1 Introduction

The regularity of solutions to second order elliptic equation has been extensively studied by many authors, see the book [5]. But for degenerate elliptic equation, there is few work about the regularity results until now. In this paper, we focus on a special degenerate elliptic equation-like Grushin elliptic equation. The main reason why we can deal with it is that we can get the expression of heat kernel by using the probability method.

For Grushin operator, many authors studied it. Beckner [2] obtained the Sobolev estimates for the Grushin operator in low dimensions by using hyperbolic symmetry and conformal geometry. Riesz transforms and multipliers for the Grushin operator was considered by Jotsaroop et al. [11]. Tri [16] studied the generalized Grushin equation. \(L^p\)-estimates for the wave equation associated to the Grushin operator was studied by Jotsaroop and Thangavelu [12]. The fundamental solution for a degenerate parabolic pseudo-differential operator covering the Grushin operator was obtained by Tsutsumi [17, 18]. Furthermore, Tsutsumi [19] constructed a left parametrix for a pesudo-differential operator. We remark that Tsutsumi did not give the exact expression of heat kernel. In the book [1], they gave the expression of heat kernel (see Page 191), but he expression is hard to use because they used the inverse Fourier transform. Yu [24] established Liouville type theorem for nonlinear elliptic equation involving Grushin operators. Chang and Li [3] studied the heat kernel asymptotic expansions for the Heisenberg sub-Laplacian and the Grushin operator. Similar degenerate elliptic equation was studied by Robinson and Sikora [15] and the Hardy inequalities for Grushin operator was considered by [23]. The blowup problem of Grushin operator was considered by Lv et al. [10]. Recently, Wang et al. [21, 22] studied the regularity for sub-elliptic systems.

About the regularity of degenerate elliptic operator, Di Fazio et al. [4] considered the regularity of a class of strongly degenerate quasilinear operators, also see [14]. Guo et al. [6, 8] studied a degenerate elliptic equations and obtained the well-posedness of the degenerate elliptic equations, also see [7]. In this paper, in view of probability point, we give a new expression and then get the regularity of solution by using the properties of heat kernel.

This paper is arranged as follows. In next section, some preliminaries are given and the main results will be proved in Sect. 3. Throughout this paper, we write C as a general positive constant and \(C_i\), \(i=1,2,\cdots \) as a concrete positive constant.

2 Main Results

Consider the Grushin operator

$$\begin{aligned} {\mathcal {L}}=\frac{1}{2}(\partial _{x}^2+x^2\partial _{y}^2), \end{aligned}$$

which is the generator of the diffusion process \((X_t,Y_t)\), where \((X_t,Y_t)\) satisfies

$$\begin{aligned}\left\{ \begin{array}{llll} dX_t=dW^1_t,\\ dY_t=X_tdW^2_t,\\ X_0=\mu _1,\ \ \ Y_0=\mu _2. \end{array}\right. \end{aligned}$$

Here \(W^i_t\), \(i=1,2\) are standard Brownian motion. It is easy to see that the process \((X_t,Y_t)\) is a two-dimensional Gaussian stochastic process. In this short paper, we consider the following operator

$$\begin{aligned} {\mathcal {L}}=\frac{1}{2}(\partial _{x}^2+x^2\partial _{y}^2) +\frac{1}{3}y\partial _x^2\partial _y, \end{aligned}$$

It is easy to get the heat kernel of the operator \({\mathcal {L}}\) is

$$\begin{aligned} K(t,x,y)=\frac{3}{\sqrt{2}\pi t^{3/2}}\exp \left\{ -\frac{3x^2}{2t}-\frac{3y^2}{t^2}\right\} , \end{aligned}$$

which yields that

$$\begin{aligned} \nabla _xK(t,x,y)= & {} -\frac{3x}{t}K(t,x,y),\\ \nabla _yK(t,x,y)= & {} -\frac{6y}{t^2}K(t,x,y). \end{aligned}$$

In [20], the authors considered a degenerate elliptic equation on a manifold. Thus the heat kernel K is different from that in [20]. In paper [9], Li studied the Grushin operator \(\Delta _x+|x|^2\Delta _u\) on \({\mathbb {R}}_x^n\times {\mathbb {R}}_u^{n'}\) with \(n\ge 3\) and \(n'=1\), which is different from our cases. We only consider the operator on \({\mathbb {R}}\times {\mathbb {R}}\). Meanwhile, we remark that Tsutsumi [19] considered the existence of left parameters of the operator

$$\begin{aligned} A=\sum _{|\alpha |=m,(\sigma ,\gamma )\ge (\tau ,\alpha )-m}a_{\alpha ,\gamma } x_1^{\gamma _1}y^{\gamma _3}D_{x_1}^{\alpha _1}D_{x_2}^{\alpha _2} D_y^{\alpha _3}, \end{aligned}$$

where \(a_{\alpha ,\gamma }\) are constants, \(m\in {\mathbb {Z}}\), \(D_j=\frac{\partial }{\partial x_j}\), \(\alpha =(\alpha _1,\alpha _2,\alpha _3)\), \(\tau =(\tau _1,\tau _2,1)\), \(\tau _j\ge 1\) \((j=1,2)\), \(\gamma =(\gamma _1,0,\gamma _3)\), \(\sigma =(\sigma _1,0,1)\), \(\min (\tau _1,\tau _2)>\sigma _1>0\) and \((\tau ,\alpha )=\sum _{j=1}^3\tau _j\alpha _j\). It is remarked that the operator A defined on \({\mathbb {R}}^3\). Moreover, there is no concrete representation of heat kernel.

We first want to solve the following elliptic equation

$$\begin{aligned} {{\mathcal {L}}^\lambda }_b:=(\lambda -{\mathcal {L}})u+b\cdot \nabla u=f. \end{aligned}$$
(1)

We say u is a solution of (1) if u satisfies

$$\begin{aligned} u=(\lambda -{\mathcal {L}})^{-1}(f-b\cdot \nabla u). \end{aligned}$$

Theorem 2.1

Assume that \(b\in C^\beta _b({\mathbb {R}}^2)\). There exists a \(\lambda _0>0\) such that for any \(f\in C^\beta _b({\mathbb {R}}^2)\) and \(\lambda >\lambda _0\), there is a unique solution \(u\in C_b^{1+\beta }({\mathbb {R}}^2)\) to Eq. (1) such that

$$\begin{aligned} \lambda ^{\delta }\Vert u\Vert _{C_b^{1+\beta }({\mathbb {R}}^2)}\le \Vert f\Vert _{ C_b^{\beta }({\mathbb {R}}^2)}, \end{aligned}$$
(2)

where \(0<\delta <(\beta \wedge (1-\beta ))\).

Assume further that \(\nabla _yh(x,\cdot )\in C_b^\beta ({\mathbb {R}})\) for any \(x\in {\mathbb {R}}\), where \(h=b\) or f. Then there is a unique solution \(u\in C_b^{2+\beta }({\mathbb {R}}^2)\) to Eq. (1) such that

$$\begin{aligned} \lambda ^{\delta }\Vert u\Vert _{C_b^{2+\beta }({\mathbb {R}}^2)}\le \Vert f\Vert _{ C_b^{\beta }({\mathbb {R}};C^{1+\beta }({\mathbb {R}}))}. \end{aligned}$$

Let

$$\begin{aligned} v(x,y):=(\lambda -{\mathcal {L}})^{-1}f(x,y)=\int _0^\infty e^{-\lambda t}K(t,\cdot )*f(x,y)dt. \end{aligned}$$
(3)

Next, we consider the \(L^p\)-regularity of v(x, y).

Theorem 2.2

Assume that \(f(\cdot ,y)\in L^{p}({\mathbb {R}})\) for every \(y\in {\mathbb {R}}\) with and \(f(x,\cdot )\in W^{s,q}({\mathbb {R}},{\mathbb {R}})\) for every \(y\in {\mathbb {R}}\), \(0<s<1\) and \(q>1\). Then we have the following estimates:

$$\begin{aligned} \Vert v\Vert _{L^r({\mathbb {R}}^2)}\le & {} C\lambda ^{-1-\frac{3}{2r}+\frac{1}{2p}+\frac{1}{q}}\Vert f\Vert _{L^p({\mathbb {R}},L^q({\mathbb {R}}))},\\ \Vert \nabla _xv\Vert _{L^r({\mathbb {R}}^2)}\le & {} C\lambda ^{-\frac{1}{2}-\frac{3}{2r}+\frac{1}{2p}+\frac{1}{q}}\Vert f\Vert _{L^p({\mathbb {R}},L^q({\mathbb {R}}))},\\ \Vert \nabla _yv\Vert _{L^r({\mathbb {R}}^2)}\le & {} C\lambda ^{-s-\frac{3}{2r}+\frac{1}{2p}}\Vert f\Vert _{L^p({\mathbb {R}},W^{s,q}({\mathbb {R}}))}.\\ \end{aligned}$$

Moreover, if we take \(0<s<1\) and \(p,q,r>1\) such that \(-\frac{3}{2r}+\frac{1}{2p}+\frac{1}{q}\le 0\) and \(-s-\frac{3}{2r}+\frac{1}{2p}+1\le 0\), we have

$$\begin{aligned} \Vert \nabla ^2_xv\Vert _{L^r({\mathbb {R}}^2)}\le C,\ \ \ \Vert \nabla ^2_yv\Vert _{L^r({\mathbb {R}}^2)}\le C. \end{aligned}$$

3 Proof of Main Results

Denote \([\cdot ]_\beta \) by the semi-norm of \(C^\beta \). Set

$$\begin{aligned} {\mathcal {A}}:=\left\{ f(x,y):\begin{array}{lllll} f(\cdot ,y)\in L^\infty ({\mathbb {R}})\ for\ any \ y\in {\mathbb {R}}, f(x,\cdot )\in C_b^\beta ({\mathbb {R}})\\ for \ any \ x\in {\mathbb {R}}\ with \ 0<\beta <1\end{array}\right\} \end{aligned}$$

Lemma 3.1

Assume that \(f\in {\mathcal {A}}\) and u is defined as in (3). Then

$$\begin{aligned} \Vert v\Vert _{L^\infty ({\mathbb {R}}^2)}\le & {} C\lambda ^{-\frac{1}{2}}\Vert f\Vert _{L^\infty ({\mathbb {R}}^2)},\ \qquad \ \ \Vert \nabla _xv\Vert _{L^\infty ({\mathbb {R}}^2)}\le C\lambda ^{-\frac{1}{2}}\Vert f\Vert _{L^\infty ({\mathbb {R}}^2)},\\ \Vert \nabla _yv\Vert _{L^\infty ({\mathbb {R}}^2)}\le & {} C\lambda ^{-\beta }\Vert f\Vert _{L^\infty ({\mathbb {R}},C_b^\beta ({\mathbb {R}}))},\ \ \ \ [\nabla _xv]_\beta \le C\lambda ^{-\frac{1}{2}+\beta }\Vert f\Vert _{L^\infty ({\mathbb {R}},C_b^\beta ({\mathbb {R}}))}. \end{aligned}$$

Moreover, if \(f(\cdot ,y)\in C_b^\beta ({\mathbb {R}})\) for any \(y\in {\mathbb {R}}\), it holds that

$$\begin{aligned}{}[\nabla _yv]_\beta \le C\lambda ^{-\delta }\Vert f\Vert _{C^\beta ({\mathbb {R}}^2)}, \end{aligned}$$

where \(0<\delta <(\beta \wedge (1-\beta ))\). That is to say, \(\nabla v\in C_b^\beta ({\mathbb {R}}^2)\).

Proof

Simply calculations show that

$$\begin{aligned} \Vert v\Vert _{L^\infty ({\mathbb {R}}^2)}= & {} \Bigg \Vert \int _0^\infty e^{-\lambda t}K(t,\cdot )*f(x,y)dt\Bigg \Vert _{L^\infty ({\mathbb {R}}^2)}\\\le & {} \int _0^\infty e^{-\lambda t}\Vert K(t,\cdot )\Vert _{L^1({\mathbb {R}})}\Vert f\Vert _{L^\infty ({\mathbb {R}}^2)}dt\\\le & {} C\lambda ^{-1}\Vert f\Vert _{L^\infty ({\mathbb {R}}^2)},\\ \Vert \nabla _xv\Vert _{L^\infty ({\mathbb {R}}^2)}= & {} \Bigg \Vert \int _0^\infty e^{-\lambda t}\nabla _xK(t,\cdot )*f(x,y)dt\Bigg \Vert _{L^\infty ({\mathbb {R}}^2)}\\\le & {} \int _0^\infty e^{-\lambda t}\Vert \nabla _xK(t,\cdot )\Vert _{L^1({\mathbb {R}})}\Vert f\Vert _{L^\infty ({\mathbb {R}}^2)}dt\\\le & {} C\lambda ^{-\frac{1}{2}}\Vert f\Vert _{L^\infty ({\mathbb {R}}^2)}, \end{aligned}$$

and

$$\begin{aligned}&\Vert \nabla _yv\Vert _{L^\infty ({\mathbb {R}}^2)}\\&\quad =\Bigg \Vert \int _0^\infty e^{-\lambda t}\nabla _yK(t,\cdot )*f(x,y)dt\Bigg \Vert _{L^\infty ({\mathbb {R}}^2)}\\&\quad =\Bigg \Vert \int _0^\infty e^{-\lambda t}\int _{{\mathbb {R}}^2}\nabla _yK(t,x-u,y-v)(f(u,v)-f(u,y))dudvdt\Bigg \Vert _{L^\infty ({\mathbb {R}}^2)}\\&\quad \le \Vert f\Vert _{L^\infty ({\mathbb {R}},C_b^\beta ({\mathbb {R}}))}\Vert \int _0^\infty e^{-\lambda t}\int _{{\mathbb {R}}^2}|\nabla _yK(t,x-u,y-v)|\cdot |y-v|^\beta dudv dt\Vert _{L^\infty ({\mathbb {R}}^2)}\\&\quad \le C\Vert f\Vert _{L^\infty ({\mathbb {R}},C_b^\beta ({\mathbb {R}}))}\int _0^\infty e^{-\lambda t}t^{-1+\beta }dt\\&\quad \le C\lambda ^{-\beta }\Vert f\Vert _{L^\infty ({\mathbb {R}},C_b^\beta ({\mathbb {R}}))}, \end{aligned}$$

where we used the fact that

$$\begin{aligned} \int _{{\mathbb {R}}^2}\nabla _yK(t,x-u,y-w)f(u,y)dudw =\int _{{\mathbb {R}}}\left( \int _{{\mathbb {R}}}\nabla _yK(t,x-u,y-w)dw\right) f(u,y)du=0. \end{aligned}$$

We also remark that the meaning of \(\Vert f\Vert _{L^\infty ({\mathbb {R}},C_b^\beta ({\mathbb {R}}))}\) is that we take infinity norm for the first variable and take Hölder norm for the second variable.

Recall the following interpolation inequality

$$\begin{aligned}{}[u]_{\sigma \alpha +(1-\sigma )\gamma }\le ([u]_{\alpha })^\sigma ([u]_\gamma )^{1-\sigma }, \ \ 0\le \alpha <\gamma \le 1,\ \ \sigma \in (0,1). \end{aligned}$$

Now, if \(0<\beta <1\), applying the above inequality with \(\alpha =0,\gamma =1\) and \(\beta =\gamma (1-\sigma )\), we have

$$\begin{aligned}{}[\nabla _xv]_\beta= & {} \left[ \int _0^\infty e^{-\lambda t}\nabla _xK(t,\cdot )*f(x,y)dt\right] _\beta \\\le & {} \int _0^\infty e^{-\lambda t}\int _{{\mathbb {R}}^2}\left[ \nabla _xK(t,\cdot ,\cdot )\right] _\beta (x,y)dxdy\Vert f\Vert _{L^\infty ({\mathbb {R}}^2)}dt\\\le & {} \Vert f\Vert _{L^\infty ({\mathbb {R}}^2)}\int _0^\infty t^{-\frac{1}{2}-\beta }e^{-\lambda t}dt\\\le & {} C\lambda ^{-\frac{1}{2}+\beta }\Vert f\Vert _{L^\infty ({\mathbb {R}}^2)}, \end{aligned}$$

where we used the definition of semi-norm. Next, we consider the derivative of the second variable.

$$\begin{aligned}{}[\nabla _yv]_\beta= & {} \left[ \int _0^\infty e^{-\lambda t}\nabla _yK(t,\cdot )*f(x,y)dt\right] _\beta \\\le & {} \int _0^\infty e^{-\lambda t}\left( \sup _{x,{\hat{x}},y,\hat{y}\in {\mathbb {R}},x\ne \hat{x},y\ne \hat{y}}\frac{1}{|x-\hat{x}|^\beta +|y-\hat{y}|^\beta }\right. \\&\times \Big |\int _{{\mathbb {R}}^2}\nabla _yK(t,x-u,w)(f( u,y-w)-f(u,\hat{y}-w))dudw\\&\left. + \int _{{\mathbb {R}}^2}\nabla _yK(t,u,\hat{y}-w)(f( x-u,w)-f(\hat{x}-u,w))dudw\Big |\right) dt\\= & {} \int _0^\infty e^{-\lambda t}\left( \sup _{x,\hat{x},y,{\hat{y}}\in {\mathbb {R}},x\ne \hat{x},y\ne \hat{y}}\frac{1}{|x-{\hat{x}}|^\beta +|y-\hat{y}|^\beta }|I_1+I_2|\right) dt. \end{aligned}$$

By dividing the real line into two parts, we have

$$\begin{aligned} I_1= & {} \int _{{\mathbb {R}}^2}\nabla _yK(t,x-u,y-w)(f( u,w)-f(u,y)dudw\\&+\int _{{\mathbb {R}}^2}\nabla _yK(t,x-u,\hat{y}-w)(f(u,\hat{y})-f( u,w))dudw\\= & {} \int _{{\mathbb {R}}^2}\nabla _yK(t,u,y-w)(f(x-u,w)-f(x-u,y ))dudw\\&+\int _{{\mathbb {R}}^2}\nabla _yK(t,u,\hat{y}-w)(f(x-u,\hat{y})-f(x-u,w))dudw\\= & {} \int _{{\mathbb {R}}}\int _{|y-w|\le 2|y-\hat{y}|}\nabla _yK(t,u, y-w)(f(x-u,w)-f(x-u, y)dudw\\&+\int _{{\mathbb {R}}}\int _{|y-w|\le 2|y-\hat{y}|}\nabla _yK(t,u,\hat{y}-w)(f(x-u,\hat{y})-f(x-u,w))dudw\\&+\int _{{\mathbb {R}}}\int _{|y-w|>2|y-\hat{y}|}(\nabla _yK(t,u,y-w)-\nabla _yK(t,u,\hat{y}-w))\\&(f(x-u,w)-f(x-u,y))dudw\\&+\int _{{\mathbb {R}}}\int _{|y-w|>2|y-\hat{y}|}\nabla _yK(t,u,\hat{y}-w) (f(x-u,\hat{y})-f(x-u,y))dudw\\=: & {} I_{11}+\cdots +I_{14}. \end{aligned}$$

Let us estimate \(I_{11}\)–\(I_{14}\). By using the form of heat kernel, we get

$$\begin{aligned} |I_{11}|= & {} \Big |\int _{{\mathbb {R}}}\int _{|y-w|\le 2|y-\hat{y}|}\nabla _yK(t,u, y-w)\\&(f(x-u,w)-f(x-u,y)dudv\Big |\\\le & {} Ct^{-\frac{3}{2}}\int _{{\mathbb {R}}} e^{-\frac{u^2}{t}}\left( \int _{|y-w|\le 2|y-\hat{y}|}\frac{|y-w|}{t^2} e^{-\frac{(y-w)^2}{t^2}}|y-w|^\beta dw\right) du\\= & {} Ct^{-\frac{3}{2}}\int _{{\mathbb {R}}} e^{-\frac{u^2}{t}}\left( \int _{|z|\le 2|y-\hat{y}|}\frac{|z|}{t^2} e^{-\frac{z^2}{t^2}}|z|^\beta dz\right) du\\\le & {} Ct^{-1+\beta }|y-\hat{y}|^\beta . \end{aligned}$$

Similarly, we can obtain

$$\begin{aligned} I_{12}\le Ct^{-1+\beta }|y-\hat{y}|^\beta . \end{aligned}$$

Next, we consider \(I_{13}\).

$$\begin{aligned} |I_{13}|= & {} \Big |\int _{{\mathbb {R}}}\int _{|y-w|>2|y-\hat{y}|}(\nabla _yK(t,u,y-w)-\nabla _yK(t,u,\hat{y}-w))\\&(f(x- u,w)-f(x-u,y))dudw\Big |\\\le & {} C\int _{{\mathbb {R}}}\int _{|y-w|>2|y-\hat{y}|}|\nabla _yK(t,u,y-w)-\nabla _yK(t,u,\hat{y}-w)|y-w|^\beta dudw\\\le & {} Ct^{-\frac{3}{2}}\int _{{\mathbb {R}}}e^{-\frac{u^2}{t}} \int _{|y-w|>2|y-{\hat{y}}|}\Big |\frac{y-w}{t^2}e^{-\frac{(y-w)^2}{t^2}}- \frac{{\hat{y}}-w}{t^2}e^{-\frac{(\hat{y}-w)^2}{t^2}}\Big ||y-w|^\beta dudw. \end{aligned}$$

Notice that \(|y-w|>2|y-\hat{y}|\). So for every \(\xi \in [y,\hat{y}]\),

$$\begin{aligned} \frac{1}{2}|y-w| \le |\xi -w|\le 2|y-w|. \end{aligned}$$

We recall the following fractional mean value formula (see (4.4) of [13])

$$\begin{aligned} f(x+h)=f(x)+\Gamma ^{-1}(1+\beta )h^\gamma f^{(\gamma )}(x+\theta h), \end{aligned}$$

where \(0<\gamma <1\) and \(\theta >0\) depends on h satisfying

$$\begin{aligned} \lim \limits _{h\downarrow 0}\theta ^\gamma =\frac{\Gamma ^2(1+\gamma )}{\Gamma (1+2\gamma )}. \end{aligned}$$

Denote

$$\begin{aligned} {\tilde{K}}(t,w)=\frac{v}{t^2}e^{-\frac{w^2}{t^2}}. \end{aligned}$$

By using the above fractional mean value formula with \(\gamma >\beta \) and the interpolation inequality in Hölder space

$$\begin{aligned} |I_{13}|\le & {} Ct^{-\frac{3}{2}}|y-{\hat{y}}|^\beta \int _{{\mathbb {R}}}e^{-\frac{u^2}{t}} \int _{|y-w|>2|y-\hat{y}|}[{\tilde{K}}(t,y-w)]_\gamma |y-\hat{y}|^{\gamma -\beta }|y-w|^\beta dudw\\\le & {} Ct^{-\frac{3}{2}}|y-{\hat{y}}|^\beta \int _{{\mathbb {R}}}e^{-\frac{u^2}{t}}du \int _{|y-w|>2|y-\hat{y}|}[{\tilde{K}}(t,y-w)]_\gamma |y-w|^{\gamma } dw\\\le & {} Ct^{-1+\gamma -\beta }|y-\hat{y}|^\beta \int _{{\mathbb {R}}}e^{-u^2}du\int _0^\infty |w|^\gamma e^{-\frac{1}{2}w^2}dw\\\le & {} Ct^{-1+\gamma -\beta }|y-\hat{y}|^\beta . \end{aligned}$$

Lastly, by using the properties of heat kernel K, it is easy to see that

$$\begin{aligned}&\int _{|u^2-(y-w)|>2|y-\hat{y}|}\nabla _yK(t,u,y-w)dw\\&\quad = \int _{|w|>2|y-\hat{y}|}\nabla _yK(t,u,u^2-w)dw\\&\quad = (2\pi )^{-1}t^{-\frac{3}{2}} e^{\frac{u^2}{t}}e^{-\frac{w^2}{t^2}}\Big |_{w=2|y-\hat{y}|}^{w=-2|y-\hat{y}|}\\&\quad =0. \end{aligned}$$

Using the above equality and similar to the operation of \(I_{13}\), we have

$$\begin{aligned} |I_{14}|= & {} \Big |\int _{{\mathbb {R}}}\int _{|y-w|>2|y-\hat{y}|}\nabla _yK(t,u,\hat{y}-w)(f( u,\hat{y})-f(u,y))dudw\Big |\\= & {} \Big |\int _{{\mathbb {R}}}\int _{|y-w|>2|y-\hat{y}|}(\nabla _yK(t,u,\hat{y}-w)-\nabla _yK(t,u,y-w))\\&\times (f( u,\hat{y})-f(u,y))dudw\Big |\\= & {} \Big |\int _{{\mathbb {R}}}(f(x- u,\hat{y})-f(x-u,y))du\int _{|y-w|>2|y-\hat{y}|}\\&\times (\nabla _yK(t,u,\hat{y}-w)-\nabla _yK(t,u,y-w))dw\Big |\\\le & {} C\Vert f\Vert _{L^\infty ({\mathbb {R}},C_b^\beta ({\mathbb {R}}))}|y-\hat{y}|^\beta \int _{{\mathbb {R}}}du\\&\times \int _{|y-w|>2|y-\hat{y}|} |\nabla _yK(t,u,\hat{y}-w)-\nabla _yK(t,u,y-w)|dw\\\le & {} C\Vert f\Vert _{L^\infty ({\mathbb {R}},C_b^\beta ({\mathbb {R}}))}t^{-1+\gamma -\beta }|y-{\hat{y}}|^\beta . \end{aligned}$$

Substituting \(I_{11}{-}I_{14}\) into \(I_1\), we get

$$\begin{aligned}&\int _0^\infty e^{-\lambda t}\left( \sup _{x,\hat{x},y,{\hat{y}}\in {\mathbb {R}},x\ne \hat{x},y\ne \hat{y}} \frac{1}{|x-\hat{x}|^\beta +|y-\hat{y}|^\beta }|I_1|\right) dt\\&\quad \le C\int _0^\infty e^{-\lambda t}\left( t^{-1+\beta }+t^{-1+\gamma -\beta }\right) dt\\&\quad \le C\lambda ^{-(\beta \wedge (\gamma -\beta ))}. \end{aligned}$$

Similarly, we can prove that if \(f(\cdot ,y)\in C_b^\beta ({\mathbb {R}})\) for any \(y\in {\mathbb {R}}\),

$$\begin{aligned} \int _0^\infty e^{-\lambda t}\left( \sup _{x,\hat{x},y,{\hat{y}}\in {\mathbb {R}},x\ne \hat{x},y\ne \hat{y}} \frac{1}{|x-{\hat{x}}|^\beta +|y-\hat{y}|^\beta }|I_2|\right) dt\le C\lambda ^{-(\beta \wedge (\gamma -\beta ))}. \end{aligned}$$

Summing the above discussion, we obtain

$$\begin{aligned}{}[\nabla _yv]_\beta \le C\lambda ^{-(\beta \wedge (\gamma -\beta ))}\Vert f\Vert _{C^\beta ({\mathbb {R}}^2)}. \end{aligned}$$

Noting that the above inequality holds for \(0<\gamma <1\). The proof of this lemma is complete. \(\square \)

Remark 3.1

It is well known that if \(f\in C^\alpha ({\mathbb {R}}^n)\), then the solution u of the following equation

$$\begin{aligned} u_t-\Delta u=f,\ \ u_0=0, \end{aligned}$$

belongs to \(C^{2+\alpha }({\mathbb {R}}^n)\), which is the Schauder theory. Noting that the heat kernel of above equation is Gauss heat kernel, that is

$$\begin{aligned} K(t,x)=\frac{1}{(\pi t)^{\frac{n}{2}}}e^{-\frac{x^2}{t}}. \end{aligned}$$

It is easy to see that \(x\sim \sqrt{t}\). But in our case, different axis has different scaling, that is,

$$\begin{aligned} x\sim \sqrt{t},\ \ \ y\sim t. \end{aligned}$$

Thus when we take derivative for variable x, we can get \(t^{-\frac{1}{2}}\), and take double derivative for variable x, we will get \(t^{-1}\). But if we take derivative for variable y, we shall get \(t^{-1}\), which is different from the classical case. In Schauder theory, we can get \(C^{2+\alpha }({\mathbb {R}}^n)\) estimates, but in our case the \(C^{1+\alpha }({\mathbb {R}}^n)\) should be optimal.

Like the classical case, we can get the \(C^{2+\alpha }\)-estimate for the x-axis if \(f(\cdot ,y)\in C_b^\beta ({\mathbb {R}})\) for any \(y\in {\mathbb {R}}\), but we can not get the same estimate for y-axis. If we want to get the \(C^{2+\alpha }\)-estimate for the y-axis, we need more regularity about the second variable. In other words, we have the following results.

Corollary 3.1

Assume that \(f(\cdot ,y)\in C_b^\beta ({\mathbb {R}})\) for any \(y\in {\mathbb {R}}\) and \(\nabla _yf(x,\cdot )\in C_b^\beta ({\mathbb {R}})\) for any \(x\in {\mathbb {R}}\). Then

$$\begin{aligned}{}[\nabla ^2u]_\beta \le C\lambda ^{-\delta }\Vert f\Vert _{C^\beta ({\mathbb {R}};C^{1+\beta }({\mathbb {R}}))}, \end{aligned}$$

where \(0<\delta <(\beta \wedge (1-\beta ))\). That is to say, \(\nabla ^2 u\in C_b^\beta ({\mathbb {R}}^2)\).

Proof of Theorem 2.1

We use Picard’s iteration to solve (1). Let \(u_0=0\) and define for \(n\in {\mathbb {N}}\),

$$\begin{aligned} u_n:=(\lambda -{\mathcal {L}})^{-1}(f-b\cdot \nabla u_{n-1}). \end{aligned}$$
(4)

It follows from Lemma 3.1 that

$$\begin{aligned} \lambda ^{\delta }\Vert u_n\Vert _{C_b^{1+\beta }({\mathbb {R}}^2)}\le & {} C \Vert f-b\cdot \nabla u_{n-1}\Vert _{ C_b^{\beta }({\mathbb {R}}^2)}\nonumber \\\le & {} C[\Vert f\Vert _{ C_b^{\beta }({\mathbb {R}}^2)} +\Vert b\Vert _{L^\infty ({\mathbb {R}}^2)}\cdot \Vert \nabla u_{n-1}\Vert _{ C_b^{\beta }({\mathbb {R}}^2)}\nonumber \\&+\Vert b\Vert _{ C_b^{\beta }({\mathbb {R}}^2)}\cdot \Vert \nabla u_{n-1}\Vert _{L^\infty ({\mathbb {R}}^2)}], \end{aligned}$$
(5)

and

$$\begin{aligned} \lambda ^{\delta }\Vert u_n-u_m\Vert _{C_b^{1+\beta }({\mathbb {R}}^2)}\le & {} C \Vert b\Vert _{L^\infty ({\mathbb {R}}^2)}\cdot \Vert \nabla u_{n-1}-\nabla u_{m-1}\Vert _{ C_b^{\beta }({\mathbb {R}}^2)}\nonumber \\&+C\Vert b\Vert _{ C_b^{\beta }({\mathbb {R}}^2)}\cdot \Vert \nabla u_{n-1}-\nabla u_{m-1}\Vert _{L^\infty ({\mathbb {R}}^2)}. \end{aligned}$$
(6)

Choosing \(\lambda _0\) be large enough so that \(C\lambda ^{-\delta }\Vert b\Vert _{C_b^\beta ({{\mathbb {R}}^2})}<1/4\) for all \(\lambda \ge \lambda _0\), we get

$$\begin{aligned} \Vert u_n\Vert _{C_b^{1+\beta }({\mathbb {R}}^2)}\le \lambda ^{-\delta }(\Vert f\Vert _{ C_b^{\beta }({\mathbb {R}}^2)} +2\Vert b\Vert _{ C_b^{\beta }({\mathbb {R}}^2)}) +\frac{1}{2}\Vert u_{n-1}\Vert _{ C_b^{1+\beta }({\mathbb {R}}^2)} \end{aligned}$$

and for all \(n\ge m\),

$$\begin{aligned} \Vert u_n-u_m\Vert _{C_b^{1+\beta }({\mathbb {R}}^2)}\le \frac{1}{2}\Vert u_{n-1}-u_{m-1}\Vert _{C_b^{1+\beta }({\mathbb {R}}^2)}. \end{aligned}$$

Substituting them into (5) and (6), we obtain

$$\begin{aligned} \lambda ^{-\delta }\Vert u_n\Vert _{C_b^{1+\beta }({\mathbb {R}}^2)}\le C\Vert f\Vert _{ C_b^{\beta }({\mathbb {R}}^2)} \end{aligned}$$

and for all \(n\ge m\),

$$\begin{aligned} \lambda ^{-\delta }\Vert u_n-u_m\Vert _{C_b^{1+\beta }({\mathbb {R}}^2)}\le \frac{C}{2^m}. \end{aligned}$$

Hence there is a \(u\in C_b^{1+\beta }({\mathbb {R}}^2)\) such that (2) holds and

$$\begin{aligned} \lambda ^{-\delta }\Vert u-u_m\Vert _{C_b^{1+\beta }({\mathbb {R}}^2)}\le \frac{C}{2^m}, \end{aligned}$$

and u solves Eq. (1) by taking limits for (4). The second result can be obtained similarly. The proof is complete. \(\square \)

Proof of Theorem 2.2

For simplicity, we only consider a special case, that is, \(f(x,y)=f_1(x)f_2(y)\). Assume that \(f_1\in L^p({\mathbb {R}})\) and \(f_2\in L^q({\mathbb {R}})\). Denote

$$\begin{aligned} K_1(t,x,y)=\int _{{\mathbb {R}}}K(t,x-u,y)f_1(u)du. \end{aligned}$$

By using the above inequality, Minkowski’s inequality and the properties of the heat kernel, we have

$$\begin{aligned} \Vert u\Vert _{L^r({\mathbb {R}}^2)}= & {} \Bigg \Vert \int _0^\infty e^{-\lambda t}K(t,\cdot )*f(x,y)dt \Bigg \Vert _{L^r({\mathbb {R}}^2)}\\\le & {} \int _0^\infty e^{-\lambda t}\left( \int _{{\mathbb {R}}^2}|\int _{{\mathbb {R}}}K_1(t,x,y-v) f_2(v)dv|^rdxdy\right) ^{\frac{1}{r}}dt\\= & {} \int _0^\infty e^{-\lambda t}\left( \int _{{\mathbb {R}}}\Vert K_1(t,x,\cdot )*f_2\Vert ^r_{L^r({\mathbb {R}})}dx\right) ^{\frac{1}{r}}dt\\\le & {} \Vert f_2\Vert _{L^q({\mathbb {R}})}\int _0^\infty e^{-\lambda t}\left( \int _{{\mathbb {R}}}\Vert K_1(t,x,\cdot )\Vert ^r_{L^m({\mathbb {R}})}dx\right) ^{\frac{1}{r}}dt\\= & {} \Vert f_2\Vert _{L^q({\mathbb {R}})}\int _0^\infty e^{-\lambda t}\left( \int _{{\mathbb {R}}}\Big |\int _{\mathbb {R}}|K_1(t,x,y)|^mdy\Big |^{\frac{r}{m}}dx\right) ^{\frac{1}{r}}dt\\\le & {} \Vert f_2\Vert _{L^q({\mathbb {R}})}\int _0^\infty e^{-\lambda t}\left( \int _{{\mathbb {R}}}\Big |\int _{\mathbb {R}}|\int _{{\mathbb {R}}}K(t,x-u,y)f_1(u)du|^rdx\Big |^{\frac{m}{r}}dy\right) ^{\frac{1}{m}}dt\\= & {} \Vert f_2\Vert _{L^q({\mathbb {R}})}\int _0^\infty e^{-\lambda t}\left( \int _{{\mathbb {R}}}\Big |\int _{\mathbb {R}}\Vert K(t,\cdot ,y)*f_1\Vert ^m_{L^r({\mathbb {R}})}dy\right) ^{\frac{1}{m}}dt\\\le & {} \Vert f_1\Vert _{L^p({\mathbb {R}})}\Vert f_2\Vert _{L^q({\mathbb {R}})}\int _0^\infty e^{-\lambda t}\Vert K(t,\cdot ,\cdot )\Vert _{L^{n}({\mathbb {R}},L^{m}({\mathbb {R}}))}dt\\\le & {} C_1\int _0^\infty e^{-\lambda t}t^{-\frac{3}{2}+\frac{1}{2n}+\frac{1}{m}}dt\\\le & {} C\lambda ^{-1-\frac{3}{2r}+\frac{1}{2p}+\frac{1}{q}}, \end{aligned}$$

where

$$\begin{aligned}&1+\frac{1}{r}=\frac{1}{n}+\frac{1}{p}, \ \ \ \ 1+\frac{1}{r}=\frac{1}{m}+\frac{1}{q},\\&C_1=\Vert f_1\Vert _{L^p({\mathbb {R}})}\Vert f_2\Vert _{L^q({\mathbb {R}})} \left( \int _{{\mathbb {R}}}\left( \int _{{\mathbb {R}}}e^{-p'x^2-p'(y-x^2)^2}dx\right) ^{\frac{q'}{p'}}dy\right) ^{\frac{1}{q'}}.\nonumber \end{aligned}$$
(7)

We remark that

$$\begin{aligned}&\left( \int _{\mathbb {R}}|\nabla _xK(t,x,y)|^rdx\right) ^{\frac{1}{r}}\le Ct^{-\frac{1}{2}+\frac{1}{2r}},\\&\left( \int _{\mathbb {R}}|\nabla _yK(t,x,y)|^rdy\right) ^{\frac{1}{r}}\le Ct^{-1+\frac{1}{r}}. \end{aligned}$$

Similarly, we obtain

$$\begin{aligned} \Vert \nabla _xu\Vert _{L^r({\mathbb {R}}^2)}\le & {} C\lambda ^{-\frac{1}{2}-\frac{3}{2r}+\frac{1}{2p}+\frac{1}{q}}. \end{aligned}$$

Furthermore, we can get

$$\begin{aligned} \Vert \nabla ^2_xu\Vert _{L^r({\mathbb {R}}^2)}\le C\lambda ^{-\frac{3}{2r}+\frac{1}{2p}+\frac{1}{q}}. \end{aligned}$$

Thus if we take \(p=q=r\), then we have \(\Vert \nabla ^2_xu\Vert _{L^r({\mathbb {R}}^2)}\le C\). However, if we deal with the second variable, it is difficult to get the decay estimate. More precisely, we have for \(\frac{3}{2r}<\frac{1}{2p}+\frac{1}{q}\),

$$\begin{aligned} \Vert \nabla _yu\Vert _{L^r({\mathbb {R}}^2)}\le & {} C\lambda ^{-\frac{3}{2r}+\frac{1}{2p}+\frac{1}{q}}\rightarrow \infty , \ \ \mathrm{as}\ \ \lambda \rightarrow \infty . \end{aligned}$$

Hence we must add more regularity on the second variable. Meanwhile, we recall that if \(h\in W^{s,p}({\mathbb {R}}^n)\) with \(0<s<1\), then

$$\begin{aligned} \Vert h\Vert _{W^{s,p}({\mathbb {R}}^n)}=\left( \Vert h\Vert ^p_{L^p({\mathbb {R}}^n)}+ \int _{{\mathbb {R}}^n}\int _{{\mathbb {R}}^n}\frac{|h(x)-h(y)|^p}{|x-y|^{n+sp}}dxdy\right) ^{\frac{1}{p}}. \end{aligned}$$

If we assume that \(f_2\in W^{s,q}({\mathbb {R}})\) and let

$$\begin{aligned} K_2(t,x,y)=\int _{{\mathbb {R}}}\nabla _yK(t,x,y-v)f_2(v)dv. \end{aligned}$$

we get

$$\begin{aligned} \Vert \nabla _yu\Vert _{L^r({\mathbb {R}}^2)}= & {} \Bigg \Vert \int _0^\infty e^{-\lambda t}\nabla _yK(t,\cdot )*f(x,y)dt\Bigg \Vert _{L^r({\mathbb {R}}^2)}\\\le & {} \int _0^\infty e^{-\lambda t}\left( \int _{{\mathbb {R}}^2}|\int _{{\mathbb {R}}}K_2(t,x-u,y) f_1(u)du|^rdxdy\right) ^{\frac{1}{r}}dt\\\le & {} \Vert f_1\Vert _{L^p({\mathbb {R}})}\int _0^\infty e^{-\lambda t}\left( \int _{{\mathbb {R}}}\Vert K_2(t,\cdot ,y)\Vert ^n_{L^r({\mathbb {R}})}dy\right) ^{\frac{1}{n}}dt\\= & {} \Vert f_1\Vert _{L^p({\mathbb {R}})}\int _0^\infty e^{-\lambda t}\\&\times \left( \int _{\mathbb {R}}\left( \Big |\int _{\mathbb {R}}\nabla _yK(t,x,v)|v|^{s+\frac{1}{q}}\frac{f_2(y-v)-f_2(y)}{|v|^{s+\frac{1}{q}}}dv\Big |^rdx \right) ^{\frac{n}{r}}dy\right) ^{\frac{1}{n}}dt\\\le & {} \Vert f_1\Vert _{L^p({\mathbb {R}})}\Vert f_2\Vert _{W^{s,q}({\mathbb {R}})}\int _0^\infty e^{-\lambda t}\left( \int _{\mathbb {R}}\left( |\nabla _yK(t,x,y)|^m|y|^{sm+\frac{m}{q}}dy\right) ^{\frac{n}{m}}dx\right) ^{\frac{1}{n}}dt\\\le & {} C_2\int _0^\infty e^{-\lambda t}t^{-\frac{5}{2}+s+\frac{1}{q}+\frac{1}{2n}+\frac{1}{m}}dt\\\le & {} C\lambda ^{-s-\frac{3}{2r}+\frac{1}{2p}}, \end{aligned}$$

where m, n satisfy (7) and

$$\begin{aligned} C_2=2\Vert f_1\Vert _{L^p({\mathbb {R}})}\Vert f_2\Vert _{W^{s,q}({\mathbb {R}})} \left( \int _{{\mathbb {R}}}\left( \int _{{\mathbb {R}}}|y-x^2|^p|y|^{sp+\frac{p}{q}}e^{-p'x^2-p'(y-x^2)^2}dx\right) ^{\frac{q'}{p'}}dy\right) ^{\frac{1}{q'}}. \end{aligned}$$

Moreover, under the condition that \(f_2\in W^{s,p}({\mathbb {R}})\), we can similarly get

$$\begin{aligned} \Vert \nabla ^2_yu\Vert _{L^r({\mathbb {R}}^2)}\le & {} C\lambda ^{-s-\frac{3}{2r}+\frac{1}{2p}+1}. \end{aligned}$$

Hence it is easy to see that we can take suitable \(s\in (0,1),p>1\) and \(r>1\) such that \(-s-\frac{3}{2r}+\frac{1}{2p}+1\le 0\). That is to say, we have

$$\begin{aligned} \Vert \nabla ^2_yu\Vert _{L^r({\mathbb {R}}^2)}\le & {} C \end{aligned}$$

under the condition that \(f_2\in W^{s,p}({\mathbb {R}})\). The proof is complete. \(\square \)

Remark 3.2

It is well known that if \(f\in L^p({\mathbb {R}}^n)\), then the solution u of the following equation

$$\begin{aligned} u_t-\Delta u=f,\ \ u_0=0, \end{aligned}$$

belongs to \(W^{2,p}({\mathbb {R}}^n)\), which is the \(L^p\)-theory. Noting that the heat kernel of above equation is Gauss heat kernel, and similar to Remark 3.1, it is easy to find the difference from the classical Laplacian operator. Due to the singularity of the variable y, we must give two different assumptions. Comparing the classical \(L^p\)-theory, in our case the regularity of Theorem 2.2 should be optimal.