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I have consulted other sources, but the relevant ones have used notation that I am not entirely comfortable with. I'm told that generalizing this procedure is straightforward, but I find myself getting confused/being unsure about what I am doing.

I want to get the general solution of the PDE $u_{x} + yu_{y} + zu_{z} = 0$

Attempt:

The characteristic curves should be subject to the relations $\frac{dx}{1} = \frac{dy}{y} = \frac{dz}{z}$. To figure out what the characteristic curves/surfaces are here, I believe I should consider two separate relations among these.

I have first considered $dx = \frac{dy}{y}$ to get that $\ln{y} = x + C$ for some constant C. Since the function $u$ should be constant on this characteristic curve, I solve for $C = \ln{y} - x$.

Next I consider the relation $\frac{dy}{y} = \frac{dz}{z}$ and similarly get that $C_{2} = \ln{(y-z)}$.

From this, I believe that I can deduce $u(x,y,z) = f(\ln{(y)}-x, \ln{(y-z)})$ is a general solution.

Do I have the right idea?

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1 Answer 1

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Note that your second relation is false. The equation $dy/y = dz/z$ implies that $C_2 = y/z$ and hence:

$$ u = f(C_1,C_2) = f(\log{y} -x , y/z),$$

since the characteristics method also tells us that the fractions are also equal to $ du/0$ and hence $u = \text{const}$.

You can check that this solution satisfies the PDE taking into account that:

$$ \frac{\partial f}{\partial x_i} = \frac{\partial f}{\partial C_1} \frac{\partial C_1}{\partial x_i} + \frac{\partial f}{\partial C_2} \frac{\partial C_2}{\partial x_i}, \quad x_i = \{x,y,z\}$$

Hope this helps.

Cheers!

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