How prove there exists a point $(x_{0},y_{0})$, such $\Delta f|_{(x_{0},y_{0})}\ge 0$
Question:
Assume that the function $f(x,y)$ is twice continuously differentiable on $\mathbb R^2$, and $$f\big|_{\partial\Sigma}=0,\quad \text{where}\,\,\,\partial\Sigma=\{(x,y)\in\mathbb R^2:x^2+y^2=1\}, $$ and $$ \lim_{x\to\infty}f(x,0)=1. $$ Show that there exists a point $(x_{0},y_{0})$, such $\Delta f(x_{0},y_{0})\ge 0$, where $\Delta$ is Laplace operator. See http://en.wikipedia.org/wiki/Laplace_operator
This problem is from (problem 8):http://www.doc88.com/p-98539219502.html
My try: since $$ \Delta f=\dfrac{\partial^2f}{\partial x^2}+\dfrac{\partial^2f}{\partial y^2}, $$ so assume that $$\Delta f=\dfrac{\partial^2f}{\partial x^2}+\dfrac{\partial^2f}{\partial y^2}<0,\,\forall(x,y)\in \mathbb R^2,$$ then I can't , This problem is hard? Thank you
Solution 1:
Despite what the other answer suggests, the assertion is false. There exist functions $f$ in $C^2(\mathbb R^2)$ such that
- $\Delta f(x,y)<0$ for all $(x,y)\in\mathbb R^2$,
- $f(x,y) = 0$ if $x^2 + y^2 = 1$, and
- $\lim_{x\to\infty}f(x,0) = 1$.
Here's how to make one. Pick a smooth function $\eta$ on $\mathbb R$ such that $\eta(x) = 1$ for $x\leq 2$ and $\eta(x) = 0$ for $x\geq 3$. Let $\rho$ be a twice continuously differentiable function on $\mathbb R$ such that $\rho(x)=1$ for $x\leq 1$. (The function $\eta$ will remain unchanged, but I'll determine what else we need to ask of $\rho$ in a moment.) Define \begin{align*} f(x,y) & = 1-\eta(x)x^2 - \rho(x)y^2 \end{align*} First, $f$ is twice continuously differentiable. Second, $f$ satisfies (2) because for $x\leq1$, we have $f(x,y) = 1-x^2 - y^2$. Also, $f$ satisfies (3) because $\eta(x) = 0$ for $x\geq 3$ and so for those $x$ we have $f(x,0) = 1-\eta(x)x^2 = 1$. Finally, for $x\leq 1$, we have $$\Delta(f(x,y)) = \Delta(1-x^2-y^2) = -4 < 0.$$ The question is thus whether we can make $\Delta f<0$ when $x>1$.
For $x>1$ \begin{align*} \Delta(f(x,y)) & = \Delta(1-\eta(x)x^2 - \rho(x)y^2) \\ & = -\Delta(\eta(x)x^2) - {\partial^2\over \partial x^2}(\rho(x)y^2) - {\partial^2\over \partial y^2}(\rho(x)y^2) \\ & = -\Delta(\eta(x)x^2) - \rho''(x)y^2 - 2\rho(x). \tag{1} \end{align*} Now for $1<x\leq 2$, we have $\eta(x) = 1$, so $-\Delta(\eta(x)x^2) = -\Delta(x^2) = -2$. Thus if we can ensure that $\rho''$ and $\rho$ are positive, then $(1)$ will be negative for $x\in(1,2]$. On the other hand, $\eta$ has compact support in $[2,\infty)$, so $\Delta(\eta(x)x^2)$ does as well. This means in particular that $\Delta(\eta(x)x^2)$ is bounded in $[2,\infty)$, say $|\Delta(\eta(x)x^2)|<M$. Therefore, if we can ensure that $$ M - \rho''(x)y^2 - 2\rho(x) <0, \tag{2} $$ then $(1)$ will be negative for all $x\in [2,\infty)$. Actually, if we make sure that $\rho''(x)\geq 0$ for $x>1$, then we need only have $M-2\rho(x)<0$ for $x\geq 2$ to satisfy $(2)$.
In sum, we will be done if we find $\rho\in C^2(\mathbb R)$ such that (i) $\rho(x)=1$ for $x\leq 1$, (ii) $\rho(x)>0$ and $\rho''(x)\geq 0$ for $x\geq 1$, and (iii) $M- 2\rho(x)<0$ for $x\geq 2$. But it's easy to come up with $\rho$ satisfying (i), (ii), and (iii). For example, we could take \begin{align*} \rho(x) = \left\{\begin{array}{ll} 1& x\leq 1, \\ 1+M(x-1)^3 & x>1. \end{array}\right. \end{align*} With that, we are done.