# Solving linear first order PDE

Let $v = v(x,y)$ be a scalar fields on $\mathbb R^2$. Find the general solution for the following equation and determine the curves along which $v$ = constant $$(x\partial_x + y\partial_y)v = 0$$

For the first part where finding the general solution for the equation I have tried:

$(x\partial_x + y\partial_y)v$ = $(1 + 1)v$ = $2v$

$(x\partial_x + y\partial_y)v = 2v = 0$

$2u(x,y)=0$

General solution in $D = \mathbb R^2$ is $u = f(x) + f(y)$

For the second part where I have to find the curves along I have tried:

Rate of change of $u(x,y)$ with $\alpha$ as we move along the curve is

$\frac{du}{d\alpha} = u_x\frac{dx}{d\alpha} + u_y\frac{dy}{d\alpha}$ = $a(x,y)u_x +b(x,y)u_y$ which is the directional derivative in the direction of $(a,b)$ at $(x,y)$.

I am very new to PDE, having lots of trouble solving the basic questions. Any

help will be appreciated.

First, let us rewrite the above linear transport equation as \begin{align} \partial_x\nu+\frac{y}{x}\partial_y\nu=0 \ \ (\ast) \end{align} Using the method of characteristic, we could further rewrite $(\ast)$ as follows \begin{align} \frac{d}{dx}\nu(x, y(x))= \partial_x\nu+ y'(x)\partial_y \nu = \partial_x\nu+\frac{y}{x}\partial_y\nu=0, \end{align} i.e. we have the ode \begin{align} y' = \frac{y}{x} \ \ \Rightarrow \ \ y = Cx. \end{align} Let us impose the artifical initial condition $y(1) = y_0$ and $u(1, y_0) = f(y_0)$ . Hence it follows \begin{align} y = y_0x \ \ \Rightarrow \ \ y_0 = \frac{y}{x}. \end{align} which means \begin{align} \nu(x, y) = \nu\left(x, y_0x\right) = f(y_0) = f\left(\frac{y}{x} \right). \end{align}