I'm having a really hard time understanding the concept behind the mass matrix in the discretization of PDEs (I get the stiffness one, but not this one), also I know that is related to the identity (like the identity "moved"). Can someone make light on this doubts ? Thank you
EDIT: Since I have not put any background on it, generally speaking could someone explain the difference between mass matrix and stiffness matrix ? Like, why in a lot of examples there is only the stiffness matrix and not the mass matrix ? When we need it ?
Given a base $\{b_j\}$ of some discrete subspace $V_h\subset V$ where $V$ is usually some Sobolev space on a domain $\Omega$, the mass matrix is the matrix $M=(m_{j,k})$ with $$m_{j,k}=\int_\Omega b_j(x)b_k(x)\,dx.$$ It can be considered as the discrete version of the $L_2(\Omega)$ inner product on $V_h$.
Edit: Consider the simple problem: find $u\in V_h$ (with $V_h=\mathrm{span}\{b_j:j=1,\ldots,n\}$ as before) such that $$u_h=f$$ for some given function $f\in L_2(\Omega)$. Of course for $f\notin V_h$ we there is no solution in strong sense. But we can go over to the weak form. To this end we multiply with testfunctions $v_h$ and integrate to find the variational formulation $$\int_\Omega u_h(x) v_h(x)\, dx=\int_\Omega f(x) v_h(x)\, dx,\quad\forall v_h\in V_h$$ Now plug in $u_h=\sum_{j=1}^n \alpha_j b_j$ and all basis functions $b_j$ for $v_h$ and you end up with $$M \alpha=F,$$ with $\alpha=(\alpha_1,\ldots,\alpha_n)^T$, $F=(f_1,\ldots,f_n)^T$ and $$f_j=\int_\Omega b_j(x)f(x)\,dx.$$ Thus, the mass matrix appears naturally when you want to approximate a given function $f$ in a discrete space.
The stiffness matrix consists of entries $$s_{j,k}=\int_\Omega\nabla b_j(x)\cdot\nabla b_k(x)\,dx.$$ It is the discrete version of the $H^1(\Omega)$ semi-norm.
These matrices appear in finite element discretizations of PDEs. For example a discretization of Poisson's equation with homogenous Neumann boundary data and right hand side $f$ i.e. \begin{align} -\Delta u(x)&=f(x),&x&\in\Omega\\ \frac{\partial u}{\partial n}(x)&=0,&x&\in\partial\Omega\\ \end{align} would be $$S\alpha=F$$ with $\alpha,F$ as before.