Half-Space Solutions with 7/2 Frequency in the Thin Obstacle Problem

Abstract

For the thin obstacle problem in ${\mathbb {R}}^3$, we show that half-space solutions form an isolated family in the space of $\frac{7}{2}$-homogeneous solutions. For a general solution with one blow-up profile in this family, we establish the rate of convergence to this profile. As a consequence, we obtain the regularity of the free boundary near such contact points.

Appendices

Appendix A. Fourier expansion in spherical caps

In this appendix, we study the decay of a harmonic function in a slit domain near the boundary of a spherical cap if some of its Fourier coefficients vanish along a smaller cap.

Recall that we use $(x_1,r,\theta )$ as the coordinate system for ${\mathbb {R}}^3$, where $r\ge 0$ and $\theta \in (-\pi ,\pi ]$ are the polar coordinates for the $(x_2,x_3)$-plane. For small $r_0>0$, the $r_0$-spherical cap is defined as

$$\begin{aligned} {\mathcal {C}}_{r_0}:=\{r<r_0, x_1>0\}\cap {\mathbb {S}}^{2}. \end{aligned}$$

The main result of this appendix is

Lemma A.1

For two small parameters $\eta ,\varepsilon$ with $\varepsilon \ll \eta$, suppose that v is a bounded solution to

$$\begin{aligned} {\left\{ \begin{array}{ll} (\Delta _{{\mathbb {S}}^{2}}+\lambda _\frac{7}{2})v=0 &{}\text { in } \widehat{{\mathcal {C}}_\eta \backslash {\mathcal {C}}_\varepsilon },\\ v=0 &{}\text { on } \partial {\mathcal {C}}_\eta \cup \widetilde{{\mathcal {C}}_\eta \backslash {\mathcal {C}}_\varepsilon }. \end{array}\right. } \end{aligned}$$

If we have, for $n=0,1,\dots ,m-1$,

$$\begin{aligned} \int _{\partial {\mathcal {C}}_\varepsilon }v\cdot \cos ((n+\frac{1}{2})\theta )=0, \end{aligned}$$

then

$$\begin{aligned} \sup _{{\mathcal {C}}_\eta \backslash {\mathcal {C}}_{\eta /2}}|v|\le C^m\varepsilon ^{m+\frac{1}{2}}\cdot \sup _{\partial {\mathcal {C}}_\varepsilon }|v| \end{aligned}$$

for a constant C depending only on $\eta .$

Recall the notations for slit domains and homogeneous harmonic functions in slit domains from (2.10) and (2.13).

Proof

With the functions from (2.17), we define, for $n=0,1,\dots$,

$$\begin{aligned} f_n(x_1,r,\theta ):=u_{-(n+\frac{1}{2})}\cdot \sum _{0\le k\le k^*}a_kx_1^{n+4-2k}r^{2k} \end{aligned}$$

where $k^*$ satisfies $(n-2k^*+4)(n-2k^*+3)=0$, $a_0=1$, and

$$\begin{aligned} a_k(n-2k+4)(n-2k+3)=a_{k+1}(2n-2k-1)(2k+2). \end{aligned}$$

(A.1)

It is elementary to verify that $f_n$ is $\frac{7}{2}$-homogeneous and harmonic in $\widehat{{\mathbb {R}}^3}$.

By the iterative relation (A.1), we can find a universal large constant M such that

$$\begin{aligned} |a_k|\le M^n \text { for }k=0,1,\dots , k^*. \end{aligned}$$

As a result, by taking M larger if necessary, we have that

$$\begin{aligned} |f_n|\le r^{-n-\frac{1}{2}}[1+r^2M^n] \text { in }{\mathcal {C}}_\eta . \end{aligned}$$

On the other hand, we have $f_n(x_1,r,0)\ge r^{-n-\frac{1}{2}}[1-r^2M^n]$ in ${\mathcal {C}}_\eta$, which gives that

$$\begin{aligned} f_n(x_1,\varepsilon ,0)\ge \frac{1}{2}\varepsilon ^{-n-\frac{1}{2}} \end{aligned}$$

if $\varepsilon$ is small and $n\le 5.$ For $n\ge 6,$ the same comparison follows directly from the fact that $a_k\ge 0$ for all k if $n\ge 6.$

Consequently, the ratio $f_n(r,\theta )/f_n(\varepsilon ,0)$ satisfies that

$$\begin{aligned} f_n(\varepsilon ,\theta )/f_n(\varepsilon ,0)=\cos ((n+\frac{1}{2})\theta ) \end{aligned}$$

and

$$\begin{aligned} |f_n(r,\theta )/f_n(\varepsilon ,0)|\le \varepsilon ^{n+\frac{1}{2}}r^{-n-\frac{1}{2}}M^n \text { in }{\mathcal {C}}_\eta \end{aligned}$$

by choosing M larger if necessary.

For each n, let $\varphi _n$ denote the solution to

$$\begin{aligned} {\left\{ \begin{array}{ll} (\Delta _{{\mathbb {S}}^{2}}+\lambda _\frac{7}{2})\varphi _n=0 &{}\text { in }\widehat{{\mathcal {C}}_\eta \backslash {\mathcal {C}}_\varepsilon ,}\\ \varphi _n=\cos ((n+\frac{1}{2})\theta ) &{}\text { along }\partial {\mathcal {C}}_\varepsilon ,\\ \varphi _n=0 &{}\text { along } \partial {\mathcal {C}}_\eta \cup \widetilde{{\mathcal {C}}_\eta \backslash {\mathcal {C}}_\varepsilon }. \end{array}\right. } \end{aligned}$$

With the maximum principle, we have that

$$\begin{aligned} |\varphi _n-\frac{f_n(r,\theta )}{f_n(\varepsilon ,0)}|\le \varepsilon ^{n+\frac{1}{2}}\eta ^{-n-\frac{1}{2}}M^n \text { in }{\mathcal {C}}_\eta \backslash {\mathcal {C}}_\varepsilon , \end{aligned}$$

which implies that

$$\begin{aligned} |\varphi _n|\le C\varepsilon ^{n+\frac{1}{2}}r^{-n-\frac{1}{2}}M^n \text { in }{\mathcal {C}}_\eta \backslash {\mathcal {C}}_\varepsilon . \end{aligned}$$

Now with $\{\cos ((n+\frac{1}{2})\theta )\}$ being a basis for $L^2(\partial {\mathcal {C}}_\varepsilon )$, for v as in the statement of the lemma, we can write $v=\sum c_n\varphi _n$, where

$$\begin{aligned} c_n=\frac{\int _{\partial {\mathcal {C}}_\varepsilon }v\cdot \cos ((n+\frac{1}{2})\theta )}{\int _{\partial {\mathcal {C}}_\varepsilon }\cos ^2((n+\frac{1}{2})\theta )}. \end{aligned}$$

For $r\ge \eta /2$, this implies that

$$\begin{aligned} |v(r,\theta )|\le \left( \sum c_n^2\right) ^{\frac{1}{2}}\left( \sum _{c_n\ne 0}\varphi _n(r,\theta )^2\right) ^{\frac{1}{2}}\le C\sup _{\partial {\mathcal {C}}_\varepsilon }|v|\cdot \left( \sum _{c_n\ne 0}\varepsilon ^{2n+1}M^{2n}\right) ^{1/2} \end{aligned}$$

for a constant C depending on $\eta$.

With our assumption on v, we have $c_n=0$ for $n\le m-1$. The conclusion follows by observing that

$$\begin{aligned} \sum _{n\ge m}\varepsilon ^{2n+1}M^{2n}\le C\varepsilon ^{2m+1}M^{2m}. \end{aligned}$$

$\square$

Appendix B. The thin obstacle problem in ${\mathbb {R}}^2$

Our treatment of solutions near $u_{\frac{7}{2}}=r^{\frac{7}{2}}\cos (\frac{7}{2}\theta )$ relies on a fine analysis of the thin obstacle problem in tiny spherical caps around ${\mathbb {S}}^{2}\cap \{r=0\}$. In the limit, this problem leads to the thin obstacle problem in ${\mathbb {R}}^2$ with prescribed expansion at infinity.

In this section, we use $(r,\theta )$ to denote the polar coordinates of ${\mathbb {R}}^2=\{(x_1,x_2)\}$. The notations for slit domains from (2.9) and (2.10) carry over with straightforward modifications. We will also take advantage of the functions from (1.6) and (2.17). Similarly to the functions in (2.16), in this appendix, we denote the derivatives of $u_{\frac{7}{2}}$ by the following⁵:

$$\begin{aligned} w_{\frac{5}{2}}:=\frac{\partial }{\partial x_1}u_{\frac{7}{2}}, w_{\frac{3}{2}}:=\frac{\partial }{\partial x_1}w_{\frac{5}{2}}, \text { and } w_{\frac{1}{2}}:=\frac{\partial }{\partial x_1}w_{\frac{3}{2}}.^5 \end{aligned}$$

The following two derivatives are singular near $\{r=0\}$:

$$\begin{aligned} w_{-\frac{1}{2}}:=\frac{\partial }{\partial x_1}w_{\frac{1}{2}}, \text { and }w_{-\frac{3}{2}}:=\frac{\partial }{\partial x_1}w_{-\frac{1}{2}}. \end{aligned}$$

(B.1)

Let $p=u_{\frac{7}{2}}+a_1u_{\frac{5}{2}}+a_2u_{\frac{3}{2}}+a_3u_{\frac{1}{2}}=u_{\frac{7}{2}}+\tilde{a}_1w_{\frac{5}{2}}+\tilde{a}_2w_{\frac{3}{2}}+\tilde{a}_3w_{\frac{1}{2}}$, then p solves the thin obstacle problem in ${\mathbb {R}}^2$ if and only if

$$\begin{aligned} a_2\ge 0, a_3=0, \text { and } a_1^2\le \frac{84}{25}a_2; \text { equivalently, } \tilde{a}_2\ge 0, \tilde{a}_3=0, \text { and }\tilde{a}_1^2\le \frac{12}{5}\tilde{a}_2.\nonumber \\ \end{aligned}$$

(B.2)

For $\tau \in {\mathbb {R}}$, the translation operator ${\text {U}}_\tau$ is defined by its action on points, sets, and functions in the following manner:

$$\begin{aligned} {\text {U}}_\tau (x_1,x_2)=(x_1+\tau ,x_2), \quad {\text {U}}_\tau (E)=\{x:{\text {U}}_{-\tau }x\in E\}, \quad {\text {U}}_\tau (f)(x)=f({\text {U}}_{-\tau }x). \end{aligned}$$

In this appendix, for $p=u_{\frac{7}{2}}+a_1u_{\frac{5}{2}}+a_2u_{\frac{3}{2}}+a_3u_{\frac{1}{2}}$, we study solutions to the thin obstacle problem in ${\mathbb {R}}^2$ with data p at infinity:

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}u \text { solves (}1.1\text {) in } {\mathbb {R}}^2,\\ &{}\sup _{{\mathbb {R}}^2}|u-p|<+\infty . \end{array}\right. } \end{aligned}$$

(B.3)

The starting point is the following proposition:

Proposition B.1

For $|a_j|\le 1$, there is a unique solution to (B.3).

For this solution, there is a universal constant $A>0$ such that

$$\begin{aligned} \sup _{{\mathbb {R}}^2}|u-p|\le A; \Delta u=0 \text { in }\widehat{\{r>A\}}; \text { and }u=0 \text { in }\widetilde{\{r>A\}}. \end{aligned}$$

Moreover, we can find $b_1,b_2$ satisfying $|b_j|\le A$ such that

$$\begin{aligned} |u-(p+b_1u_{-\frac{1}{2}}+b_2u_{-\frac{3}{2}})|\le A|x|^{-2}u_{-\frac{1}{2}} \text { for all } x\in {\mathbb {R}}^2. \end{aligned}$$

Recall the harmonic functions with negative homogeneities from (2.17).

Remark B.1

For simplicity, we will denote the coefficients $b_j$ by $b_j^{{\mathbb {R}}^2}[a_1,a_2,a_3]$ or simply $b_j[a_1,a_2,a_3]$ when there is no ambiguity.

Proof

Step 1: Uniqueness.

Suppose that $u_1$ and $u_2$ are two solutions to (1.1) in ${\mathbb {R}}^2$ with $\sup |u_j-p|<+\infty .$ With a similar argument as in Lemma 3.1, we find $R>0$ such that

$$\begin{aligned} \Delta u_j=0 \text { in }\widehat{\{r>R\}}, \text { and } u_j=0 \text { in }\widetilde{\{r>R\}}. \end{aligned}$$

Let $w(x):=(u_1-u_2)(R^2x/|x|^2)$ be the Kelvin transform of $(u_1-u_2)$ with respect to $\partial B_R$. Then w is a harmonic function in the slit domain $\widehat{B_R}$, as defined in (2.11). Applying Theorem 2.1, we have that $|w|\le Cu_{\frac{1}{2}}\text { in } B_R,$ which implies that

$$\begin{aligned} |u_1-u_2|\le Cu_{-\frac{1}{2}} \text { in }{\mathbb {R}}^2. \end{aligned}$$

From here we have $u_1=u_2$ bythe maximum principle.

Step 2: A barrier function.

Rewrite p in the basis $\{u_{\frac{7}{2}},w_{\frac{5}{2}},w_{\frac{3}{2}},w_{\frac{1}{2}}\}$ as $p=u_{\frac{7}{2}}+\tilde{a}_1w_{\frac{5}{2}}+\tilde{a}_2w_{\frac{3}{2}}+\tilde{a}_3w_{\frac{1}{2}}$.

For $\tau >0$ to be chosen, if we let $(\alpha _1,\alpha _2)$ denote the solution to

$$\begin{aligned} \alpha _1+\tau =\tilde{a}_1, \text { and }\alpha _2+\alpha _1\tau +\frac{1}{2}\tau ^2=\tilde{a}_2, \end{aligned}$$

and define

$$\begin{aligned} q=u_{\frac{7}{2}}+\alpha _1w_{\frac{5}{2}}+\alpha _2w_{\frac{3}{2}}, \end{aligned}$$

then Taylor’s Theorem gives that

$$\begin{aligned} {\text {U}}_{-\tau }(q)-p\ge \left( \frac{1}{6}\tau ^3-\frac{1}{2}\tilde{a}_1\tau ^2+\tilde{a}_2\tau \right) u_{\frac{1}{2}}-C\tau ^4u_{-\frac{1}{2}}. \end{aligned}$$

Choosing $\tau$ large universally,

$$\begin{aligned} {\text {U}}_{-\tau }(q)-p\ge 0 \text { on } \{r\ge A\} \end{aligned}$$

for a universal large A.

By choosing $\tau$ larger, if necessary, it is elementary to verify that $(\alpha _1,\alpha _2)$ satisfies condition (B.2), and consequently, $Q:={\text {U}}_{-\tau }q$ solves the thin obstacle problem in ${\mathbb {R}}^2.$

Step 3: Existence, universal boundedness, and localization of contact set.

For large $n\in {\mathbb {N}}$, let $u_n$ be the solution to the thin obstacle problem (1.1) in $B_n$ with $u_n=p$ along $\partial B_n$.

By the maximum principle, we have that

$$\begin{aligned} u_n\ge p \text { in }B_n, \text { and } u_n\le Q \text { in }B_n \end{aligned}$$

(B.4)

if n is large. Consequently, this family $\{u_n\}$ is locally uniformly bounded. Therefore, we can extract a subsequence converging to some $u_\infty$ locally uniformly on ${\mathbb {R}}^2$. This limit $u_\infty$ solves the thin obstacle problem in ${\mathbb {R}}^2$.

With (B.4), we have $u_n=0 \text { in }B_n\cap \{x_1\le -A, x_2=0\}$ and $u_n\ge 1 \text { in }B_n\cap \{x_1\ge A, x_2=0\}$ for a universal $A>0$. Thus we have that

$$\begin{aligned} \Delta u_\infty =0 \text { in }\widehat{\{r>A\}}; \text { and }u_\infty =0 \text { in }\widetilde{\{r>A\}}. \end{aligned}$$

Along $\{r=A\}$, we have $0\le u_\infty -p\le Q-p\le C$. Thus the maximum principle, applied in the domain$\{r>A\}$, gives that

$$\begin{aligned} |u_\infty -p|\le C \end{aligned}$$

for a universal constant C. In particular, $u_\infty$ is the unique solution to (B.3), according to Step 1.

Step 4: Finer expansion.

Let $w(x):=(u-p)(A^2x/|x|^2)$ be the Kelvin transform of $(u-p)$ with respect to $\partial B_A$. Results from the previous step implies that w is a harmonic function in the slit domain $\widehat{B_A}$. An application of Theorem 2.1 gives universally bounded $b_1$ and $b_2$ such that

$$\begin{aligned} |w-(b_1u_{\frac{1}{2}}+b_2u_{\frac{3}{2}})|\le C|x|^2u_{\frac{1}{2}}\text { in }B_A. \end{aligned}$$

Inverting the Kelvin transform, we have that

$$\begin{aligned} |u-(p+b_1u_{-\frac{1}{2}}+b_2u_{-\frac{3}{2}})|\le C|x|^{-2}u_{-\frac{1}{2}} \text { in }{\mathbb {R}}^2. \end{aligned}$$

$\square$

For the solution from the previous proposition, we have precise information on its first two Fourier coefficients along big circles.

Corollary B.1

With the same assumptions and notations from Proposition B.1, we have that

$$\begin{aligned} \int _{\partial B_R}[u-(p+b_1u_{-\frac{1}{2}}+b_2u_{-\frac{3}{2}})]\cdot \cos \left( \frac{1}{2}\theta \right) =0 \end{aligned}$$

and

$$\begin{aligned} \int _{\partial B_R}[u-(p+b_1u_{-\frac{1}{2}}+b_2u_{-\frac{3}{2}})]\cdot \cos \left( \frac{3}{2}\theta \right) =0 \end{aligned}$$

for all $R\ge A.$

Proof

For simplicity, let us denote that

$$\begin{aligned} p_{ext}:=p+b_1u_{-\frac{1}{2}}+b_2u_{-\frac{3}{2}}. \end{aligned}$$

With Proposition B.1, we have $\Delta (u-p_{ext})=0 \text { in } \widehat{\{r>A\}}, \text { and } u-p_{ext}=0 \text { in } \widetilde{\{r>A\}}.$

For $R>A$, define $v:=(r^{\frac{1}{2}}-Rr^{-\frac{1}{2}})\cos (\frac{1}{2}\theta )$. Then

$$\begin{aligned} \Delta v=0 \text { in }\widehat{{\mathbb {R}}^2}, \text { and } v=0 \text { along } \widetilde{\{r>0\}}. \end{aligned}$$

With these properties, we have, for $L>R$, that

$$\begin{aligned}&0=\int _{B_L\backslash B_R}(u-p_{ext})\cdot \Delta v-\Delta (u-p_{ext})\\&\quad \cdot v=\int _{\partial (B_L\backslash B_R)}(u-p_{ext})_\nu \cdot v-(u-p_{ext})\cdot v_\nu . \end{aligned}$$

Along $\partial B_L$, we have $|u-p_{ext}|=O(L^{-\frac{5}{2}})$, $|(u-p_{ext})_\nu |=O(L^{-\frac{7}{2}})$, $|v|=O(L^{\frac{1}{2}})$ and $|v_\nu |=O(L^{-\frac{1}{2}})$, thus

$$\begin{aligned} \int _{\partial B_L}(u-p_{ext})_\nu \cdot v-(u-p_{ext})\cdot v_\nu =O(L^{-2}). \end{aligned}$$

Along $\partial B_R$, we have $v=0$ and $v_\nu =-R^{-\frac{1}{2}}\cos (\frac{1}{2}\theta ).$ Combining all of these, we have that

$$\begin{aligned} \int _{\partial B_R}(u-p_{ext})\cdot \cos (\frac{1}{2}\theta )=O(R^{\frac{1}{2}}L^{-2}). \end{aligned}$$

Sending $L\rightarrow \infty$ gives the first conclusion. The second follows from a similar argument. $\quad \square$

The following lemma is one of the main reasons for the restriction to 3d in the main part of this work:

Lemma B.1

Given functions

$$\begin{aligned} p=u_{\frac{7}{2}}+a_1u_{\frac{5}{2}}+a_2u_{\frac{3}{2}}+a_3u_{\frac{1}{2}}, \text { and } q=u_{\frac{7}{2}}-a_1u_{\frac{5}{2}}+a_2u_{\frac{3}{2}}-a_3u_{\frac{1}{2}}\end{aligned}$$

with $|a_j|\le 1,$ suppose that u and v are solutions to (B.3) with p and q as data at infinity, respectively.

Assume $b_1[a_1,a_2,a_3]=b_1[-a_1,a_2,-a_3]$ and $b_2[a_1,a_2,a_3]=-b_2[-a_1,a_2,-a_3]$, then we can find universally bounded constants $\alpha _1$, $\alpha _2$ and $\tau$ such that

$$\begin{aligned} u={\text {U}}_\tau (u_{\frac{7}{2}}+\alpha _1u_{\frac{5}{2}}+\alpha _2u_{\frac{3}{2}}), \text { and }v={\text {U}}_{-\tau }(u_{\frac{7}{2}}-\alpha _1u_{\frac{5}{2}}+\alpha _2u_{\frac{3}{2}}). \end{aligned}$$

Recall the definition of $b_j$’s from Remark B.1.

Proof

Step 1: Two auxiliary polynomials.

For simplicity, let us define $b_j=b_j[a_1,a_2,a_3]$ for $j=1,2$, and

$$\begin{aligned} p_{ext}:=p+b_1u_{-\frac{1}{2}}+b_2u_{-\frac{3}{2}}, \text { and } q_{ext}:=q+b_1u_{-\frac{1}{2}}-b_2u_{-\frac{3}{2}}. \end{aligned}$$

With Proposition B.1, we have that

$$\begin{aligned} |u-p_{ext}|+|v-q_{ext}|\le A|x|^{-\frac{5}{2}} \text { in }{\mathbb {R}}^2, \end{aligned}$$

which implies that

$$\begin{aligned} |\nabla u-\nabla p_{ext}|+|\nabla v-\nabla q_{ext}|\le C|x|^{-\frac{7}{2}} \text { for }|x|\ge 1.\end{aligned}$$

(B.5)

Since u is an entire solution to the thin obstacle problem of order $O(|x|^\frac{7}{2})$ at infinity, we see that $(\partial _{x_1}u-i\partial _{x_2}u)^2$ is a polynomial of degree 5. Meanwhile, a direct computation gives that

$$\begin{aligned} (\partial _{x_1}p_{ext}-i\partial _{x_2}p_{ext})^2={\mathcal {P}}(x_1+ix_2)+\sum _{k=1}^{5}{\mathcal {R}}_k(x_1+ix_2), \end{aligned}$$

where ${\mathcal {P}}$ is a polynomial of degree 5, and ${\mathcal {R}}_k$ is a $(-k)$-homogeneous rational function for $k=1,2,\dots ,5$.

With (B.5), it follows that

$$\begin{aligned} (\partial _{x_1}u-i\partial _{x_2}u)^2={\mathcal {P}} \text { in }{\mathbb {R}}^2. \end{aligned}$$

If we define

$$\begin{aligned} P(t):={\text {Re}}(\partial _{x_1}u-i\partial _{x_2}u)^2(t,0)=[(\partial _{x_1}u)^2-(\partial _{x_2}u)^2](t,0)={\text {Re}}{\mathcal {P}}(t), \end{aligned}$$

then P is a real polynomial of degree 5.

Similarly, corresponding to v and $q_{ext}$, we have that

$$\begin{aligned} (\partial _{x_1}q_{ext}-i\partial _{x_2}q_{ext})^2={\mathcal {Q}}(x_1+ix_2)+\sum _{k=1}^{5}{\mathcal {S}}_k(x_1+ix_2), \end{aligned}$$

where ${\mathcal {Q}}$ is a polynomial of degree 5, and ${\mathcal {S}}_k$ is a $(-k)$-homogeneous rational function for $k=1,2,\dots ,5$. Moreover, we have that

$$\begin{aligned} Q(t):={\text {Re}}(\partial _{x_1}v-i\partial _{x_2}v)^2(t,0)=[(\partial _{x_1}v)^2-(\partial _{x_2}v)^2](t,0)={\text {Re}}{\mathcal {Q}}(t), \end{aligned}$$

also a real polynomial of degree 5.

With $b_1[a_1,a_2,a_3]=b_1[-a_1,a_2,-a_3]$ and $b_2[a_1,a_2,a_3]=-b_2[-a_1,a_2,-a_3]$, a direct computation gives that

$$\begin{aligned} P(t)=-Q(-t). \end{aligned}$$

(B.6)

Step 2: Half-space solutions.

With (B.6), we show that up to a translation, u must be a half-space solution. Since $u=0$ in $\widetilde{\{r>A\}}$ according to Proposition B.1, it suffices to show that ${\text {spt}}(\Delta u)$ has only one component.

Suppose, on the contrary, that

$$\begin{aligned} (-\infty ,a]\cup [b,+\infty )\supset {\text {spt}}(\Delta u)\supset (-\infty ,a]\cup [b,c] \text { with }b>a, \end{aligned}$$

Note that the second component has to terminate in finite length since $\Delta u=0$ in $\widehat{\{r>A\}}$.

On $(-\infty ,a]\cup [b,c]$, we have $\partial _{x_1}u=0$. Thus $P(t)=-(\partial _{x_2}u)^2\le 0$ for $t\in (-\infty ,a]\cup [b,c]$. On the contrary, on (a, b), $\partial _{x_2}u=0$ and $P(t)=(\partial _{x_1}u)^2\ge 0$. Moreover, since $u(a)=u(b)=0$ and $u>0$ on (a, b), we must have $\partial _{x_1}u(d)=0$ at some point $d\in (a,b)$. Thus $P(d)=0$. Note that d is a root of multiplicity at least 2. Together with the roots a, b, c, this implies that P cannot have other roots; see Fig. 4.

Fig. 4

P and Q along the $x_1$-axis

With the symmetry described in (B.6), if we let $b'=-b$ and $c'=-c$, then $Q(b')=Q(c')=0$ while $Q>0$ on $(b',c')$. This implies that $v>0$ on $(b',c')$, while $v(b')=v(c')=0$. However, this implies that $\partial _{x_1}v$ must vanish at some point on $(b',c'),$ and so does Q. This is a contradiction.

As a result, ${\text {spt}}(\Delta u)$ must be a half line. A similar result holds for ${\text {spt}}(\Delta v).$ With (B.6), we see that if ${\text {spt}}(\Delta u)=(-\infty ,a]$, then ${\text {spt}}(\Delta v)=(-\infty , -a].$

Step 4: Conclusion.

After the previous step, we can apply Theorem 2.1 to get that

$$\begin{aligned} {\text {U}}_{-a}u=a_0'u_{\frac{7}{2}}+\alpha _1u_{\frac{5}{2}}+\alpha _2u_{\frac{3}{2}}+a_3'u_{\frac{1}{2}}. \end{aligned}$$

We must have $a_3'=0$ by (B.2). With $|u-(u_{\frac{7}{2}}+a_1u_{\frac{5}{2}}+a_2u_{\frac{3}{2}})|$ being bounded in ${\mathbb {R}}^2$, we conclude $a_0'=1.$ Therefore,

$$\begin{aligned} {\text {U}}_{-a}u=u_{\frac{7}{2}}+\alpha _1u_{\frac{5}{2}}+\alpha _2u_{\frac{3}{2}}. \end{aligned}$$

Similarly, we have that

$$\begin{aligned} {\text {U}}_{a}v=u_{\frac{7}{2}}+\beta _1u_{\frac{5}{2}}+\beta _2u_{\frac{3}{2}}. \end{aligned}$$

From here, we use (B.6) to conclude $\alpha _1=\beta _1$ and $\alpha _2=-\beta _2$. The conclusion follows. $\quad \square$

A perturbation of the previous lemma leads to the following corollary (recall notations from (B.1) and Remark B.1):

Corollary B.2

Given $p=u_{\frac{7}{2}}+a_1u_{\frac{5}{2}}+a_2u_{\frac{3}{2}}+a_3u_{\frac{1}{2}}=u_{\frac{7}{2}}+\tilde{a}_1w_{\frac{5}{2}}+\tilde{a}_2w_{\frac{3}{2}}+\tilde{a}_3w_{\frac{1}{2}}$ with $|a_j|\le 1$, we set

$$\begin{aligned} b_j^+:=b_j[a_1,a_2,a_3], b_j^-:=b_j[-a_1,a_2,-a_3] \text { for }j=1,2, \end{aligned}$$

and

$$\begin{aligned} p_{ext}=p+b_1^+u_{-\frac{1}{2}}+b_2^+u_{-\frac{3}{2}}=p+\tilde{b}_1^+w_{-\frac{1}{2}}+\tilde{b}_2^+w_{-\frac{3}{2}}. \end{aligned}$$

Then there is a universal modulus of continuity, $\omega$, such that

$$\begin{aligned}&|\tilde{a}_1-(\alpha _1+\tau )|+\left| \tilde{a}_2-\left( \alpha _2+\alpha _1\tau +\frac{1}{2}\tau ^2\right) \right| + \left| \tilde{a}_3-\left( \alpha _2\tau +\frac{1}{2}\alpha _1\tau ^2+\frac{1}{6}\tau ^3\right) \right| \\&\qquad +\left| \tilde{b}_1^+-\left( \frac{1}{2}\alpha _2\tau ^2+\frac{1}{6}\alpha _1\tau ^3+\frac{1}{24}\tau ^4\right) \right| +\left| \tilde{b}_2^+-\left( \frac{1}{6}\alpha _2\tau ^3+\frac{1}{24}\alpha _1\tau ^4+\frac{1}{120}\tau ^5\right) \right| \\&\quad \le \omega (|b_1^+-b_1^-|+|b_2^++b_2^-|) \end{aligned}$$

for universally bounded $\alpha _j$ and $\tau$ satisfying

$$\begin{aligned} \alpha _2\ge 0 \text { and }\alpha _1^2\le \frac{12}{5}\alpha _2. \end{aligned}$$

(B.7)

Proof

Suppose there is no such $\omega$, we find a sequence $(a_j^n)$ such that the corresponding $(b_j^{\pm ,n})$ satisfy

$$\begin{aligned} |b_1^{+,n}-b_1^{-,n}|+|b_2^{+,n}+b_2^{-,n}|\rightarrow 0, \end{aligned}$$

(B.8)

but for any bounded $\alpha _j$ and $\tau$ satisfying (B.7), we have that

$$\begin{aligned}&|\tilde{a}^n_1-(\alpha _1+\tau )|+\left| \tilde{a}^n_2-\left( \alpha _2+\alpha _1\tau +\frac{1}{2}\tau ^2\right) \right| +\left| \tilde{a}^n_3-\left( \alpha _2\tau +\frac{1}{2}\alpha _1\tau ^2+\frac{1}{6}\tau ^3\right) \right| \nonumber \\&\quad +\left| \tilde{b}_1^{+,n}-\left( \frac{1}{2}\alpha _2\tau ^2+\frac{1}{6}\alpha _1\tau ^3+\frac{1}{24}\tau ^4\right) \right| +\left| \tilde{b}_2^{+,n}-\left( \frac{1}{6}\alpha _2\tau ^3+\frac{1}{24}\alpha _1\tau ^4+\frac{1}{120}\tau ^5\right) \right| \nonumber \\&\quad \ge \varepsilon >0 \end{aligned}$$

(B.9)

Up to a subsequence, we have that

$$\begin{aligned} a_j^n\rightarrow a_j^\infty \text { and }b_j^{\pm ,n}\rightarrow b_j^{\pm ,\infty }. \end{aligned}$$

If we take that

$$\begin{aligned} p_n^+=u_{\frac{7}{2}}+a^n_1u_{\frac{5}{2}}+a^n_2u_{\frac{3}{2}}+a^n_3u_{\frac{1}{2}}=u_{\frac{7}{2}}+\tilde{a}^n_1w_{\frac{5}{2}}+\tilde{a}^n_2w_{\frac{3}{2}}+\tilde{a}^n_3w_{\frac{1}{2}}\end{aligned}$$

and denote by $u^+_n$ the solution to (B.3) with data $p^+_n$ at infinity, then, by Proposition B.1, we have that

$$\begin{aligned} |u^+_n-p^+_n|\le A \text { in }{\mathbb {R}}^2. \end{aligned}$$

Up to a subsequence, we have $u^+_n$ locally uniformly converge to $u^+_\infty$, a solution to the thin obstacle problem in ${\mathbb {R}}^2$. Moreover, we have

$$\begin{aligned} |u^+_\infty -[u_{\frac{7}{2}}+a^\infty _1u_{\frac{5}{2}}+a^\infty _2u_{\frac{3}{2}}+a^\infty _3u_{\frac{1}{2}}]|\le A \text { in }{\mathbb {R}}^2. \end{aligned}$$

Thus $u^+_\infty$ is the solution to (B.3) with data $p^+_\infty =u_{\frac{7}{2}}+a^\infty _1u_{\frac{5}{2}}+a^\infty _2u_{\frac{3}{2}}+a^\infty _3u_{\frac{1}{2}}$ at infinity.

With Corollary B.1, we see that $b_j^{+,\infty }:=b_j[a_j^\infty ]=\lim b_j^{+,n}.$ A similar argument applied to $p_n^-=u_{\frac{7}{2}}-a^n_1u_{\frac{5}{2}}+a^n_2u_{\frac{3}{2}}-a^n_3u_{\frac{1}{2}}$ leads to $b_j^{-,\infty }:=b_j[-a^\infty ,a_2^\infty ,-a_3^\infty ]=\lim b_j^{-,n}.$ With (B.8), we conclude that

$$\begin{aligned} b_1[a_1^\infty ,a_2^\infty , a_3^\infty ]= & {} b_1[-a_1^\infty , a_2^\infty , -a_3^\infty ] \text { and } b_2[a_1^\infty , a_2^\infty , a_3^\infty ]\\= & {} -b_2[-a_1^\infty , a_2^\infty , -a_3^\infty ]. \end{aligned}$$

Lemma B.1 gives that

$$\begin{aligned} u_\infty ^+={\text {U}}_\tau (u_{\frac{7}{2}}+\alpha _1w_{\frac{5}{2}}+\alpha _2w_{\frac{3}{2}}) \end{aligned}$$

for $\alpha _j$ satisfying (B.7).

Consequently, we have that

$$\begin{aligned}&\left| \tilde{a}_1^\infty -(\alpha _1+\tau )|+|\tilde{a}^\infty _2-\left( \alpha _2+\alpha _1\tau +\frac{1}{2}\tau ^2\right) \right| +\left| \tilde{a}^\infty _3-\left( \alpha _2\tau +\frac{1}{2}\alpha _1\tau ^2+\frac{1}{6}\tau ^3\right) \right| \\&\quad +\left| \tilde{b}_1^{\infty ,+}-\left( \frac{1}{2}\alpha _2\tau ^2+\frac{1}{6}\alpha _1\tau ^3+\frac{1}{24}\tau ^4\right) \right| \\&\quad +\left| \tilde{b}_2^{\infty ,+}-\left( \frac{1}{6}\alpha _2\tau ^3+\frac{1}{24}\alpha _1\tau ^4+\frac{1}{120}\tau ^5\right) \right| =0. \end{aligned}$$

With convergence of $\tilde{a}_j^n\rightarrow \tilde{a}_j^\infty$ and $\tilde{b}_j^{+,n}\rightarrow \tilde{b}_j^{+,\infty }$, this contradicts (B.9). $\quad \square$