Finite-dimensional parabolic Galerkin ODEs with measurable coefficients

If $V_h=\{0\}$, the only possible solution is $u_h=0$ and the equation and estimate hold. Otherwise choose a finite $H$-orthonormal basis $({\varphi }_1,\dots ,{\varphi }_m)$ of $V_h$. In this basis the variational equation is $c^{\mathrm{\prime }}(t)+B(t)c(t)=F(t)$, where $B_{ij}(t)=a(t;{\varphi }_j,{\varphi }_i)$ and $F_i(t)=⟨f(t),{\varphi }_i⟩$. The entries of $B$ are measurable and bounded because the basis is finite and the form is uniformly bounded; $F$ is in $L^2(0,T;{\mathbb{R}}^m)\subset L^1(0,T;{\mathbb{R}}^m)$ by finite dimensionality and Cauchy–Schwarz.

Choose a finite partition of $[0,T]$ with each interval length $\delta$ satisfying $\delta \mathrm{\parallel }B{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}<1$ (any partition works if $B=0$). On each interval $J=[s,s+\delta ]$, fix the incoming vector $c_s$ independently of the trial curve: initially it is the coefficient vector of $u_h^0$, and subsequently the endpoint of the preceding interval solution. The map $\mathrm{\Phi }(c)(t)=c_s+{\int }_s^t(F(r)-B(r)c(r))dr$ maps $C(J,{\mathbb{R}}^m)$ into itself: $F$ is integrable and $B$ is essentially bounded, so its integrand is integrable for every continuous $c$. The space is nonempty and complete by [F2], and $\mathrm{\parallel }\mathrm{\Phi }(c)-\mathrm{\Phi }(d){\mathrm{\parallel }}_{\mathrm{\infty }}\le \delta \mathrm{\parallel }B{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}\mathrm{\parallel }c-d{\mathrm{\parallel }}_{\mathrm{\infty }}$. Thus $\mathrm{\Phi }$ is a contraction, and [F3] gives its unique fixed point (the limit of the Picard iterates). This fixed point is absolutely continuous, equals $c_s$ at $s$, and solves the ODE almost everywhere. Concatenating the finitely many intervals gives a continuous absolutely continuous global solution with the prescribed initial value, and uniqueness follows successively on those intervals. Its continuous coefficient vector is bounded on $[0,T]$, so $c^{\mathrm{\prime }}=F-Bc\in L^2(0,T;{\mathbb{R}}^m)$; hence the solution belongs to the stated $H^1(0,T;V_h)$.

In the variational equation choose $v_h=u_h(t)$. Absolute continuity gives $\frac{1}{2}\frac{d}{dt}\mathrm{\parallel }{u}_h(t){\mathrm{\parallel }}_H^2=((u_h{)}^{\mathrm{\prime }}(t),u_h(t){)}_H$. Coercivity and Young's inequality yield $\frac{1}{2}\frac{d}{dt}\mathrm{\parallel }{u}_h{\mathrm{\parallel }}_H^2+\frac{\alpha }{2}\mathrm{\parallel }{u}_h{\mathrm{\parallel }}_V^2\le \frac{1}{2\alpha }\mathrm{\parallel }f{\mathrm{\parallel }}_{V_h^{\ast }}^2\mathrm{.}$ Integrating from $0$ to $t$ proves the stated energy estimate.