Example 2.7 Solve the equation $u_x + 3y^{2/3}u_y = 2$ subject to the initial condition $u(x,1) = 1+x$.
It's an example in "An introduction to partial differential equations" book by Yehuda Pinchover, to be solved by the method of characteristics. And I'm trying to solve it by Lagrange's method. I know that I need to find 2 orthogonal vectors so I can find $Ψ$ and $Φ$. But my struggles is finding the 2 orthogonal vectors.
$$u_x + 3y^{2/3}u_y = 2$$ The Charpit-Lagrange equations are : $$\frac{dx}{1}=\frac{dy}{ 3y^{2/3}}=\frac{du}{2}$$ A first characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{ 3y^{2/3}}$ : $$x-y^{1/3}=c_1$$ A second characteristic equation comes from solving $\frac{dx}{1}=\frac{du}{2}$ : $$u-2x=c_2$$ The general solution of the PDE is : $\quad c_2=F(c_1)=u-2x=F(x-y^{1/3})$ : $$\boxed{u=2x+F(x-y^{1/3})}$$ $F$ is an arbitrary function in the general case.
In the particular case with condition $u(x,1)=1+x$ the function F is non longer arbitrary. $$1+x=2x+F(x-1^{1/3})=2x+F(x-1)$$ Let $X=x-1\quad\implies\quad x=X+1$ $$1+(X+1)=2(X+1)+F(X)$$ $$F(X)=-X$$ Now the function $F(X)$ is known. We put it into the above general solution where $X=x-y^{1/3}$ : $$u=2x-(x-y^{1/3})$$ $$\boxed{u(x,y)=x+y^{1/3}}$$