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I am working on this question:

Suppose that $\Delta u\leqslant0$ in $\Omega$ and $u=f$ on $\partial\Omega$, and $\Delta v\leqslant0$ in $\Omega$ and $v=f$ on $\partial\Omega$. Prove that $v(x)\leqslant u(x)$ for $x\in\Omega$ using the maximum principle.

I do not want an answer to it. The context of it is partial differential equations, and I do not know what the symbols mean.

I assume that

$$\Delta=\nabla^2=\frac{\partial}{\partial x}+\frac{\partial}{\partial y}+\frac{\partial}{\partial z}$$

is the Laplace operator, $\Omega$ is some subset of $\mathbb{R}^3$, and I have no clue what $\partial\Omega$ is; maybe it is the surface of $\Omega$? Also, how could I go about answering this question? I mean, I am thinking in terms of three dimensions, but both $u$ and $v$ are functions of one variable.

I am so confused!

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Either read the book's symbols index or ask whoever gave this problem what that $\,\triangle\,$ thing means. – DonAntonio Oct 27 '12 at 1:23
My assumptions were correct, but there was a typo in the problem. I solved it now. – EsX_Raptor Oct 29 '12 at 23:27

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