0

I'm trying to understand what my professor wrote, maybe it's a stupid question..

I lost him in the step from $1D$ diffusion equation:

$\frac{d}{dx}(g\frac{du}{dx})=f(x)$

where $f$ is a given forcing function, $g$ diffusion coefficient and $u$ is the solution we are seeking.

As a next step he wrote:

$f(u)=\int[\frac{1}{2}g(x)|\frac{du}{dx}|^2+f(x)u(x)]dx$

Maybe I didn't understand something he write, but basically I can't follow how this step done.

I see that it's look a bit similar to $\frac{d}{dx}(g\frac{du}{dx})=\frac{dg}{dx}\frac{du}{dx}+g\frac{d^2u}{dx^2}$ from here.

Any good tutorial of FEM is welcome as well.

ChaosPredictor
  • 303
  • 4
  • 13
  • There is clearly a typo on the left, since two different things are denoted as f. – uranix May 24 '22 at 07:37
  • 1
    The integral is a quadratic functional that attains its minimum at $u(x)$ that satisfies Euler-Lagrange equation. For this particular case the E-L equation is exactly the same as the original diffusion equation – uranix May 24 '22 at 07:42

0 Answers0