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?
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$\begingroup$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:
In PDE (from Pinchover and Rubinstein):
In our case we have:
$$F(x, y, t, u, v,$$
$$u_x, u_y, u_t, u_{xx}, u_{yy}, u_{tt}, u_{xt}, u_{tx}, u_{xy}, u_{yx}, u_{yt}, u_{ty},$$
$$v_x, v_y, v_t, v_{xx}, v_{yy}, v_{tt}, v_{xt}, v_{tx}, v_{xy}, v_{yx}, v_{yt}, v_{ty}) = 0$$
Definition of a multivariate linear function from Wiki:
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 $(2)$
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