Nonlinear parabolic compactness and monotone evolution — Examples

Write $\alpha =1\mathrm{/}(m-1)$, $y=x{t}^{-\beta }$, and $F(y)=(C-k{y}^2{)}_{+}^{\alpha }$. Then $u=t^{-\beta }F(y)$ and $u^m=t^{-m\beta }G(y)$, where $G(y)=(C-k{y}^2{)}_{+}^{\alpha +1}$. Inside the positivity set, $u_t=t^{-\beta -1}(-\beta F-\beta y{F}^{\mathrm{\prime }}),(u^m{)}_{xx}=t^{-(m+2)\beta }{G}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{.}$ Equality of the time powers requires $\beta +1=(m+2)\beta$, giving $\beta =1\mathrm{/}(m+1)$.

In the positivity set put $q=C-k{y}^2>0$. Direct differentiation gives $F^{\mathrm{\prime }}=-2k\alpha y{q}^{\alpha -1},G^{\mathrm{\prime }\mathrm{\prime }}=-2k(\alpha +1)q^{\alpha -1}(q-2k\alpha y^2)=-2k(\alpha +1)(F+y{F}^{\mathrm{\prime }})\mathrm{.}$ Choosing $2k(\alpha +1)=\beta$ makes $G^{\mathrm{\prime }\mathrm{\prime }}=-\beta (F+y{F}^{\mathrm{\prime }})$, exactly the remaining profile equation. Since $\alpha +1=m\mathrm{/}(m-1)$, this choice is $k=(m-1)\mathrm{/}(2m(m+1))$. Thus the PDE holds pointwise wherever $u>0$; both sides vanish in the open set where $u=0$.

The interface is $q=0$. The profile $u$ tends to zero there because $\alpha >0$. The spatial flux $G$ and its first derivative $G^{\mathrm{\prime }}=-2k(\alpha +1)y{q}_{+}^{\alpha }$ also tend to zero at the interface, so distributional differentiation of $G$ creates no boundary delta. Its interior second derivative is locally integrable: its only possible singularity is proportional to $q^{\alpha -1}$ and $\alpha -1>-1$. Likewise, $u_t$ has only a locally integrable $q^{\alpha -1}$ singularity; the moving interface contributes no time delta because the jump of $u$ across it is zero. Hence the pointwise identity of step 2.1 extends distributionally across the interface.

The positivity condition is $C-k{x}^2{t}^{-2\beta }>0$, equivalent to $\mathrm{\mid }x\mathrm{\mid }<\sqrt{C\mathrm{/}k}{t}^{\beta }$. Therefore the support is the stated closed interval, with radius proportional to $t^{\beta }=t^{1\mathrm{/}(m+1)}$. This proves finite propagation and the claimed scaling.

By [F2], $V_h\subset V$ and $v\in V$. By [F3], $\mathrm{\mid }a(f,g)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^1}\mathrm{\parallel }g{\mathrm{\parallel }}_{H^1}$ and $a(f,f)=\mathrm{\parallel }{f}^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2\ge \frac{1}{2}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^1}^2$, so $a$ is bounded, symmetric and coercive as required by [F1]. The displayed hat has slopes $2$ and $-2$. Also $v^{\mathrm{\prime }}(x)=1-2x$, and direct integration gives $a(v,w)=2{\int }_0^{1\mathrm{/}2}(1-2x)dx-2{\int }_{1\mathrm{/}2}^1(1-2x)dx=1,a(w,w)=4.$ Since $R_{h}v=cw$, [F1] with test $w$ gives $4c=1$, so $R_{h}v=w\mathrm{/}4$ and its slopes are $1\mathrm{/}2$ and $-1\mathrm{/}2$.

On each half interval, $\mathrm{\mid }(R_{h}v{)}^{\mathrm{\prime }}\mathrm{\mid }(R_{h}v{)}^{\mathrm{\prime }}{w}^{\mathrm{\prime }}=1\mathrm{/}2$, so its integral is $1\mathrm{/}2$. On the left half, $v^{\mathrm{\prime }}\ge 0$ and $\mathrm{\mid }{v}^{\mathrm{\prime }}\mathrm{\mid }{v}^{\mathrm{\prime }}{w}^{\mathrm{\prime }}=2(1-2x{)}^2$, whose integral is $1\mathrm{/}3$. On the right half, $v^{\mathrm{\prime }}\le 0$ and the same integrand $2(1-2x{)}^2$ also integrates to $1\mathrm{/}3$. Therefore ${\int }_0^1\mathrm{\mid }{v}^{\mathrm{\prime }}\mathrm{\mid }{v}^{\mathrm{\prime }}{w}^{\mathrm{\prime }}=2\mathrm{/}3$.

The two residual pairings are $1\mathrm{/}2$ and $2\mathrm{/}3$, so they are unequal. This explicit conforming Ritz space refutes any general claim that a linear Ritz projection preserves every nonlinear $p$-Laplacian Galerkin residual.

Since each interval $[0,T]$ contains exactly $n$ periods of $\sin (2\pi nt\mathrm{/}T)$, a change of variable gives $\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_{L^p(0,T;V)}^p=\mathrm{\parallel }v{\mathrm{\parallel }}_V^p\frac{T}{2\pi }{\int }_0^{2\pi }\mathrm{\mid }\sin s{\mathrm{\mid }}^{p}ds,$ independent of $n$. Thus the family is uniformly bounded in each stated $L^p(V)$.

For distinct positive integers $n\mathrm{\ne }m$, sine orthogonality over $[0,T]$ gives $\mathrm{\parallel }{u}_n-u_m{\mathrm{\parallel }}_{L^2(0,T;H)}^2=\mathrm{\parallel }iv{\mathrm{\parallel }}_H^2{\int }_0^T(\sin (2\pi nt\mathrm{/}T)-\sin (2\pi mt\mathrm{/}T){)}^{2}dt=T\mathrm{\parallel }iv{\mathrm{\parallel }}_H^2>0,$ where positivity follows from injectivity in [F1]. Every two distinct members therefore stay a fixed positive distance apart, so no subsequence is Cauchy and none converges strongly in the complete space $L^2(0,T;H)$.

Scalar integration by parts against each $w\in V$ shows that the pivot weak derivative is $u_n^{\mathrm{\prime }}(t)=(2\pi n\mathrm{/}T)\cos (2\pi nt\mathrm{/}T)J(iv)$. Its $L^q(V^{\ast })$ norm is $(2\pi n\mathrm{/}T)\mathrm{\parallel }J(iv){\mathrm{\parallel }}_{V^{\ast }}{(\frac{T}{2\pi }{\int }_0^{2\pi }\mathrm{\mid }\cos s{\mathrm{\mid }}^{q}ds)}^{1\mathrm{/}q}$ for finite $q$, and $(2\pi n\mathrm{/}T)\mathrm{\parallel }J(iv){\mathrm{\parallel }}_{V^{\ast }}$ for $q=\mathrm{\infty }$. Since $⟨J(iv),v⟩=\mathrm{\parallel }iv{\mathrm{\parallel }}_H^2>0$ by [F1], these norms grow linearly in $n$. Hence no positive exponent derivative bound is present.

Let $J_p(\xi )=\mathrm{\mid }\xi {\mathrm{\mid }}^{p-2}\xi$ and $\mathrm{\Phi }(\xi )=\mathrm{\mid }\xi {\mathrm{\mid }}^p\mathrm{/}p$. The function $\mathrm{\Phi }$ is convex: the triangle inequality gives $\mathrm{\mid }\theta \xi +(1-\theta )\eta \mathrm{\mid }\le \theta \mathrm{\mid }\xi \mathrm{\mid }+(1-\theta )\mathrm{\mid }\eta \mathrm{\mid }$, and $r\mapsto r^p$ is increasing and convex for $r\ge 0$. It is differentiable with $\mathrm{\nabla }\mathrm{\Phi }=J_p$, including at zero because $p>1$. For a differentiable convex function, restriction to the segment from $\eta$ to $\xi$ gives the tangent inequalities $\mathrm{\Phi }(\xi )\ge \mathrm{\Phi }(\eta )+J_p(\eta )\cdot (\xi -\eta ),\mathrm{\Phi }(\eta )\ge \mathrm{\Phi }(\xi )+J_p(\xi )\cdot (\eta -\xi )\mathrm{.}$ Adding them yields $(J_p(\xi )-J_p(\eta ))\cdot (\xi -\eta )\ge 0$. Also $J_p(\xi )\cdot \xi =\mathrm{\mid }\xi {\mathrm{\mid }}^p$ and $\mathrm{\mid }{J}_p(\xi )\mathrm{\mid }=\mathrm{\mid }\xi {\mathrm{\mid }}^{p-1}$, giving monotonicity, coercivity with $\alpha =1,h=0$, and growth with $C=1,g=0$.

For $p\ge 2$, the derivative of $J_p$ has operator norm at most $(p-1)\mathrm{\mid }z{\mathrm{\mid }}^{p-2}$. Integrating it along the segment from $\eta$ to $\xi$ gives $\mathrm{\mid }{J}_p(\xi )-J_p(\eta )\mathrm{\mid }\le (p-1)(\mathrm{\mid }\xi \mathrm{\mid }+\mathrm{\mid }\eta \mathrm{\mid }{)}^{p-2}\mathrm{\mid }\xi -\eta \mathrm{\mid }\mathrm{.}$ For $1<p<2$, put $d_0=\mathrm{\mid }\xi -\eta \mathrm{\mid }$ and $M_0=\max (\mathrm{\mid }\xi \mathrm{\mid },\mathrm{\mid }\eta \mathrm{\mid })$. If $d_0\ge M_0\mathrm{/}2$, then $\mathrm{\mid }{J}_p(\xi )-J_p(\eta )\mathrm{\mid }\le 2M_0^{p-1}\le 2^p{d}_0^{p-1}$. If $0<d_0<M_0\mathrm{/}2$, every point on the segment has norm at least $M_0\mathrm{/}2$; there the derivative norm is at most $(M_0\mathrm{/}2{)}^{p-2}$. Hence $\mathrm{\mid }{J}_p(\xi )-J_p(\eta )\mathrm{\mid }\le (M_0\mathrm{/}2{)}^{p-2}{d}_0\le C_p{d}_0^{p-1},$ because $(d_0\mathrm{/}{M}_0{)}^{2-p}\le 1$. The case $d_0=0$ is immediate. Thus for $1<p<2$ the global estimate $\mathrm{\mid }{J}_p(\xi )-J_p(\eta )\mathrm{\mid }\le C_p\mathrm{\mid }\xi -\eta {\mathrm{\mid }}^{p-1}$ holds.

If $z_n\to z$ in $V$, then $\mathrm{\nabla }{z}_n\to \mathrm{\nabla }z$ in $L^p$. For $p\ge 2$, Hölder applied to the first estimate of step 1.2 gives $\mathrm{\parallel }{J}_p(\mathrm{\nabla }{z}_n)-J_p(\mathrm{\nabla }z){\mathrm{\parallel }}_{L^{p^{\mathrm{\prime }}}}\le (p-1)(\mathrm{\parallel }\mathrm{\nabla }{z}_n{\mathrm{\parallel }}_p+\mathrm{\parallel }\mathrm{\nabla }z{\mathrm{\parallel }}_p{)}^{p-2}\mathrm{\parallel }\mathrm{\nabla }{z}_n-\mathrm{\nabla }z{\mathrm{\parallel }}_p\to 0.$ For $1<p<2$, the second estimate and $(p-1)p^{\mathrm{\prime }}=p$ give $\mathrm{\parallel }{J}_p(\mathrm{\nabla }{z}_n)-J_p(\mathrm{\nabla }z){\mathrm{\parallel }}_{L^{p^{\mathrm{\prime }}}}\le C_p\mathrm{\parallel }\mathrm{\nabla }{z}_n-\mathrm{\nabla }z{\mathrm{\parallel }}_p^{p-1}\to 0.$ For any $w\in V$, Hölder bounds $\mathrm{\mid }⟨A(z_n)-A(z),w⟩\mathrm{\mid }\le \mathrm{\parallel }{J}_p(\mathrm{\nabla }{z}_n)-J_p(\mathrm{\nabla }z){\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\mathrm{\parallel }w{\mathrm{\parallel }}_V$. Therefore $A:V\to V^{\ast }$ is norm-continuous, an additional property of this example.

The threshold and [F2] supply the required dense compact $V⋐L^2(\mathrm{\Omega })$ inclusion, and step 1.1 verifies the pointwise flux conditions. Step 2.1 additionally proves norm continuity, without requiring it as a flux hypothesis. Hence the flux meets all hypotheses of [F1].

On $\mathrm{\Omega }=(0,1)$ in $d=1$ the threshold is $p>1$, while on the unit cube in $d\ge 2$ it is exactly the stated strict threshold. For either domain, [F4] supplies an explicit fixed-degree spatial family and time partitions satisfying the definition in [F3]. Applying the convergence theorem [F3] to the flux verified in step 3.1 gives the concrete subsequential FEM convergence assertion in the Statement, with all original data and exponent hypotheses retained.