I have the following P.D.E.: $$-yu_x + xu_y = 0 \quad\text{where } u(0, y) = f(y).$$ I derived a solution as follows: \begin{align} -yu_x + xu_y ={}& 0 \\ \iff{}& \nabla u(x,y) \cdot \langle -y, x\rangle = 0 \\ \implies{}& \frac{\Bbb dy}{\Bbb dx} = \frac{-x}{y} \\ \iff{}& y\,\Bbb dy = -x\,\Bbb dx \\ \iff{}& \int y\,\Bbb dy = \int -x\,\Bbb dx \\ \iff{}& y^2 = -x^2 + c_2 \\ \iff{}& y^2 + x^2 = c_2. \end{align} Since $u(x,y)$ is constant along the O.D.E. ${\Bbb dy}/{\Bbb dx}$, we have: \begin{align} u(x,y) &= c_1 \\ &= f(c_2) \quad\text{for some function $f$} \\ &= f(y^2+x^2). \end{align} I want to check that this satisfies the P.D.E. Specifically, any function $f$ should satisfy the P.D.E. My calculus is a bit rusty, and I am not exactly sure how to do this. Here is my reasoning:
Since $u(x,y) = -yu_x + xu_y = 0$ we have to substitute in for $f$ which yields $$u(x, y) = -yf_x2x + xf_y2y = 2xyf_y - 2xyf_x.$$ We need the above to equal $0$. The $2xy$ and $-2xy$ give me evidence that it should cancel, and that my calculus is off$\ldots$ What is not clear to me is that, we don’t know what $f$ is, hence I can’t find out what $f_x$ or $f_y$ are. Though, the problem does look symmetrical, and I could see a potential solution involving polar coordinates, but I’m not quite sure how to solve it in this way either.
- How can I verify that the above is indeed equal to $0$ and hence satisfies the P.D.E.?
Thanks for all the help!
Since we conjecture $u(x,y)=f(x^2+y^2)$, let us (carefully) apply calculus and verify $-yu_x+xu_y=0$.
Note (by chain rule), $$ u_x = f'(x^2+y^2)*2x, \qquad u_y = f'(x^2+y^2) * 2y. $$
Thus, $-yu_x+xu_y=-2xyf'+2xyf'=0$.
So, you had it, except a small detail with the chain rule. Another way to have caught the mistake -- $f$ is a function of a single variable, so what does $f_x$ or $f_y$ mean?