partial a(u)h = \partial_x u * \partial_y h + \partial_y u * \partial_x h.$$

$\endgroup$
  • $\begingroup$ Thank you for the detailed explanation, I think I follow but shouldn't $a(u) = (\nabla u)^2$ then result in $$ \int_\Omega 2\nabla u(x) \nabla\phi_i(x)*(\nabla u(x) \cdot \nabla \phi_j(x))\,dx.$$ instead of $$ \int_\Omega 2u(x) *(\nabla \phi_i(x) \cdot \nabla \phi_j(x))\,dx.$$ where $2\nabla u(x) \nabla\phi_i(x)$ is due to $\partial_u a(u)$ and $(\nabla u(x) \cdot \nabla \phi_j(x))$ from the original bilinear form (now evaluated explicitly in $u$)? $\endgroup$ – B Ring Jan 11 '18 at 3:32
  • $\begingroup$ Yes, sorry, I forgot the gradient there. Fixing it now! $\endgroup$ – Christian Clason Jan 11 '18 at 7:47
  • $\begingroup$ ...and actually using your example and not the full gradient for $a(u)$. (What threw me initially is that I usually see the the nonlinearity as containing the full dependence on $u$, so I mixed up the gradients.) $\endgroup$ – Christian Clason Jan 11 '18 at 8:43
3
$\begingroup$

You need to understand how to actually compute derivatives when you want to take the derivative with respect to a function. I've recorded a lengthy example in lecture 31.55 here: http://www.math.colostate.edu/~bangerth/videos.html

In short, you need to consider the derivative as the limit of a variation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.