 Solution

Take linearly independent basis functions ${\phi }_i\in H_0^1(0,1)$ and write $u_h(x,t)={\sum }_j{U}_j(t){\phi }_j(x)$. The weak formulation of the heat equation, tested against ${\phi }_i$, isThus the semidiscrete finite element heat equation isThe vector $U$ contains the basis coefficients. A natural initial vector $G$ is the $L^2$ projection of the initial data: $MG=b$, where $b_i={\int }_0^{1}g{\phi }_i$, and $U(0)=G$. Other consistent initial approximations can be specified, but exact representation of an arbitrary $g$ in the trial space is not implicit.

For a nonzero coefficient vector $a$, the function $v_h={\sum }_i{a}_i{\phi }_i$ is nonzero by basis independence. Hence $a^{T}Ma=\Vert v_h{\Vert }_{L^2}^2>0$, proving that the mass matrix is symmetric positive definite. Also $a^{T}Ka=-\Vert v_h^{\prime }{\Vert }_{L^2}^2<0$: if the derivative vanished, $v_h$ would be a constant, and its zero endpoint traces would make it zero. Therefore $K$ is symmetric negative definite, not merely semidefinite, for these homogeneous Dirichlet conditions.

The squared spatial norm is $U^{T}MU$. Differentiating and using the semidiscrete system gives the energy identityUsing the trapezoidal rule in time gives the Crank-Nicolson methodThe left matrix is positive definite for every $\Delta t>0$, so every step exists uniquely. To prove stability, set $d=U^{k+1}-U^k$ and $s=U^{k+1}+U^k$. The update says $Md=(\Delta t/2)Ks$. Multiplying by $s^T$ givesConsequentlyThe same inequality applied to the difference of two discrete solutions proves nonamplification of initial perturbations. This is unconditional stability in the mass-matrix norm, exactly the spatial $L^2$ norm of the finite element functions.