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I'm trying to solve the following equation $$\dfrac{\partial}{\partial x}\left(e^{au}\dfrac{\partial u}{\partial x}\right) = 0$$

Of course, this equation can be solved analytically. I am trying to understand the way of working with exponential coefficients.

For the approximation, I use the finite volume method which leads to

$$F_{i+1/2} - F_{i-1/2} = 0,\qquad F_{i-1/2} = \left.e^{au}\dfrac{\partial u}{\partial x}\right|_{i+1/2} $$

For the further linearization, I'm trying the following two methods:

  • simple iteration, where $e^u$ is calculated based on the previous iteration
  • Newton's method (with the series expansion, dropping terms that are proportional to $\delta u^2$ and smaller)

$$u^{k+1} = u^k + \delta u$$

The first one worked well until some value of $a$, and the second one didn't work at all (I have not found the mistake yet)

Is there any way to deal with exponential coefficient? Will be glad for any advice. Thanks!

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  • $\begingroup$ After a bunch of test i come to working Newton method, but it has the same field of correctnes as a simple iteration. And didnt solve equation with strong exponenta. $\endgroup$ – Sergey Konoplev Jan 1 '17 at 13:25
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    $\begingroup$ If $a$ is too large, your problem is strongly nonlinear, and you probably need to use a method such as line search to stabilize the Newton iteration. The same is true for a fixed point method like you do in the first approach, where it would be called "damping". $\endgroup$ – Wolfgang Bangerth Jan 3 '17 at 14:27
  • $\begingroup$ Thanks for your reply. I honestly suppose that methods such as line search needed to improve convergence? I using some simple dumping method, but it seems like i need some special current approximation method to handle with strong nonlinearity. Or using a better approximation for derivatives. (I know about exponential scheme but still in search for deeper understanding of the problem) $\endgroup$ – Sergey Konoplev Jan 3 '17 at 20:34
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    $\begingroup$ For strongly nonlinear problems, there is no way around damping or line search. $\endgroup$ – Wolfgang Bangerth Jan 4 '17 at 1:41

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