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Question:

let $D$ is Bounded closed region,and Assume $f$ is Continuous differentiable on $D$,if such $$\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y}+\dfrac{\partial f}{\partial z}=f,f|_{\partial D}=0$$

show that $f=0$

this problem is from compition,and I consider sometimes,and I want use this contradiction methods,But I can't,Thank you

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    First show the solution is unique. Since $f=0$ is a solution, then it is that one. – OR. Dec 07 '13 at 16:03
  • Assume that $f$ and $g$ are two solutions, then $f-g$ is a solutions of the homogeneous equation $d_x(f-g)+d_y(f-g)+d_z(f-g)=0$, $(f-g)_{\partial D}=0$. So, it is enough to show the homogeneous equation has a unique solution. – OR. Dec 07 '13 at 16:04
  • why $f=g$? Thank you –  Dec 07 '13 at 16:10
  • Assume $h$ is a solution of $d_xh+d_yh+d_zh=0$, $h|_{\partial D}=0$. Then look at the function $r(t)=h(a+t[1,1,1])$. – OR. Dec 07 '13 at 16:12
  • I would purpose that if $f=0$,then $\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}=0$ – Claude Leibovici Dec 07 '13 at 16:49

1 Answers1

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Simply follow the line...

For every $(x,y,z)$ in $D$, consider $g:t\mapsto g(t)=f(x+t,y+t,z+t)$, then $$g'(t)=(\partial_xf+\partial_yf+\partial_zf)(x+t,y+t,z+t)=f(x+t,y+t,z+t)=g(t)$$ hence $g(t)=g(0)\mathrm e^t$ for every $t$ such that $(x+s,y+s,z+s)$ is in $D$ for every $s$ between $0$ and $t$.

There exists some finite $T$ depending on $(x,y,z)$ such that $(x+T,y+T,z+T)$ is in $\partial D$ while $(x+t,y+t,z+t)$ is in $D$ for every $t$ in $[0,T)$. Since $f(x+T,y+T,z+T)=0$, $g(t)\to0$ when $t\to T$, $t\lt T$. This is only possible if $g(0)=0$, that is, $f(x,y,z)=0$.

Did
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