Here are some thoughts on quasi linear first order PDEs which can be expressed as a(x, y, u)u_x+b(x, y, u)u_y=c(x, y, u), where u_x is the partial derivative of the dependent variable u with respect to x and likewise for u_y. Problems of this type can be solved by the method of characteristics. Mathematica software can be useful for finding such solutions. One feature of quasi-linear PDES is that their solution exhibits rapidly steepening solution fronts, such as shocks. I have attached a PDF tutorial on this subject which also illustrated how Mathematica can be used so solve/display such solutions . More examples, including Mathematica notebooks can be found at my google site https://sites.google.com/site/chemengwithmathematica/home/reaction-engineering

I have no information about this.

I have no information about this.

Dear A. M. Abdallah,

(Lagrange Method)

The general form of the quasi-linear partial differential equation is p(x,y,u)(∂u/∂x)+q(x,y,u)(∂u/∂y)=R(x,y,u), where u = u(x,y).

Or the form

p(x,y,u).(∂u/∂x)+q(x,y,u)(∂u/∂y)+R(x,y,u)(∂u/∂z)=0, where u=u(x,y,z).

It is linear in ((∂u)/(∂x)),((∂u)/(∂y)) and ((∂u)/(∂z))

but p(x,y,z),q(x,y,z) and R(x,y,u) are not necessarily linear,

(all are differentiable over some domain D).

that's why it is called quasi- linear pde.

The solution can be determined using the Lagrange Method

by solving the auxiliary equations:

dx/p(x,y,z)= dy/q(x,y,z) = du/R(x,y,u)

A simple example:

Solve: x(∂u)/(∂x))+y(∂u/∂y)=u

using the auxiliary equations: (dx/x)=(dy/y)=(du/u)

By integration, we obtain ln(x)=ln(y)+ constant

which is equivalent to x = cy where c is constant

similarly, ky = u where k is constant.

the general solution is given by f(c, k )= 0

where f is any arbitrary function or

k=φ(c), φ is arbitrary function.

i.e, (u/y)=φ((x/y))

Finally, u(x,u)=yφ((x/y)), φ is arbitrary function.

If one choose φ(t)= sin t

you can check u(x,u)= ysin(x/y)

satisfies x((∂u)/(∂x))+y((∂u)/(∂y))=u

((∂u)/(∂x)) = ycos((x/y))((1/y)) = cos((x/y))

((∂u)/(∂y)) = sin(x/y)+y(cos(x/y))(-x/y²)=sin((x/y))+(cos((x/y)))(-x)/y)

substitute in the pde, we obtain:

x((∂u)/(∂x))+y((∂u)/(∂y))=

x(cos((x/y)))+y(sin((x/y))+(cos((x/y)))(((-x)/y)))=y(sin((x/y))=u.

Best regards