Mathematics Asked by Still_waters on December 1, 2020
I have an PDE $dfrac{df}{d xi}-xidfrac{d x_1}{d xi}=0$ homogeneous for $xi$, where $f:mathbb{R}^ntomathbb{R}$ is function of $x_1,cdots,x_n:mathbb{R}tomathbb{R}$, which in turn are functions of $xi$ (so, $dfrac{df}{d xi}$ is a total derivative).
I also have $xi=y/zinmathbb{R}$, where $y,zinmathbb{R}$, and one said that once the PDE is homogeneous in $xi$ I have $dfrac{df}{d y}-xidfrac{d x_1}{d y}=0$.
I missed this gap. I imagine that I can multiply the first ODE by $dfrac{dxi}{d y}=dfrac{1}{z}$, so
$$dfrac{df}{d xi}dfrac{dxi}{d y}-xidfrac{d x_1}{d xi}dfrac{dxi}{d y}=sum_{i=1}^n dfrac{partial f}{partial x_i}dfrac{d x_i}{dxi}dfrac{dxi}{d y}-xidfrac{d x_1}{d y}=sum_{i=1}^n dfrac{partial f}{partial x_i}dfrac{d x_i}{d y}-xidfrac{d x_1}{d y}=dfrac{d f}{d y}-xidfrac{d x_1}{d y}=0.$$
But I am not sure. Indeed, I’d like to know how state this only from the fact of homogeneous.
Thanks so much.
Assume that $f:textbf{R}^nrightarrowtextbf{R}$ and $f=f(x_1,x_2,ldots,x_n)$ is a function of $n$ variables. By saying that $x_i=x_i(xi)$, then $C:overline{x}={x_1(xi),x_2(xi),ldots,x_n(xi)}$, $xiintextbf{R}$, then $C$ is one dimentional object in $textbf{R}^n$ and hence $C$ is a curve of $textbf{R}^n$. Then $$ frac{df}{dxi}=sum^{n}_{k=1}frac{partial f}{partial x_k}frac{dx_k}{dxi} $$ is the derivative of $f$ allong $C$ (or total derivative of $f$ allong the curve $C$). You also have the equation: $$ frac{df}{dxi}-xifrac{dx_1}{dxi}=0Leftrightarrow sum^{n}_{k=1}frac{partial f}{partial x_k}frac{dx_k}{dxi}=xifrac{dx_1}{dxi} tag 1 $$ If $xi=u y$, then $frac{dxi}{dy}=u$. Hence $$ frac{df}{dxi}-xifrac{dx_1}{dxi}=0Leftrightarrow frac{df}{dy}frac{dy}{dxi}-xifrac{dx_1}{dy}frac{dy}{dxi}=0Leftrightarrow frac{df}{dy}frac{1}{u}-xifrac{dx_1}{dy}frac{1}{u}=0Leftrightarrow $$ $$ frac{df}{dy}-xifrac{dx_1}{dy}=0.tag 2 $$ This answer your first question about the change of variables.
About the homogenicity
However if $f$ is homogeneous function then we have even more
If the function $f$ is homogeneous of degree $lambda$. Then setting $x_i=uy_i$ in equation (1) we have, (knowing that $f(x_1,x_2,ldots,x_n)$ and $(x_1,x_2,ldots,x_n)rightarrow x_1$ are homogeneous i.e. $f(uy_1,uy_2,ldots,uy_n)=u^{lambda}f(y_1,y_2,ldots,y_n)$ and $(ux_1)=ux_1$ of degree 1): $$ sum^{n}_{k=1}frac{partial f}{partial x_k}(uy_1,uy_2,ldots,uy_n)left(ufrac{dy_k}{dxi}right)-xileft(ufrac{dy_1}{dxi}right)=0Leftrightarrow $$ $$ u^{lambda-1}sum^{n}_{k=1}frac{partial f}{partial y_k}(y_1,y_2,ldots,y_n)left(ufrac{dy_k}{dxi}right)-xi ufrac{dy_1}{dxi}=0 $$ $$ u^{lambda-1}sum^{n}_{k=1}frac{partial f}{partial y_k}(y_1,y_2,ldots,y_n)frac{dy_k}{dxi}-xi frac{dy_1}{dxi}=0.tag 3 $$ (That is because when $f(x_1,x_2,ldots ,x_n)$ is homogeneous of degree $lambda$, then $frac{partial f}{partial x_{j}}$ is homogeneous of degree $lambda-1$ i.e. $frac{partial f}{partial x_j}(uy_1,uy_2,ldots,uy_j,ldots,uy_n)=u^{lambda-1}frac{partial f}{partial x_j}(y_1,y_2,ldots,y_n)$). Hence when $lambda=1$, then (3) becomes: $$ sum^{n}_{k=1}frac{partial f}{partial y_k}frac{dy_k}{dxi}-xifrac{dy_1}{dxi}=0.tag 4 $$ Hence if $f=f(x_1,x_2,ldots,x_n)$ is homogeneous of degree 1, then equation (1) is homogeneous PDE (invariant under any transformation of variables of the form $x_i=uy_i$, $i=1,2,ldots,n$).
Correct answer by Nikos Bagis on December 1, 2020
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