# Textbook classification of linear, semi-linear, quasi-linear, and fully-nonlinear PDEs

My textbook has classified the following PDEs accordingly:

$$u_{xxx} - 4u_{xxyy} + u_{yyzz} = f(x, y, z) \ \ \ \text{Linear}$$

$$u^2_{x}u_{tt} - \dfrac{1}{2}u = 1 - u^2 \ \ \ \text{Semi-Linear}$$

$$u_{tt}u_{xxx} - u_{x}u_{ttt} = f(u, x, t) \ \ \ \text{Quasi-Linear}$$

$$\exp(u_{xtt}) - u_{xt}u_{xxx} + u^2 = 0 \ \ \ \text{Fully Non-Linear}$$

$$2\cos(xt)u_t - xe^tu_x - 9u = e^t \sin(x) \ \ \ \text{Linear}$$

$$x\cos(t)u_t + t u_x + u^2 u = \dfrac{x}{t} u \sin(u) \ \ \ \text{Semi-Linear}$$

$$uu_t + u^2 u_x + u = e^x \ \ \ \text{Quasi-Linear}$$

$$u_t + \dfrac{1}{2} u_x^2 - u = \cos(xt) \ \ \ \text{Fully Non-Linear}$$

The textbook gives the following explanations of linear, semi-linear, quasi-linear, and fully non-linear PDEs:

Linear: In effect, all coefficients of $$u$$ and any of its derivatives must depend only on the independent variables.

Semi-linear: The coefficients of the highest derivatives of $$u$$ do not depend on $$u$$ or any derivatives of $$u$$.

Quasi-linear: The coefficients of the highest derivatives of $$u$$ depend only on lower derivatives of $$u$$.

Otherwise the PDE is fully nonlinear.

Reading through the classification of the aforementioned PDEs, I have a suspicion that there are some errors. I would greatly appreciate it if people could please review the author's classification of these PDEs and comment on its correctness.

• I don't get the point of missing the classical rule:: the PDE that is not linear... it is non-linear, there is no any other possibility. The PDE $Lu=b$ is linear and the PDE $L(u)\,u=b$ is non-linear.
– HBR
Aug 27, 2018 at 12:56

I agree with you that this classification is not easy to follow. For example for $$u^2_{x}u_{tt} - \dfrac{1}{2}u = 1 - u^2 \ \ \ \text{Semi-Linear}$$ What is the highest derivative and what is its coefficient? If you say it is u_{tt}, then the coefficient depends on derivative of $u$ so it is not semi-linear.