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

Using the exponential definition of the hyperbolic cosine and Euler's formula,Therefore

For a short string element of length $\Delta x$, the small-slope approximation makes the upward components of tension at its two ends $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$. Newton's second law givesDivide by $\Delta x$ and take the limit to obtain the wave equation on a string,Here $\rho$ is linear mass density, and the derivation assumes small transverse slopes, uniform tension, and a flexible string.

At the right endpoint the string pulls the ring toward the interior. Its vertical force is therefore $-T{y}_x(L,t)$ to first order; the frictionless wire supplies no vertical force. The ring obeysThis is the dynamic endpoint law for a string with an endpoint mass.

For a separated oscillatory solution, write $y=X(x)q(t)$ with $q^{\prime \prime }=-{\omega }^{2}q$ and $X^{\prime \prime }+(\omega /c{)}^{2}X=0$. The fixed left end removes the spatial cosine, so the nonzero normal modes haveThe endpoint law gives $M{\omega }_n^2\sin ({\omega }_{n}L/c)=T({\omega }_n/c)\cos ({\omega }_{n}L/c)$. Since $\sin ({\omega }_{n}L/c)=0$ would also force the cosine to vanish, it can be divided out for a nonzero mode. Using $T=\rho c^2$ givesFor physical $M\ge 0$, these give all positive modal frequencies. There is no nonzero static linear mode: the left value and right slope conditions would force it to vanish. Positivity also follows by integration by parts from

For $M=0$, the right endpoint is free, with the Neumann boundary condition $y_x(L,t)=0$. At $L=\pi$, the allowed wave numbers are $k_n=n+1/2$, $n\ge 0$, and ${\omega }_n=c{k}_n$. Zero initial displacement removes every cosine-in-time coefficient. The initial velocity contains only the modes $k=1/2$ and $k=5/2$, each with coefficient one. Dividing those velocity coefficients by their angular frequencies givesAt $x=0$ each term vanishes, at $x=\pi$ each spatial derivative vanishes, and at $t=0$ the displacement and its time derivative reproduce the two initial conditions.