Find the parametric form of the solution of the following PDE using method of characteristics: $$-yu_x+xu_y=0\qquad u(x,x^2)=x^3$$
I assume that $(x,y)$ are functions of a parameter, say, $t$, i.e. $(x(t), y(t))$. In that case, the chain rule gives
$$\frac{du}{dt}=\frac{\partial u}{\partial x}.\frac{d x}{dt}+\frac{\partial u}{\partial y}.\frac{dy}{dt}=0$$
Comparing the equations, we need to solve the ODEs to find the characteristic curves.
$$\frac{du}{dt}=0,\quad\frac{d x}{dt}=-y \qquad\&\quad\frac{d y}{dt}=x$$
I get the solution, $x(t)=C_{11}\cos(t)+C_{12}\sin(t)$, $y(t)=C_{21}\cos(t)+C_{22}\sin(t)$ and $u(t)=C$.
Using initial conditions $x(0,s)=s$, $y(0,s)=s^2$ and $u(0,s)=s^3$ we get, $x(t)=s\cos(t)+C_{12}\sin(t)$, $y(t)=s^2\cos(t)+C_{22}\sin(t)$ and $u(t)=s^3$. But how to remove $C_{12},C_{22}$ constants without initial conditions?
I noticed they used one more extra parameter, $s$. Why it was needed? Like why, we re-interpret the $x(t)$ as $x(t,s)$?
I am new to this method. It will be a great help if anyone provide some insight along the solution.
Thanks in advance.
Best Answer
$$-yu_x+xu_y=0$$ The charpit-Lagrange characteristic ODEs are : $$\frac{dx}{-y}=\frac{dy}{x}=\frac{du}{0}$$ A first characteristic equation comes from $du=0$ $$u=c_1$$ which is an obvious particular solution of the PDE.
A second characteristic equation comes from solving $\frac{dx}{-y}=\frac{dy}{x}$: $$x^2+y^2=c_2$$ The general solution of the PDE$\quad c_1=F(c_2)\quad$is : $$\boxed{u(x,y)=F(x^2+y^2)}$$ $F$ is an arbitrary function until no boundary condition is taken into account.
The most interesting part of the problem is to find the particular solution(s) which satisfy the condition $u(x,x^2)=x^3$. $$u(x,x^2)=F(x^2+(x^2)^2)=x^3$$ Let $X=x^2+(x^2)^2$. Solving the quadratic eq. $(x^2)^2+(x^2)-X=0$ for $(x^2)$ then for $x$ :
$x=\pm \left(\frac{-1\pm \sqrt{1+4X}}{2} \right)^{1/2}\quad$ with $\quad X(0)=0\quad\implies\quad x=\pm \left(\frac{-1+ \sqrt{1+4X}}{2} \right)^{1/2}$
$$F(X)=x^3\quad\implies\quad\begin{cases} F(X)=+\left(\frac{-1+\sqrt{1+4X}}{2} \right)^{3/2}\quad \text{if}\quad x>0 \\ F(X)=0\quad \text{if}\quad x=0 \\ F(X)=-\left(\frac{-1+\sqrt{1+4X}}{2} \right)^{3/2}\quad \text{if}\quad x<0 \end{cases}$$
Now the function $F(X)$ is determined. We put it into the above general solution where $X= x^2+y^2$ :
$$\begin{cases} u(x,y)=\left(\frac{-1+\sqrt{1+4(x^2+y^2)}}{2} \right)^{3/2}\quad \text{if}\quad x>0 \\ u(x,y)=0\quad \text{if}\quad x=0 \\ u(x,y)=-\left(\frac{-1+\sqrt{1+4(x^2+y^2)}}{2} \right)^{3/2}\quad \text{if}\quad x<0 \end{cases}$$