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A useful version of the Aubin-Lions lemma concerns Banach spaces with embeddingswhere the first embedding is compact and the second continuous. For $1<p<\infty$ and $1<q\le \infty$, a family bounded in $L^p(0,T;X_0)$ whose distributional time derivatives are bounded in $L^q(0,T;X_1)$ is relatively compact in $L^p(0,T;X)$. The exponents and embedding hypotheses must be stated; an unspecified suitable derivative bound does not identify a version of the theorem.

For a bounded Lipschitz domain $\Omega$, Rellich-Kondrachov compactness theorem and continuous dual embeddings giveConsequently boundedness of $u_j$ in $L^2(0,T;H^1(\Omega ))$ and of ${\partial }_t{u}_j$ in $L^2(0,T;H^{-1}(\Omega ))$ yields a strongly convergent subsequence in $L^2(0,T;L^2(\Omega ))$. The spatial bound prevents fine-scale spatial oscillations, while the derivative bound controls time translations. This is why the lemma lets a Galerkin method pass to nonlinear terms, for which weak convergence alone is often insufficient. On an unbounded spatial domain, the displayed spatial compact embedding can fail through translations.

On the zero-boundary Sobolev space $H_0^1(0,1)$ with norm $\Vert u^{\prime }{\Vert }_{L^2}$, assume measurable coefficients $0<p_0\le p\le p_1$ and $0\le q\le q_1$. The Poincaré inequality and Cauchy-Schwarz inequality give $|a(u,v)|\le (p_1+q_1/{\pi }^2)\Vert u^{\prime }{\Vert }_2\Vert v^{\prime }{\Vert }_2$, while nonnegativity of $q$ gives $a(u,u)\ge p_0\Vert u^{\prime }{\Vert }_2^2$. Thus the form is a bounded bilinear form and a coercive bilinear form. For a source in $H^{-1}$, the Lax-Milgram theorem yields a unique weak solution and every conforming Galerkin method approximation. No derivative of the bounded coefficient $p$ is needed for this weak formulation.

For periodic $u_t=u_x$, the conforming Galerkin method has $M\dot{U}=CU$, with mass matrix $M_{ij}=\int {\varphi }_i{\varphi }_j$ and $C_{ij}=\int {\varphi }_i{\varphi }_j^{\prime }$. Integration by parts and periodicity give $C^T=-C$. ThereforeThe conserved quantity is the squared L2 norm of the finite element function. On a uniform one-dimensional mesh of spacing $h$, the piecewise-linear hat functions give $M_{ii}=2h/3$, $M_{i,i\pm 1}=h/6$, $C_{i,i+1}=1/2$ and $C_{i,i-1}=-1/2$.

The finite element method replaces an infinite-dimensional weak formulation by finitely many unknowns attached to a finite element mesh. Simple local functions, often piecewise polynomial functions, join to form a global approximation. The attraction is that each finite element can be integrated separately, while the assembled equations enforce a global balance. A piecewise-linear hat function only meets its immediate neighbors, so a large problem can have a sparse matrix rather than a dense one. Gilbert Strang's MIT lecture develops this connection from the one-dimensional weak formulation through local hats to the stiffness matrix and load integrals.^[ref]

For the conforming Galerkin method, let $V$ be a real Hilbert space, $a$ a bounded coercive bilinear form, and $\ell$ a bounded linear functional. Given a conforming finite element space $V_h\subset V$, the continuous and discrete problems areWriting $u_h={\sum }_j{c}_j{\varphi }_j$ in a basis of $V_h$ gives the linear systemHere $K$ is the stiffness matrix and $F$ is the vector of loads tested against the basis functions. The Galerkin method uses the same trial and test space in this construction. Finite element methods also use the Petrov-Galerkin method, with different trial and test spaces; testing against the trial space is therefore a property of this construction, rather than a definition of every finite element method.

The finite element solution of the one-dimensional Poisson equation below gives every entry explicitly. Refining the finite element mesh then improves the approximation by adding local degrees of freedom. For a symmetric coercive bilinear form, Galerkin orthogonality makes the discrete solution the best approximation in the energy norm; the geometry of orthogonal projection explains why this local construction yields a globally meaningful answer.

Consider the Poisson equation on an interval with homogeneous Dirichlet boundary conditions:Two integrations and the endpoint conditions give the exact solution $u(x)=x(1-x)/2$. To approximate it, multiply by a test function $v$ that vanishes at both endpoints. Integration by parts givesThe boundary term is zero. Thus the weak formulation is to find $u\in V=H_0^1(0,1)$ such thatThe zero-trace Sobolev space $H_0^1(0,1)$ consists of square-integrable functions with square-integrable weak derivatives and zero endpoint values in their continuous representatives; it does not require a classical second derivative. This is why continuous piecewise linear functions are admissible despite corners. The Poincaré inequality makes $a$ a coercive bilinear form on $V$. Strang derives this one-dimensional weak formulation, the piecewise-linear hat functions, and their load and derivative-product integrals in the cited lecture; its endpoint example uses a free left endpoint, while both endpoints here are fixed.^[ref]

Take the uniform finite element mesh $x_j=jh$, $j=0,\dots ,n$, with $h=1/n$. The continuous $P_1$ trial space isHere $P_1$ means polynomials of degree at most one on each interval. This is a conforming finite element space in $V$, with $n-1$ interior piecewise-linear hat functions ${\varphi }_i$ as a basis. Their endpoint values enforce the Dirichlet boundary conditions automatically. Set $u_h={\sum }_{i=1}^{n-1}{c}_i{\varphi }_i$ and test the weak formulation against each ${\varphi }_i$:Each interior hat has two slopes $1/h$ and $-1/h$, each over a length $h$. Its squared derivative therefore integrates to $2/h$. Adjacent hats have opposite slopes on their one shared interval, giving $-1/h$. All other derivative products vanish. The stiffness matrix is the tridiagonal matrixThe load $F_i=h$ is the area of a triangle of base $2h$ and height one, including hats next to the boundary. Boundary hats are omitted because the boundary values are prescribed. The piecewise-linear hat function article explains how the same entries arise by adding local two-node contributions.

With four mesh intervals, $h=1/4$, there are three unknowns:For example, the first row gives $8(3/32)-4(1/8)=1/4$ and the middle gives $-4(3/32)+8(1/8)-4(3/32)=1/4$. The coefficients are the nodal values $u_h(x_i)=c_i$. At $x=3/8$, linear interpolation gives $u_h(3/8)=7/64=0.109375$, whereas $u(3/8)=15/128=0.1171875$: exactness at the nodes still leaves a visible error between them.

In this model the Galerkin method solution equals the nodal approximation $I_{h}u$ obtained by linear interpolation on any finite element mesh. Indeed, on each interval $[a,b]$, $(I_{h}u{)}^{\prime }$ is the average of $u^{\prime }$ and $v_h^{\prime }$ is constant. Therefore ${\int }_a^b(u^{\prime }-(I_{h}u{)}^{\prime })v_h^{\prime }dx=0$. Summing proves Galerkin orthogonality for $I_{h}u$, and uniqueness gives $u_h=I_{h}u$. This argument relies on the constant diffusion coefficient, conformity and exact integration; nodal exactness is not a general promise for finite element methods with other operators or integration rules.

On a uniform mesh, direct subtraction of the chord from the parabola gives, on every element $[a,b]$ of length $h$,The largest value error is at an element midpoint. Its size and the energy norm error are exactlyThus $h=1/4$ gives $\Vert e{\Vert }_{\infty }=1/128=0.0078125$ and $\Vert e{\Vert }_a=1/(4\sqrt{12})\approx 0.0721688$. With eight elements, $h=1/8$, these become $1/512=0.001953125$ and $1/(8\sqrt{12})\approx 0.0360844$. Halving the mesh spacing quarters the maximum value error but only halves the energy norm error: values and slopes converge at different rates. The figure shows the small value error at its own scale, where the four-to-one reduction becomes clear.

A Galerkin method retaining one stream function mode, two temperature modes and two transverse-velocity modes produces five coupled amplitudes. A useful normalization gives $a^{\prime }=-\sigma a+\sigma rb-{\sigma }^{2}qd$, $b^{\prime }=a-b-ac$, $c^{\prime }=h(-c+ab)$, $d^{\prime }=a-\sigma d-ae$, $e^{\prime }=-(4-h)\sigma e+had$. Here $\sigma >0$ is the Prandtl number, $0<h<4$, and $q$ measures squared rotation. The trivial equilibrium point has characteristic polynomial $(s+\sigma {)}^2(s+1)-\sigma r(s+\sigma )+{\sigma }^{2}q(s+1)$ in its three coupled linear modes; the remaining two modes decay. Stationary bifurcations and oscillatory onset meet at $q=(1+\sigma )/(1-\sigma )$, $r=2/(1-\sigma )$ when $\sigma <1$.

A Galerkin method approximation can retain one roll streamfunction mode, two temperature modes and two magnetic-flux perturbation modes in vertical-field magnetoconvection. Projection onto these five modes, rather than pointwise equality of the truncated fields, gives a closed system of five ordinary differential equations. Products generate higher spatial harmonics that are discarded by orthogonal projection. The system has symmetry $(a,b,c,d,e)\mapsto (-a,-b,c,-d,e)$ and hence admits paired steady roll branches.

Let $V_h\subset V$ be a conforming finite element space. Suppose the exact solution satisfies $a(u,v)=\ell (v)$ for every $v\in V$ and its Galerkin method approximation satisfies $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$, using the same bilinear form and linear functional. Subtracting the two identities gives Galerkin orthogonality:This identity itself does not require symmetry. Its interpretation as an orthogonal projection in an energy norm does require a bounded symmetric coercive bilinear form: then $a$ is an inner product, $e=u-u_h$ is perpendicular to every available discrete correction, andConsequently $u_h$ is the unique closest point in $V_h$ in the energy norm. This Pythagorean theorem in a function space explains the beauty of the Galerkin method: integrated local equations compute a global best approximation. The Céa lemma supplies the corresponding estimate in a chosen reference norm.

For the finite element solution of the one-dimensional Poisson equation, $a(w,v)={\int }_0^1{w}^{\prime }{v}^{\prime }dx$ and $\Vert w{\Vert }_a=\Vert w^{\prime }{\Vert }_{L^2(0,1)}$. Testing the error against the $i$th piecewise-linear hat function on a uniform mesh givesIn this model $e$ vanishes at every node, so each of these integrals is individually zero, although $e$ is positive inside every element. Galerkin orthogonality controls the derivative-weighted balance; it does not say that the pointwise error vanishes.

The exact identity assumes conformity and exact evaluation of the continuous forms, including any numerical integration that is used. If assembly instead uses $a_h$ and ${\ell }_h$, subtraction gives the consistency defectThus inexact numerical integration need not preserve the projection. In the Petrov-Galerkin method the error is tested against a different test space, and for a nonsymmetric bilinear form the displayed identity need not describe perpendicularity in any energy norm.

A five-mode Galerkin method model retains the roll velocity $a$, its correlated temperature and concentration modes $b,d$, and the mean-profile distortions $c,e$. Its parameters include the Prandtl number $\sigma$, diffusivity ratio $\tau$, thermal forcing $r$, stabilizing solutal forcing $r_s$, and a positive geometric decay ratio $\varpi$. The velocity equation is $\dot{a}=\sigma (-a+rb-r_{s}d)$, while the scalar equations are $\dot{b}=-b+a(1-c)$, $\dot{c}=\varpi (-c+ab)$, $\dot{d}=-\tau d+a(1-e)$ and $\dot{e}=\varpi (-\tau e+ad)$. The solutal forcing is normalized using the thermal diffusion scale, so its steady stabilizing effect is $r_s/\tau$.

The first-order Sobolev space is $H^1(\Omega )=\{v\in L^2(\Omega ):{\partial }_{1}v,{\partial }_{2}v\in L^2(\Omega )\}$, with derivatives understood as weak derivatives and norm squared ${\int }_{\Omega }(v^2+|\nabla v{|}^2)$.

The printed normal boundary condition must be distinguished from a conormal condition. Take $K$ symmetric positive definite, as required for the symmetric matrix requested later. Integration by parts givesThe natural boundary term contains the conormal derivative $\mathbf{n}\cdot K\nabla u$, not the ordinary normal derivative ${\partial }_{n}u$. On a smooth boundary, writing $\mathbf{t}$ for its unit tangent, the literal condition ${\partial }_{n}u=0$ leaves the term $b{\partial }_{\mathbf{t}}u$, where $b=\mathbf{n}\cdot K\mathbf{t}$. Its weak form therefore retains a boundary pairing:Here the tangential derivative of the Sobolev trace can be interpreted in $H^{-1/2}$ and paired against $H^{1/2}$ on a smooth boundary. This form is generally nonsymmetric; the requested usual symmetric energy construction does not follow from the literal boundary condition.

For an explicit failure of the claimed mean-zero compatibility, take the unit disk, $K=\mathrm{diag}(2,1)$ and $u=(x^2-y^2)[1-(x^2+y^2)/2]$. It has ${\partial }_{n}u=0$ on the circle, butThus replacing the printed normal condition by a zero conormal flux is not harmless.

For the intended zero-conormal derivative problem, the requested variational and finite element conclusions are as follows. Seek $u\in H^1(\Omega )$ withPutting $v=1$ gives ${\int }_{\Omega }f=0$. On a connected domain this condition permits a solution unique modulo constants; the Poincare-Wirtinger inequality gives coercivity after fixing the mean.

The energy is $J(v)=\frac{1}{2}a(v,v)-\ell (v)$. A minimizer satisfies $\frac{d}{dt}J(u+tv){|}_{t=0}=a(u,v)-\ell (v)=0$. Conversely, a weak solution satisfiesso it is a minimizer. Testing by $u$ gives the minimum valueThis value is independent of the arbitrary additive constant when $\int f=0$. If the compatibility fails, constant shifts make $J$ unbounded below.

Write $u_h={\sum }_j{c}_j{\varphi }_j$. The Galerkin method givesSymmetry of $K$ gives $A^T=A$. For $v_h={\sum }_j{z}_j{\varphi }_j$, $z^{T}Az={\int }_{\Omega }\nabla v_h\cdot K\nabla v_h\ge 0$. Equality forces $v_h$ constant by positive definiteness and connectedness. Since the basis sums to one, the coefficient vector $1=(1,\dots ,1{)}^T$ represents that constant, givingThus the stiffness matrix is symmetric positive semidefinite and singular; the displayed compatibility condition is necessary and sufficient for the finite system. Positive definiteness of $K$ controls every gradient direction, preventing additional zero-energy directions and supplying ellipticity. A degenerate coefficient can introduce further kernel modes, while an indefinite one destroys the minimizing-energy argument.

A real bilinear form is a coercive bilinear form on $V$ when $a(v,v)\ge \alpha \Vert v{\Vert }_V^2$ for all $v\in V$ and some fixed $\alpha >0$. It is a bounded bilinear form, equivalently continuous, when $|a(v,w)|\le M\Vert v{\Vert }_V\Vert w{\Vert }_V$ for one finite $M$. A continuous linear functional satisfies $|\ell (v)|\le C_{\ell }\Vert v{\Vert }_V$.

The Galerkin method seeks $u_h\in S$ with $a(u_h,v_h)=\ell (v_h)$ for all $v_h\in S$. Subtraction from the continuous equations gives Galerkin orthogonality: $a(u-u_h,v_h)=0$ for every $v_h\in S$. Under the stated continuity and coercivity assumptions, Céa's lemma saysIndeed, with $e=u-u_h$ and any $v_h\in S$, Galerkin orthogonality eliminates $v_h-u_h$:If $e=0$ the claim is immediate; otherwise cancel $\Vert e{\Vert }_V$ and take the infimum. No symmetry of $a$ was used. These statements define an approximation when it exists; a finite-dimensional $S$ gives existence and uniqueness by the Lax-Milgram theorem, as does a closed vector subspace.

For the interval example write $L=c-b$ and $R=\max (|b|,|c|)$. The one-dimensional Sobolev representative of $v\in V$ satisfiesby the Cauchy-Schwarz inequality. The representation follows by taking the absolutely continuous function primitive of the weak derivative; its difference from $v$ has zero weak derivative, hence is constant, and $v(c)=0$ fixes that constant. Integrating the estimate gives the Endpoint Poincaré inequalityThus the nonnegative potential $x^2$ givesFor continuity, $|x|\le R$, the Cauchy-Schwarz inequality for the two integrals, and then its two-dimensional version giveSimilarly $\boxed{|\ell (v)|\le \Vert f{\Vert }_2\Vert v{\Vert }_2\le \Vert f{\Vert }_2\Vert v{\Vert }_V}$, proving continuity of the linear functional.

For the last displayed integral take $\Omega =(-1,1)$ and retain the endpoint condition $v(1)=0$, the domain-consistent interpretation of the changed formula. The changed form is still coercive. A negative potential on part of the interval does not by itself destroy coercivity. The sharper pointwise estimate above givesDiscarding only the positive part of the potential yieldsThe proved Endpoint Poincaré inequality with $L=2$ gives $\Vert v{\Vert }_V^2\le 3\Vert v^{\prime }{\Vert }_2^2$, and thereforeThis is a concrete application of negative-potential domination by endpoint control. The constant is sufficient, without claiming to be optimal.

Take $V_h={\mathcal{U}}_h\times {\mathcal{W}}_h\subset V$. The Galerkin method asks for $U_h=(u_h,w_h)\in V_h$ such thatSince a finite-dimensional subspace is closed, $V_h$ is itself a Hilbert space. The bounded bilinear form and coercive bilinear form estimates in part (a) restrict to $V_h$ with the same constants. The Lax-Milgram theorem therefore gives exactly one $U_h$ for every $h$.

Put $E=U-U_h=(e_u,e_w)$. Subtracting the continuous and discrete weak formulations gives Galerkin orthogonality:For any $Y_h=(v_h,z_h)\in V_h$, the difference $Y_h-U_h$ also belongs to $V_h$, soThe last step uses the constant-one Cauchy-Schwarz inequality estimate from part (a). Bounding the four terms separately by a sum would obscure the precise constant requested here.

The two minimization problems over ${\mathcal{U}}_h$ and ${\mathcal{W}}_h$ are independent. Each has a best approximation because its vector subspace is finite-dimensional. Consequently, ifthen the preceding inequality gives exactlyBoth errors belong to the zero-boundary Sobolev space. Applying the Poincaré inequality componentwise now yieldsWhen $U=U_h$ the desired estimate is automatic; otherwise divide by the nonzero norm. ThusThe constant $C=1+K$ is independent of $h$. This is the Céa lemma estimate for this particular skew-coupled elliptic system.

For a conforming triangular finite element mesh, takeThis is the quadratic conforming finite element space. Across a common edge of a polygon, the restrictions from neighboring triangles are univariate quadratic polynomials. Their agreement at the two endpoints and the midpoint makes their difference a degree-at-most-two polynomial with three distinct roots of a polynomial, so it vanishes on the entire common edge of a polygon. Thus matching the local finite element degrees of freedom produces the required continuity. Similarly, setting the three nodal values on every boundary edge carrying the Dirichlet boundary condition to zero makes its entire restriction vanish.

Continuity across element interfaces means the piecewise gradients give the global weak gradient, with no jump contribution, so $V_h\subset V$. There are finitely many local finite element degrees of freedom, hence $V_h$ is finite-dimensional. The Galerkin method seeksThe bounded bilinear form and coercive bilinear form estimates from part (b) remain valid on $V_h$. Since a finite-dimensional subspace is closed, the Lax-Milgram theorem gives exactly one $u_h$. Alternatively, the stiffness matrix plus the mass matrix is a positive-definite matrix in any basis of $V_h$. The same existence and uniqueness conclusion holds for any finite-dimensional vector subspace of this full space.

Use the Galerkin method, rather than pointwise equality: products generate harmonics outside the retained five-dimensional space. Put $X=\alpha x$, $Z=\pi z$, and denote the five prefactors byThe retained streamfunction is $P\sin X\sin Z$, the retained temperature perturbation is $T\cos X\sin Z+C\sin 2Z$, and the retained magnetic-flux perturbation is $D\sin X\cos Z+E\sin 2X$. These satisfy all the stated boundary conditions. Let $J(f,g)=f_x{g}_z-f_z{g}_x$, the streamfunction advection bracket. Orthogonality of the sine and cosine modes gives the following coefficients of the retained harmonics:For example the $C$ contribution to the temperature bracket is $\alpha \pi PC[\cos X\sin 3Z-\cos X\sin Z]$. The magnetic bracket follows by writing $A=A_D+A_E$: $J(A,\Delta A)=({\beta }^2-4{\alpha }^2)J(A_D,A_E)$ and $J(A_D,A_E)=\alpha \pi DE(\sin 3X-\sin X)\sin Z$. These identities exhibit both the retained terms and discarded harmonics.

Before rescaling time, the projected coefficient equations areSubstitute the prefactors and divide time derivatives by ${\beta }^2$. With $\tau ={\beta }^{2}t$, $\varpi =4{\pi }^2/{\beta }^2$ and $q={\pi }^{2}Q/{\beta }^4$, the resulting five-mode vertical-field magnetoconvection equations areHere primes denote differentiation in $\tau$. The factor $3-\varpi$ is $(3{\alpha }^2-{\pi }^2)/{\beta }^2$, which checks the magnetic nonlinear sign. There is a terminology qualification in the source: on this thermal diffusion time scale, its $\zeta$ is the magnetic-to-thermal diffusivity ratio $\eta /\kappa$. The conventional magnetic Prandtl number is $\nu /\eta =\sigma /\zeta$, where $\sigma$ is the Prandtl number. The displayed equations and subsequent calculation use the source's $\zeta$ unchanged.

Use the Galerkin method: substitute the retained trigonometric modes and project onto them with the $L^2$ inner product over the rectangle. Each mode satisfies the printed boundary conditions. Write $X=\alpha x$, $Z=\pi z$, and abbreviate the four mode prefactors by $C=2\sqrt{2}\beta /\alpha$, $D=2\sqrt{2}/\beta$, $V_1=2\sqrt{2}\sigma \pi R_{\Omega }/(\alpha \beta )$ and $V_2=\sigma R_{\Omega }/\alpha$. The Jacobian determinant in the advection terms is $J(f,g)=f_x{g}_z-f_z{g}_x$.

The stream function is a single Laplacian eigenfunction, so ${\nabla }^2\psi =-{\beta }^2\psi$ and $J(\psi ,{\nabla }^2\psi )=0$ exactly. For the temperature modes, direct multiplication givesFor the transverse-velocity modes, the retained products areFor example, $\sin Z\cos 2Z=(\sin 3Z-\sin Z)/2$ and $\sin X\cos 2X=(\sin 3X-\sin X)/2$ supply the signs of the two retained $ac$ and $ae$ couplings. The third harmonics are orthogonal to the retained modes and are discarded by this Galerkin method; they are not claimed to vanish in the full partial differential equations.

Before changing time, the five coefficient equations areDividing by ${\beta }^2$ gives the stated five-mode rotating-convection truncation, withThe squared rotation parameter is $q=r_{\Omega }^2\ge 0$. The printed $R_{\Omega }$ is proportional to rotation rate, so it is a square-root-type rotation parameter rather than the more usual Taylor number proportional to squared rotation rate; the algebra uses the coefficient as printed. At zero rotation the physical transverse velocity modes vanish; the normalized $d,e$ equations can still be understood through their continuous zero-rotation limit.

The truncated fields must be interpreted as a Galerkin method approximation: project each equation onto its retained spatial modes and discard orthogonal higher harmonics. They are not an exact pointwise solution of the full partial differential equations, because the nonlinear terms generate additional harmonics. Write $X=kx$, $Z=\pi z$, and abbreviate the printed normalizations byThen $\psi =Sa\sin X\sin Z$, $\theta =Tb\cos X\sin Z-c\sin 2Z/\pi$, and $\chi =Dd\sin X\cos Z+Ee\sin 2X$. All five amplitudes a,b,c,d,e depend on $\tau =pt$. Use the spatial Jacobian determinant $J(F,G)=F_x{G}_z-F_z{G}_x$ and the inner product over the stated rectangle. Orthogonality of the trigonometric modes makes projection equivalent to retaining their coefficients. In particular, ${\nabla }^2\psi =-p\psi$, so $J(\psi ,{\nabla }^2\psi )=0$.

For the temperature equation, direct products giveThe other temperature contribution is $-2kSac\cos X\sin Z\cos 2Z$. Since $\sin Z\cos 2Z=(\sin 3Z-\sin Z)/2$, its retained component is $kSac\cos X\sin Z$. Thus the projected temperature equation isHere the dot denotes $d/d\tau$. Since $kS/(pT)=1$ and $\phi =4{\pi }^2/p$, these equations give $\dot{b}=-b+a(1-c)$ and $\dot{c}=\phi (-c+ab)$.

Similarly, the two magnetic-advection products arewhile $J(\psi ,Ee\sin 2X)=-2\pi Sae\sin X\cos Z\cos 2X$ has retained component $\pi Sae\sin X\cos Z$. The projected induction equation is thereforeUsing $\pi S/(pD)=1$ and $4k^2/p=4-\phi$ gives $\dot{d}=-\zeta d+a(1-e)$ and $\dot{e}=-(4-\phi )\zeta e+\phi ad$.

It remains to project the Lorentz force in the vorticity equation. The linear term is $J(x,{\nabla }^2\chi )=p\pi Dd\sin X\sin Z$. The nonlinear cross term isThe Jacobian on the right is $2\pi kDEde\sin X\sin Z\cos 2X$. Its retained component is $-\pi kDEde\sin X\sin Z$, so the total retained magnetic term isMeanwhile the projected acceleration, viscous term and temperature forcing have coefficients $-p^{2}S\dot{a}$, $p^{2}Sa$, and $-RkTb$. Divide the projected vorticity equation by $-p^{2}S$ to obtain $\dot{a}=\sigma [-a+rb+\zeta qd\{(\phi -3)e-1\}]$, using $r=k^{2}R/p^3$ and $q={\pi }^{2}Q/p^2$. This derives the complete five-mode vertical-field magnetoconvection system:The Rayleigh number, Chandrasekhar number, Prandtl number and magnetic-to-thermal diffusivity ratio enter through r,q,$\sigma$ and $\zeta$ respectively.