Characteristic Curve of a PDE

Question

Question from a Exam:

Consider the pde $xu_{xx}+2x^2u_{xy}=u_x-1$. Find the characteristic curves of the above.

Can someone please tell me how are these types of problems handled?

I dont want exact solution.Please tell me how to solve these type of questions.

Answer 1

Your first step should be noting that part of the PDE has a $u_y$ or $u_yy$, so first define $v=u_x$. Your PDE translates to $xv_x+2x^2v_y=v-1$.

You have a constant term, and none of your coefficients are dependent on y. Therefore, $v=fy+g$ where f and g are functions of x. Substituting $v=fy+g, v_x=f'y+g', v_y=f$, you get $xf'y+xg'+2x^2f=fy+g-1$. Dividing this up into two equations (one with y and one without) gives that $xf'=f$ and that $xg'+2x^2f=g-1$. Solving the first of these equations gives that $f=Ax$. Then, solving the second equation gives: $$xg'+2Ax^3=g-1$$ $$g-xg'=2Ax^3+1$$ $$g=-Ax^3+Bx+1$$ Together, this gives that $u_x=v=Axy-Ax^3+Bx+1$. Integrating with respect to x gives $u=\frac{A}{2}x^2y-\frac{A}{4}x^4+\frac{B}{2}x^2+x+C$. Redefining our constants of integration gives $u=C_1(x^4-2x^2y)+C_2x^2+x+C_3$

Answer 2

$$xu_{xx}+2x^2u_{xy}=u_x-1 \tag 1$$ $v=u_x$

$$xv_{x}+2x^2v_{y}=v-1 \tag 2$$ Charpit-Lagrange equations: $$\frac{dx}{x}=\frac{dy}{2x^2}=\frac{dv}{v-1}$$ First family of characteristic curves from $\frac{dx}{x}=\frac{dy}{2x^2}$ : $$y-x^2=c_1$$ Second family of characteristic curves from $\frac{dx}{x}=\frac{dv}{v-1}$ $$\frac{v-1}{x}=c_2$$ General solution of the PDE Eq.$(2)$: $$\frac{v-1}{x}=F(y-x^2)$$ where $F$ is an arbitrary function. $$v(x,y)=1+xF(y-x^2)$$ $u=\int \left(1+xF(y-x^2)\right)dx$ $$u(x,y)=x+\int xF(y-x^2) dx+G(y)$$ $F$ and $G$ are arbitrary functions. They have to be determined according to some boundary conditions.

Since no boundary condition is specified in the wording of the question it is not possible to fully solve the problem.