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I was solving a problem and I encountered the following partial differential equation:

$$\sum_nA_n\frac{\partial f}{\partial x_n}=B$$ where $A_n$ and B are constants and $f:\mathbb{R}^n \longrightarrow \mathbb{R}$ is the function we are trying to find. I have some knowledge on Ordinary differential equations, so I have no idea how to solve this or if it's even possible to solve it. Does such function $f$ exists? If so how do I solve the equasion?

Eduardo Magalhães
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    The left hand side is a directional derivative. You may want to check the method of characteristics. In your particular case, $f$ is any function that grows linearly in the direction of the vector of $(A_1,A_2,\dots)$. You need some initial data, the value of $f$ on some hyper surface. – GReyes May 02 '20 at 01:48
  • one of my initial conditions is that $f(0,...,0)=0$, so that would mean that B=0, simplifying the equation to: $\sum_nA_n\frac{\partial f}{\partial x_n}=0$ – Eduardo Magalhães May 02 '20 at 11:33
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    To get a unique solution, you need to specify $f$ not just at a point but on a whole hypersurface, for example on some $x_i=0$ (a coordinate hyperplane). – GReyes May 03 '20 at 07:30
  • @GReyes But without specifying is it possible to get a general solution? – Eduardo Magalhães May 03 '20 at 16:14
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    $f(0)=0$ does not mean that $B=0$. for example, in 1 dimension, i.e. $n=1$, your differential equation simplifies to $Af'(x)=B$, i.e. $f'(x)=B/A$, so $f(x)=Bx/A+C$ for some constant $C$. – Botond May 03 '20 at 21:06
  • You're right @Botond – Eduardo Magalhães May 03 '20 at 21:24

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This is a particular case of the method of characteristics, which works for very general equations of order one, even non-linear. We can assume that $A=(A_1,A_2,\dots A_n)\neq 0$, Otherwise any differentiable function would be a solution if $B=0$ or there is no solution if $B\neq 0$. Suppose $A_1\neq 0$. The left hand side of your equation is just the derivative of $f$ along a straight line with direction vector $A$ through the given point. This is just the chain rule, $$ \frac{df(x(t))}{dt}=\nabla f\cdot \frac{dx}{dt}=\nabla f\cdot A. $$ Therefore, given a point $x$, you consider a line through it with direction vector $A$ and you move along it until you hit a point where $f$ is known and integrate the resulting ODE from there. Under the assumption $A_1\neq 0$, if you know the value of $f$ on the coordinate plane $x_1=0$, the intersection of that line with the coordinate plane is $(0,y_2,y_3,\dots y_n)$ such that $$ x-(0,y_2,y_3\dots y_n)=At $$ for some $t$, which gives you $t=x_1/A_1$ and $y_i=A_it=A_ix_1/A_1$. According to your equation, $f$ increases linearly in $t$, with rate $B$. That is, $$ f(x)=f(0,y_2,y_3\dots y_n)+Bt=f(0,A_2x_1/A_1,A_3x_1/A_1,\dots A_nx_1/A_1)+ $$ $$ +bx_1/A_1 $$ which is an explicit formula for the solution.

If you only know $f(0)=0$ you have infinitely many degrees of freedom for your solution. You have as many as (differentiable) functions $g$ of $n-1$ variables satisfying $g(0)=0$.

The same argument applies in general, choosing a hyperplane transversal to the vector $A$ (such hyperplanes are called non-characteristic in the general theory, and are the ones where the boundary values determine a unique solution).

GReyes
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