3

Consider the following second order PDEs

\begin{align} u_{t} + v_{t} + x^{4}u_{xx} + v_{yy} &=0\,\, (1)\\ u_{t} + v_{t} + xu_{xx} + v_{yy} + x^{2}v_{y}&=0\,\,(2) \end{align}

I want to classify them into linear and nonlinear. i believe that both equations are linear. Is that correct?

Paul
  • 31
  • What's your reasoning? –  Nov 17 '15 at 08:18
  • 1
    The classification of the equation being linear or non-linear comes from the dependent variable $u$. It has nothing to do with whether $x$ to $t$ appear linearly. – okrzysik Nov 17 '15 at 08:24
  • So it only matters if there appear products of the partial derivatives and the functions, e.g. $uu_{xx}$ or powers, e.g. $u_{xx}^{3}$, to classify as nonlinear? I read it before but have not understant it completely. So, both equations are linear? – Paul Nov 17 '15 at 08:26
  • @Paul that is correct. The same concept as an ODE, just because it has variable coefficients doesn't mean that it is non-linear; for the equation to be non-linear it has to be so in the dependent variable. Which in this case is $u(x,t)$. – okrzysik Nov 17 '15 at 08:31
  • @okrzysik thank you for making that clear for me. – Paul Nov 17 '15 at 08:33

1 Answers1

2

Intuitively, the equations are linear because all the u's and v's don't have exponents, aren't the exponents of anything, don't have logarithms or any non-identity functions applied on them, aren't multiplied w/ each other and the like.

Precisely, just go back to the definition of linear.

In ODE:

enter image description here

enter image description here

In PDE (from Pinchover and Rubinstein):

enter image description here

In our case we have:

$$F(x, y, t, u, v,$$

$$u_x, u_y, u_t, u_{xx}, u_{yy}, u_{tt}, u_{xt}, u_{tx}, u_{xy}, u_{yx}, u_{yt}, u_{ty},$$

$$v_x, v_y, v_t, v_{xx}, v_{yy}, v_{tt}, v_{xt}, v_{tx}, v_{xy}, v_{yx}, v_{yt}, v_{ty}) = 0$$

enter image description here

Definition of a multivariate linear function from Wiki:

enter image description here

enter image description here

In our case, define $f$ s.t.

$$F(x, y, t, u, v,$$

$$u_x, u_y, u_t, u_{xx}, u_{yy}, u_{tt}, u_{xt}, u_{tx}, u_{xy}, u_{yx}, u_{yt}, u_{ty},$$

$$v_x, v_y, v_t, v_{xx}, v_{yy}, v_{tt}, v_{xt}, v_{tx}, v_{xy}, v_{yx}, v_{yt}, v_{ty})$$

$$:= f(u, v,$$

$$u_x, u_y, u_t, u_{xx}, u_{yy}, u_{tt}, u_{xt}, u_{tx}, u_{xy}, u_{yx}, u_{yt}, u_{ty},$$

$$v_x, v_y, v_t, v_{xx}, v_{yy}, v_{tt}, v_{xt}, v_{tx}, v_{xy}, v_{yx}, v_{yt}, v_{ty})$$

It should be clear that $f$ is linear w/rt its arguments for both $(1)$ and $(2)$

BCLC
  • 13,459