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Solving another pde

  1. Nov 27, 2015 #1
    1. The problem statement, all variables and given/known data

    Consider the following pde: ##\sum_{i=1}^n c_i f_{x_i} = 0##,

    where all the ##c_i## are real valued and ##c_1 \neq 0##, and ##f## is the unknown defined from ##\mathbb{R}^n\to \mathbb{R}## and of class ##{\cal C}^1(\mathbb{R}^n,\mathbb{R})##

    Show there exists an invertible ##n\times n## real matrix ##M## such that such that the change of variable ## X = M x## simplifies the pde to ##F_{X_1} = 0##, where ##F(X) = f(x)##. Then what is the general solution to this pde.


    2. Relevant equations


    3. The attempt at a solution

    Assuming there exists such a solution, and such a matrix. Then the change of variable is of class ##{\cal C}^1(\mathbb{R}^n,\mathbb{R}^n)##, bijective, and its inverse has same regularity. So ##F## has the same regularity as ##f## and its partial derivatives are defined. Wa have:

    ## 0 = \sum_{i=1}^n c_i f_{x_i}(x) = \sum_{i=1}^n c_i (\sum_{k=1}^n \frac{\partial X_k}{\partial x_i} F_{X_k}(X)) = \sum_{i=1}^n c_i (\sum_{k=1}^n m_{ki} F_{X_k}(X)) = \sum_{k=1}^n (\sum_{i=1}^n c_i m_{ki}) F_{X_k}(X) ##

    So we must choose ##M## such that ## M_{1,\bullet} \ \vec c =1## and ##M_{k,\bullet}\ \vec c = 0## for ##k\neq 1##. So we can choose

    ##M = \begin{pmatrix}
    \frac{1}{c_1} & 0 & \ldots & 0 \\
    \frac{-c_2}{c_1} & 1 & 0 & \vdots \\
    \vdots &0 & \ddots & \ldots \\
    \frac{-c_n}{c_1} & \ldots & \ldots & 1
    \end{pmatrix} ##

    It's determinant is ##\frac{1}{c_1}\neq 0 ## so ##M## is invertible and with such a matrix, ##F_{X_1} = 0##.

    The solutions to this pde have the form ## F(X) = \lambda(X_2,...,X_n) ##, where ##\lambda \in {\cal C}^1(\mathbb{R}^ {n-1},\mathbb{R})##, so returning to ##f##, then

    ##f(x) = \lambda(M_{2,\bullet}\ x,...,M_{n,\bullet}\ x) = \lambda( x_2 - \frac{c_2}{c_1} x_1 , ..., x_n - \frac{c_n}{c_1} x_1)##

    Are you ok with that proof?
     
  2. jcsd
  3. Nov 28, 2015 #2

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    Nobody?

    I checked the math and it looks correct. A second opinion would be appreciated though.
     
  4. Nov 29, 2015 #3
    Thank you Samy for looking at it.
     
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