Petrov–Galerkin method

An example of differential equation containing a term with odd order is as follows:

${\displaystyle a(x){\dfrac {\mathrm {d} u}{\mathrm {d} x}}+b(x){\dfrac {\mathrm {d} ^{2}u}{\mathrm {d} x^{2}}}=f(x),\quad x\in (0,L)\quad \&\quad u(0)=u_{o},\left.{\dfrac {\mathrm {d} u}{\mathrm {d} x}}\right|_{x=L}=u_{L}'}$

If a test function ${\displaystyle v(x)}$ is used to obtain the weak form, after integration by parts the final Galerkin formulation will be given as follows:

${\displaystyle \int _{0}^{L}a(x)v(x){\dfrac {\mathrm {d} u}{\mathrm {d} x}}\mathrm {d} x-\int _{0}^{L}b(x){\dfrac {\mathrm {d} v}{\mathrm {d} x}}{\dfrac {\mathrm {d} u}{\mathrm {d} x}}\mathrm {d} x+\left[b(x)v{\dfrac {\mathrm {d} u}{\mathrm {d} x}}\right]_{0}^{L}=\int _{0}^{L}v(x)f(x)\,\mathrm {d} x}$

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 ${\displaystyle H_{0}^{1}}$. However, there is no way the first term on LHS can be made this way. In this case the solution space ${\displaystyle H_{0}^{1}}$ and test function space ${\displaystyle L^{2}}$ are different and hence the usually employed Bubnov Galerkin method cannot be used.

• Bubnov-Galerkin method