# How to construct a function in $H^3(\Omega)$, but not in $H^4(\Omega)$?

As the title mentioned, some suggestions and references are needed to help me to construct a function in $H^3(\Omega)$, but not in $H^4(\Omega)$？The domain $\Omega=[0,1]\times[0,1]$ and has zero trace. I'm very glad to receive some suggestions.

More precisely, I want to construct a function satisfying \begin{align} &-\Delta u=f\quad in~~\Omega=[0,1]\times[0,1],\\ &~~u=0,~~~~~~~~~on ~~~~\partial\Omega. \end{align} It may be like $r^{2/3}\sin(2/3\theta)$ on a L-shape domain $[-1,1]\times [-1,1]\backslash[0,1]\times[-1,0]$, which has singularity in $(0,0)$.

• Do you need an explicit expression or is it ok if I define the function as a solution to some PDE? – knl Mar 1 '17 at 13:54
• $x(1-x)y(1-y)|x-0.5|^{3.5}$ with the idea that $|x-0.5|^{3.5}$ is in $C^3$ but has unbounded 4th derivative and will generate a point singularity of type $x^{-1}$ at the line $x=0.5$? – VorKir Mar 1 '17 at 22:50
• The definition of classical derivative and weak derivative is different, so the function you provided in $C^3$, not in $C^4$. However, i need a function in $H^3$, but not in $H^4$.@Vorkir – Feng Young Mar 2 '17 at 1:26
• Yes, i want to find an explicit expression to verify my prior estimate (e.g. $\|u-u_h\|_0$). @knl – Feng Young Mar 2 '17 at 1:29
• Take the radial function $f(r)=r^{5/2}$ in the unit disc. This is in $H^3$ but not in $H^4$. – cpraveen Mar 2 '17 at 3:52

There are fundamentally two options you have:

• You construct a function based on the radial distance $r=\sqrt{x^2+y^2}$ and make the ansatz $w(x,y)=r^\alpha$ as suggested by Praveen Chandrashekar in the comments. Then compute the $H^s$ norm of this function on the unit disk via $$\|w\|^2_{H^s} = 2\pi \int_0^1 |\partial^s_r w(r)|^2 \, r\, dr,$$ which yields, up to constants $$\|w\|^2_{H^s} \propto \int_0^1 r^{2(\alpha-s)} \, r\, dr.$$ This integral is only finite if $2(\alpha-s)+1 > -1$, or in other words if $\alpha > -1+s$. That is, the function $w=r^\alpha$ is in $H^3$ if $\alpha>2$ but it is not in $H^4$ if $\alpha\le 3$. Praveen C.'s choice of $\alpha=\frac 52$ is a valid choice then.

• You construct a one dimensional function $w(x,y)=|x|^\beta$ that simply doesn't depend on the $y$ coordinate at all. Consequently, all two-dimensional norms really just come down to one dimensional ones. By a similar argument as above, the norm of $w$ is given by $$\|w\|^2_{H^s} = \int_{-1}^1 |\partial^s_x w(x)|^2\, dr,$$ which yields again, up to constants $$\|w\|^2_{H^s} \propto \int_0^1 |x|^{2(\beta-s)}\, dr.$$ This is again finite if $2(\beta-s)>-1$, or $\beta>-\frac 12 + s$. So $w\in H^3$ but $w\not\in H^4$ if $\frac 52 < \beta \le \frac 72$.

In both cases, $w$ does not satisfy the boundary conditions you ask for. But that can be addressed in the following way: Let $u_0$ satisfy the equation $$-\Delta u_0 = 0 \qquad \text{on \Omega=(-1,1)^2},$$ with boundary values $u_0|_{\partial\Omega}=-w|_{\partial\Omega}$, then $u=w+u_0$ does satisfy the equation you are looking for.

It is, of course, not a priori clear that $u_0$ actually has at least $H^3$ regularity, to ensure that $u$ is also in $H^3$. But a bit of functional analysis shows that since $w\in H^3$ by construction, $w|_{\partial\Omega}\in H^{5/2}(\partial\Omega)$, which in turn again implies that $u_0\in H^3$.

• This is a very nice explanation. Thank you very much. @Wolfgang Bangerth – Feng Young Mar 2 '17 at 5:01
• If $\beta=1$ and $s=2$, then $w(x)=|x|$ and $\partial^2_x w = 2\delta(x)$. Clearly, $\int_{-1}^1 |\delta(x)|^2 dx$ is still a divergent integral, but this doesn't look like your integral $\int_0^1 |x|^{2(\beta-s)} dx$. Am I missing something? Am I being overly picky about the details? BTW, I really appreciated your explanation. – LedHead Mar 2 '17 at 9:31
• @led23head your case is too singular. I should have specifically excluded the case where $\beta$ is an integer and $s>\beta$ because in that case you are indeed left with only the delta functions. In all other cases, you can exclude the origin from the integration without loss, because the remainder of the integral is already divergent. – Wolfgang Bangerth Mar 3 '17 at 4:42
• @WolfgangBangerth That makes sense. Thanks for the clarification. – LedHead Mar 3 '17 at 5:01
• @led23head -- I think I should have just restricted my derivation to cases where $\beta\ge s$. If $\beta<s$, then we're already in the regime where the solution is not in $H^s$ and we're not interested in that. – Wolfgang Bangerth Mar 3 '17 at 19:31