A fundamental system $\Delta$ is a basis made of roots such that every root is a linear combination of $\Delta$ whose nonzero coefficients all have the same sign. Its members are the simple roots.

The $G_2$ root system has six short and six long roots. For a short simple root ${\alpha }_1$ and a long simple root ${\alpha }_2$, its positive roots are

These are the root data associated with G2.

A Dynkin diagram records the angles and relative lengths of the simple roots. A multiple-edge arrow points toward the shorter root.

The reflection associated with a root is $s_{\alpha }(x)=x-⟨x,{\alpha }^{\vee }⟩\alpha$.

The weight lattice is the set of vectors $\lambda$ for which $⟨\lambda ,{\alpha }^{\vee }⟩$ is an integer for every root $\alpha$.

The root lattice is the integer span of the roots, equivalently of the simple roots.

The coroot associated with a nonzero root is ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$.

A choice of positive roots selects exactly one of $\alpha$ and $-\alpha$ and is closed under addition whenever the sum is a root.

For a graded ring $S$, $\mathrm{Proj}S$ consists of homogeneous prime ideals not containing the irrelevant ideal $S_{+}$. Its standard affine opens satisfy $D_{+}(f)\cong \mathrm{Spec}(S_f{)}_0$.

For each $\nu >0$, choose the solution from parts (a)--(c). The maximum estimate gives a $\nu$-independent boundThus a sequence ${\nu }_j\downarrow 0$ has ${\omega }_{{\nu }_j}{\rightharpoonup }^{\ast }\omega$ in $L^{\infty }$. The periodic elliptic estimates forbound ${\Psi }_{{\nu }_j}$ in $W^{2,p}$ and $u_{{\nu }_j}$ in $W^{1,p}$ for every finite $p$. Taking $p>2$ and using compact Sobolev embedding gives, after a further subsequence, uniform convergence $u_{{\nu }_j}\to u$. It follows that $u_{{\nu }_j}{\omega }_{{\nu }_j}\rightharpoonup u\omega$ distributionally.

The first energy estimate givesTherefore the viscous term vanishes in $H^{-1}$, and the weak formulation passes to the limit asThe elliptic relations pass to the limit as well and giveSince $\omega \in L^{\infty }\subset L^2$, periodic elliptic regularity gives $\Psi \in H_{\mathrm{per}}^2$ and $u\in H_{\mathrm{per}}^1\cap H$. Moreover $u\omega \in L^2$, so its divergence lies in $H_{\mathrm{per}}^{-1}$. This constructs the required weak solution of the damped-driven Euler system by a vanishing-viscosity limit.

Let $q=2p+2$ and multiply the vorticity equation by ${\omega }^{q-1}$, first for an odd power as in the hint. Periodic integration by parts and $\nabla \cdot u=0$ giveThe diffusion term is nonnegative, while the Holder inequality bounds the right-hand side by $\Vert g{\Vert }_q\Vert \omega {\Vert }_q^{q-1}$. ConsequentlyOn the finite-volume torus, $L^q$ norms increase to the essential supremum as $q\to \infty$. Hence the damped-vorticity maximum estimate gives

Part (c)(i) already gives weak continuity of $t\mapsto \theta (t)$ in $L^2$. In the assumed energy equality, the dissipation integralis absolutely continuous. The forcing integrand is in $L^1(0,T)$ because $u\in L^{\infty }(0,T;L^2)$ and $\theta \in L^{\infty }(0,T;L^2)$. The equality therefore makes $t\mapsto \Vert \theta (t){\Vert }_2^2$ continuous.

Whenever $t_n\to t$, weak continuity gives $\theta (t_n)\rightharpoonup \theta (t)$ and the energy equality gives convergence of their norms. The Radon-Riesz theorem, or directly the strong continuity from weak continuity and an energy equality, now gives $\theta (t_n)\to \theta (t)$ in $L^2$. Thus

Let $(u_1,{\theta }_1)$ and $(u_2,{\theta }_2)$ be two weak solutions with the same initial data, and set $U=u_1-u_2$ and $\Theta ={\theta }_1-{\theta }_2$. The diagnostic Stokes equation givesSubtracting the temperature equations yieldsPair this equation with $\Theta$. The term transported by $u_1$ vanishes by the skew-symmetry of incompressible transport, while the other nonlinear term satisfiesThe forcing difference is at most $C\Vert \Theta {\Vert }_2^2$. ConsequentlyThe coefficient is integrable on $[0,T]$ because ${\theta }_2\in L^2(0,T;H^1)$. Since $\Theta (0)=0$, the Gronwall inequality gives $\Theta =0$, and the Stokes equation then gives $U=0$. The weak solution is unique.