Solving another pde

1. Nov 27, 2015

geoffrey159

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. Nov 28, 2015

Samy_A

Nobody?

I checked the math and it looks correct. A second opinion would be appreciated though.

3. Nov 29, 2015

geoffrey159

Thank you Samy for looking at it.