Confused solving quasilinear PDE

Question

The equation to solve is:

$\displaystyle x\frac{\partial u}{\partial x}+ \displaystyle y\frac{\partial u}{\partial y}+ \displaystyle z\frac{\partial u}{\partial z}= \displaystyle xyz$

Applying the Lagrange-Charpit equations to $x$ and $y$, and $y$ and $z$ we trivially get $\frac{x}{y}=C_1$ and $\frac{z}{y}=C_2$. However I have a problem solving

$(1) \displaystyle \frac{x}{dx}=\frac{xyz}{du}$

This leads to

$\displaystyle xdu=xyzdx$, and

$\displaystyle du=yzdx$, integrating

$\displaystyle u+C_3=xyz$

This would mean that the solution to the original problem is

$F(\displaystyle \frac{x}{y},\frac{z}{y}, \displaystyle u-xyz)=0$ with an arbitrary $F$.

And indeed $\displaystyle u=xyz$ satisfies $(1)$. But the third function in the actual solution should be $3u=xyz$ which in turn does not satisfy $(1)$.

What am I missing?

Answer 1

Your Lagrange equations are statements about the relation of x,y,z,u along a curve in 4-space. This is somewhat hidden in the notation, but is brought out clearly in the solution by doraemonpaul. So when you say that $x/y=C_1$, that means along one of these curves; for the pde, x and y are still going to be independent variables. So when you say that your equation (1) leads to $du = yzdx$, that is still a statement along the same curve. So your deduction $u+C_3 = xyz$ is not correct. You could write instead $du = yzdx=\frac{x}{C_1}\frac{C_2x}{C_1}dx$, and integrate to get $u = \frac{1}{3}\frac{x}{C_1}\frac{C_2x}{C_1}x+C_3 = \frac{1}{3}xyz+C_3$. In general I think it is better to include the $t$ parameter as doraemonpaul did, because the idea of the method of characteristics is to use these parametrized curves to transfer Cauchy data into the space.

Answer 2

$$\displaystyle x\frac{\partial u}{\partial x}+ \displaystyle y\frac{\partial u}{\partial y}+ \displaystyle z\frac{\partial u}{\partial z}= \displaystyle xyz$$ Let $u(x,y,z)=U(x,y,z)+\frac{1}{3}xyz$ $$\displaystyle x\frac{\partial U}{\partial x}+ \displaystyle y\frac{\partial U}{\partial y}+ \displaystyle z\frac{\partial U}{\partial z}= \displaystyle 0$$ Let $X=e^x$ ; $Y=e^y$ ; $Z=e^z$ $$\displaystyle \frac{\partial U}{\partial X}+ \displaystyle \frac{\partial U}{\partial Y}+ \displaystyle \frac{\partial U}{\partial Z}= \displaystyle 0$$ which the wellknown solution is : $U(X,Y,Z)=F(aX+bY+cZ)$ with $a+b+c=0$

$a, b, c$ are constant and $F$ is any differentiable function. $$U(x,y,z)=F(\ln(x^a)+\ln(y^b)+\ln(z^c))=F(ln(x^ay^bz^c))$$ Since $F(\space)$ is any function, $F(ln(\space))=f(\space)$ insofar $f$ is any function. $$U(x,y,z)=f(x^ay^bz^c)$$

$$u(x,y,z)=\frac{1}{3}xyz+f(x^ay^bz^c)$$ with $a+b+c=0$ , $a,b,c$ are constant anf $f$ is any differentiable function.

Answer 3

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 $z=z_0e^t=z_0x$

$\dfrac{du}{dt}=xyz=y_0z_0e^{3t}$ , we have $u(x,y,z)=\dfrac{y_0z_0e^{3t}}{3}+f(y_0,z_0)=\dfrac{xyz}{3}+f\left(\dfrac{y}{x},\dfrac{z}{x}\right)$