[proofplan]
We show that the Backlund system relates solutions of the sine-Gordon equation: $(\varphi_1)_{\xi\tau} = \sin\varphi_1$ if and only if $(\varphi_2)_{\xi\tau} = \sin\varphi_2$. The strategy is to differentiate one equation of the Backlund system with respect to the complementary variable, substitute the other equation to eliminate the mixed combination, and apply the product-to-sum trigonometric identity $2\cos A\sin B = \sin(A+B) - \sin(A-B)$ to decouple $\varphi_1$ and $\varphi_2$.
[/proofplan]
[step:Differentiate the first Backlund equation with respect to $\tau$]
The first equation of the Backlund system is
\begin{align*}
\frac{\partial}{\partial\xi}(\varphi_1 - \varphi_2) = 2\varepsilon\sin\!\left(\frac{\varphi_1 + \varphi_2}{2}\right).
\end{align*}
Differentiating both sides with respect to $\tau$, and using the equality of mixed partial derivatives $\partial_\tau \partial_\xi = \partial_\xi \partial_\tau$ on the left-hand side:
\begin{align*}
(\varphi_1)_{\xi\tau} - (\varphi_2)_{\xi\tau} = 2\varepsilon\cos\!\left(\frac{\varphi_1 + \varphi_2}{2}\right) \cdot \frac{1}{2}\frac{\partial}{\partial\tau}(\varphi_1 + \varphi_2).
\end{align*}
[/step]
[step:Substitute the second Backlund equation to eliminate $\partial_\tau(\varphi_1 + \varphi_2)$]
The second equation of the Backlund system states
\begin{align*}
\frac{\partial}{\partial\tau}(\varphi_1 + \varphi_2) = \frac{2}{\varepsilon}\sin\!\left(\frac{\varphi_1 - \varphi_2}{2}\right).
\end{align*}
Substituting this into the right-hand side:
\begin{align*}
(\varphi_1)_{\xi\tau} - (\varphi_2)_{\xi\tau} &= 2\varepsilon\cos\!\left(\frac{\varphi_1 + \varphi_2}{2}\right) \cdot \frac{1}{2} \cdot \frac{2}{\varepsilon}\sin\!\left(\frac{\varphi_1 - \varphi_2}{2}\right) \\
&= 2\cos\!\left(\frac{\varphi_1 + \varphi_2}{2}\right)\sin\!\left(\frac{\varphi_1 - \varphi_2}{2}\right).
\end{align*}
The parameter $\varepsilon$ cancels: the factor $\varepsilon$ from the first equation and the factor $1/\varepsilon$ from the second multiply to 1.
[/step]
[step:Apply the product-to-sum identity to decouple $\varphi_1$ and $\varphi_2$]
We apply the trigonometric identity
\begin{align*}
2\cos A\sin B = \sin(A + B) - \sin(A - B)
\end{align*}
with $A = (\varphi_1 + \varphi_2)/2$ and $B = (\varphi_1 - \varphi_2)/2$. Then $A + B = \varphi_1$ and $A - B = \varphi_2$, giving
\begin{align*}
2\cos\!\left(\frac{\varphi_1 + \varphi_2}{2}\right)\sin\!\left(\frac{\varphi_1 - \varphi_2}{2}\right) = \sin\varphi_1 - \sin\varphi_2.
\end{align*}
Substituting into the equation from the previous step:
\begin{align*}
(\varphi_1)_{\xi\tau} - (\varphi_2)_{\xi\tau} = \sin\varphi_1 - \sin\varphi_2.
\end{align*}
[/step]
[step:Rearrange to obtain the equivalence of the two sine-Gordon equations]
Rewriting the identity as
\begin{align*}
\bigl((\varphi_1)_{\xi\tau} - \sin\varphi_1\bigr) - \bigl((\varphi_2)_{\xi\tau} - \sin\varphi_2\bigr) = 0,
\end{align*}
we see that $(\varphi_1)_{\xi\tau} - \sin\varphi_1 = (\varphi_2)_{\xi\tau} - \sin\varphi_2$ at every point. In particular:
- If $(\varphi_1)_{\xi\tau} = \sin\varphi_1$, then $(\varphi_2)_{\xi\tau} - \sin\varphi_2 = 0$, so $\varphi_2$ satisfies the sine-Gordon equation.
- Conversely, if $(\varphi_2)_{\xi\tau} = \sin\varphi_2$, then $(\varphi_1)_{\xi\tau} - \sin\varphi_1 = 0$, so $\varphi_1$ satisfies the sine-Gordon equation.
This establishes that the Backlund system is an auto-Backlund transformation for the sine-Gordon equation.
[guided]
The structure of this proof is characteristic of Backlund transformation verifications: one differentiates one equation of the pair with respect to the "other" independent variable, substitutes the second equation of the pair to close the system, and then uses algebraic identities to decouple the two unknown functions. The remarkable feature is that the nonlinear sine function, which would ordinarily prevent any such decoupling, cooperates precisely because of the product-to-sum identity. This is not a coincidence --- it reflects the integrability of the sine-Gordon equation. For a generic nonlinear PDE, no such parameter-dependent system of first-order equations exists, and no trigonometric identity would produce the required decoupling.
Note also that the proof is symmetric in $\varphi_1$ and $\varphi_2$: the identity $(\varphi_1)_{\xi\tau} - \sin\varphi_1 = (\varphi_2)_{\xi\tau} - \sin\varphi_2$ treats both functions on equal footing. This reflects the fact that the Backlund transformation is invertible: given $\varphi_2$, one can recover $\varphi_1$ by solving the same system.
[/guided]
[/step]
