1
$\begingroup$

\begin{cases} xu_x+yu_y+zu_z=4u\\ u(x,y,1)=xy\\ \end{cases}

Using Lagrange method we get:

$$\frac{x}{y}=c_1, \frac{y}{z}=c_2, \frac{z}{\sqrt[4]{u}}=c_3$$

So the general solution is $$u=\frac{z^4}{c_3}$$?

$\endgroup$
  • 2
    $\begingroup$ Any homogeneous function of degree four satisfies that PDE, so how about $xyz^2$? $\endgroup$ – Lord Shark the Unknown Sep 9 '19 at 20:22
  • $\begingroup$ @LordSharktheUnknown a second thought is $F(\frac{x}{y},\frac{y}{z})=\frac{z^4}{u}$ and to use $u(x,y,1)=xy$ $\endgroup$ – newhere Sep 9 '19 at 20:25
2
$\begingroup$

Converting to spherical coordinates, we get

$$ru_r = 4u \implies u = f(\theta,\phi)r^4$$

Then plugging in our boundary condition at $r\cos\theta = 1$ and $xy=r^2\sin^2\theta\sin\phi\cos\phi$, we can get

$$f(\theta,\phi)r^4 = r^2\sin^2\theta\sin\phi\cos\phi\cdot(1)=r^2\sin^2\theta\sin\phi\cos\phi\cdot (r^2\cos^2\theta)$$

$$\implies f(\theta,\phi) = \cos^2\theta\sin^2\theta\sin\phi\cos\phi$$

by canceling out the $r^4$ on both sides. In other words when we convert back to Cartesian we get the solution

$$u(x,y,z) = xyz^2$$

$\endgroup$

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.