- #1
geoffrey159
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Homework Statement
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.
Homework Equations
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?