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I have the following PDE: \begin{equation} -yu_x + xu_y = 0 \quad\text{where } u(0, y) = f(y) \end{equation} I derived a solution as follows: \begin{align} -yu_x + xu_y =& 0 \\ \iff& \nabla u(x,y)\dot \langle -y, x\rangle = 0 \\ \implies& \frac{dy}{dx} = \frac{-x}{y} \\ \iff& ydy = -xdx \\ \iff& \int ydy = \int -xdx \\ \iff& y^2 = -x^2 + c_2 \\ \iff& y^2 + x^2 = c_2 \end{align} Since $u(x,y)$ is constant along the ODE $\frac{dy}{dx}$, 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 PDE. Specifically, any function $f$ should satisfy the pde. 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 \begin{equation} u(x, y) = -yf_x2x + xf_y2y = 2xyf_y - 2xyf_x \end{equation} We need the above to equal $0$. The $2xy$ and $-2xy$ give me evidence that it should cancel, and that my calculus is off... 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 PDE?

Thanks for all the help!

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  • $\begingroup$ Did you look at my question at the bottom? I tried to make this more clear by making the question bold. I think this question is no more than a partial derivative question... $\endgroup$ – CodeKingPlusPlus Apr 13 '14 at 6:01
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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?

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