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## 数学代写|非线性偏微分方程代写Nonlinear Partial Differential Equation代考|COMPATIBILITY IN （2+1） DIMENSIONS

Up until now, we have considered compatibility of PDEs in $(1+1)$ dimensions. We now extend this idea and consider PDEs in $(2+1)$ dimensions. In this section we consider the compatibility between the $(2+1)$ dimensional reaction-diffusion equation
$$u_t=u_{x x}+u_{y y}+Q\left(u, u_x, u_y\right)$$
and the first-order partial differential equation
$$u_t=F\left(t, x, y, u, u_x, u_y\right) .$$
This section is based on the work of Arrigo and Suazo [39]. We will assume that $F$ in (2.222) is nonlinear in the first derivatives $u_x$ and $u_y$. The case where $F$ is linear in the first derivatives $u_x$ and $u_y$, is left as an exercise to the reader.

Compatibility between (2.221) and (2.222) gives rise to the compatibility equation constraints
Eliminating the $x$ and $y$ derivatives in (2.223b) and (2.223c) by $(i)$ cross differentiation and (ii)
$\begin{array}{ll}2 F_{u p}+\left(F_p-Q_p\right) F_{p p}+\left(F_q-Q_q\right) F_{p q}=0, & \quad(2.224 \mathrm{a}) \ 2 F_{u q}+\left(F_p-Q_p\right) F_{p q}+\left(F_q-Q_q\right) F_{q q}=0 . & (2.224 \mathrm{~b})\end{array}$

## 数学代写|非线性偏微分方程代写Nonlinear Partial Differential Equation代考|COMPATIBILITY IN （2+1） DIMENSIONS

\begin{aligned} &\text { Further, eliminating } F_{u p} \text { and } F_{u q} \text { by again }(i) \text { cross differentiation and (ii ) imposing (2.223a) } \ &\text { gives rise to } \ &\qquad \begin{array}{r} \left(2 F_{p p}-Q_{p p}+Q_{q q}\right) F_{p p}+2\left(F_{p q}-Q_{p q}\right) F_{p q}=0, \ \left(Q_{p p}-Q_{q q}\right) F_{p q}+2 Q_{p q} F_{q q}=0 . \end{array} \end{aligned}
Solving (2.223a), (2.225a) and (2.225b) for $F_{p p}, F_{p q}$ and $F_{q q}$ gives rise to two cases:
(i) $F_{p p}=F_{p q}=F_{q q}=0$,
(ii) $\quad F_{p p}=\frac{1}{2}\left(Q_{p p}-Q_{q q}\right), \quad F_{p q}=Q_{p q}, \quad F_{q q}=\frac{1}{2}\left(Q_{q q}-Q_{p p}\right)$
As we are primarily interested in compatible equations that are more general than quasilinear, we omit the first case. If we require that the three equations in (2.226b) be compatible, then to within equivalence transformations of the original equation, $Q$ satisfies
$$Q_{p p}+Q_{q q}=0 .$$
Using (2.227), we find that (2.226b) becomes
$$F_{p p}=Q_{p p}, \quad F_{p q}=Q_{p q}, \quad F_{q q}=Q_{q q},$$
from which we find that
$$F=Q(u, p, q)+X(t, x, y, u) p+Y(t, x, y, u) q+U(t, x, y, u),$$
where $X, Y$ and $U$ are arbitrary functions. Substituting (2.229) into (2.224a) and (2.224b) gives
$$\begin{gathered} 2 Q_{u p}+X Q_{p p}+Y Q_{p q}+2 X_u=0 \ 2 Q_{u q}+X Q_{p q}+Y Q_{q q}+2 Y_u=0 \end{gathered}$$
while (2.223b) and (2.223c) become (using (2.227) and (2.230))
\begin{aligned} &(X p+Y q+2 U) Q_{p p}+(X q-Y p) Q_{p q}+2\left(X_x-Y_y\right)=0 \ &(X q-Y p) Q_{p p}-(X p+Y q+2 U) Q_{p q}-2\left(X_y+Y_x\right)=0 \end{aligned}

## 数学代写非线性偏微分方程代写Nonlinear Partial Differential Equation代 考|COMPATIBILITY IN $(2+1)$ DIMENSIONS

$$u_t=u_{x x}+u_{y y}+Q\left(u, u_x, u_y\right)$$

$$u_t=F\left(t, x, y, u, u_x, u_y\right) .$$

(2.221) 和 $(2.222)$ 之间的相容性昌致相容性方程

$$2 F_{u p}+\left(F_p-Q_p\right) F_{p p}+\left(F_q-Q_q\right) F_{p q}=0, \quad(2.224 \mathrm{a}) 2 F_{u q}+\left(F_p-Q_p\right) F_{p q}+\left(F_q-Q_q\right) F_{q q}=0 . \quad(2.224 \mathrm{~b})$$

## 数学代写非线性偏微分方程代写Nonlinear Partial Differential Equation代考|COMPATIBILITY IN (2+1) DIMENSIONS

Further, eliminating $F_{u p}$ and $F_{u q}$ by again $(i)$ cross differentiation and (ii ) imposing (2.223a) $\quad$ gives rise to $\quad\left(2 F_{p p}-Q_{p p}+Q_{q q}\right) F_{p p}+2($

(i) $F_{p p}=F_{p q}=F_{q q}=0$
(ii) $\quad F_{p p}=\frac{1}{2}\left(Q_{p p}-Q_{q q}\right), \quad F_{p q}=Q_{p q}, \quad F_{q q}=\frac{1}{2}\left(Q_{q q}-Q_{p p}\right)$

$$Q_{p p}+Q_{q q}=0 .$$

$$F_{p p}=Q_{p p}, \quad F_{p q}=Q_{p q}, \quad F_{q q}=Q_{q q},$$

$$F=Q(u, p, q)+X(t, x, y, u) p+Y(t, x, y, u) q+U(t, x, y, u),$$

$$2 Q_{u p}+X Q_{p p}+Y Q_{p q}+2 X_u=02 Q_{u q}+X Q_{p q}+Y Q_{q q}+2 Y_u=0$$

$$(X p+Y q+2 U) Q_{p p}+(X q-Y p) Q_{p q}+2\left(X_x-Y_y\right)=0 \quad(X q-Y p) Q_{p p}-(X p+Y q+2 U) Q_{p q}-2\left(X_y+Y_x\right)=0$$

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