Quantitative uniqueness of solutions to second order elliptic equations with singular potentials in two dimensions

Abstract

In this article, we study the vanishing order of solutions to second order elliptic equations with singular lower order terms in the plane. In particular, we derive lower bounds for solutions on arbitrarily small balls in terms of the Lebesgue norms of the lower order terms for all admissible exponents. Then we show that a scaling argument allows us to pass from these vanishing order estimates to estimates for the rate of decay of solutions at infinity. Our proofs rely on a new $L^p - L^q$ Carleman estimate for the Laplacian in $\mathbb {R}^2$.

Appendix

In this section, we first present the proof of Lemma 3 for the case $1<p<2$. The eigenfunction estimates in Lemma 2 play an important role in the argument. Then we prove a quantitative Caccioppoli inequality in dimension $n=2$.

Proof of Lemma 3

We define a conjugated operator $L^-_\tau$ of $L^-$ by

$$\begin{aligned} L^-_\tau u=e^{-\tau \varphi (t)}L^-(e^{\tau \varphi (t)}u). \end{aligned}$$

With $v=e^{\tau \varphi (t)}u$, it is equivalent to prove

$$\begin{aligned} \Vert t^{-{1}}u \Vert _{L^2(dtd\omega )}\le C \tau ^\beta \Vert t L^-_\tau u\Vert _{L^p(dtd\omega )}. \end{aligned}$$

(A.1)

From the definition of $\varLambda$ and $L^-$ in (11) and (12), the operator $L^-_\tau$ can be written as

$$\begin{aligned} L^-_\tau =\sum _{k\ge 0} (\partial _t+\tau \varphi '(t)-k)P_k. \end{aligned}$$

(A.2)

Let $M=\lceil 2\tau \rceil$. Since $\displaystyle \sum _{k\ge 0} P_k v= v$, we split $\displaystyle \sum _{k \ge 0} P_k v$ into

$$\begin{aligned} P^+_\tau =\sum _{k> M}P_k, \quad \quad P^-_\tau =\sum _{k=0}^{M}P_k. \end{aligned}$$

Then (A.1) is reduced to showing that

$$\begin{aligned} \Vert t^{-1} P^+_\tau {u} \Vert _{L^2(dtd\omega )} \le \tau ^{\beta }\Vert t{L_\tau ^- u} \Vert _{L^{{p}}(dtd\omega )} \end{aligned}$$

(A.3)

and

$$\begin{aligned} \Vert t^{-1} P^-_\tau {u}\Vert _{L^2(dtd\omega )}\le \tau ^\beta \Vert t {L_\tau ^- u} \Vert _{L^{{p}}(dtd\omega )} \end{aligned}$$

(A.4)

hold for all $u \in C^\infty _c\left( (-\infty , \ t_0)\times S^{1} \right)$ and $1< p < 2$. We first establish (A.3). From (A.2) and properties of the projection operator $P_k$, it follows that

$$\begin{aligned} P_k L^-_\tau u= (\partial _t+\tau \varphi '(t)-k)P_k u. \end{aligned}$$

(A.5)

For $u\in C^\infty _{0}\left( (-\infty , \ t_0)\times S^{1} \right)$, the solution $P_k u$ of this first order differential equation is given by

$$\begin{aligned} P_k u(t, \omega ) =-\int _{-\infty }^{\infty } H(s-t)e^{k(t-s)+\tau \left[ \varphi (s)-\varphi (t)\right] } P_k L^-_\tau u (s,\omega )\, ds, \end{aligned}$$

(A.6)

where $H(z)=1$ if $z\ge 0$ and $H(z)=0$ if $z<0$.

For $k\ge M \ge 2 \tau$, it can be shown that

$$\begin{aligned} H(s-t)e^{k(t-s)+\tau \left[ \varphi (s)-\varphi (t)\right] } \le e^{-\frac{1}{2}k|t-s|} \end{aligned}$$

for all $s, t\in (-\infty , \ t_0)$. Taking the $L^2\left( S^{1} \right)$-norm in (A.6) yields that

$$\begin{aligned} \Vert P_k u(t, \cdot )\Vert _{L^2(S^{1})} \le \int _{-\infty }^{\infty } e^{-\frac{1}{2}k|t-s|} \Vert P_k L^-_\tau u(s, \cdot )\Vert _{L^2(S^{1})} \,ds. \end{aligned}$$

Applying the eigenfunction estimates (19) gives that

$$\begin{aligned} \Vert P_k u(t, \cdot )\Vert _{L^2(S^{1})} \le C \int _{-\infty }^{\infty } e^{-\frac{1}{2}k|t-s|} \Vert L^-_\tau u(s, \cdot )\Vert _{L^p(S^{1})} \,ds \end{aligned}$$

for all $1\le p\le 2$. Furthermore, the application of Young’s inequality for convolution yields that

$$\begin{aligned} \Vert P_k u\Vert _{L^2(dt d\omega )} \le C \left( \int _{-\infty }^{\infty } e^{-\frac{\sigma }{2}k|z|} dz \right) ^{\frac{1}{\sigma }}\Vert L^-_\tau u\Vert _{L^p(dtd\omega )} \end{aligned}$$

with $\frac{1}{\sigma }=\frac{3}{2}-\frac{1}{p}$. Therefore,

$$\begin{aligned} \Vert P_k u\Vert _{L^2(dt d\omega )} \le C k^{\frac{1}{p} - \frac{3}{2}} \Vert L^-_\tau u\Vert _{L^p(dtd\omega )}, \end{aligned}$$

where we have used the fact that

$$\begin{aligned} \left( \int _{-\infty }^{\infty } e^{-\frac{\sigma }{2}k|z|} dz \right) ^{\frac{1}{\sigma }}\le C k^{\frac{1}{p} - \frac{3}{2}}. \end{aligned}$$

Squaring and summing up $k> M$ shows that

$$\begin{aligned} \sum _{k> M} \Vert P_k u\Vert ^2_{L^2(dt d\omega )} \le C \sum _{k> M} k^{\frac{2}{p} - 3} \Vert L^-_\tau u\Vert ^2_{L^p(dtd\omega )}. \end{aligned}$$

$\displaystyle \sum _{k> M} k^{\frac{2}{p} - 3}$ converges if $p>1$. If $p=1$, $\displaystyle \sum _{k> M} k^{\frac{2}{p}- 3}$ diverges. Therefore,

$$\begin{aligned} \Vert P^+_\tau u\Vert _{L^2(dtd\omega )} \le C\tau ^{\frac{1-p}{p}}\Vert L^-_\tau u\Vert _{L^p(dtd\omega )}, \end{aligned}$$

which implies estimate (A.3) since $|t|\ge |t_0|$, where $\left| t_0\right|$ is large.

Set $N=\lceil \tau \varphi ^\prime (t)\rceil$. Recall that $\varphi (t)=t+\log t^2$. By Taylor’s theorem, for all $s, t \in (-\infty , \ t_0)$, we have

$$\begin{aligned} \varphi (s)-\varphi (t)&= \varphi '(t)(s-t)-\frac{1}{(s_0)^2}(s-t)^2, \end{aligned}$$

(A.7)

where $s_0$ is some number between s and t. If $s>t$, then

$$\begin{aligned} S_k(s, t) = e^{k(t-s) + \tau \left[ \varphi \left( s - \varphi \left( t \right) \right) \right] } \le e^{-(k-\tau \varphi '(t))(s-t)-\frac{\tau }{t^2}(s-t)^2}. \end{aligned}$$

Hence

$$\begin{aligned} H(s-t) S_k(s, t)\le e^{-|k- N ||s-t|-\frac{\tau }{t^2}(s-t)^2}. \end{aligned}$$

(A.8)

Furthermore, we consider the case $N\le k\le M$. The summation of (A.6) over k shows that

$$\begin{aligned} \Vert \sum ^M_{k=N} P_k u(t, \cdot )\Vert _{L^2(S^{1})} \le \int _{-\infty }^{\infty } \Vert \sum ^M_{k=N} H(s-t)S_k(s,t) P_k L^-_\tau u(s, \cdot ) \Vert _{L^2(S^{1})}\, ds. \end{aligned}$$

(A.9)

Let $c_k= H(s-t)S_k(s,t)$. It is clear that $|c_k|\le 1$. Now we make use of Lemma 2. An application of estimate (14) shows that for all $1< p < 2$

$$\begin{aligned}&\Vert \sum ^M_{k=N}H(s-t)S_k(s,t) P_k L^-_\tau u(s, \cdot )\Vert _{L^2(S^{1})}\le \nonumber \\&C \left( \sum ^M_{k=N} H(s-t)|S_k(s,t)|^2 \right) ^{\frac{1}{p}-\frac{1}{2}} \Vert L^-_\tau u(s, \cdot ) \Vert _{L^p(S^{1})}. \end{aligned}$$

(A.10)

From (A.8), we have

$$\begin{aligned} \sum ^M_{k=N} H(s-t)|S_k(s,t)|^2&\le \left( \sum ^M_{k=N+1} e^{-2|k- N||s-t|}+1 \right) e^{ -\frac{2\tau }{t^2}(s-t)^2} \nonumber \\&\le C \left( \frac{1}{|s-t|}+1 \right) e^{ -\frac{\tau }{t^2}(s-t)^2} . \end{aligned}$$

(A.11)

Therefore, the last two inequalities imply that

$$\begin{aligned} \Vert \sum ^M_{k=N}H(s-t)S_k(s,t) P_k L^-_\tau u(s,\cdot )\Vert _{L^2(S^{1})}&\le C (|s-t|^{-\alpha _2}+1)e^{-\frac{\alpha _2\tau }{t^2}(s-t)^2}\Vert L^-_\tau u(s, \cdot ) \Vert _{L^p(S^{1})}, \end{aligned}$$

where $\alpha _2=\frac{(2-p)}{2p}$. It can be shown that

$$\begin{aligned} e^{-\frac{\alpha _2\tau }{t^2}(s-t)^2}\le C|t|\left( 1+\sqrt{\tau }|s-t| \right) ^{-1}. \end{aligned}$$

(A.12)

Thus,

$$\begin{aligned} \Vert \sum ^M_{k=N}H(s-t)S_k(s,t) P_k L^-_\tau u(s,\cdot )\Vert _{L^2(S^{1})} \le \frac{C|t|(|s-t|^{-\alpha _2}+1)\Vert L^-_\tau u(s, \cdot ) \Vert _{L^p(S^{1})}}{1+\sqrt{\tau }|s-t|}. \end{aligned}$$

It follows from (A.9) that

$$\begin{aligned} |t|^{-1}\Vert \sum ^M_{k=N} P_k u(t, \cdot )\Vert _{L^2(S^{1})} \le C \int _{-\infty }^{\infty }\frac{(|s-t|^{-\alpha _2}+1)\Vert L^-_\tau u(s, \cdot ) \Vert _{L^p(S^{1})}}{1+\sqrt{\tau }|s-t|}. \end{aligned}$$

(A.13)

For the case $k\le N-1$, solving the first order differential Eq. (A.5) gives that

$$\begin{aligned} P_k u(t, \omega )=\int _{-\infty }^{\infty } H(t-s)S_k(s, t) P_k L^-_\tau u\left( s, \omega \right) \, ds . \end{aligned}$$

(A.14)

The estimate (A.7) shows that for any s, t

$$\begin{aligned} H(t-s) S_k(s, t)\le e^{-|N- 1 - k ||s-t|-\frac{\tau }{s^2}(t-s)^2}. \end{aligned}$$

(A.15)

Using (A.15) and performing the calculation as before, we conclude that

$$\begin{aligned} \Vert \sum _{k=0}^{N-1} P_k u(t, \cdot )\Vert _{L^2(S^{1})} \le C \int _{-\infty }^{\infty } \frac{|s|(|s-t|^{-\alpha _2}+1)\Vert L^-_\tau u(s, \cdot ) \Vert _{L^p(S^{1})}}{1+\sqrt{\tau }|s-t|}. \end{aligned}$$

(A.16)

Since s, t are in $(-\infty , \ t_0)$ with $|t_0|$ large enough, the combination of estimates (A.13) and (A.16) gives

$$\begin{aligned} |t|^{-1}\Vert P_\tau ^- u(t, \cdot )\Vert _{L^2(S^{1})} \le C \int _{-\infty }^{\infty } \frac{|s|(|s-t|^{-\alpha _2}+1)\Vert L^-_\tau u(s, \cdot ) \Vert _{L^p(S^{1})}}{1+\sqrt{\tau }|s-t|}. \end{aligned}$$

Applying Young’s inequality for convolution, we obtain

$$\begin{aligned} \Vert t^{-1} P_\tau ^- u (t, \cdot ) \Vert _{L^2(dtd\omega )} \le C \left[ \int _{-\infty }^{\infty } \left( \frac{ |z|^{-\alpha _2}+1}{1+\sqrt{\tau }|z|} \right) ^\sigma \, dz\right] ^{\frac{1}{\sigma }} \Vert tL^-_\tau u \Vert _{L^p(dtd\omega )}, \end{aligned}$$

where $\frac{1}{\sigma }=\frac{3}{2}-\frac{1}{p}$. Therefore,

$$\begin{aligned} \Vert t^{-1} P_\tau ^- u (t, \cdot ) \Vert _{L^2(dtd\omega )}\le C\tau ^{-\frac{1}{2\sigma }+\frac{\alpha _2}{2}} \Vert t L^-_\tau u \Vert _{L^p(dtd\omega )}, \end{aligned}$$

where we have used the fact

$$\begin{aligned} \left[ \int _{-\infty }^{\infty } \left( \frac{ |z|^{-\alpha _2}+1}{1+\sqrt{\tau }|z|} \right) ^\sigma \, dz\right] ^{\frac{1}{\sigma }} \le C\tau ^{-\frac{1}{2\sigma }+\frac{\alpha _2}{2}} \end{aligned}$$

with $\alpha _2 \in \left( 0, \frac{1}{2} \right)$ and $\sigma \in \left( 1, 2 \right)$. This completes (A.4) since $-\frac{1}{2\sigma }+\frac{\alpha _2}{2}=\frac{1-p}{p}$. Finally, the proof is complete. $\square$

We state and prove a Caccioppoli inequality for the second order elliptic Eq. (5) with singular lower order terms. Because our lower order terms are assumed to be singular, we must employ a Sobolev embedding in $\mathbb {R}^2$, and this forces the right hand side to be relatively larger than it was in Lemma 5 from [6], the corresponding result for $n \ge 3$.

Lemma 7

Assume that for some $s \in \left( 2, \infty \right]$ and $t \in \left( 1, \infty \right]$, $\left| \left| W\right| \right| _{L^s\left( B_{R} \right) } \le K$ and $\left| \left| V\right| \right| _{L^t\left( B_{R} \right) } \le M$. Let u be a solution to Eq. (5) in $B_R$. Then for any $\delta > 0$, there exists a constant C, depending only on s, t and $\delta$, such that for any $r<R$,

$$\begin{aligned} \Vert \nabla u\Vert ^2_{L^2(B_r)}\le C\left[ \frac{1}{(R-r)^2}+M^{\frac{t}{t-1}+ \delta } +K^{\frac{2s}{s-2} + \delta }\right] \Vert u\Vert ^2_{L^2(B_R)}. \end{aligned}$$

Proof

We start by decomposing V and W into bounded and unbounded parts. For some $M_0, K_0$ to be determined, let

$$\begin{aligned} V(x)=\overline{V}_{M_0}+V_{M_0}, \quad W(x)=\overline{W}_{K_0}+W_{K_0}, \end{aligned}$$

where

$$\begin{aligned} {\overline{V}}_{M_0}=V(x)\chi _{\left\{ |V(x)|\le \sqrt{M_0}\right\} }, \quad {V}_{M_0}=V(x)\chi _{\left\{ |V(x)|>\sqrt{M_0}\right\} }, \end{aligned}$$

and

$$\begin{aligned} {\overline{W}}_{K_0}=W(x)\chi _{\left\{ |W(x)|\le \sqrt{K_0}\right\} }, \quad {W}_{K_0}=W(x)\chi _{\left\{ |W(x)|>\sqrt{K_0}\right\} }. \end{aligned}$$

For any $q \in \left[ 1, t\right]$,

$$\begin{aligned} \Vert V_{M_0}\Vert _{L^q}\le M_0^{-\frac{t-q}{2q}}\Vert V_{M_0}\Vert _{L^{t}}^{\frac{t}{q}}\le M_0^{-\frac{t-q}{2q}}\Vert V\Vert _{L^{t}}^{\frac{t}{q}} \le M_0^{-\frac{t-q}{2q}}M^{\frac{t}{q}}. \end{aligned}$$

(A.17)

Similarly, for any $q \in \left[ 1, s\right]$, we have

$$\begin{aligned} \Vert W_{K_0}\Vert _{L^q}\le K_0^{-\frac{s-q}{2q}}\Vert W_{K_0}\Vert _{L^{s}}^{\frac{s}{q}}\le K_0^{-\frac{s-q}{2q}}\Vert W\Vert _{L^{s}}^{\frac{s}{q}} \le K_0^{-\frac{s-q}{2q}}K^{\frac{s}{q}}. \end{aligned}$$

(A.18)

Let $B_R\subset B_1$. Choose a smooth cut-off function $\eta \in C^\infty _0\left( B_R \right)$ such that $\eta (x) \equiv 1$ in $B_r$ and $|\nabla \eta |\le \frac{C}{|R-r|}$. Multiplying both sides of Eq. (5) by $\eta ^2 u$ and integrating by parts, we obtain

$$\begin{aligned} \int |\nabla u|^2 \eta ^2 = \int V\eta ^2 u^2 +\int W\cdot \nabla u \, \eta ^2 u - 2\int \nabla u\cdot \nabla \eta \, \eta \, u. \end{aligned}$$

(A.19)

We estimate the terms on the right side of (A.19). For the first term, we see that

$$\begin{aligned} \int V \eta ^2 u^2 \le \int |{\overline{V}_{M_0}}|\eta ^2 u^2 + \int |{{V}_{M_0}}|\eta ^2 u^2. \end{aligned}$$

It is clear that

$$\begin{aligned} \int |{\overline{V}_{M_0}}|\eta ^2 u^2\le M_0^{\frac{1}{2}}\int \eta ^2 u^2. \end{aligned}$$

(A.20)

Fix $\delta > 0$. Let $\delta _0 = \delta \frac{\left( t-1 \right) ^2}{t+\delta \left( t-1 \right) }$. By Hölder’s inequality, (A.17) with $q = 1 + \delta _0$, and Sobolev embedding with $2q^\prime := 2\left( 1 + \delta _0^{-1} \right) > 2$, we get

$$\begin{aligned} \left| \int {V}_{M_0}\eta ^2 u^2 \right|&\le \left( \int |{{V}_{M_0}}|^{q} \right) ^{\frac{1}{q}} \left( \int \left| \eta ^2u^2\right| ^{q^\prime } \right) ^{\frac{1}{q^\prime }} \nonumber \\&\le C_{\delta _0} {M_0}^{-\frac{t-q}{2q}}M^{\frac{t}{q}} \int |\nabla (\eta u)|^2. \end{aligned}$$

(A.21)

Taking $C_{\delta _0} {M_0}^{-\frac{t-q}{2q}}M^{\frac{t}{q}}=\frac{1}{16}$, i.e $M_0 = C_{\delta _0, t} M^{\frac{2t}{t-q}}$, from (A.20) and (A.21), we get

$$\begin{aligned} \int V \eta ^2 u^2\le & {} C M^{\frac{t}{t-q}}\int |\eta u|^2+\frac{1}{16}\int |\nabla (\eta u)|^2 \nonumber \\\le & {} C M^{\frac{t}{t-1} + \delta }\int |\eta u|^2+\frac{1}{8}\int |\nabla \eta |^2 u^2 +\frac{1}{8}\int |\nabla u|^2 \eta ^2. \end{aligned}$$

(A.22)

Now we estimate the second term in the righthand side of (A.19). We have that

$$\begin{aligned} \int W \cdot \nabla u \eta ^2 u = \int |{\overline{W}_{K_0}}| |\nabla u| \eta ^2 u + \int |{W}_{K_0}| | \nabla u| \eta ^2 u. \end{aligned}$$

(A.23)

By Young’s inequality for products,

$$\begin{aligned} \int |{\overline{W}_{K_0}}| |\nabla u| \eta ^2 u \le \frac{1}{8} \int |\nabla u|^2\eta ^2 +CK_0 \int \eta ^2 u^2. \end{aligned}$$

(A.24)

This time, set $\delta _0 =\delta \frac{\left( s-2 \right) ^2}{2s+\delta \left( s-2 \right) }$. By Hölder’s inequality, (A.17) with $q = 2 + \delta _0$, and Sobolev embedding with $q^\prime = 2\left( 1 + 2 \delta _0^{-1} \right) > 2$, we get

$$\begin{aligned} \int {W}_{K_0} \cdot \nabla u \eta ^2 u&\le \left( \int |{{W}_{K_0}}|^{{q}} \right) ^{\frac{1}{q}} \left( \int |\nabla u\cdot \eta |^{2} \right) ^{\frac{1}{2}} \left( \int |u \eta |^{q^{\prime }} \right) ^{\frac{1}{q^\prime }} \nonumber \\&\le C_{\delta _0} K_0^{-\frac{s-q}{2q}}K^{\frac{s}{q}}\Vert |\nabla u|\eta \Vert _{L^2} \Vert \nabla (\eta u)\Vert _{L^2} \nonumber \\&\le C_{\delta _0} K_0^{-\frac{s-q}{2q}}K^{\frac{s}{q}}\Vert |\nabla u|\eta \Vert _{L^2}\left( \Vert |\nabla \eta | u \Vert _{L^2} + \Vert |\nabla u|\eta \Vert _{L^2} \right) \nonumber \\&\le 2 C_{\delta _0} K_0^{-\frac{s-q}{2q}}K^{\frac{s}{2}}\Vert |\nabla u|\eta |\Vert ^2_{L^2} + \frac{1}{2} C_{\delta _0} K_0^{-\frac{s-q}{2q}}K^{\frac{s}{2}} \Vert |\nabla \eta | u \Vert _{L^2}^2. \end{aligned}$$

(A.25)

We choose $C_{\delta _0} K_0^{-\frac{s-q}{2q}}K^{\frac{s}{q}}=\frac{1}{16}$, that is, $K_0=C_{\delta _0, s} K^{\frac{2s}{s-q}}$. The combination of (A.23), (A.24) and (A.25) gives that

$$\begin{aligned} \int W \cdot \nabla u \, \eta ^2 u\le & {} \frac{1}{4} \Vert |\nabla u| \eta \Vert ^2_{L^2}+CK^{\frac{2s}{s-2} + \delta }\Vert u\eta \Vert ^2_{L^2} +\frac{1}{32}\Vert |\nabla \eta | u\Vert ^2_{L^2} . \end{aligned}$$

(A.26)

Finally, Young’s inequality for products implies that

$$\begin{aligned} 2\int \nabla u\cdot \nabla \eta \, \eta \, u \le \frac{1}{8} \Vert |\nabla u| \eta \Vert ^2_{L^2} + 8 \Vert |\nabla \eta | u \Vert ^2_{L^2} \end{aligned}$$

Together with (A.19), (A.22) and (A.26), we obtain

$$\begin{aligned} \int |\nabla u|^2 \eta ^2 \le C \left( M^{\frac{t}{t-1} + \delta } +K^{\frac{2s}{s-2} + \delta } \right) \int |u\eta |^2\,dx+ C \int |\nabla \eta |^2u^2\,dx. \end{aligned}$$

From the assumptions on $\eta$, this completes the proof in the lemma. $\square$