0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.