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

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.

$\dfrac{dx}{dt}=x$ , 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 $u(x,y,z)=f(y_0,z_0)=f\left(\dfrac{y}{x},\dfrac{z}{x}\right)$

• I wish I knew that shortcut...too bad I am stuck with this crazy book...otherwise I would be done... hey what would be the solution if we have the pde $u_x-u_y+u=z$? @doraemonpaul then I can practice and master the shortcut instead of the bs from the textbook. Sep 21, 2014 at 9:38

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 $\mathbb R\smallsetminus\{0\}$.

• whoa hold it! I don't know any of that. I have to use whatever this crazy book says. I can link it to you if you want to see the example I can scan the book page. I have to do it that way -____-! Sep 21, 2014 at 8:37
• hmmm $xdx+ydy+zdz=0$.... I have another question... Can we just use separation of variables on this ? Sep 21, 2014 at 8:41

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 $$u=f(\eta-\xi,\zeta-\xi)=f(\log(y)-\log(x),\log(z)-\log(x))\\= f\left(\log(\frac{y}{x}),\log(\frac{z}{x})\right)\\=F\left(\frac{y}{x},\frac{z}{x}\right)$$