The "Wrong Minimal Surface Equation" does not have the Bernstein property

TL;DR

This paper demonstrates that a specific nonlinear PDE, similar to the minimal surface equation, does not possess the Bernstein property, meaning not all solutions over the entire plane are affine linear functions.

Contribution

It provides a simple proof that the equation ef{toll} does not have the Bernstein property, contrasting with classical minimal surface equations.

Findings

The equation ef{toll} admits non-affine solutions over the entire plane.

The classical Bernstein theorem does not extend to this nonlinear PDE.

A straightforward argument shows the failure of the Bernstein property for this equation.

Abstract

A celebrated result of S. Bernstein states that every solution of the minimal surface equation over the entire plane has to be an affine linear function. Since the paper of Bernstein appeared in 1927, many different proofs and generalizations of this beautiful theorem were given, namely to higher dimensions and to more general equations, for a careful account we refer to the paper by Simo and to the monograph by Dierkes-Hildebrandt-Tromba. In his paper Simon posed the question whether the equation \begin{equation} (1+{u_x}^2)u_{xx}+2u_x u_y u_{xy}+ (1+{u_y}^2)u_{yy} = 0 \label{toll} \end{equation} has the Bernstein property i.e. whether every $C^2$-solution defined over the entire plane necessarily has to be affine. We here show by a very simple argument that this is not the case.

Equations

$$(1+{u_x}^2)u_{xx}+2u_x{u}_y{u}_{xy}+(1+{u_y}^2)u_{yy}=0$$

$$u(x,y)=h(x)+g(y),$$

$$(1+(h^{\prime }(x){)}^2)h^{\prime \prime }(x)+(1+(g^{\prime }(y){)}^2)g^{\prime \prime }(y)=0.$$

$$h^{\prime }(x)=f(x)=\frac{1}{\sqrt[3]{2}}\left(\sqrt[3]{\sqrt{9x^2+4}+3x}-\sqrt[3]{\sqrt{9x^2+4}-3x}\right).$$

$$\begin{array}{rl}h(x)=& \frac{-1}{\sqrt[3]{1024}}\{9x\left(\sqrt[3]{\sqrt{9x^2+4}-3x}-\sqrt[3]{\sqrt{9x^2+4}+3x}\right)\\ & +\sqrt{9x^2+4}\left(\sqrt[3]{\sqrt{9x^2+4}-3x}+\sqrt[3]{\sqrt{9x^2+4}+3x}\right)\}.\end{array}$$

$$\begin{array}{rl}g(y)=& \frac{1}{\sqrt[3]{1024}}\{9y\left(\sqrt[3]{\sqrt{9y^2+4}-3y}-\sqrt[3]{\sqrt{9y^2+4}+3y}\right)\\ & +\sqrt{9y^2+4}\left(\sqrt[3]{\sqrt{9y^2+4}-3y}+\sqrt[3]{\sqrt{9y^2+4}+3y}\right)\}=-h(y).\end{array}$$

$$u(x,y)=h(x)-h(y)$$

$$(1+{u_x}^2)u_{xx}-2u_x{u}_y{u}_{xy}+(1+{u_y}^2)u_{yy}=0.$$

$$f_1(u_x)\cdot u_{xx}+2B\cdot u_{xy}+f_2(u_y)\cdot u_{yy}=0,$$

$$f_1(h^{\prime }(x))h^{\prime \prime }(x)+f_2(g^{\prime }(y))g^{\prime \prime }(y)=0.$$

$$F_1(h^{\prime }(x))=x\text{ and }{F}_2(g^{\prime }(y))=-y.$$

$$u(x,y)=\int {F_1}^{-1}(x)dx+\int {F_2}^{-1}(-y)dy,$$

$$u_{xx}+u_{xy}+u_{yy}=0.$$

$$\frac{u_{xx}}{\sqrt{1+{u_x}^2}}+\frac{u_{yy}}{\sqrt{1+{u_y}^2}}=0$$

$$\sqrt{1+{u_y}^2}\cdot u_{xx}+u_{xy}+\sqrt{1+{u_x}^2}\cdot u_{yy}=0$$

$$\sqrt{1+{u_y}^2}\cdot u_{xx}+2\tilde{B}{u}_{xy}+\sqrt{1+{u_x}^2}\cdot u_{yy}=0,$$

$$\text{with an arbitary B~ such that}|\tilde{B}|<\sqrt[4]{(1+{u_y}^2)(1+{u_x}^2)}.$$

$$u=u(x,y)=\int {F_1}^{-1}(x)dx+\int {F_2}^{-1}(-y)dy$$

$$\frac{u_{xx}}{f_2(u_y)}+2\tilde{B}{u}_{xy}+\frac{u_{yy}}{f_1(u_x)}=0,$$

$$\text{the minimal surface equation }(1+{u_y}^2)u_{xx}-2u_x{u}_y{u}_{xy}+(1+{u_x}^2)u_{yy}=0$$

$${u_x}^2{u}_{xx}+2u_x{u}_y{u}_{xy}+{u_y}^2{u}_{yy}=0.$$

Full text

In his paper [Simon] Simon posed the question whether the equation

[TABLE]

has the Bernstein property i.e. whether every $C^{2}$-solution defined on all of $\mathds{R}^{2}$ necessarily has to be affine.

We here show by a very simple argument that this is not the case.

Keywords: Bernstein property, wrong minimal surface equation, entire non-linear solutions

AMS 2000 MSC: 35A01, 35B08, 35B65, 35D05, 35J15

To start with we consider $u\in C^{2}(\mathds{R}^{2})$ to be a solution of the elliptic equation (1) with $u_{xy}\equiv 0$ in the whole of $\mathds{R}^{2}$. Then $u$ has the form

[TABLE]

We put $(1+(h^{\prime}(x))^{2})h^{\prime\prime}(x)=c$ and hence $(1+(g^{\prime}(y))^{2})g^{\prime\prime}(y)=-c$, $c\in\mathds{R}$, and choose $c=1$. (To get the linear solutions take $c=0$.)

By separation of variables we solve the equation  $(1+f^{2})f^{\prime}=1$ with $f=h^{\prime}(x)$  and obtain   $(1+f^{2})df=dx$,  or  $f+\frac{\displaystyle f^{3}}{\displaystyle 3}=x$. By Cardano’s formulae:

[TABLE]

An integration yields:

[TABLE]

Similarly, we get

[TABLE]

Thus, the non-linear $C^{2}-$function

[TABLE]

solves (1) in the whole plane $\mathds{R}^{2}$.

[TABLE]

More generally, we have the following

[TABLE]

*with an arbitary $B$ (depending on $x,y,u,u_{x},u_{y},u_{xx},u_{xy},u_{yy}$) has non-linear entire $C^{2}-$solutions in $\mathds{R}^{2}$, i.e. (2) does not have the Bernstein property.

(For the ellipticity of (2) assume $|B|<\sqrt{f_{1}(u_{x})f_{2}(u_{y})}$.)*

[TABLE]

Put $f_{1}(h^{\prime}(x))h^{\prime\prime}(x)=c$ and $f_{2}(g^{\prime}(y))g^{\prime\prime}(y)=-c$, with an arbitrary constant $c\in\mathds{R}$.

For the linear solutions take $c=0$. Since we are interested in non-linear ones, let us choose $c=1$:

By separation of variables we get:

[TABLE]

Since $F_{i}$ ($i=1,2$) is bijective in the whole of $\mathds{R}$, a non-linear entire $C^{2}-$solution is given by

[TABLE]

wherein ${F_{i}}^{-1}$ is the bijective continuous inverse of $F_{i}$ ($i=1,2$). ∎

Example 1.

Taking $F_{i}(t)=t$ we find that $u(x,y)=x^{2}-y^{2}$ solves the elliptic equation

[TABLE]

Example 2.

With $f_{i}(t)=1+t^{2}$  and $F_{i}(t)=t+\frac{t^{3}}{3}$ we obtain the equation (1).

Example 3.

Take  $f_{i}(t)=\frac{\displaystyle 1}{\displaystyle\sqrt{1+t^{2}}}$  and  $F_{i}(t)=\operatorname{arsinh}t$  respectively,

  then  $u(x,y)=\cosh(x)-\cosh(y)$ solves

[TABLE]

The function*

[TABLE]

also solves the equation

[TABLE]

*with an arbitary $\tilde{B}$, i.e. this equation does not have the Bernstein property in $\mathds{R}^{2}$.

(For the ellipticity of our last equation assume $|\tilde{B}|<\frac{\displaystyle 1}{\displaystyle\sqrt{f_{2}(u_{y})f_{1}(u_{x})}}$.)*

In fact we have the following counterexamples:

Example

Take $f_{i}(t)=\frac{\displaystyle 1}{\displaystyle 1+t^{2}}$ and $F_{i}(t)=\arctan(t)$ respectively (so $F_{i}:\mathds{R}\to\mathds{R}$ is not bijective), then we get:

$u(x,y)=\ln(\cos y)-\ln(\cos x)\in C^{2}\left(\left(-\frac{\pi}{2};\frac{\pi}{2}\right)\times\left(-\frac{\pi}{2};\frac{\pi}{2}\right)\right)$ which solves

[TABLE]

and the equation $\frac{\displaystyle u_{xx}}{\displaystyle 1+{u_{x}}^{2}}+\frac{\displaystyle u_{yy}}{\displaystyle 1+{u_{y}}^{2}}=0$ respectively, clearly not on all of $\mathds{R}^{2}$.

The condition $F_{i}^{\prime}>0$ cannot be replaced by the strong monotonicity of $F_{i}$ (for $i=1$ or $i=2$). Otherwise the solution $u$ can develop singularities:

Example

Take $f_{i}(t)=t^{2}$ and $F_{i}(t)=\frac{1}{3}t^{3}$ respectively, then we obtain:

$u(x,y)=\frac{9}{4}\left(|x|^{4/3}-|y|^{4/3}\right)$ which solves the equation

[TABLE]

Aronsson presented this $C^{1,1/3}(\mathds{R}^{2})-$ ”singular solution” of (3) in [Aronsson2]:

$u$ is $C^{1}$ in $\mathds{R}^{2}$, $C^{\infty}$ in each open quadrant and the coordinate axes are lines of singularity for $u$.

Interestingly, the equation (3) has the Bernstein-property, see [Aronsson].