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could any one just tell me in a short what is the geometric idea of this theorem?

"let $u:\Omega\subseteq\mathbb{R}^2\rightarrow \mathbb {R}$,$p=(x_0,y_0)\in\Omega$, $u_x,u_y$ exist at every point in a neighborhood of $p$ and continous at $p$ Then for sufficiently small $s,t$ in $\mathbb{R}$,

$$u(x_0+s,y_0+t)-u(x_0,y_0)=su_x(x_0+s^*,y_0+t^*)+tu_y(x_0,y_0+t^*)$$ with $|s^*|<|s|,|t^*|<|t|$

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Isn't this just a generalization of the Mean Value Theorem? –  echoone Aug 24 '12 at 20:36
    
how mow khau?... –  miosaki Aug 24 '12 at 20:39
    
You go from $(x,y)$ to $(x,y+t)$ applying the mean value theorem on $t$. Then you go from $(x,y+t)$ to $(x+s,y+t)$ applying the mean value theorem on $s$. The latter doesn't quite fit because $u_x$ is evaluated at $(x+s^*,y+t^*)$ and not $(x+s^*,y+t)$; are you sure that's not a typo? –  Rahul Aug 24 '12 at 21:38
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1 Answer 1

Indeed as echoone mentions, this is a generalization of the mean value theorem from single variable calculus.

Consider the rectangle $[x_0,\ x_0 + s]\times[y_0,\ y_0 + t]$. The idea is to apply the mean value theorem component wise to $u(x,y)$.

By taking $x$ fixed, we can consider $u(x,y)$ as a function of $y$. We can then write $$u(x_0, y_0 + t) - u(x_0, y_0) = tu_y(x_0, y_0 + t^*)$$ Now we apply the mean value theorem again to $x$ $$u(x_0 + s, y_0 + t) - u(x_0, y_0 + t) = su_x(x_0 + s^*, y_0 + t)$$ Combining the two yields the desired expression. There appears to be a small error in that the statement should read $$u(x_0 + s, y_0+t) - u(x_0, y_0) = su_x(x_0 + s^*, y_0 + t) - tu_y(x_0, y_0 + t^*)$$ and not $$u(x_0 + s, y_0+t) - u(x_0, y_0) = su_x(x_0 + s^*, y_0 + t^*) - tu_y(x_0, y_0 + t^*)$$ The $t$ in the first term on the right is not $t^*$.

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