Solve this PDE $u_x+u_y+2u_z=0$,$u(1,y,z)=yz$ (problem with initial conditions/particular solution)

Question

Find the particular solution of the part differential equation $u_x+u_y+2u_z=0$ that satisfies the data $u(1,y,z)=yz$

So:

$\frac {dx}{1}=\frac {dy}{1} = \frac{dz}{2}$

$\int dx=\int dy \Rightarrow x=y+C_1 \Rightarrow C_1=x-y$

$2\int dx=\int dz \Rightarrow 2x=z+C_2 \Rightarrow C_2=2x-z$

General solution: $u(x,y,z)=F(x-y,2x-z)$

I use a initial condition: $yz=F(1-y,2-z) \Rightarrow $ Now, I don't know what to do next?

How can I get particular solution for this condition?

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PS: We only did this example in class earlier: $\\ u(x,y)=F(2x-3y),u(0,y)=81y^{4}$ $81y^{4}=F(2*0-3y) \Rightarrow 81y^{4}=F(-3y) \Rightarrow t=-3y \Rightarrow F(t)=t^{4} \Rightarrow u=(2x-3y)^{4}$

I try use wolfram alpha, which can give general solution with this prompt D[u[x,y,z],x]+D[u[x,y,z],y]+2*D[u[x,y,z],z]=0, but when I write this prompt D[u[x,y,z],x]+D[u[x,y,z],y]+2*D[u[x,y,z],z]=0,u(1,y,z)=y*z, doesn't return a particular solution, there is another way to do it (with wolfram)?

Answer 1

Hint: Let $y'=1-y,z'=2-z$. Now rewrite $F(1-y,2-z)=yz$ in terms of $y',z'$.

Answer 2

Here is another solution. The characteristic curves satisfy the equations $$ \dot{x}=1,\qquad\dot{y}=1,\qquad\dot{z}=2,\qquad\dot{u}=0, \tag{1} $$ which have as general solution \begin{align} x&=x_0+t, \tag{2} \\ y&=y_0+t, \tag{3} \\ z&=z_0+2t, \tag{4}\\ u&=u_0. \tag{5} \end{align} The particular solution $u(1,y,z)=yz$ corresponds to the initial conditions $x_0=1$ and $u_0=y_0z_0$. The latter condition and Eqs. $(3)$-$(5)$ imply $u=(y-t)(z-2t)$; the former and Eq. $(2)$ imply $t=x-1$. Combining these results we finally obtain $$ u(x,y,z)=(y-x+1)(z-2x+2). \tag{6} $$