## 5 Test Problem for Numerical Experiments  
The test problem is defined in terms of the parabolic Equation (1), such that \(\Omega=(-1,1)\times(-1,1)\), \(T=0.1\), and an initial condition \(u^{0}(x)=0\). In all numerical experiments, \(\alpha=0.5\), which is considered as a representative case to prove the concept of the analysed methods and algorithms.  
We denote by \(\tilde{f}(x)\) the checker-board function on \(\Omega\setminus\partial\Omega=(-1,1)\times(-1,1)\)  
\[\tilde{f}(x)=\left\{\begin{array}{cc}1&\quad\text{if}\;\;x_{1}x_{2}>0\\ -1&\quad\text{otherwise}\end{array}\right.. \tag{10}\]  
Then the right-hand side of \(f(x,t)\in L^{2}(\Omega\times(0,T))\) has the form  
\[f(x,t)=\left\{\begin{array}{cc}100\tilde{f}(x)&\quad\text{if}\;\;t\in(0, \tilde{t}_{1}]\\ 200\tilde{f}(x)&\quad\text{if}\;\;t\in(\tilde{t}_{2},\tilde{t}_{3}]\\ 0&\quad\text{otherwise}\end{array}\right., \tag{11}\]  
where \(\tilde{t}_{1}=0.01\), \(\tilde{t}_{2}=0.05\) and \(\tilde{t}_{3}=0.06\). For the adaptive time stepping scheme, we set the parameters \(\gamma=1.5\), \(\delta=0.01\), and \(\theta=0.1\). We also apply successive refinement of FEM meshes, resulting in an increase in degrees of freedom in space with \(N=2131,\;4167,\;8030,\;12,805,\;16,184,\;24,892,\;32,302\).  
In the case of adaptive time stepping, the accuracy in time is controlled by the initial time-step \(\tau_{0}\), which must be aligned with the FEM accuracy in space. Furthermore, in the context of our test problem, \(\tau_{j}\) must agree on jump points of \(f(t)\).  
In the performance analysis, we use also results for the backward Euler algorithm with uniform step \(\tau\). For this purpose, the relative error in the norm \(l_{2}\) is used  
\[R_{j}=\frac{\left\|\mathbf{u}^{j}-\mathbf{u}^{j+1}\right\|_{l_{2}}}{\left\| \mathbf{u}^{j}\right\|_{l_{2}}}=\frac{\sqrt{\sum_{i=1}^{N}(u_{i}^{j+1}-u_{i}^ {j})^{2}}}{\sqrt{\sum_{i=1}^{N}(u_{i}^{j})^{2}}},\]  
where the solution \(\mathbf{u}^{j}\) is obtained with \(m=200\times 2^{j},j=0,1,2,\dots\). In the considered example, there are several points where \(f\) has jumps. To fit them, we add additional checks that will not allow the time step to go beyond these two points. If \(t_{i}<0.01\) and \(\tilde{t}_{i+1}>0.05\), we lower \(\tilde{\tau}_{i}\) such that \(\tilde{t}_{i}+\tilde{\tau}_{i}=0.05\). In a similar way, we treat the interval \([0.05,0.06]\).  
In Figure 1, we present the relative errors for \(N=2131\) with decreasing constant time step \(\tau\) (the behaviour is quite similar for all seven variants of \(N\)). The results shown are for \(t=0.025,\;0.05,\;0.075,\;0.1\). The relative error between two consecutive values of the time steps becomes smaller than \(10^{-3}\) at \(m=6400\) and smaller than \(10^{-4}\) at \(m=204800\). Based on this simple observation, in the numerical experiments for the adaptive scheme, we will consider the cases \(m_{\max}=6400\) and \(m_{\max}=204,800\), which correspond to \(\tau_{0}=1.5625\times 10^{-5}\) and \(\tau_{0}=4.8828125\times 10^{-7}\). As you will see in the next section, the two chosen options for \(\tau_{0}\) turn out to be very representative for the purposes of the presented performance analysis.  
Figure 1: Backward Euler method with uniform time step: reduction in relative error \(R_{j}\) versus decrease in \(\tau\) for \(N=2131\).  
Finally, for the absolute and relative threshold used in the HSS compression implemented in STRUMPACK, we will use a fixed \(\varepsilon_{\text{abs}}=10^{-8}\) and vary \(\varepsilon_{\text{rel}}=10^{-2},10^{-4},10^{-6}\) and \(10^{-8}\).