[proofplan] We verify by direct substitution that $u(x,t) = -2\chi_1^2\operatorname{sech}^2(\chi_1(x - 4\chi_1^2 t))$ satisfies $u_t + u_{xxx} - 6uu_x = 0$. Setting $\eta = \chi_1(x - 4\chi_1^2 t)$, we compute $u_t$, $u_x$, and $u_{xxx}$ using the chain rule and hyperbolic identities, then collect all terms and verify cancellation. [/proofplan] [step:Set up the substitution and compute derivatives of $\operatorname{sech}^2$] Write $u = -2\chi_1^2 \operatorname{sech}^2\eta$ where $\eta = \chi_1(x - 4\chi_1^2 t)$. The partial derivatives of $\eta$ are \begin{align*} \eta_x = \chi_1, \qquad \eta_t = -4\chi_1^3. \end{align*} We need the derivatives of $\operatorname{sech}^2\eta$ with respect to $\eta$. Using $\frac{d}{d\eta}\operatorname{sech}\eta = -\operatorname{sech}\eta\tanh\eta$: \begin{align*} \frac{d}{d\eta}\operatorname{sech}^2\eta &= -2\operatorname{sech}^2\eta\tanh\eta, \\ \frac{d^2}{d\eta^2}\operatorname{sech}^2\eta &= -2\operatorname{sech}^2\eta + 6\operatorname{sech}^2\eta\tanh^2\eta = -2\operatorname{sech}^2\eta(1 - 3\tanh^2\eta), \\ \frac{d^3}{d\eta^3}\operatorname{sech}^2\eta &= 16\operatorname{sech}^2\eta\tanh\eta - 24\operatorname{sech}^2\eta\tanh^3\eta = 8\operatorname{sech}^2\eta\tanh\eta(2 - 3\tanh^2\eta), \end{align*} where we used $\operatorname{sech}^2\eta + \tanh^2\eta = 1$ to simplify. [/step] [step:Compute $u_t$, $u_x$, and $u_{xxx}$ via the chain rule] Applying the chain rule with $u = -2\chi_1^2\operatorname{sech}^2\eta$: \begin{align*} u_x &= -2\chi_1^2 \cdot (-2\operatorname{sech}^2\eta\tanh\eta) \cdot \chi_1 = 4\chi_1^3\operatorname{sech}^2\eta\tanh\eta, \\ u_t &= -2\chi_1^2 \cdot (-2\operatorname{sech}^2\eta\tanh\eta) \cdot (-4\chi_1^3) = -16\chi_1^5\operatorname{sech}^2\eta\tanh\eta, \\ u_{xxx} &= -2\chi_1^2 \cdot 8\operatorname{sech}^2\eta\tanh\eta(2 - 3\tanh^2\eta) \cdot \chi_1^3 = -16\chi_1^5\operatorname{sech}^2\eta\tanh\eta(2 - 3\tanh^2\eta). \end{align*} [/step] [step:Substitute into the KdV equation and verify cancellation] We also need $uu_x$: \begin{align*} uu_x &= (-2\chi_1^2\operatorname{sech}^2\eta)(4\chi_1^3\operatorname{sech}^2\eta\tanh\eta) = -8\chi_1^5\operatorname{sech}^4\eta\tanh\eta. \end{align*} Now substitute into $u_t + u_{xxx} - 6uu_x$: \begin{align*} u_t + u_{xxx} - 6uu_x &= -16\chi_1^5\operatorname{sech}^2\eta\tanh\eta - 16\chi_1^5\operatorname{sech}^2\eta\tanh\eta(2 - 3\tanh^2\eta) + 48\chi_1^5\operatorname{sech}^4\eta\tanh\eta. \end{align*} Factor out $-16\chi_1^5\operatorname{sech}^2\eta\tanh\eta$: \begin{align*} &= -16\chi_1^5\operatorname{sech}^2\eta\tanh\eta\left[1 + (2 - 3\tanh^2\eta) - 3\operatorname{sech}^2\eta\right]. \end{align*} The bracket simplifies using $\operatorname{sech}^2\eta = 1 - \tanh^2\eta$: \begin{align*} 1 + 2 - 3\tanh^2\eta - 3(1 - \tanh^2\eta) = 3 - 3\tanh^2\eta - 3 + 3\tanh^2\eta = 0. \end{align*} Therefore $u_t + u_{xxx} - 6uu_x = 0$ identically, confirming $u$ is a solution. [/step]