\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 $$u=\frac{z^4}{c_3}$$?

  • 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

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

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


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