# Homework Help: Variable Coefficient PDEs and Continuity of the General Solution

1. Sep 23, 2012

### Tsunoyukami

Variable Coefficient PDEs

My homework question:

"Find the general solution of $xu_{x} + 4yu_{y} = 0$ in ${(x,y)\neq(0,0})$; when is this solution continuous at (0,0)?"

$\frac{dx}{dy} = \frac{x}{4y}$

$\frac{dx}{x} = \frac{dy}{4y}$

Integrating both sides, we find:

$lnx + c = \frac{1}{4} lny + d$

$lnx + c = lny^{\frac{1}{4}} + d$

$lnx + b= lny^{\frac{1}{4}}$, where $b = c-d$

$e^(lnx + b) = e^(lny^{\frac{1}{4}})$

$x e^b = y^{\frac{1}{4}}$, but $e^b$ is a constant, so we say $e^b = C$

$Cx = y^{\frac{1}{4}}$

$(Cx)^4 = y$

$u(x,y) = u(x,[Cx]^4)$
$\frac{du(x, [Cx]^4)}{dx} = \frac{du}{dx} + \frac{du}{dy}\frac{dy}{dx} = \frac{du}{dx} + \frac{du}{dy}\frac{4y}{x} = xu_{x} + 4yu_{y} = 0$; this is the initial question, so we know that this works. Now we know:

$u(x,y) = u(x, [Cx]^4) = f(C)$, but we can solve for C in terms of x and y to find $C = \frac{y^(\frac{1}{4})}{x}$ so we have the general solution:

$u(x,y) = f(\frac{y^(\frac{1}{4})}{x})$

In terms of continuity is this just asking when the argument of the general solution f is equal to 0? If so, then the general solution would be discontinuous at any point $(x,y) = (0,y)$ or any point along the vertical line described by $x = 0$. How can I make this function continuous at $(x,y) = (0,0)$, or better - how can I make it continuous at this point given that f is an arbitrary function; do I use the notion that f is continuous at this point as a boundary condition?

EDIT

I think it seems reasonable that I would have to take the limit of f as $(x,y)$ approaches $(0,0)$. If I find that the limit along two separate paths do not agree then the function is not continuous at this point...(or that just means that the limit does not exist - just because the limit at a point exists does not necessarily mean that the function is continuous, such as at a point discontinuity, correct?)

Last edited: Sep 23, 2012