Overlarge streamline stabilization suppresses a two-cell solution

The function $u(x)=x-(e^x-1)\mathrm{/}(e-1)$ satisfies the endpoint conditions. Its derivatives are $u^{\mathrm{\prime }}=1-e^x\mathrm{/}(e-1)$ and $u^{\mathrm{\prime }\mathrm{\prime }}=-e^x\mathrm{/}(e-1)$, so $-u^{\mathrm{\prime }\mathrm{\prime }}+u^{\mathrm{\prime }}=1$. Hence $u(1\mathrm{/}2)=u_{\ast }:=1\mathrm{/}2-1\mathrm{/}(\sqrt{e}+1)>0$.

Let $\varphi$ be the midpoint hat function and $U$ its coefficient. Its derivative is $2$ on the left cell and $-2$ on the right. Thus ${\int }_0^1\mathrm{\mid }{\varphi }^{\mathrm{\prime }}{\mathrm{\mid }}^{2}dx=4$, ${\int }_0^1{\varphi }^{\mathrm{\prime }}\varphi dx=\frac{1}{2}[{\varphi }^2{]}_0^1=0$, and ${\int }_0^1\varphi dx=1\mathrm{/}2$. Also ${\int }_0^1{\varphi }^{\mathrm{\prime }}dx=\varphi (1)-\varphi (0)=0$, so the stabilization load vanishes. The single discrete equation is therefore $4(1+M\tau )U=1\mathrm{/}2$, so $U=1\mathrm{/}[8(1+M\tau )]$.

Since $\tau >0$ by [F3], $U\to 0$ as $M\to \mathrm{\infty }$, whereas the exact value $u_{\ast }>0$ from step 1.1 is fixed. In particular, every $M$ satisfying $M\tau >1\mathrm{/}(4u_{\ast })-1$ gives $U<u_{\ast }\mathrm{/}2$. Oversizing the parameter can therefore smear this discrete solution.