Take the spatial interval $0<x<L$ with $L>0$, and write $v=|c|>0$. A separated nonzero solution satisfiesThe zero-endpoint Dirichlet boundary conditions rule out nonzero solutions for $\lambda \le 0$: linear or hyperbolic spatial factors cannot vanish at both ends. For $\lambda >0$, they give $X_n=\sin (n\pi x/L)$ and ${\lambda }_n=(n\pi /L{)}^2$, $n\ge 1$. The temporal factors solve $T_n^{\prime \prime }+{\omega }_n^2{T}_n=0$, with the fixed-end standing wave frequencies ${\omega }_n=n\pi v/L$. Superposition givesThe constants are arbitrary subject to the convergence and regularity required of the solution. For initial displacement $f$ and initial velocity $g$, their Fourier sine series give $A_n=(2/L){\int }_0^{L}f(x)\sin (n\pi x/L)dx$ and $B_n=(2/(L{\omega }_n)){\int }_0^{L}g(x)\sin (n\pi x/L)dx$. Smooth compatible data give classical solutions; weaker data have the corresponding energy interpretation. A negative printed $L$ describes the interval with endpoints reversed and gives the same modes after using its positive length.

If $\psi =f(\xi )+g(\eta )$ with twice differentiable $f,g$, its mixed partial derivative vanishes because each summand depends on only one coordinate. Conversely ${\psi }_{\xi \eta }=0$ means ${\psi }_{\xi }$ is independent of $\eta$; integration in $\xi$ supplies $f(\xi )$ and an integration function $g(\eta )$. Hence, on a characteristic rectangle, the general solution isThese terms travel in opposite directions without changing profile, as in D'Alembert's formula.

Assume $c\ne 0$, so $(\xi ,\eta )=(x-ct,x+ct)$ is an invertible coordinate change. The chain rule givesFor a twice continuously differentiable solution the mixed partial derivatives commute. Squaring the two operators therefore givesThus the one-dimensional wave equation becomes

Let $a=f(b,c)$ and work where both $f_b$ and $f_c$ are nonzero, so the required inverse partial derivatives exist. At fixed $a$, the total differential gives $0=f_{b}db+f_{c}dc$, hence $(\partial b/\partial c{)}_a=-f_c/f_b$. At fixed $b$, $(\partial c/\partial a{)}_b=1/f_c$. Their product with $f_b$ is $-1$: this is the cyclic partial derivative identity.

For explicit verification in this example, put $P=b^4-6b^2{c}^2+c^4$ and $H=(b^2+c^2{)}^3$. Direct differentiation givesThereforeThe identity is local to regular branches: for these explicit formulas require $bcP\ne 0$. At singular points one cannot justify cancellation of nonexistent inverse derivatives merely by writing a formal product.

For an exact differential $Pdx+Qdy$, equality of mixed partial derivatives requires $P_y=Q_x$. Here this givesOn an open two-dimensional domain the coefficient relation is $x{\mu }^{\prime }+3\mu =0$, so $\mu =C{x}^{-3}$ on a connected interval avoiding zero. The normalization gives $C=1$ on the positive interval through $x=1$. Integrating $f_x=2x^{-3}\sin y$ and checking $f_y$ yieldsThere is no smooth extension of this normalized integrating factor across $x=0$; an independently specified factor on $x<0$ is not fixed by data at one.

Regard $u=u(s,t(s,v))$. Applying the multivariable chain rule while holding $v$ fixed givesEach subscript records the variable held fixed. A differentiable local coordinate change from $(s,t)$ to $(s,v)$ requires $(\partial v/\partial t{)}_s\ne 0$; the question's notation $t=t(s,v)$ presupposes such a branch.

The product rule givesThus this is an exact differential, and its zero level differential means $y{\sin }^{2}x=C$. The initial data set $C=1$, givingThe displayed interval is the maximal connected real interval containing $\pi /2$ on which this finite-valued solution is defined.

The integrand in the preceding expression is nonnegative. Tonelli's theorem permits squaring as a quadrant integral; then polar coordinates giveThe defining integral is positive, fixing the sign of the square root. ThereforeThis is the positive-quadrant proof of the Gaussian integral.

In the Gamma function integral at $p=1/2$, use $x=t^2$, $dx=2tdt$, with $t>0$. Then $x^{-1/2}dx=2dt$, soThe substitution can first be made on finite positive intervals and then extended by monotone limits, justifying the endpoint passage.

Set $t=\log (1/x)$, so $x=e^{-t}$ and $dx=-e^{-t}dt$. The improper integral becomesThe Gamma function recurrence givesThe exponential-transformed integrand also proves convergence at the logarithmically singular endpoint.

For $p>0$, integration by parts gives the Gamma function recurrence:The endpoint terms vanish because $x^p\to 0$ at zero and the exponential dominates the power at infinity. Also $\Gamma (1)={\int }_0^{\infty }{e}^{-x}dx=1$. Iterating for nonnegative integers $n$ therefore yieldsNegative integers are outside this integral definition and factorial conclusion.

At an intersection, $a+tu=b+t^{\prime }v$, so $b-a=tu-t^{\prime }v$. Alternating multilinearity of the scalar triple product then gives $[v,b-a,u]=t[v,u,u]-t^{\prime }[v,v,u]=0$, establishing $[v,b,u]=[v,a,u]$.

For the location, write $b=a+tu-t^{\prime }v$ and take its scalar triple product with $a$ and $v$:Under the additional stated assumption $[a,u,v]\ne 0$, division givesThese are intersection parameters from scalar triple products. The formula presupposes an intersection; the first coplanarity condition by itself also holds for distinct parallel lines. Intersections can exist with the displayed denominator zero, in which case a different auxiliary vector or a component calculation is needed.

By the vector triple product identity,The last inner product vanishes by perpendicularity. Taking the inner product with $d$ givesThis is the identity for cross products with a common vector, and holds without nondegeneracy assumptions.

The exponential definitions giveThus the double-angle identity follows directly, without any restriction beyond the domain of the exponential.