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\frac{\partial u}{\partial t} = \nu \nabla^{2} u + q
$$
1.integrate and introduce weight functions
1.Integrate and introduce weight functions
$$
\int_{\Omega} w \frac{\partial u}{\partial t} d\Omega = \int_{\Omega} w \nu \nabla^2 u d\Omega + \int_{\Omega} wq d\Omega
$$
2. Integration by parts: changing the second derivative, changing sign and introducing boundary condition
2. Integration by parts: changing the second derivative, changing sign and introducing boundary condition (not necessary for the term with time-derivative)
$$
\int_{\Omega} w \frac{\partial u}{\partial t} d\Omega = - \int_{\Omega} \nu \nabla w \cdot \nabla u d\Omega + \int_{\Gamma} w \mu \nabla u \cdot \bar{n} d\Gamma + \int_{\Omega} w q d\Omega
$$
3.substitute boundary condition (w= 0 on $\Gamma_D$ and $\mu \nabla u \cdot \mathbf{n}$ on $\Gamma_N$):
3.Substitute boundary condition (w= 0 on $\Gamma_D$ and $\mu \nabla u \cdot \mathbf{n}$ on $\Gamma_N$) (not necessary for the term with time-derivative):
$$\int_{\Omega} w \frac{\partial u}{\partial t} d\Omega + \int_{\Omega} \nu \nabla w \cdot \nabla u d\Omega = \int_{\Gamma N} w h d\Gamma_N + \int_{\Omega} w q d\Omega$$
4.discretization:
4.Discretization:
$$
u^h = N\mathbf{u}
u^h = \mathbf{N}\mathbf{u}
$$
$$w^h = N\mathbf{w}$$
$$w^h = \mathbf{N}\mathbf{w}$$
$$\nabla u^h = B\mathbf{u}$$
$$\nabla w^h = B\mathbf{w}$$
$$\nabla u^h = \mathbf{B}\mathbf{u}$$
$$\nabla w^h = \mathbf{B}\mathbf{w}$$
5. Substitute and take $\mathbf{u,w}$ out of the integral as they don't depend on x and y
5. Substitute and take $\mathbf{u,w}$ out of the integral as they don't depend on $x$ and $y$
$$
\mathbf{w^T} \int_{\Omega} N^T N d\Omega \frac{\partial u}{\partial t} + \mathbf{w^T} \int_{\Omega} \nu B^T B d\Omega \mathbf{u} = \mathbf{w^T} \int_{\Gamma N} N^T h d\Gamma_N + w^T \int_{\Omega} N^T q d\Omega
\mathbf{w^T} \int_{\Omega} N^T N d\Omega \frac{\partial \mathbf{u}}{\partial t} + \mathbf{w^T} \int_{\Omega} \nu B^T B d\Omega \mathbf{u} = \mathbf{w^T} \int_{\Gamma N} N^T h d\Gamma_N + w^T \int_{\Omega} N^T q d\Omega
$$
6."Devide" by $w^T$
6.Eliminate $w^T$
$$ \int_{\Omega} N^T N d\Omega \frac{\partial u}{\partial t}+ \int_{\Omega} \nu B^T B d\Omega \mathbf{u} = \int_{\Gamma N} N^T h d\Gamma_N + \int_{\Omega} N^T q d\Omega
$$ \int_{\Omega} N^T N d\Omega \frac{\partial \mathbf{u}}{\partial t}+ \int_{\Omega} \nu B^T B d\Omega \mathbf{u} = \int_{\Gamma N} N^T h d\Gamma_N + \int_{\Omega} N^T q d\Omega
