Talk:Petrov–Galerkin method

Wrong math
Right now this article has several flaws:
 * The derivation of the weak form is wrong due to the existence of the factor $$b(x)$$ and its dependency on $$x$$:
 * $$\begin{align}

\int_0^L bu''\cdot v\,\mathrm{d}x & = \left[bu'\cdot v\right]_0^L - \int_0^L u'\left(b'\cdot v + b\cdot v'\right) \,\mathrm{d}x \\ & \neq \left[b(x)\dfrac{\mathrm{d}u}{\mathrm{d}x}\cdot v(x)\right]_0^L -\int_0^L b(x)\dfrac{\mathrm{d}v}{\mathrm{d}x}\dfrac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x \end{align}$$
 * The second line is quoted from the article.


 * The following statement needs some correction:
 * $$\int_0^L a(x)v(x)\dfrac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x-\int_0^L b(x)\dfrac{\mathrm{d}v}{\mathrm{d}x}\dfrac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x + \left[b(x)v\dfrac{\mathrm{d}u}{\mathrm{d}x}\right]_0^L = \int_0^L v(x)f(x) \, \mathrm{d}x $$
 * The term with even order (2nd term in LHS) is now symmetric, as the test function and solution function both have same order of differentiation and they both belong to $$H_0^1$$. However, there is no way the first term on LHS can be made this way. In this case the solution space $$H_0^1$$ and test function space $$L^2$$ are different and hence the usually employed Bubnov Galerkin method cannot be used.
 * The test function $$v(x)$$ lives in $$H_0^1$$, otherwise the second term is not defined. The test function doesn't change the space its coming from within the same equation, thus the first term is the product of two functions living in $$H_0^1$$. Actually taking a second-order PDE as an example seems wrong in this case.


 * The article mentions the usually employed Bubnov Galerkin method and links to Galerkin method.
 * I have never heard Bubnov-Galerkin before and it's not explained in the linked article, but it seems to be the Russian name for Ritz-Galerkin.

I leave this here as a reference. I will incorporate the corrections when I have time for it. --Hoelzlw (talk) 10:30, 15 November 2017 (UTC)