SQGE.Error.tex

The main goal of this section is to develop a rigorous error analysis for the FE
discretization of the SQGE \eqref{eqn:SQGE_Psi} by using conforming FEs. In
particular, we develop error bounds for the FE discretiztion of
\eqref{eqn:SQGE_Psi} with Argyris element in \autoref{thm:Errors}. First, in
\autoref{thm:EnergyNorm} we prove error estimates in the $H^2$ norm by using an
approach similar to that used in \cite{Cayco86}.  Second, in
\autoref{thm:Errors}, we prove error estimates in the $L^2$ and $H^1$ norms by
using a duality argument.

The following lemma will introduce some useful bounds for the forms introduced
in \autoref{sec:QGEStrong}.
\begin{lemma} \label{lma:ContinuousForms}
  The linear form $(F,\chi)$, the bilinear forms $(\nabla \psi, \nabla \chi)$,
  $(\Delta \psi, \Delta \chi)$, $(\psi_x, \chi)$ and the trilinear form $b(\psi;
  \psi, \chi)$ are continuous \cite{Cayco86}: There exist $\Gamma_0,\,
   \Gamma_2 > 0$ such that for all $\psi, \chi, \varphi\in X$
  \begin{align}
    (\nabla \psi, \nabla \chi) &\le \Gamma_0\, |\psi|_2 |\chi|_2,
      \label{eqn:a0cont} \\
    (\Delta \psi, \Delta \chi) &\le |\psi|_2\, |\chi|_2, \label{eqn:a1Cont} \\
    b(\psi;\varphi,\chi) &\le \Gamma_1 \|\psi\|_2\, \|\varphi\|_2\, \|\chi\|_2,
      \label{eqn:BH2Bounds} \\
    (\psi_x,\chi) &\le \Gamma_2 \, |\psi|_2 \, |\chi|_2, \label{eqn:a3Cont} \\
    (F,\chi) &\le \|F\|_{-2} \, |\chi|_2.
      \label{eqn:lCont}
  \end{align}
\end{lemma}
\begin{proof}
  We first establish the most straightforward estimate \eqref{eqn:a1Cont}. This
  follows directly from the H\"older inequality, \eqref{eqn:Holder}, i.e.
  \begin{equation*}
    (\Delta \psi, \Delta \chi) \le \|\Delta \psi\|_{L^p}\,\|\Delta \chi\|_{L^r},
    \quad \frac{1}{p} + \frac{1}{r} = 1
  \end{equation*}
  with $p=r=2$. \\
  The bound \eqref{eqn:lCont} can be proven by first noting that, $\forall \chi
  \in X$,
  \begin{equation*}
    \frac{(F,\chi)}{|\chi|_2} \le \sup_{\xi \in X} \frac{(F,\xi)}{|\xi|_2} =
      \|F\|_{-2}.
  \end{equation*}
  Multiplying both sides of the above inequality by $|\chi|_2$ gives the desired
  result:
  \begin{equation*}
    (F,\chi) \le \|F\|_{-2}|\chi|_2.
  \end{equation*}
  Next, we prove \eqref{eqn:a0cont} by first applying H\"older inequality, \eqref{eqn:Holder} with
  $p=r=2$. Therefore
  \begin{equation*}
    (\nabla \psi, \nabla \chi) \le \|\nabla \psi\|\, \|\nabla \chi\|.
  \end{equation*}
  Now, we apply the Poincar\'e inequality, \eqref{eqn:Poincare} to get the
  desired result:
  \begin{equation*}
    (\nabla \psi, \nabla \chi) \le \Gamma_0 |\psi|_2, |\chi|_2,
  \end{equation*}
  where $\Gamma_0$ is the square of the Poincar\'e constant. \\
  Next, we will obtain \eqref{eqn:a3Cont} by first applying the H\"older
  inequality to get
  \begin{equation*}
    (\psi_x, \chi) \le \|\psi_x\|\, \|\chi\|.
  \end{equation*}
  Noting that $\|\psi_x\| \le \|\nabla \psi\|$ and applying the Poincar\'e
  inequality once to the term involving $\psi$ and twice to the term involving
  $\chi$ gives the desired result
  \begin{equation*}
    (\psi_x, \chi) \le \Gamma_2\,|\psi|_2\, |\chi|_2,
  \end{equation*}
  where $\Gamma_2$ is the cube of the Poincar\'e constant. \\
  Finally, for \eqref{eqn:BH2Bounds} we start by applying the H\"older inequality, \eqref{eqn:Holder}
  \begin{equation}
    b(\psi;\varphi,\chi) \le \|\Delta \psi\|_{L^p} \|\nabla \varphi\|_{L^q}
      \|\nabla \chi\|_{L^r},\quad \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1,
    \label{eqn:HolderB}
  \end{equation}
  where $1\le p,\,q,\,r\le \infty$. Thus, setting $p = 2$, and $q = r = 4$ gives
  \begin{equation*}
    b(\psi;\varphi,\chi) \le \|\Delta \psi\| \|\nabla \varphi\|_{L^4} \|\nabla
      \chi\|_{L^4}.
  \end{equation*}
  Then applying the Ladyzhenskaya inequality, \eqref{eqn:Ladyzhenskaya} results in
  \begin{equation}
    b(\psi;\varphi,\chi) \le \Gamma \|\Delta \psi\|
      \|\nabla \varphi\|^{\nicefrac{1}{2}} \|\Delta \varphi\|^{\nicefrac{1}{2}}
      \|\nabla \chi\|^{\nicefrac{1}{2}} \|\Delta \chi\|^{\nicefrac{1}{2}},
    \label{eqn:bLady}
  \end{equation}
  where $\Gamma$ is the square Ladyzhenskaya constant, from
  \eqref{eqn:Ladyzhenskaya}.  Using the Poincar\'e inequality,
  \eqref{eqn:Poincare} gives
  \begin{equation*}
    b(\psi;\varphi,\chi) \le \Gamma_1 \|\Delta \psi\|\, \|\Delta \varphi\|\,
      \|\Delta \chi\|,
  \end{equation*}
  where $\Gamma_1$ is the square of the Poincar\'e constant time $\Gamma$ from
  \eqref{eqn:bLady}.
\end{proof}

\begin{thm}
\label{thm:EnergyNorm}
  Let $\psi$ be the solution of \eqref{eqn:SQGEWF} and $\psi^h$ be the solution
  of \eqref{eqn:SQGEFEF}.
  Furthermore, assume that the following small data condition is satisfied:
  \begin{equation}
    Re^{-2} \, Ro \geq \Gamma_1 \, \| F \|_{-2} ,
    \label{eqn:small_data_condition}
  \end{equation}
  where $Re$ is the Reynolds number defined in \eqref{eqn:reynolds_number}, $Ro$
  is the Rossby number defined in \eqref{eqn:rossby_number}, $\Gamma_1$ is the
  continuity constant of the trilinear form $b$ in \eqref{eqn:BH2Bounds}, and $F$
  is the forcing term. Then the following error estimate holds:
  \begin{equation}
    |\psi - \psi^h|_2 \le C(Re, Ro, \Gamma_1, \Gamma_2, F) \,
      \inf_{\chi^h \in X^h} |\psi - \chi^h|_2 ,
    \label{eqn:EnergyNorm}
  \end{equation}
  where
  \begin{equation}
    C(Re, Ro, \Gamma_1, \Gamma_2, F)
      := \frac{ Ro^{-1} \Gamma_2 + 2 \, Re^{-1} + \Gamma_1 \, Re \, Ro^{-1} \,
        \| F \|_{-2} }
        { Re^{-1} - \Gamma_1 \, Re \, Ro^{-1} \, \| F \|_{-2} }
    \label{eqn:constant_definition}
  \end{equation}
  is a generic constant that can depend on $Re$, $Ro$, $\Gamma_1$, $\Gamma_2$, $F$, but
  \emph{not} on the meshsize $h$.
\end{thm}

\begin{remark}
  Note that the small data condition in \autoref{thm:EnergyNorm} involves
  \emph{both} the Reynolds number and the Rossby number, the latter quantifying
  the rotation effects in the QGE.

  Furthermore, note that the standard small data condition $Re^{-2} \geq
  \Gamma_1 \, \| F \|_{-2}$ used to prove the uniqueness for the steady-state 2D
  NSE \cite{Girault79,Girault86,Layton08} is significantly more restrictive for
  the QGE, since \eqref{eqn:small_data_condition} has the Rossby number (which
  is small when rotation effects are significant) on the left-hand side.  This
  is somewhat counterintuitive, since in general rotation effects are expected
  to help in proving the well-posedness of the system.  We think that the
  explanation for this puzzling situation is the following: Rotation effects do
  make the mathematical analysis of 3D flows more amenable by giving them a 2D
  character.  We, however, are concerned with 2D flows (the QGE).  In this case,
  the small data condition \eqref{eqn:small_data_condition} (needed in proving
  the uniqueness of the solution) indicates that rotation effects make the
  mathematical analysis of the (2D) QGE more complicated than that of the 2D
  NSE. %We emphasize that there is no contradiction, since, although the
  %rotation effects do help in the three-dimensional case, they seem to
  %complicate the situation if one compare the QGE with the two-dimensional (not
  %the three-dimensional) NSEi.
\end{remark}
\begin{proof}
  Since $X^h \subset X$, \eqref{eqn:SQGEWF} holds for all $\chi = \chi^h\in X^h$.
  Subtracting \eqref{eqn:SQGEFEF} from \eqref{eqn:SQGEWF} with $\chi=\chi^h \in
  X^h$ gives
  \begin{equation}
    \begin{split}
      Re^{-1}\left(\Delta \left[\psi - \psi^h\right],\Delta \chi^h \right)
        + b(\psi;\psi,\chi^h) - b(\psi^h;\psi^h,\chi^h) \\
        - Ro^{-1} \left(\left[\psi-\psi^h\right]_x,\chi^h\right) = 0 \qquad \forall \chi^h \in
    X^h.
  \end{split}
    \label{eqn:ErrorEq}
  \end{equation}
  Next, adding and subtracting $b(\psi^h;\psi,\chi^h)$ to \eqref{eqn:ErrorEq},
  we get:
  \begin{equation}
    \begin{split}
      & Re^{-1} (\Delta \left[\psi - \psi^h\right], \Delta \chi^h)
        + b(\psi;\psi,\chi^h) - b(\psi^h;\psi,\chi^h) %\\
        + b(\psi^h;\psi,\chi^h) - b(\psi^h;\psi^h,\chi^h) \\
      & \hspace*{3.0cm} - Ro^{-1} (\left[\psi-\psi^h\right]_x,\chi^h) = 0
        \qquad \forall \chi^h \in X^h.
    \end{split}
    \label{eqn:interError}
  \end{equation}
  The error $e$ can be decomposed as
  \begin{equation}
    e:= \psi-\psi^h = (\psi-\lambda^h)+(\lambda^h-\psi^h):= \eta + \Phi^h,
    \label{eqn:ErrorTrick}
  \end{equation}
  where $\lambda^h\in X^h$ is arbitrary.
  Thus, equation \eqref{eqn:interError} can be
  rewritten as
  \begin{equation}
    \begin{split}
      Re^{-1}(\Delta \eta + \Delta \Phi^h, \Delta \chi^h)
        + b(\eta+\Phi^h,\psi,\chi^h) + b(\psi^h;\eta+\Phi^h,\chi^h) \\
      + Ro^{-1} (\eta_x+\Phi^h_x,\chi^h) = 0 \qquad \forall \chi^h \in X^h.
    \end{split}
    \label{eqn:ErrorEta}
  \end{equation}
  Letting $\chi^h := \Phi^h$ in \eqref{eqn:ErrorEta}, we obtain
  \begin{equation}
    \begin{split}
      Re^{-1} (\Delta \Phi^h, \Delta \Phi^h) - Ro^{-1} (\Phi^h_x,\Phi^h)
        = -Re^{-1} (\Delta \eta, \Delta \Phi^h)
        - b(\eta;\psi,\Phi^h) - b(\psi^h;\psi,\Phi^h) \\
      - b(\psi^h;\eta,\Phi^h) - b(\psi^h;\Phi^h,\Phi^h)
        + Ro^{-1} (\eta,\Phi^h).
    \end{split}
    \label{eqn:ErrorVarphi}
  \end{equation}
  Note that, since $(\Phi^h_x,\Phi^h)=-(\Phi^h,\Phi^h_x)\; \forall
  \Phi^h \in X^h \subset X = H^2_0$, it follows that
  \begin{equation}
    (\Phi^h_x,\Phi^h)=0 .
    \label{eqn:a30}
  \end{equation}
  Also, it follows immediately from \eqref{eqn:b} that
  \begin{equation}
    b(\psi^h;\Phi^h,\Phi^h)=0 .
    \label{eqn:b0}
  \end{equation}
  Combining \eqref{eqn:b0}, \eqref{eqn:a30}, and \eqref{eqn:ErrorVarphi}, we get:
  \begin{equation}
    \begin{split}
      Re^{-1} (\Delta \Phi^h, \Delta \Phi^h) = -Re^{-1} (\Delta \eta, \Delta \Phi^h)
        - b(\Phi^h;\psi,\Phi^h) - b(\eta;\psi,\Phi^h) \\
      - b(\psi^h;\eta,\Phi^h) + Ro^{-1} (\eta_x,\Phi^h).
    \end{split}
    \label{eqn:ErrorZeroed}
  \end{equation}
  Using
  \begin{equation*}
    (\Delta \Phi^h, \Delta \Phi^h) = |\Phi^h|^2_2
  \end{equation*}
  and inequalities \eqref{eqn:a1Cont} -- \eqref{eqn:a3Cont} in equation
  \eqref{eqn:ErrorZeroed} gives
  \begin{equation}
    \begin{split}
      Re^{-1} \, |\Phi^h|^2_2 \le Re^{-1} \,  |\eta|_2 \, |\Phi^h|_2 + \Gamma_1
        \biggl( |\eta|_2 \, |\psi|_2 \, |\Phi^h|_2 + |\psi^h|_2 \, |\eta|_2 \,
          |\Phi^h|_2 \biggr) \\
      + \Gamma_1 \, |\Phi^h|^2_2 \, |\psi|_2 + Ro^{-1}\, \Gamma_2 \, |\eta|_2 \, |\Phi^h|_2 .
    \end{split}
    \label{eqn:varphiIneq}
  \end{equation}
  Simplifying and rearranging terms in \eqref{eqn:varphiIneq} gives
  \begin{equation}
      |\Phi^h|_2 \le \left( Re^{-1} - \Gamma_1 \, | \psi |_2 \right)^{-1} \,
        \left( Re^{-1} + \Gamma_1 \, |\psi|_2 + \Gamma_1 \, |\psi^h|_2 +
        Ro^{-1}\, \Gamma_2 \right) \, |\eta|_2 .
    \label{eqn:phihIneq}
  \end{equation}
  Using \eqref{eqn:phihIneq} and the triangle inequality along with the
  stability estimates \eqref{eqn:stability_sqge} and
  \eqref{eqn:stability_fem_sqge} gives
  \begin{align}
    |e|_2 &\le |\eta|_2 + |\Phi^h|_2 \nonumber \\[0.2cm]
    &\le \left[ 1 + \frac{Re^{-1} + \Gamma_1 \, |\psi|_2 + \Gamma_1 \,
      |\psi^h|_2 + Ro^{-1}\, \Gamma_2} {Re^{-1} - \Gamma_1 \, |\psi|_2} \right] \, |\eta|_2
      \nonumber \\[0.2cm]
    &\le \left[ 1 + \frac{Re^{-1} + \Gamma_1 \, \left( Re \, Ro^{-1} \, \| F
      \|_{-2} \right) + \Gamma_1 \, \left( Re \, Ro^{-1} \, \| F \|_{-2} \right)
      + Ro^{-1}\, \Gamma_2} {Re^{-1} - \Gamma_1 \, \left( Re \, Ro^{-1} \, \| F \|_{-2}
      \right) } \right] \, |\eta|_2 \nonumber \\
    &= \left[ \frac{ Ro^{-1}\, \Gamma_2 + 2 \, Re^{-1} + \Gamma_1 \, Re \, Ro^{-1} \,
      \| F \|_{-2} } { Re^{-1} - \Gamma_1 \, Re \, Ro^{-1} \, \| F \|_{-2} }
        \right] \, | \psi-\lambda^h |_2 ,
    \label{eqn:EnergyError}
  \end{align}
  where $\lambda^h \in X^h$ is arbitrary.  Taking the infimum over $\lambda^h \in
  X^h$ in \eqref{eqn:EnergyError} proves the error estimate
  \eqref{eqn:EnergyNorm}.
\end{proof}

In \autoref{thm:EnergyNorm}, we proved an error estimate in the $H^2$ norm.
In \autoref{thm:Errors}, we will prove error estimates in the $L^2$ and $H^1$
norms by using a duality argument. To this end, we first notice that the QGE
\eqref{eqn:QGE_psi} can be written as
\begin{equation}
  \mathcal{N} \, \psi = Ro^{-1} \, F ,
  \label{eqn:qge_operator_formulation}
\end{equation}
where the nonlinear operator $\mathcal{N}$ is defined as
\begin{equation}
  \mathcal{N} \, \psi := Re^{-1} \, \Delta^2 \psi + J(\psi , \Delta \psi)
    - Ro^{-1} \, \frac{\partial \psi}{\partial x} .
  \label{eqn:nonlinear_operator}
\end{equation}
The linearization of $\mathcal{N}$ around $\psi$, a solution of
\eqref{eqn:QGE_psi}, yields the following \emph{linear} operator:
\begin{equation}
  \mathcal{L} \, \chi := Re^{-1} \, \Delta^2 \chi + J(\chi , \Delta \psi) +
    J(\psi, \Delta \chi) - Ro^{-1} \, \frac{\partial \chi}{\partial x} .
  \label{eqn:linear_operator}
\end{equation}
To find the dual problem associated with the QGE
\eqref{eqn:qge_operator_formulation}, we first define the \emph{dual operator}
$\mathcal{L}^*$ of $\mathcal{L}$:
\begin{equation}
  (\mathcal{L} \, \chi , \psi^*) = ( \chi , \mathcal{L}^* \, \psi^*)
    \qquad \forall \, \psi^* \in X .
  \label{eqn:dual_operator}
\end{equation}
To find $\mathcal{L}^*$, we use the standard procedure:
In \eqref{eqn:dual_operator}, we use the definition of $\mathcal{L}$ given in \eqref{eqn:linear_operator} and we ``integrate by parts" (i.e., use Green's theorem):
\begin{align}
  (\mathcal{L} \, \chi , \psi^*) =& \left( Re^{-1} \, \Delta^2 \chi
    + J(\chi , \Delta \psi) + J(\psi, \Delta \chi)
    - Ro^{-1} \, \frac{\partial \chi}{\partial x} \, , \, \psi^* \right) \nonumber \\
  =& \left( \chi \, , \, Re^{-1} \, \Delta^2 \, \psi^* - J(\psi , \Delta \psi^* )
    + Ro^{-1} \, \frac{\partial \psi^*}{\partial x} \right)
    + \biggl( J(\chi , \Delta \psi) , \psi^* \biggr) ,
  \label{eqn:dual_operator_1}
\end{align}
where to get the first term on the right-hand side of
\eqref{eqn:dual_operator_1} we used the skew-symmetry of the trilinear form
$b$ in the last two variables and Green's theorem (just as we did in the proof
of \autoref{thm:stability_sqge}).
Next, we apply Green's theorem to the second term on the right-hand side of
\eqref{eqn:dual_operator_1}:
\begin{align}
  \biggl( J(\chi , \Delta \psi) , \psi^* \biggr) =& \int_{\Omega}\! \chi_x \,
    \Delta \psi_y \, \psi^* - \chi_y \, \Delta \psi_x \, \psi^* \nonumber\,
    d\mathbf{x} \\
  \stackrel{Green}{=}& \cancelto{0}{\int_{\partial \Omega}\!
      \left(\chi \, \Delta \psi_{x y} \, \psi^* - \chi \, \Delta \psi_{y x} \,
      \psi^* \right) \cdot \mathbf{n} \, dS}
    + \int_{\Omega}\! \chi \, \Delta \psi_x \, \psi^*_y
      - \chi \, \Delta \psi_y \, \psi^*_x\, d\mathbf{x} \nonumber \\[1em]
  =& \biggl( \chi , J(\Delta \psi , \psi^*) \biggr) .
  \label{eqn:dual_operator_2}
\end{align}
Equations \eqref{eqn:dual_operator_1}-\eqref{eqn:dual_operator_2} imply:
\begin{align}
  (\mathcal{L} \, \chi , \psi^*) =& \left( \chi \, , \, Re^{-1} \, \Delta^2 \, \psi^*
    - J(\psi , \Delta \psi^* ) + Ro^{-1} \, \frac{\partial \psi^*}{\partial x} \right)
    + \biggl( \chi , J(\Delta \psi , \psi^*) \biggr) \nonumber \\
  =& ( \chi , \mathcal{L}^* \, \psi^*) .
  \label{eqn:dual_operator_3}
\end{align}
Thus, the \emph{dual operator} $\mathcal{L}^*$ is given by
\begin{equation}
  \mathcal{L}^* \, \psi^* = Re^{-1} \, \Delta^2 \, \psi^* - J(\psi , \Delta \psi^* )
    + J(\Delta \psi , \psi^* ) + Ro^{-1} \, \frac{\partial \psi^*}{\partial x} .
  \label{eqn:dual_operator_4}
\end{equation}
For any given $g \in L^2(\Omega)$, the weak formulation of the \emph{dual
problem} is:
\begin{equation}
  ( \mathcal{L}^* \, \psi^* , \chi ) = (g , \chi)
    \qquad \forall \, \chi \in X = H_0^2(\Omega) .
  \label{eqn:dual_operator_5}
\end{equation}
We assume that $\psi^*$, the solution of \eqref{eqn:dual_operator_5}, satisfies
the following elliptic regularity estimates:
\begin{align}
  & \psi^* \in H^4(\Omega) \cap H^2_0(\Omega), \label{eqn:dual_operator_6a} \\[0.2cm]
  & \| \psi^* \|_4 \le C \, \|g\|_{0}, \label{eqn:dual_operator_6b} \\[0.2cm]
  & \| \psi^* \|_3 \le C \, \|g\|_{-1}, \label{eqn:dual_operator_6c}
\end{align}
where $C$ is a generic constant that can depend on the data, but not on the
mesh size $h$.
\begin{remark}
  We note that this type of elliptic regularity was also assumed in \cite{Cayco86}
  for the streamfunction formulation of the 2D NSE.  In that report, it was also
  noted that, for a polygonal domain with maximum interior vertex angle $\theta <
  126^{\circ}$, the assumed elliptic regularity was actually proved by Blum and
  Rannacher \cite{Blum1980}.  We note that the theory developed in
  \cite{Blum1980} carries over to our case as well.  In Section 5 in
  \cite{Blum1980} it is proved that, for weakly nonlinear problems that
  involve the biharmonic operator as linear main part and that satisfy certain
  growth restrictions, each weak solution satisfies elliptic regularity results of
  the form \eqref{eqn:dual_operator_6a}-\eqref{eqn:dual_operator_6c}.  Assuming
  that $\Omega$ is a bounded polygonal domain with inner angle $\omega$ at each
  boundary corner satisfying $\omega < 126.283696\ldots^{\circ}$, Theorem 7 in
  \cite{Blum1980} with $k = 0$ and $k = 1$ implies
  \eqref{eqn:dual_operator_6a}-\eqref{eqn:dual_operator_6c}.  Using an argument
  similar to that used in Section 6(b) in \cite{Blum1980} to prove that
  the streamfunction formulation of the 2D NSE satisfies the restrictions in
  Theorem 7, we can prove that $\psi^*$, the solution of our dual problem
  \eqref{eqn:dual_operator_5}, satisfies the elliptic regularity results in
  \eqref{eqn:dual_operator_6a}-\eqref{eqn:dual_operator_6c}.  Indeed, the main
  point in Section 6(b) in \cite{Blum1980} is that the corner
  singularities arising in flows around sharp corners are essentially determined
  by the linear main part $\Delta^2$ in the streamfunction formulation of the 2D
  NSE, which is the linear main part of our dual problem
  \eqref{eqn:dual_operator_5} as well.
\end{remark}

We emphasize that although, the error estimates in the $L^2$ and $H^1$ norms
proven in \autoref{thm:Errors} are derived for the particular space $X^h\subset
H^2_0(\Omega)$ consisting of Argyris elements, the same results can be derived
for other conforming $C^1$ finite element spaces.
\begin{thm} \label{thm:Errors}
  Let $\psi$ be the solution of \eqref{eqn:SQGEWF} and $\psi^h$ be the solution
  of \eqref{eqn:SQGEFEF}.  Assume that the same small data condition as in
  \autoref{thm:EnergyNorm} is satisfied:
  \begin{equation}
    Re^{-2} \, Ro \geq \Gamma_1 \, \| F \|_{-2} .
    \label{eqn:small_data_condition_dual}
  \end{equation}
  Furthermore, assume that $\psi\in H^6(\Omega) \cap H^2_0(\Omega)$.  Then there
  exist positive constants $C_0, \, C_1 \text{ and } C_2$ that can depend on
  $Re$, $Ro$, $\Gamma_1$, $\Gamma_2$, $F$, but \emph{not} on the meshsize $h$, such that
  \begin{align}
    |\psi - \psi^h|_2 &\le C_2 \, h^4\, |\psi|_6, \label{eqn:H2Error} \\
    |\psi - \psi^h|_1 &\le C_1 \, h^5\, |\psi|_6, \label{eqn:H1Error} \\
    \|\psi - \psi^h\|_0 &\le C_0 \, h^6\, |\psi|_6. \label{eqn:L2Error}
  \end{align}
\end{thm}
\begin{proof}
  Estimate \eqref{eqn:H2Error} follows immediately from
  \eqref{eqn:argyris_approximation_0} and \autoref{thm:EnergyNorm}.
  Estimates \eqref{eqn:L2Error} and \eqref{eqn:H1Error} follow from a duality
  argument.

  The error in the primal problem \eqref{eqn:SQGEWF} and the interpolation error
  in the dual problem  \eqref{eqn:dual_operator_5} are denoted as
  \begin{equation}
    e := \psi - \psi^h \qquad e^* : = \psi^* - {\psi^*}^h ,
    \label{eqn:theorem_dual_1}
  \end{equation}
  respectively.

  We start by proving the $L^2$ norm estimate \eqref{eqn:L2Error}.
  \begin{align}
    |e|^2 = (e, e) &= (e , \mathcal{L}^* \, \psi^*)
      = (\mathcal{L} \, e , \psi^*) \nonumber \\
    &= (e , \mathcal{L}^* \, e^*) + (e , \mathcal{L}^* \, {\psi^*}^h)
      = (\mathcal{L} \, e , e^*) + (\mathcal{L} \, e , {\psi^*}^h) .
    \label{eqn:theorem_dual_2}
  \end{align}
  The last term on the right-hand side of \eqref{eqn:theorem_dual_2} is given by
  \begin{equation}
    (\mathcal{L} \, e , {\psi^*}^h) = \left( Re^{-1} \, \Delta^2 e
      + J(e , \Delta \, \psi) + J(\psi , \Delta \, e)
      - Ro^{-1} \, \frac{\partial e}{\partial x} \, , \, {\psi^*}^h \right) .
    \label{eqn:theorem_dual_3}
  \end{equation}
  To estimate this term, we consider the error equation obtained by subtracting
  \eqref{eqn:SQGEFEF} (with $\psi^h= {\psi^*}^h$) from \eqref{eqn:SQGEWF} (with
  $\chi = {\psi^*}^h$):
  \begin{equation}
    \left( Re^{-1} \, \Delta^2 e - Ro^{-1} \, \frac{\partial e}{\partial x} \, ,
      \, {\psi^*}^h \right) + \left( J(\psi , \Delta \, \psi)
      - J(\psi^h , \Delta \, \psi^h) \, , \, {\psi^*}^h \right) = 0 .
    \label{eqn:theorem_dual_4}
  \end{equation}
  Using \eqref{eqn:theorem_dual_4}, equation \eqref{eqn:theorem_dual_3} can be
  written as follows:
  \begin{equation}
    (\mathcal{L} \, e , {\psi^*}^h) = \left( J(e , \Delta \, \psi)
      + J(\psi , \Delta \, e) - J(\psi , \Delta \, \psi)
      + J(\psi^h , \Delta \, \psi^h) \, , \, {\psi^*}^h \right) .
    \label{eqn:theorem_dual_5}
  \end{equation}
  Thus, by using \eqref{eqn:theorem_dual_5} equation \eqref{eqn:theorem_dual_2}
  becomes:
  \begin{align}
    |e|^2 =& (\mathcal{L} \, e , e^*) + (\mathcal{L} \, e , {\psi^*}^h) \nonumber \\
    =& Re^{-1} \, (\Delta e , \Delta e^*)
      - Ro^{-1} \, \left( \frac{\partial e}{\partial x} , e^* \right)
      + \left( J(e , \Delta \psi) + J(\psi , \Delta e) , e^* \right) \nonumber \\
    & + \left( J(e , \Delta \, \psi) + J(\psi , \Delta \, e)
      - J(\psi , \Delta \, \psi) + J(\psi^h , \Delta \, \psi^h)
      \, , \, {\psi^*}^h \right) \nonumber \\
    =& Re^{-1} (\Delta e ,\Delta e^*) - Ro^{-1}(e_x , e^*)
      + b(e , \psi , e^*) + b(\psi , e, e^*) + b(e , \psi , {\psi^*}^h) \nonumber \\
    & - b(\psi , \psi , e^*) + b(\psi^h , \psi^h , e^*) \nonumber \\
    =& Re^{-1} (\Delta , \Delta e^*) - Ro^{-1}(e_x , e^*)
      + b(e , \psi , e^*) + b(\psi , e, e^*) \nonumber \\
    & - b(e , \psi , e^*) + b(e , \psi^h , e^*) + b(e , e , \psi^*).
    \label{eqn:theorem_dual_6}
  \end{align}
  Using the bounds in \eqref{eqn:a1Cont}-\eqref{eqn:a3Cont},
  \eqref{eqn:theorem_dual_6} yields:
  \begin{align}
    |e|^2 =& Re^{-1} (\Delta e ,\Delta e^*) - Ro^{-1} (e_x , e^*)
      + b(e , \psi , e^*) + b(\psi , e, e^*) \nonumber \\
    & - b(e , \psi , e^*) + b(e , \psi^h , e^*) + b(e , e , \psi^*) \nonumber \\
    \leq& Re^{-1} \, | e |_2 \, |e^* |_2 + Ro^{-1} \, \Gamma_2 | e |_2 \, |e^* |_2
      + \Gamma_1 \, | e |_2 \, | \psi |_2 \, | e^* |_2
      + \Gamma_1 \, | \psi |_2 \, | e |_2 \, | e^* |_2 \nonumber \\
    & + \Gamma_1 \, | e |_2 \, | \psi |_2 \, | e^* |_2
      + \Gamma_1 \, | e |_2 \, | \psi^h |_2 \, | e^* |_2
      + \Gamma_1 \, | e |_2 \, | e |_2 \, | \psi^* |_2 \nonumber \\
    =& | e |_2 \, |e^* |_2 \, \left( Re^{-1} + Ro^{-1}\, \Gamma_2 + \Gamma_1 \, | \psi |_2
      + \Gamma_1 \, | \psi |_2
      + \Gamma_1 \, | \psi |_2
      + \Gamma_1 \, | \psi^h |_2 \right) \nonumber \\
    & + | e |_2^2 \, \left( \Gamma_1 \, | \psi^* |_2 \right) .
    \label{eqn:theorem_dual_7}
  \end{align}
  We start bounding the terms on the right-hand side of
  \eqref{eqn:theorem_dual_7}.  First, we note that using the stability estimates
  \eqref{eqn:stability_sqge} for $\psi$ and \eqref{eqn:stability_fem_sqge} for
  $\psi^h$, the right-hand side of \eqref{eqn:theorem_dual_7} can be bounded as
  follows:
  \begin{equation}
    |e|^2 \leq C \, | e |_2 \, |e^* |_2 + | e |_2^2 \,
      \left( \Gamma_1 \, | \psi^* |_2 \right) ,
    \label{eqn:theorem_dual_8}
  \end{equation}
  where $C$ is a generic constant that can depend on $Re$, $Ro$, $\Gamma_1$,
  $\Gamma_2$, $F$, but \emph{not} on the mesh size $h$.  By using the
  approximation results \eqref{eqn:argyris_approximation_1}, we get:
  \begin{equation}
    |e^* |_2 \leq C \, h^2 \, | \psi^* |_4 .
    \label{eqn:theorem_dual_9}
  \end{equation}
  By using \eqref{eqn:dual_operator_6a} and \eqref{eqn:dual_operator_6b}, the
  elliptic regularity results of the dual problem \eqref{eqn:dual_operator_5}
  with $g := e$, we also get:
  \begin{equation}
    | \psi^* |_4 \leq C \, | e | ,
    \label{eqn:theorem_dual_10}
  \end{equation}
  which obviously implies
  \begin{equation}
    | \psi^* |_2 \leq C \, | e | .
    \label{eqn:theorem_dual_11}
  \end{equation}
  Inequalities \eqref{eqn:theorem_dual_9}-\eqref{eqn:theorem_dual_10} imply:
  \begin{equation}
    |e^* |_2 \leq C \, h^2 \, | e | .
    \label{eqn:theorem_dual_12}
  \end{equation}
  Inserting \eqref{eqn:theorem_dual_11} and \eqref{eqn:theorem_dual_12}  in
  \eqref{eqn:theorem_dual_8}, we get:
  \begin{equation}
    |e|^2 \leq C \, h^2 \, | e |_2 \, | e | + C \, | e |_2^2 \, | e | .
    \label{eqn:theorem_dual_13}
  \end{equation}
  Using the obvious simplifications and the $H^2$ error estimate
  \eqref{eqn:H2Error} in \eqref{eqn:theorem_dual_13} yields:
  \begin{equation}
    |e| \leq C \, h^2 \, | e |_2 + C \, | e |_2^2 \leq C \, h^6 |\psi|_6 + C \,
      h^8 |\psi|_6^2 = C_0 \, h^6\, |\psi|_6 ,
    \label{eqn:theorem_dual_14}
  \end{equation}
  which proves the $L^2$ error estimate \eqref{eqn:L2Error}.

  Next, we prove the $H^1$ norm estimate \eqref{eqn:H1Error}.  Since the duality
  approach we use is similar to that we used in proving the  $L^2$ norm estimate
  \eqref{eqn:L2Error}, we only highlight the main differences.  We start again
  by writing the $H^1$ norm of the error in terms of the dual operator
  $\mathcal{L}^*$:
  \begin{align}
    |e|_1^2 =& (\nabla e , \nabla e) = ( e , - \Delta e) =
      (e , \mathcal{L}^* \, \psi^*) = (\mathcal{L} \, e , \psi^*) \nonumber \\
    =& (e , \mathcal{L}^* \, e^*) + (e , \mathcal{L}^* \, {\psi^*}^h)
      = (\mathcal{L} \, e , e^*) + (\mathcal{L} \, e , {\psi^*}^h) .
    \label{eqn:theorem_dual_15}
  \end{align}
  Thus, the second and fourth equalities in \eqref{eqn:theorem_dual_15} clearly
  indicate that in the dual problem \eqref{eqn:dual_operator_5}, one should
  choose $g = - \Delta e$, and not $g = e$, as we did in
  \eqref{eqn:theorem_dual_10}, when we proved the $L^2$ error estimate
  \eqref{eqn:L2Error}.  Using \eqref{eqn:dual_operator_6a} and
  \eqref{eqn:dual_operator_6c}, the elliptic regularity results of the dual
  problem \eqref{eqn:dual_operator_5} with $g :=  - \Delta e$, we also get:
  \begin{equation}
    | \psi^* |_3 \leq C \, | - \Delta e |_{-1} \leq C \, | e |_1 ,
    \label{eqn:theorem_dual_16}
  \end{equation}
  where in the last inequality we used the fact that $e \in H_0^2(\Omega)$.
  Inequality \eqref{eqn:theorem_dual_16} obviously implies
  \begin{equation}
    | \psi^* |_2 \leq C \, | e |_1 .
    \label{eqn:theorem_dual_17}
  \end{equation}
  All the results in \eqref{eqn:theorem_dual_3}-\eqref{eqn:theorem_dual_7} carry
  over to our setting.  Thus, we get:
  \begin{equation}
    |e|_1^2 \leq C \, | e |_2 \, |e^* |_2 + C \, | e |_2^2 \, | \psi^* |_2 ,
    \label{eqn:theorem_dual_18}
  \end{equation}
  where $C$ is a generic constant that can depend on $Re$, $Ro$, $\Gamma_1$,
  $\Gamma_2$,  $F$, but \emph{not} on the mesh size $h$.  By using the
  approximation result \eqref{eqn:argyris_approximation_2}, we get:
  \begin{equation}
    |e^* |_2 \leq C \, h \, | \psi^* |_3 .
    \label{eqn:theorem_dual_19}
  \end{equation}
  Inequalities \eqref{eqn:theorem_dual_19}-\eqref{eqn:theorem_dual_16} imply:
  \begin{equation}
    |e^* |_2 \leq C \, h \, | e |_1 .
    \label{eqn:theorem_dual_20}
  \end{equation}
  Inserting \eqref{eqn:theorem_dual_17} and \eqref{eqn:theorem_dual_20}  in
  \eqref{eqn:theorem_dual_18}, we get:
  \begin{equation}
    |e|_1^2 \leq C \, h \, | e |_2 \, | e |_1 + C \, | e |_2^2 \, | e |_1 .
    \label{eqn:theorem_dual_21}
  \end{equation}
  Using the obvious simplifications and the $H^2$ error estimate
  \eqref{eqn:H2Error} in \eqref{eqn:theorem_dual_21} yields:
  \begin{equation}
    |e|_1 \leq C \, h \, | e |_2 + C \, | e |_2^2 \leq C \, h^5\, |\psi|_6 + C
      \, h^8\, |\psi|_6 = C_1 \, h^5 |\psi|_6,
    \label{eqn:theorem_dual_22}
  \end{equation}
  which proves the $H^1$ error estimate \eqref{eqn:H1Error}.
\end{proof}