PDE , n-variables

Dec 2015
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169
Earth
(1) \(\displaystyle \dfrac{\partial Z}{\partial t_1 }+\dfrac{\partial Z}{\partial t_{1}^{2}} +\dotsc + \dfrac{\partial Z}{\partial t_{n}^{n}}=0\).

(2) \(\displaystyle \partial_x f(x,y) = \partial_y f(x,y)\).

(3) \(\displaystyle \dfrac{\partial^2 z(x,y)}{\partial x \partial y }=\dfrac{\partial z}{\partial x} + \dfrac{\partial z}{\partial y } \).
 

romsek

Math Team
Sep 2015
2,969
1,676
USA
You need to look at (1). The way you have it written can't be correct.

(2) is trivially solved by any function $f(x,y) \ni f(x,y)=f(y,x)$
 
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Dec 2015
1,085
169
Earth
You need to look at (1). The way you have it written can't be correct.

(2) is trivially solved by any function $f(x,y) \ni f(x,y)=f(y,x)$
\(\displaystyle \partial_x f(x,y) = \partial_y f(x,y) \; \Rightarrow x\equiv y \; \) ; \(\displaystyle \; f(x,y)=f(x,x)\pm f(y,y)=g(x\pm y)\).

\(\displaystyle f(x,y)=x\pm y\) satisfies \(\displaystyle [x\pm y]_{x}'=1=[x\pm y]_{y}'=1\).

can we say the general solution is \(\displaystyle z=Cf(x\pm y)\) ?

\(\displaystyle \dfrac{\partial }{\partial x} Cf(x\pm y) = \dfrac{\partial }{\partial y } Cf(x\pm y)\).

For (3) I got : \(\displaystyle y\dfrac{\partial z}{\partial x} +z=0\).
 
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