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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.

Define the squared L2 norm $E(t)={\int }_0^{1}u(x,t{)}^{2}dx$. Multiplication of the semilinear partial differential equation by $u$, followed by integration by parts, givesAt $x=1$ the Dirichlet boundary condition makes the boundary term zero. At $x=0$ the Robin boundary condition is $u_x(0,t)=\alpha u(0,t)$, so subtraction of the lower endpoint gives $-\alpha u(0,t{)}^2$. In particular the sign is dissipative. Thus the exact energy identity iswhich implies

Write $w=u-v$ and $r=f-g$. Subtraction of the two boundary value problems gives $-w^{\prime \prime }+w+u^3-v^3=r$, with $w(0)=w(1)=0$. Multiply by $w$ and apply integration by parts:Consequently the required energy identity isThe nonlinear term is nonnegative pointwise, sinceThis is the monotonicity energy estimate for a semilinear Dirichlet problem for the monotone function $s\mapsto s^3$. Apply Young's inequality for products, $rw\le (r^2+w^2)/2$, inside the last integral and discard the nonnegative nonlinear term. Moving $\frac{1}{2}{\int }_0^1{w}^{2}dx$ to the left givesIn particular, equal forcing gives $w=0$, proving uniqueness and a quantitative continuous dependence on forcing estimate for this boundary value problem.

Let $f_i=f(ih)$ and $g_i=g(ih)$, and define the scaled forward difference operator, backward difference operator and second-order central difference byThe two finite difference methods are the nonlinear systemswith the prescribed Dirichlet boundary conditions $U_0=V_0=A$ and $U_N=V_N=B$. These equations determine the interior nodal values.

For clarity distinguish the interior and edge inner products and their norms:The bracketed norm includes the last edge $i=N$. Let $W=U-V$, so $W_0=W_N=0$, and put $q_i=D_x^{-}{W}_i$ for $1\le i\le N$. Direct telescoping gives discrete summation by parts:Subtract the two nonlinear systems, multiply each equation by $h{W}_i$ and sum. The resulting discrete energy identity isThe nonlinear summands are nonnegative by the factorization in (a). Using Young's inequality for products at every node givesThis also proves uniqueness of each discrete solution. Existence can be seen by minimizing, with the endpoints fixed, the strictly convex energy functionalIt is a coercive function of its finitely many interior values, so it attains a minimum. Its first derivatives are $h(-D_x^{+}{D}_x^{-}{W}_i+W_i+W_i^3-f_i)$; thus its minimizer solves the scheme. The same argument applies to $g$.