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.
Best Answer
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 $u(x,y,z)=f(y_0,z_0)=f\left(\dfrac{y}{x},\dfrac{z}{x}\right)$