# Characteristic Curve of a PDE

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.

• You can change the variable to $v=u_x$ and apply the characteristic curves method. – Dante Grevino Dec 3 '18 at 15:43
• @DanteGrevino; i obtain $v=u_x,xv_x+2x^2v_y=v-1$ – user596656 Dec 3 '18 at 15:50
• then i have if $a=x,b=2x^2,c=v-1\implies \frac{dx}{a}=\frac{dy}{2a^2}=\frac{dv}{v-1}$ – user596656 Dec 3 '18 at 15:51
• @DanteGrevino;how to solve it now – user596656 Dec 3 '18 at 15:52
• Do you have boundary conditions? – Dante Grevino Dec 3 '18 at 16:00

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 $$u=C_1(x^4-2x^2y)+C_2x^2+x+C_3$$
$$xu_{xx}+2x^2u_{xy}=u_x-1 \tag 1$$ $$v=u_x$$
$$xv_{x}+2x^2v_{y}=v-1 \tag 2$$ 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.