Quasilinear PDE definition?

Question

Here it's written that:

Definition 3: A partial differential equation is said to be quasilinear if it is linear with respect to all the highest order derivatives of the unknown function.

What does mean to be linear with respect to all the highest order derivatives?

Does it mean that the terms containing the highest order derivatives are linear, but there might be other terms that are nonlinear?

score 11 · Accepted Answer · answered Sep 11 '17 at 15:35

Yes, that's it. The term $$\frac{\partial u}{\partial y}\frac{\partial ^2u}{\partial y^2}$$ is linear with respect to $\frac{\partial ^2u}{\partial y^2}$, but the term $$\left(\frac{\partial ^2u}{\partial y^2}\right)^2$$ is not linear.

If you have a second order derivative, it does not matter if the equation has a term like $u^2$ to be quasilinear. You just need to check the highest order derivatives (second order ones in the examples of the link).

score 6 · Answer 2 · answered Apr 08 '21 at 09:04

To understand the meaning of quasilinear, it is helpful to start with the first-order case. On reference.wolfram.com, quasilinear is defined as follows:

The PDE is said to be quasilinear if it can be expressed in the form $$a(x, y, u(x, y)) \frac{\partial u(x,y)}{\partial x}+b(x, y, u(x, y)) \frac{\partial u(x, y)}{\partial y}=c(x, y, u(x, y))$$

where $a, b$ and $c$ can be nonlinear functions.

For higher order equations, we can have nonlinear functions of some of the derivates, so long as they are not the highest order. For example, the following PDE is second order and quasilinear because it is linear it terms of the highest order derivative, $u_{xx}$: $$ u_{xx} + u_y^2 = 0.$$ It is not linear, however, because the $u_y^2$ term.

score -3 · Answer 3 · answered Jul 08 '18 at 11:48

-3

The simplest definition of a quasi-linear PDE says:

A PDE in which at least one coefficient of the partial derivatives is really a function of the dependent variable(say u).

For example,

$\mathrm{\frac{\partial^2u}{\partial x_1^2} +u\frac{\partial^2u}{\partial x_2^2} =0}$

answered Jul 08 '18 at 11:48

Ariana

27

What about the order of those derivatives? – user539887 Jul 08 '18 at 12:21
Order doesn't matter, any order of any derivative containing the function of "u" as a coefficient. – Ariana Jul 08 '18 at 13:55
Is, for example, $(u_{xy})^2+uu_x=0$ quasi-linear? The coefficient of a partial derivative of the first order is a function of $u$. – user539887 Jul 08 '18 at 18:50
No, it is not, $\mathrm{\left ( \frac{\partial}{\partial y}\left( \frac{\partial u}{\partial x} \right) \right )^2 + u \frac{\partial u}{\partial x} = 0}$, although the condition is valid, the first order PD has a coefficient containing u but the exponent of the second order PD is 2, not 1. This makes this equation non-linear. – Ariana Jul 08 '18 at 19:14
$u_{xy}=(u_x)^2$? – user539887 Jul 08 '18 at 19:22
Non-linear..... first order PD has a power raised to 2, not 1. – Ariana Jul 08 '18 at 19:22
No, that is a quasi-linear equation: see Glitch's answer to difference between second order quasi linear and semi linear PDE. – user539887 Jul 08 '18 at 19:52
The definition @Ariana gave appears to be for the special case where the PDE is first-order: https://reference.wolfram.com/language/tutorial/DSolvePartialDifferentialEquations.html#18819 – Paul Wintz Apr 08 '21 at 08:52

Quasilinear PDE definition?

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