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Stationarity requires both first partial derivatives to vanish:For example, choosing $x=\pi$ and $\cos y=0$ gives two points strictly inside the prescribed rectangle:Both satisfy the two derivative equations directly.

The vectors $(1,1)$ and $(1,-1)$ are eigenvectors, with respective eigenvalues $\cosh \varphi +\sinh \varphi$ and $\cosh \varphi -\sinh \varphi$. The exponential definitions reduce these toAt $\varphi =0$ they coincide and the matrix is the identity.

Set $\xi =x-ct$. The chain rule gives $u_t=-c{f}^{\prime }(\xi )$ and $u_x=f^{\prime }(\xi )$, so $u_t+c{u}_x=0$ for any differentiable $f$. The initial profile fixes $f(\xi )=\cos \xi$, givingThis is the translated initial profile, as described by the transport equation.

Take the Cartesian partial derivatives component by component. The divergence isThe constant third component has zero derivatives. The curl consequently has only two nonzero components:

The characteristic equation of the homogeneous equation is $(r-1)(r-3)=0$. Thus the complementary function isA linear trial $y_p=ax+b$ gives $3ax+(3b-4a)=3x$, so $a=1$, $b=4/3$. One particular integral is therefore $\boxed{y_p=x+4/3}$.

Multiply the standard Taylor series through quadratic order:The linear term is $x$ and the quadratic coefficient is $1-1/2$, hence