1. Introduction

In this paper, we consider the following equation for $u$

$-\Delta u=f(u)-\frac{1}{|\Omega|}\int_{\Omega} f(u)d\mathbf{x}$

(1.1)

under periodic boundary conditions. The domain $\Omega=\mathbb{T}^N$, the $N$-dimensional torus, with $N=2, 3$. Here $f$ is a given smooth function of $u$ for $u(\mathbf{x}) \in G \subset \mathbb{R}$.

We will prove that there exists a solution $u$ to equation (1.1) which is unique if $ |\frac{d f}{d u}(u_0) | < \frac{1}{(C_0)^2}$, where $u_0 \in G$ is a given constant and where $C_0$ is the constant from Poincarés inequality. And we will prove that the solution $u$ is not unique if $\frac{df}{du}(u_0) $ is a simple eigenvalue of $-\Delta$.

In previous related work, many researchers have studied the equation $-\Delta u=f(u)+g$. Existence of a solution $u$ to the equation $-\Delta u=f(u)+g$ has been proven for a Dirichlet boundary condition $u|_{\partial \Omega}=0$ (see, e.g., ^{[1, 2, 5, 7]}) under certain conditions on $f$ and, $g$. And existence of a solution $u$ to the equation $-\Delta, ,u=f(u)+g$ has been proven for a Neumann boundary condition $\frac{\partial u}{\partial n}|_{\partial \Omega}=h$ (see, e.g., ^{[3, 4, 6]}) under certain conditions on $f$ and $g$. We have not seen any work by other researchers on the existence of a solution $u$ to equation (1.1) under periodic boundary conditions. And we have not seen any work by other researchers which contains the particular condition that $ |\frac{d f}{d u}(u_0) | < \frac{1}{(C_0)^2}$, where $C_0$ is the constant from Poincarés inequality and where $u_0$ is a given constant in the domain of the function $\frac{d f}{d u}$.

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2. Existence theorem

In the proof that follows, we use the standard notation for the $L^2(\Omega)$ norm of a function $g$, namely, $\|g\|_0^2=\int_\Omega | g|^2d \mathbf{x}$. And we denote the inner product as $(g, h)=\int_\Omega gh d \mathbf{x}$. Also, we let $Du$ denote the gradient of a function $u$. We also use the notation $ |\frac{df}{du} |_{0, \overline{G}_1}=\max\{|\frac{df}{du}(u_{*}) |: u_{*} \in \overline{G}_1\}$, where $\frac{df}{du}$ is a function of $u$ and where $ \overline{G}_1 \subset \mathbb{R}$ is a closed bounded interval.

The purpose of this article is to prove the following theorem.

Theorem 2.1. Consider the following equation for $u$

$-\Delta u=f(u)-\frac{1}{|\Omega|}\int_{\Omega} f(u)d\mathbf{x}$

(2.1)

where the domain $\Omega=\mathbb{T}^N$, the $N$-dimensional torus, with $N=2$ or $N=3$, and where $f$ is a given smooth function of $u$ for $u(\mathbf{x}) \in G \subset \mathbb{R}$. Let $u_0 \in G$ be a given constant. Then we have the following two cases:

(1) If $ |\frac{d f}{d u}(u_0) | < \frac{1}{(C_0)^2}$, where $C_0$ is the constant from Poincarés inequality, then there exists a unique classical solution $u(\mathbf{x})\in \overline{G}_1$ to equation (2.1) which satisfies the condition $u(\mathbf{x}_0)=u_0$, where $\overline{G}_1 \subset G \subset \mathbb{R}$ and where $u_0 \in \overline{G}_1$ and where $\mathbf{x}_0 \in \Omega$ is a given point. This unique classical solution is $u=u_0$.

(2) If $\frac{df}{du} (u_0) $ is a simple eigenvalue of $-\Delta$ then there exists a solution $u$ of equation (2.1) which is not the constant function $u_0$. This solution $u$ may not necessarily satisfy the condition $u(\mathbf{x}_0)=u_0$. }

Proof.

We will consider separately each of the two cases from the statement of the theorem. First, we will consider Case 1 from the statement of Theorem 2.1

Suppose that $ |\frac{d f}{d u}(u_0) | < \frac{1}{(C_0)^2}$, where $C_0$ is the constant from Poincarés inequality and where $u_0 \in G$ is a given constant. It follows that there exists a closed bounded interval $\overline{G}_1 \subset G$ such that $u_0 \in \overline{G}_1$ and such that $|\frac{d f}{d u} |_{0, \overline{G}_1} < \frac{1}{(C_{0})^2}$, where $|\frac{d f}{d u} |_{0, \overline{G}_1}=\max\{|\frac{d f}{d u} (u_{*}) |:u_{*} \in \overline{G}_1\}$. Suppose that $u$ is a classical solution of equation (2.1) such that $u(\mathbf{x}) \in \overline{G}_1$ for all $ \mathbf{x} \in \Omega$ and $u$ satisfies the condition $u(\mathbf{x}_0)=u_0$, where $\mathbf{x}_0 \in \Omega$ is a given point. We will now prove that this solution is $u=u_0$.

From equation (2.1), and from using integration by parts and Poincarés inequality, we obtain the following estimate for $ \Vert D u\Vert_{0}^2$:

$\begin{gathered} \left\| {Du} \right\|_0^2=(-\Delta u, u-\frac{1}{{|\Omega |}}\int_\Omega u d{\mathbf{x}}) \\=(f(u)-\frac{1}{{|\Omega |}}\int_\Omega f (u)d{\mathbf{x}}, u-\frac{1}{{|\Omega |}}\int_\Omega u d{\mathbf{x}}) \\ \leq {\left\| {f(u)-\frac{1}{{|\Omega |}}\int_\Omega f (u)d{\mathbf{x}}} \right\|_0}{\left\| {u-\frac{1}{{|\Omega |}}\int_\Omega u d{\mathbf{x}}} \right\|_0} \\ \leq {({C_0})^2}{\left\| {Df(u)} \right\|_0}{\left\| {Du} \right\|_0} \\ \end{gathered} $

(2.2)

where we used Poincarés inequality to obtain $\|u-\frac{1}{|\Omega|}\int_{\Omega} u d\mathbf{x} \|_0 \leq C_0 \|D u \|_0$ and $\|f(u)-\frac{1}{|\Omega|}\int_{\Omega} f(u)d\mathbf{x} \|_0 \leq C_0 \|Df(u) \|_0$.

From (2.2) we obtain the inequality

$\begin{gathered} \left\| {Du} \right\|_0^2 \leq {({C_0})^4}\left\| {Df(u)} \right\|_0^2 \\ \leq {({C_0})^4}|\frac{{df}}{{du}}|_{{L^\infty }(\Omega)}^2\left\| {Du} \right\|_0^2 \\ \leq {({C_0})^4}|\frac{{df}}{{du}}|_{0, {{\bar G}_1}}^2\left\| {Du} \right\|_0^2 \\ \end{gathered} $

(2.3)

where we used the assumption that $u(\mathbf{x})\in \overline{G}_1$ for all $\mathbf{x}\in \Omega$, and so it follows that $|\frac{d f}{d u} |_{L^{\infty}(\Omega)}\leq |\frac{d f}{d u} |_{0, \overline{G}_1}$, where $|\frac{d f}{d u} |_{0, \overline{G}_1}=\max\{|\frac{d f}{d u} (u_{*}) |:u_{*} \in \overline{G}_1\}$.

Since $ \Big|\frac{d f}{du} \Big|_{0, \overline{G}_1}^2 < \frac{1}{(C_0)^4}$, it follows from (2.3) that $\Vert D u\Vert_{0}=0$ and so the solution $u$ of equation (2.1) is a constant. Therefore the solution $u=u_0$ is the unique classical solution of equation (2.1) in $\overline{G}_1$ which satisfies the condition $u(\mathbf{x}_0)=u_0$. This completes the proof of Case 1 in the statement of Theorem 2.1.

Next, we consider Case 2 in the statement of Theorem 2.1. We now prove that if $\frac{d f}{du} (u_0) $ is a simple eigenvalue of $-\Delta$ then there exists a solution $u$ of equation (2.1) which is not the constant solution $u_0$. We remark that this solution $u$ may not necessarily satisfy the condition that $u(\mathbf{x}_0)=u_0$, where $\mathbf{x}_0 \in \Omega$ is a given point.

We begin by letting $v=u-u_0$ and write equation (2.1) equivalently as

$\begin{gathered}-\Delta v=-\Delta u=f(u)-\frac{1}{{|\Omega |}}\int_\Omega f (u)d{\mathbf{x}} \\=(f(u)-f({u_0}))-\frac{1}{{|\Omega |}}\int_\Omega {(f(} u)-f({u_0}))d{\mathbf{x}} \\=(\frac{{df}}{{du}}({u_0} + {t_1}(u-{u_0})))(u-{u_0})-\frac{1}{{|\Omega |}}\int_\Omega (\frac{{df}}{{du}}({u_0} + {t_1}(u-{u_0})))(u-{u_0})d{\mathbf{x}} \\=(\frac{{df}}{{du}}({u_0} + {t_1}v))v-\frac{1}{{|\Omega |}}\int_\Omega (\frac{{df}}{{du}}({u_0} + {t_1}v))vd{\mathbf{x}} \\ \end{gathered} $

(2.4)

where $t_{1} \in (0, 1)$. Here we used the mean value theorem.

We next obtain the identity

$\begin{gathered} \frac{{df}}{{du}}({u_0} + {t_1}v)=\frac{{df}}{{du}}({u_0} + {t_1}v)-\frac{{df}}{{du}}({u_0}) + \frac{{df}}{{du}}({u_0}) \\=(\frac{{{d^2}f}}{{d{u^2}}}({u_0} + {t_2}({t_1}v))){t_1}v + \frac{{df}}{{du}}({u_0}) \\ \end{gathered} $

(2.5)

where $t_{2} \in (0, 1)$. And we again used the mean value theorem.

Substituting (2.5) into (2.4) yields

$-\Delta v=\frac{{df}}{{du}}({u_0})v + (\frac{{{d^2}f}}{{d{u^2}}}({u_0} + {t_2}({t_1}v))){t_1}{v^2}-\frac{1}{{|\Omega |}}\int_\Omega (\frac{{df}}{{du}}({u_0} + {t_1}v))vd{\mathbf{x}}$

(2.6)

where $v=u-u_0$, where $t_{1} \in (0, 1)$, and where $t_{2} \in (0, 1)$.

We can write equation (2.6) in the form

$\Delta v+ \lambda v=g(v)$

(2.7)

where $\lambda=\frac{d f}{du}(u_{0})$ and where $g(v)=-\Big(\frac{d^2 f}{du^2}(u_0+t_{2}(t_{1}v))\Big)t_{1}v^2 $ $+\frac{1}{|\Omega|}\int_{\Omega}\Big(\frac{d f}{du}(u_{0}+t_{1}v)\Big) v d\mathbf{x}$.

Let $F(v, \lambda)=\Delta v+ \lambda v-g(v) $. We will apply the the implicit function theorem to the equation $F(v, \lambda)=0$. Note that $g(0)=0$ and $g^{\prime}(0)=0 $.

If $\lambda=\frac{d f}{du}(u_{0})$ is not an eigenvalue of $-\Delta$, it follows from the implicit function theorem that $v=0$ is the only small solution to the equation $F(v, \lambda)=0$ when $F(v, \lambda)=\Delta v+ \lambda v-g(v) $ and when $g(0)=0$ and $g^{\prime}(0)=0 $ (see, e.g., ^[7]). Therefore $u=u_0$ is the only solution of equation (2.1) in a neighborhood of $u_0$.

If $\lambda=\frac{d f}{du}(u_{0})$ is a simple eigenvalue of $-\Delta$, it follows from the implicit function theorem that there exists a non-trivial solution $v$ to the equation $F(v, \lambda)=0$ when $F(v, \lambda)=\Delta v+ \lambda v-g(v) $ and when $g(0)=0$ and $g^{\prime}(0)=0 $ (see, e.g., ^[7]). Therefore there exists a solution $u$ to equation (2.1) which is not the constant function $u_0$.

This completes the proof of Theorem 2.1.

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