Given this PDE
$$xu_x + yu_y + zu_z=0$$
It is asked, in some old paper-and-pencil notes I am reading, to solve it.
1) Firstly, finding the general solution
2) Then, finding the particular solution under the condition $u(1,y,z)=3yz$
3) Then, finding (again from scratch) the particular solution under the condition $u(1,y,1)=y^2$
Now, the first point is simple, because I've used the characteristics method, finding
$$\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{z}$$
so, the system
$$ \begin{cases} \frac{dx}{x}=\frac{dy}{y}\\ \frac{dy}{y}=\frac{dz}{z} \end{cases} $$
leads to
$$ \begin{cases} \zeta_1:=\pm e^{K_1}=\frac{x}{y}\\ \zeta_2:=\pm e^{K_2}=\frac{y}{z} \end{cases} $$
so
$$ u(x,y,z)=\Phi(\zeta_1,\zeta_2)=\Phi\left(\frac{x}{y},\frac{y}{z}\right) $$
solution that I've just verified by chain rule.
But, what about the second point?
I've tried to solve like this:
$$3yz=u(1,y,z)=\Phi\left(\frac{1}{y},\frac{y}{z}\right)$$
$$\zeta_1=\frac{1}{y} \rightarrow 3 y^2 z\zeta_1=3yz$$
$$\zeta_2=\frac{y}{z} \rightarrow 3 z^2 \zeta_2=3yz$$
obtaining something that I am not able to recombine.
Can anyone help me to see if the passages here are right and how to complete?
Thanks in advance.
Hint: For the second question:
$$u = \Phi(x/y,y/z)=F(y/x,z/y)$$
Now use the $u(1,y,z)=F(y,z/y)=3yz$
Use the substitution $v=z/y \implies z = vy \implies F(y,v)=3y^2v$
Hence, we obtain $u=3(y/x)^2(z/y)=3yz/x^2.$
Edit: The goal is to transform the function such that it is $F(a,b)$.
Let us have a look at the second part of the question. We know that $$u(x,y,z)=\Phi(x/y,y/z)=F(y/x,y/z)$$
I inverted the first argument such that the variables that we will fix are now in the denominator. Using $u(1,y,1)=y^2$ we obtain:
$$u(1,y,1)=F(y,y)=y^2=y\cdot y.$$
Hence, the function $F(\alpha, \beta)$ just multiplies is arguments and we obtain:
$$u(x,y,z)=F(y/x,y/z)=\frac{y^2}{xz}.$$