## 6 Numerical Examples
### An Illustrative Example  
The first example is given to describe the detailed steps of the adaptive procedure, showing in particular how to form the FE matrix equation, how to estimate the maximum errors, and how to adjust the element size. The problem is the linearized Problem 3 in the next example, i.e.,  
\[u^{\prime}+4 tu=0,u(0)=1,u(t)=e^{-2t^{2}},0\leq t \tag{26}\]  
The objective is to solve for the solution on the first element \((0,h)\) with the element size \(h\) being adaptively adjusted. The reduced linear element (\(m=1\)) is used as the final solution and, accordingly, the full element is a quadratic element, which can provide fourth-order nodal solutions (Equation (12)). From Equation (5), the trial and test functions are respectively as follows  
\[u^{h}=\bigg{(}1-\frac{t}{h}\bigg{)}u_{1}^{h}+6\bigg{(}1-\frac{t}{h}\bigg{)} \bigg{(}\frac{t}{h}\bigg{)}u_{3}^{h}+\bigg{(}\frac{t}{h}\bigg{)}u_{2}^{h},\;v ^{h}=v_{1}^{h}+\bigg{(}-1+\frac{2t}{h}\bigg{)}v_{2}^{h}. \tag{27}\]  
Then the bilinear form in Equation (11) is  
\[a^{e}(u^{h},v^{h})=\bigg{\{}v_{1}^{h}v_{2}^{h}\bigg{\}} \begin{bmatrix}\frac{2 \eta^{2}}{3}-1&2\bar{h}^{2}&\frac{4\bar{h}^{2}}{3}+1\\ 0&\frac{2\bar{h}^{2}}{5}-2&\frac{2\bar{h}^{2}}{3}\end{bmatrix}\begin{pmatrix}u _{0}\\ u_{3}^{h}\\ u_{2}^{h}\end{pmatrix} . \tag{28}\]  
Note that by introducing the initial condition, \(u_{1}^{h}\) has been replaced by \(u_{0}\). Then, from the arbitrariness of \(v_{1}^{h}\) and \(v_{2}^{h}\), the matrix equation can be derived based on Equation (11) as  
\[ \begin{bmatrix}2\bar{h}^{2}&\frac{4}{3}h^{2}+1\\ \frac{2\bar{h}^{2}}{5}-2&\frac{2\bar{h}^{2}}{3}\end{bmatrix}\begin{pmatrix}u _{3}^{h}\\ u_{2}^{h}\end{pmatrix} = \begin{cases}1-\frac{2\bar{h}^{2}}{3}\\ 0\end{cases} u_{0}, \tag{29}\]  
the  
Note that \(u_{2}^{h}\) is the FE solution at second end node of the element. It is easily seen that  
\[\left|\frac{2h^{4}-13h^{2}+15}{6h^{4}+17h^{2}+15}\right|<1\ \text{when}\ h>0, \tag{31}\]  
which means the element is unconditionally stable. To see the local (truncated) error of the full FE solution, the series expansion of the true error gives  
\[u(h)-u^{h}(h)=u(h)-u_{2}^{h}=\frac{2}{15}h^{6}+O(h^{7})\ \text{with}\ u_{0}=1. \tag{32}\]  
Note that the local error is of order 6, which is one order higher than it is estimated in Equation (12). This is because that the series expansion of the exact solution \(e^{-2h^{2}}=1-2h^{2}+2h^{4}-4h^{6}/3+O(h^{7})\) lacks the term of order 5. With the full FE (conventional element) solution \(u^{h}\) obtained, the reduced solution \(u^{h}_{\text{R}}\) (Equation (13)), the estimated error term \(\varepsilon_{\text{R}}^{h}\) (Equation (17)) and the estimated maximum error \(\max\left|\varepsilon_{\text{R}}^{h}\right|\) (Equation (18)) are, respectively, as follows:  
\[u^{h}_{\text{R}}=\left(1-\frac{t}{h}\right)u^{h}_{0}+\left(\frac{t}{h}\right) u^{h}_{2},\ \varepsilon_{\text{R}}^{h}=6\bigg{(}1-\frac{t}{h}\bigg{)}\bigg{(}\frac{t}{h} \bigg{)}u^{h}_{3},\ \max\left|\varepsilon_{\text{R}}^{h}\right|=\frac{3}{2}u^{h}_{3}. \tag{33}\]  
Now we are in a position to implement the adaptive FE analysis for the first element with the preset tolerance \(T_{l}=0.001\). Taking the initial element size \(h_{0}=0.5\) and setting the initial value \(u_{0}=1\), the following numerical results can be obtained from Equation (30)  
\[u^{h}_{3}=0.05307856,\ u^{h}_{2}=0.60509554. \tag{34}\]  
The true error at the right end node is \(u(h)-u^{h}_{2}=0.00143511\). It would be informative to calculate the same problem setting by using the fourth-order Runge-Kutta method, which yields the value \(0.60416667\) at \(t=0.5\) with the true error being \(0.00236399\), and turns out to be less accurate than the fourth-ordered solution of the quadratic element.