# Difference between linear and quasi linear differential equation. Which is more strong?

I am currently studying PDE for the first time. So I came across some definitions of linear differential operator and quasi-linear differential operator. What exactly is the difference? Can someone explain in simple words?

This is the definition in my script • Let $F(x,u,u')=0$ be a first-order ODE for the function $u(x)$. We say that it is linear whenever it can be written as $a(x)u+b(x)u'=0$ for functions $a$ and $b$. We say that it's quasilinear if it can be written as $G(x,u)+b(x,u)u'=0$. Note that in this case, $G$ is a nonlinear part, and $b$ is a function of $x$ and $u$ instead of just $x$. This is the "one-by-one" case, i.e. first-order in a single variable $x$. The "two-by-two" case is given in the answer I posted. – Saurabh Agrawal Feb 7 '20 at 11:58
• Check out the Chapter 3 of the book: Partial differential equations of applied mathematics, Zauder,1989 – Mark May 18 '20 at 19:07

It's helpful to consider the "two by two" case, i.e. second-order PDEs in $$x$$ and $$y$$. Recall that such a PDE always has the form $$F(x,y,u,u_x,u_y,u_{xy},u_{yx},u_{xx},u_{yy})=0$$ for some function $$F$$. This PDE is called linear whenever it can be written $$a(x,y)u+b(x,y)u_x+c(x,y)u_y+d(x,y)u_{xy}+e(x,y)u_{yx}+f(x,y)u_{xx}+g(x,y)u_{yy}=0$$ which we will abbreviate $$au+bu_x+cu_y+du_{xy}+eu_{yx}+fu_{xx}+gu_{yy}=0$$ The quasilinear case is different in two respects. First of all, the original function $$F$$ need only be linear in derivatives of the highest order (in this case $$\text{ord}=2$$). That is to say, it can be written as $$G(x,y,u,u_x,u_y)+du_{xy}+eu_{yx}+fu_{xx}+gu_{yy}=0$$ for some (possibly nonlinear) function $$G$$. The second difference is this: the remaining coefficients of the "linear part" (or principal part) can depend on $$u$$, $$u_x$$, and $$u_y$$ as well as $$x$$ and $$y$$. In particular, we should write $$G(x,y,u,u_x,u_y)+d(x,y,u,u_x,u_y)u_{xy}+e(x,y,u,u_x,u_y)u_{yx}+f(x,y,u,u_x,u_y)u_{xx}+g(x,y,u,u_x,u_y)u_{yy}=0$$