I am studying the second order PDE's and I am a bit confused with classification of quasi linear and semi linear PDEs.
Could anybody explain on examples what is a difference between them please?

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1 Answers
Let $L$ be a $k^{th}$-order linear differential operator, i.e. one which satisfies $L(\alpha u + \beta v) = \alpha L u + \beta Lv$ for all $u,v\in C^k$ and constants $\alpha,\beta$ (of course this notion can be weakened past $C^k$, but this will do for here). We say that the equation $L u =f$, for some given function $f$, is linear.
A semilinear equation is one of the form $L u = f(x,u,Du, \dotsc, D^{k-1} u)$, where $D^j j$ denotes all $j^{th}$ order derivatives. The right hand side is a generic nonlinearity that involves any possible combination of derivatives up to one order less than the order of the main linear operator. This is the key feature of semilinear equations: they are linear "at the highest order" and possibly nonlinear at lower order.
To describe a quasilinear equation we need to be more careful with naming $L$. Let's say it's of the form $$ L = \sum_{|\alpha| \le k} a_\alpha \partial_\alpha. $$ In the above treatment we have that $a_\alpha = a_\alpha(x)$ in order for the operator $L$ to be linear. Now for a quasilinear equation we allow the $a_\alpha$ coefficients to depend on the solution itself, but only up to $k-1$ order derivatives. That is, a quasilinear problem is one of the form $$ \sum_{|\alpha| \le k} a_\alpha(x,u,Du,\dotsc,D^{k-1}u) \partial_\alpha u = f(x,u,Du,\dotsc,D^{k-1}u). $$ The key feature here is that the coefficients can only depend on the lower order derivatives, so in some sense if we think of "freezing" the lower order derivatives, then the resulting problem is actually a linear one.
In some sense the main reason for making these distinctions lies in the tools available to solve such problems. Roughly speaking, linear problems are the easiest. Semilinear ones are next, and one often views a semilinear problem as a "small nonlinear perturbation" of a linear one. Quasilinear problems are next in the hierarchy; the construction of solutions is often built on the linear theory but in a more complicated way than for semilinear problems. After quasilinear comes fully nonlinear, which essentially means that the nonlinearity occurs at the highest order of differentiability.
EDIT: Some first-order examples.
A linear equation: $$ \partial_1 u + \partial_2 u + u = 0. $$
A semilinear equation: $$ \partial_1 u + \partial_2 u + u = \cos(u). $$
A quaslinear equation: $$ u\partial_1 u + u^2 \partial_2 u + u = e^u. $$
A fully nonlinear equation: $$ (\partial_1 u)^2 + (\partial_2 u)^2 = 1. $$
Some second-order examples.
A linear equation: $$ \partial_1 \partial_1 u + \partial_2 \partial_2 u + u = 0. $$
A semilinear equation: $$ \partial_1 \partial_1 u + \partial_2 \partial_2 u + u = \cos(u). $$
A quaslinear equation: $$ u\partial_1 \partial_1 u + u^2 \partial_2 \partial_2 u + u = e^u. $$
A fully nonlinear equation: $$ (\partial_1 \partial_1 u)^2 + (\partial_2 \partial_2 u)^2 = 1. $$

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could you give some example equations please? – Michal Jun 26 '16 at 23:46
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Well, in principle I've given you all possible examples above, but sure... see the edit. – Glitch Jun 27 '16 at 14:33
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does the $\partial_2 u$ in this notation refer to the second order? like $u_{xx}$? – Michal Jun 28 '16 at 10:40
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No. These are derivatives with respect to the variables $x_1$ and $x_2$, i.e. $\partial_i = \frac{\partial}{\partial x_i}$ when $u = u(x_1,x_2)$. – Glitch Jun 28 '16 at 13:59
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1Classification of the first order PDE is clear. The difference between semi linear and quasi linear for second order is not clear to me. For example $u_{xx} + u_{yy} + u u_{y} + u^2 \ u_x + u = e^u$ is it semi or quasi linear? Could you explain on the examples for the second order please? – Michal Jun 28 '16 at 19:41
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Your example is semilinear as the highest-order term (the terms involving second order derivatives) is linear. See my second edit for more examples. – Glitch Jun 28 '16 at 21:11
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How do you categorize equations with a $\lvert u\rvert^p u$ term? The Schrödinger equation with such a term is categorized as semilinear but should'nt it fall into the quasilinear category? – Rem Feb 20 '21 at 13:22