Solve the 3 Dimension first order PDE $xu_x+yu_y+zu_z=0$

Question

This is a 3D Higher Dimension first order PDE and I am kind of confused because I am taking the partial derivatives and for some reason nothing is being cancelled out.

Question: Solve the PDE $xu_x+yu_y+zu_z=0$

Attempt: Here are our characteristic equations

$ a = x, b = y,$ and $ c = z $

$\frac{dy}{dx} = \frac{y}{x}$ and $ \frac{dz}{dx} = \frac{z}{x}$

Ok. I am going to integrate both of these in terms of x and since both of them have x, I should have a problem.

$ Y_x = \frac{1}{x}y $ and $Z_x =\frac{1}{x}z$

$ Y = \ln x y + \alpha$ and $ Z = \ln x z + \beta$

So now I need my alpha and beta by itself and then assign new variables..let's make it $\bar x = \alpha$ and $\bar y = \beta $.

$ Y -\ln x y = \bar x $

$ Z -\ln x z= \bar y$

$ z = \bar z$

Ok. Now I am going to take partial derivatives on all of the terms. I should see some things cancel out

$x[u_xX_x+u_yY_x+u_zZ_x]+y[u_xx_Y+u_yY_y+u_zZ_y]+z[[u_xX_z+u_yY_z+u_zZ_z] =0$

$ Y -\ln x y = \bar x $

$X_x = \frac{-1}{x}y$

$ X_y = 1-\ln x$

$X_z=0$

---

$ Z -\ln x z= \bar y$

$Y_x =\frac{-1}{x}z $

$Y_y =0 $

$Y_z= 1 -\ln x$

---

$ z = \bar z$

$Z_x = 0 $

$Z_y =0$

$Z_z =1$

$x[u_x(\frac{-1}{x}y)+u_y(\frac{-1}{x}z )]+y[u_x(1-\ln x)]+z[u_y(1 -\ln x)+u_z] =0$

$[-u_x(y)-u_yz )]+y[u_x(1-\ln x)]+z[u_y(1 -\ln x)+u_z] =0$

But I am stuck here. Did I take the partial derivatives wrong? I thought that I when I am dealing with multiplication type derivatives whatever I leave alone and take partial derivative in terms of a variable it should still stay. Like for example.

$w = 6sinxcosy$

$w_y = -6sinxsiny$ or is it $w_y = -6siny$? I don't think it's the second one.

Answer 1

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$

$\dfrac{dy}{dt}=y$ , letting $y(0)=y_0$ , we have $y=y_0e^t=y_0x$

$\dfrac{dz}{dt}=z$ , letting $z(0)=z_0$ , we have $z=z_0e^t=z_0x$

$\dfrac{du}{dt}=0$ , letting $u(0)=f(y_0,z_0)$ , we have $u(x,y,z)=f(y_0,z_0)=f\left(\dfrac{y}{x},\dfrac{z}{x}\right)$

Answer 2

This PDE is equivalent to that $$ \frac{d}{dt}u(at,bt,ct)=0,\quad\text{for all $a,b,c\in\mathbb R$}, $$ and it means that $u:\mathbb R^3\smallsetminus\{0\}\to \mathbb R$, is constant on every straight half-line starting from the origin, and not containing it.

If fact, the solution is $$ f(\boldsymbol{x})=g\left(\frac{\boldsymbol{x}}{\|\boldsymbol{x}\|}\right), $$ where $g$ is differentiable in $\mathbb R\smallsetminus\{0\}$.

Answer 3

After change variables $$\xi=\log(x),\quad \eta=\log(y)\quad \zeta=\log(z)$$

we get equation $$u_\xi+u_\eta+u_\zeta=0$$ with solution $$u=f(\eta-\xi,\zeta-\xi)=f(\log(y)-\log(x),\log(z)-\log(x))\\= f\left(\log(\frac{y}{x}),\log(\frac{z}{x})\right)\\=F\left(\frac{y}{x},\frac{z}{x}\right) $$