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\section{Some Remarks on Lumping} 
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\label{REMARKS ON LUMPING} 
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Explicit time integration schemes (two examples are discussed later in this 
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section), require very small time steps in order to maintain numerical stability. 
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Unfortunately, these small time increments often result in a prohibitive 
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computational cost. 
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In order to minimise these costs, a technique termed lumping can be utilised. 
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Lumping is applied to the coefficient matrix, reducing it to a simple diagonal 
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matrix. This can significantly improve the computational speed, because the 
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solution updates are simplified to a simple componentbycomponent 
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vectorvector product. However, some care is required when making radical 
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approximations such as these. In this section, two commonly applied lumping 
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techniques are discussed, namely row sum lumping 
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\index{linear solver!row sum lumping}\index{row sum lumping} and HRZ 
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lumping \index{linear solver!HRZ lumping}\index{HRZ lumping}. 
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\subsection{Scalar wave equation} 
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One example where lumping can be applied to a hyperpolic problem, is 
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the scalar wave equation 
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\begin{eqnarray} \label{LUMPING WAVE} 
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u_{,tt}=c^2 u_{,ii} \; . 
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\end{eqnarray} 
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In this example, both of the lumping schemes are tested against the reference solution 
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\begin{eqnarray} \label{LUMPING WAVE TEST} 
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u=sin(5 \pi (x_0c*t) ) 
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\end{eqnarray} 
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over the 2D unit square. Note that $u_{,i}n_i=0$ on faces $x_1=0$ and $x_1=1$. 
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Thus, on the faces $x_0=0$ and $x_0=1$ the solution is constrained. 
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To solve this problem the explicit Verlet scheme\index{Verlet scheme} was used 
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with a constant time step size $dt$ given by 
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\begin{eqnarray} \label{LUMPING WAVE VALET} 
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u^{(n)}=2u^{(n1)}u^{(n2)} + dt^2 a^{(n)} 
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\end{eqnarray} 
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for all $n=2,3,\ldots$ where the upper index ${(n)}$ refers to values at 
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time $t^{(n)}=t^{(n1)}+h$ and $a^{(n)}$ is the solution of 
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\begin{eqnarray} \label{LUMPING WAVE VALET 2} 
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a^{(n)}=c^2 u^{(n1)}_{,ii} \; . 
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\end{eqnarray} 
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This equation can be interpreted as a PDE for the unknown value $a^{(n)}$, 
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which must be solved at each timestep. 
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In the notation of equation~\ref{LINEARPDE.SINGLE.1} we thus set $D=1$ and 
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$X=c^2 u^{(n1)}_{,i}$. Furthermore, in order to maintain stability, 
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the time step size needs to fullfill the Courantâ€“Friedrichsâ€“Lewy condition 
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(CFL condition). 
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\index{Courant condition} 
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\index{explicit scheme!Courant condition} 
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For this example, the CFL condition takes the form 
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\begin{eqnarray} \label{LUMPING WAVE CFL} 
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dt = f \cdot \frac{dx}{c} . 
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\end{eqnarray} 
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where $dx$ is the mesh size and $f$ is a safety factor. In this example, 
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we use $f=\frac{1}{6}$. 
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Figure~\ref{FIG LUMPING VALET A} depicts a temporal comparison between four 
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alternative solution algorithms: the exact solution; using a full mass matrix; 
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using HRZ lumping; and row sum lumping. The domain utilsed rectangular order 1 
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elements (element size is $0.01$) with observations taken at the point 
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$(\frac{1}{2},\frac{1}{2})$. 
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All four solutions appear to be identical for this example. This is not the case 
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for order $2$ elements, as illustrated in Figure~\ref{FIG LUMPING VALET B}. 
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For the order $2$ elements, the row sum lumping has become unstable. Row sum 
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lumping is unstable in this case because for order $2$ elements, a row sum can 
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result in a value of zero. HRZ lumping does not display the same problems, but 
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rather exhibits behaviour similar to the full mass matrix solution. When using 
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both the HRZ lumping and full mass matrix, the wavefront is slightly delayed 
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when compared with the analytical solution. 
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\begin{figure}[ht] 
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\centerline{\includegraphics[width=7cm]{lumping_valet_a_1}} 
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\caption{Amplitude at point $(\frac{1}{2},\frac{1}{2})$ using the accelaraton formulation~\ref{LUMPING WAVE VALET 2} of the 
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Velet scheme with order $1$ elements, element size $dx=0.01$, and $c=1$.} 
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\label{FIG LUMPING VALET A} 
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\end{figure} 
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\begin{figure}[ht] 
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\centerline{\includegraphics[width=7cm]{lumping_valet_a_2}} 
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\caption{Amplitude at point $(\frac{1}{2},\frac{1}{2})$ using the accelaraton formulation~\ref{LUMPING WAVE VALET 2} of the 
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Velet scheme with order $2$ elements, element size $0.01$, and $c=1$.} 
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\label{FIG LUMPING VALET B} 
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\end{figure} 
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\begin{figure}[ht] 
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\centerline{\includegraphics[width=7cm]{lumping_valet_u_1}} 
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\caption{Amplitude at point $(\frac{1}{2},\frac{1}{2})$ using the displacement formulation~\ref{LUMPING WAVE VALET 3} of the 
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Velet scheme with order $1$ elements, element size $0.01$ and $c=1$.} 
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\label{FIG LUMPING VALET C} 
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\end{figure} 
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Alternatively, one can directly solve for $u^{(n)}$ by inserting 
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equation~\ref{LUMPING WAVE VALET} into equation~\ref{LUMPING WAVE VALET 2}: 
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\begin{eqnarray} \label{LUMPING WAVE VALET 3} 
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u^{(n)}=2u^{(n1)}u^{(n2)} + (dt\cdot c)^2 u^{(n1)}_{,ii} \; . 
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\end{eqnarray} 
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This can also be interpreted as a PDE that must be solved at each timestep, but 
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for the unknown $u^{(n)}$. 
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As per equation~\ref{LINEARPDE.SINGLE.1} we set the general form coefficients to: 
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$D=1$; $Y=2u^{(n1)}u^{(n2)}$; and $X=(h\cdot c)^2 u^{(n1)}_{,i}$. 
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For the full mass matrix, the acceleration ~\ref{LUMPING WAVE VALET 2} and displacement formulations ~\ref{LUMPING WAVE VALET 3} 
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are identical. 
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The displacement solution is depicted in Figure~\ref{FIG LUMPING VALET C}. The 
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domain utilised order $1$ elements (for order $2$ both 
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lumping methods are unstable). The solutions for the exact and the full mass 
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matrix approximation are almost identical while the lumping solutions, whilst 
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identical to each other, exhibit a considerably faster wavefront propagration 
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and a decaying amplitude. 
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\subsection{Advection equation} 
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Consider now, a second example that demonstrates the advection equation 
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\begin{eqnarray} \label{LUMPING ADVECTIVE} 
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u_{,t}=(v_i u)_i \; . 
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\end{eqnarray} 
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where $v_i$ is a given velocity field. To simplify this example, set $v_i=(1,0)$ and 
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\begin{equation} \label{LUMPING ADVECTIVE TEST} 
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u(x,t)= 
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\left\{ 
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\begin{array}{cl} 
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1 & x_0 < t \\ 
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0 & x_0 \ge t 
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\end{array} 
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\right\}. 
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\end{equation} 
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The solution scheme implemented, is the twostep TaylorGalerkin scheme 
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\index{TaylorGalerkin scheme} 
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(which is in this case equivalent to SUPG\index{SUPG}): 
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the incremental formulation is given as 
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\begin{eqnarray} \label{LUMPING SUPG 1} 
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du^{(n\frac{1}{2})} = \frac{dt}{2} (v_i u^{(n1)})_i \\ 
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du^{(n)} = dt (v_i (u^{(n1)}+du^{(n\frac{1}{2})}) )_i \\ 
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u^{(n)} = u^{(n)} + du^{(n)} 
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\end{eqnarray} 
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This can be reformulated to calculate $u^{(n)}$ directly: 
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\begin{eqnarray} \label{LUMPING SUPG 2} 
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u^{(n\frac{1}{2})} = u^{(n1)} + \frac{dt}{2} (v_i u^{(n1)})_i \\ 
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u^{(n)} = u^{(n1)} + dt (v_i u^{(n\frac{1}{2})} )_i 
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\end{eqnarray} 
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In some cases it may be possible to combine the two equations to calculate 
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$u^{(n)}$ without the intermediate step. This approach is not discussed, because 
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it is inflexible when a greater number of terms (e.g. a diffusion term) are 
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added to the right hand side. 
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The advection problem is thus similar to the wave propagation problem, because 
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the time step also needs to satisfy the CFL condition 
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\index{Courant condition}\index{explicit scheme!Courant condition}. For the 
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advection problem, this takes the form 
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\begin{eqnarray} \label{LUMPING ADVECTION CFL} 
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dt = f \cdot \frac{dx}{\v\} . 
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\end{eqnarray} 
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where $dx$ is the mesh size and $f$ is a safty factor. 
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For this example, we again use $f=\frac{1}{6}$. 
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Figures~\ref{FIG LUMPING SUPG INC A} and~\ref{FIG LUMPING SUPG INC B} illustrate 
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the four incremental formulation solutions: the true solution; the exact mass matrix; 
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the HRZ lumping; and the row sum lumping. Observe, that for the order $1$ elements 
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case, there is little deviation from the exact solution before the wave front, 
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whilst there is a significant degree of osciallation after the wavefront has 
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passed. For the order $2$ elements example, all of the numerical techniques fail. 
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\begin{figure}[ht] 
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\centerline{\includegraphics[width=7cm]{lumping_SUPG_du_1}} 
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\caption{Amplitude at point $(\frac{1}{2},\frac{1}{2})$ using the incremental formulation~\ref{LUMPING SUPG 1} of the 
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TaylorGalerkin scheme with order $1$ elements, element size $dx=0.01$, $v=(1,0)$.} 
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\label{FIG LUMPING SUPG INC A} 
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\end{figure} 
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\begin{figure}[ht] 
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\centerline{\includegraphics[width=7cm]{lumping_SUPG_du_2}} 
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\caption{Amplitude at point $(\frac{1}{2},\frac{1}{2})$ using the incremental formulation~\ref{LUMPING SUPG 1} of the 
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TaylorGalerkin scheme with order $2$ elements, element size $0.01$, $v=(1,0)$.} 
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\label{FIG LUMPING SUPG INC B} 
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\end{figure} 
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Figure~\ref{FIG LUMPING SUPG A} depicts the results from the direct formulation 
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of the advection problem for an order $1$ mesh. Generally, the results have 
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improved when compared with the incremental formulation. The full mass matrix 
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still introduces some osciallation both before and after the arrival of the 
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wavefront at the observation point. The two lumping solutions are identical, and 
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have introduced additional smoothing to the solution. There are no oscillatory 
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effects when using lumping for this example. In Figure~\ref{FIG LUMPING SUPG Ab} 
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the mesh or element size has been reduced from 0.01 to 0.002 units. As predicted 
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by the CFL condition, this significantly improves the results when lumping is 
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applied. However, when utilising the full mass matrix, a smaller mesh size will 
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result in post wavefront oscilations which are higher frequency and slower to 
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decay. 
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Figure~\ref{FIG LUMPING SUPG B} illustrates the results when utilising elements 
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of order $2$. The full mass matrix and HRZ lumping formulations are unable to 
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correctly model the exact solution. Only the row sum lumping was capable of 
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producing a smooth and sensical result. 
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\begin{figure}[ht] 
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\centerline{\includegraphics[width=7cm]{lumping_SUPG_u_1}} 
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\caption{Amplitude at point $(\frac{1}{2},\frac{1}{2})$ using the direct formulation~\ref{LUMPING SUPG 2} of the 
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TaylorGalerkin scheme using order $1$ elements, element size $dx=0.01$, $v=(1,0)$.} 
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\label{FIG LUMPING SUPG A} 
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\end{figure} 
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\begin{figure}[ht] 
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\centerline{\includegraphics[width=7cm]{lumping_SUPG_u_1b}} 
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\caption{Amplitude at point $(\frac{1}{2},\frac{1}{2})$ using the direct formulation~\ref{LUMPING SUPG 2} of the 
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TaylorGalerkin scheme using order $1$ elements, element size $dx=0.002$, $v=(1,0)$.} 
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\label{FIG LUMPING SUPG Ab} 
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\end{figure} 
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\begin{figure}[ht] 
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\centerline{\includegraphics[width=7cm]{lumping_SUPG_u_2}} 
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\caption{Amplitude at point $(\frac{1}{2},\frac{1}{2})$ using the direct formulation~\ref{LUMPING SUPG 2} of the 
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TaylorGalerkin scheme using order $2$ elements, element size $0.01$, $v=(1,0)$.} 
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\label{FIG LUMPING SUPG B} 
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\end{figure} 
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\subsection{Sumamry} 
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The examples in this section have demonstrated the capabilities and limitations 
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of both HRZ and row sum lumping with comparisons to the exact and full mass 
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matrix solutions. Wave propagation type problems that utilise lumping, produce 
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results simular the full mass matrix at significantly 
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lower computation cost. An accelleration based formulation, with HRZ lumping 
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should be implemented for such problems, and can be appied to both order $1$ and 
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order $2$ elements. 
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In transport type problems, it is essential that row sum lumping is used to 
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achieve a smooth solution. Additionally, it is not recommended that second order 
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elements be used in advection type problems. 
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