# 3 Variable Method of Characteristics

Is it possible to solve the below 3-Dimensional Quasli-Linear partial differential equation using the method of characteristics?

$$(y-z)u_x+(z-x)u_y+(x-y)u_z=0 \quad u(x,y,0)=xy$$

So first we want to convert the above PDE into a system of ODE's to solve.

Hence the equivalent system of ODE's is given by:

$$\dot{X}=y-z \quad X(\sigma,0)=\sigma$$ $$\dot{Y}=z-x \quad Y(\sigma,0)=\sigma$$ $$\dot{Z}=x-y \quad Z(\sigma,0)=0$$ $$\dot{U}=0 \quad U(\sigma,0)=X(\sigma,0)Y(\sigma,0)=\sigma^2$$

I am not sure how to determine $$X(\sigma,s)$$, $$Y(\sigma,s)$$ and $$Z(\sigma,s)$$ however. Am I going about this the right way? I would appreciate any help/hints if possible.

As a note, I also tried solving the above PDE using the matrix exponential, but the system of equations generated was too hard to solve.

$$(y-z)u_x+(z-x)u_y+(x-y)u_z=0 \quad u(x,y,0)=xy$$ Charpit-Legendre equations : $$\frac{dx}{y-z}=\frac{dy}{z-x}=\frac{dz}{x-y}=\frac{du}{0}$$ A first characteristic equation comes from $$du=0$$ : $$u=c_1$$ A second characteristic equation comes from

$$\frac{dx}{y-z}=\frac{dy}{z-x}=\frac{dz}{x-y}=\frac{dx+dy+dz}{(y-z)+(z-x)+(x-y)}=\frac{dx+dy+dz}{0}$$ $$x+y+z=c_2$$

A third characteristic equation comes from

$$\frac{dx}{y-z}=\frac{dy}{z-x}=\frac{dz}{x-y}=\frac{ydx+xdy+zdy+ydz+xdz+zdx}{y(y-z)+x(z-x)+z(z-x)+y(x-y)+x(x-y)+z(y-z)}=\frac{d(xy+yz+zx)}{0}$$ $$xy+yz+zx=c_3$$ The general solution of the PDE expressed on the form of implicite equation $$c_1=F(c_2,c_3)$$ is : $$\boxed{u(x,y,z)=F(x+y+z \:,\: xy+yz+zx)}$$ where $$F(X,Y)$$ is an arbitrary function of two variables $$\begin{cases}X=x+y+z\\Y=xy+yz+zx\end{cases}$$

Among the infinite set of solutions they are some sub-sets such as $$u(x,y,z)=f(xy+yz+zx)$$ $$u(x,y,z)=g(x+y+z)$$ and of course many others, for example $$u(x,y,z)=h(x^2+y^2+z^2)$$ because $$(x^2+y^2+z^2)=(x+y+z)^2-2(xy+yz+zx)-1$$ .

$$f,g,h$$ are arbitrary functions.

Boundary condition :

$$u(x,y,0)=xy=F(x+y\:,\:xy)=f(xy)$$

The function $$f(X)$$ is determined : $$f(X)=X$$ that we bring back into the above general solution where $$X=xy+yz+zx$$. The function : $$\boxed{u(x,y,z)=xy+yz+zx}$$ satisfies the PDE and the boundary condition.