How to solve the characteristic equations $\frac{dx_1}{x_1} = \frac{dx_2}{x_2} = \frac{dV}{2V}$?

Question

How to solve the following $$\frac{dx_1}{x_1} = \frac{dx_2}{x_2} = \frac{dV}{2V},$$ where $V = V(x_1,x_2)$.

My effort is that I take $$\int dx_1/x_1 = \int dV/2V + C \, \Rightarrow \, V = C_1(x_2) x_1^2$$ and $$\int dx_2/x_2 = \int dV/2V + D \, \Rightarrow \, V = D_1(x_1) x_2^2.$$

Since they are the same $V$, we get $$C_1(x_2)x_1^2=D_1(x_1)x_2^2 \, \Rightarrow \, C_1(x_2) = (\frac{x_2}{x_1})^2D_1(x_1).$$ So we get $$V(x_1,x_2) = (\frac{x_2}{x_1})^2D_1(x_1) x_1^2$$

Am I on the correct way?

Thanks!

Answer 1

Probably you want to solve the PDE : $$x_1\frac{\partial V}{\partial x_1}+x_2\frac{\partial V}{\partial x_2}=2V,$$ and you correctly wrote the Charpit-Lagrange system of characteristic ODEs : $$\frac{dx_1}{x_1} = \frac{dx_2}{x_2} = \frac{dV}{2V}.$$ A first characteristic equation comes from $\frac{dx_1}{x_1} = \frac{dV}{2V}$ .

You correctly get : $V=C_1x_1^2$.

$C_1$ is an arbitrary parameter which sets a particular characteristic curve on which : $$\frac{V(x_1,x_2)}{x_1^2}=C_1$$

A second characteristic equation comes from $\frac{dx_1}{x_1} = \frac{dx_2}{x_2}$ leading to :
$$\frac{x_2}{x_1}=C_2$$ Again $C_2$ is an arbitrary parameter to which a characteristic curve corresponds for each value of the parameter.

The general solution of the PDE (see reference added at the end) is expressed on the form of implicit equation is : $$\Phi(C_1,C_2)=0$$ where $\Phi$ is an arbitrary function of two variables. Or equivalently $$C_1=F(C_2)\quad\text{or}\quad C_2=G(C_1)$$ where $F$ and $G$ are arbitrary functions.

$$C_1=F(C_2)=\frac{V}{x_1^2}=F\left( \frac{x_2}{x_1}\right)$$ $$\boxed{V(x_1,x_2)=x_1^2\:F\left( \frac{x_2}{x_1}\right)}$$ If some boundary conditions are correctly specified the function $F$ can be determined.

NOTE:

Alternatively one could consider $\frac{dx_2}{x_2} = \frac{dV}{2V}$ instead of $\frac{dx_1}{x_1} = \frac{dV}{2V}$ . This leads to

$V(x_1,x_2)=x_2^2\:H\left( \frac{x_2}{x_1}\right)$ where $H$ is an arbitrary function related to the above function $F$ :

$F\left( \frac{x_2}{x_1}\right)=\left( \frac{x_2}{x_1}\right)^2H\left( \frac{x_2}{x_1}\right)$ . Since both $F$ and $H$ are arbitrary functions, the result is the same than above.

REFERENCE: Copy from https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf

The symbols are not the same as those used in my above answer. Be carefull about possible confusion.