Chapter 15 - Differentiation in Several Variables - 15.3 Partial Derivatives - Exercises - Page 783: 83

We take the derivatives of the function $u\left( {x,t} \right) = sec{h^2}\left( {x - t} \right)$ and show that it satisfies the Korteweg-deVries equation: $4{u_t} + {u_{xxx}} + 12u{u_x} = 0$

We have $u\left( {x,t} \right) = sec{h^2}\left( {x - t} \right)$. From Section 7.9 on page 378, we have $\frac{d}{{dv}}sechv = - sechv\tanh v$ $\frac{d}{{dv}}\tanh v = sec{h^2}v$ 1. Find the derivatives with respect to $t$: ${u_t} = \left( {2sech\left( {x - t} \right)} \right)\left( { - sech\left( {x - t} \right)\tanh \left( {x - t} \right)} \right)\left( { - 1} \right)$ $ = 2sec{h^2}\left( {x - t} \right)\tanh \left( {x - t} \right)$ 2. Find the derivatives with respect to $x$: ${u_x} = \left( {2sech\left( {x - t} \right)} \right)\left( { - sech\left( {x - t} \right)\tanh \left( {x - t} \right)} \right)$ $ = - 2sec{h^2}\left( {x - t} \right)\tanh \left( {x - t} \right)$ $ = - 2u\tanh \left( {x - t} \right)$ ${u_{xx}} = - 2{u_x}\tanh \left( {x - t} \right) - 2usec{h^2}\left( {x - t} \right)$ $ = - 2{u_x}\tanh \left( {x - t} \right) - 2{u^2}$ ${u_{xxx}} = - 2{u_{xx}}\tanh \left( {x - t} \right) - 2{u_x}sec{h^2}\left( {x - t} \right) - 4u{u_x}$ $ = - 2{u_{xx}}\tanh \left( {x - t} \right) - 2{u_x}u - 4u{u_x}$ $ = - 2{u_{xx}}\tanh \left( {x - t} \right) - 6u{u_x}$ From previous results we have ${u_x} = - 2sec{h^2}\left( {x - t} \right)\tanh \left( {x - t} \right)$ ${u_{xx}} = - 2{u_x}\tanh \left( {x - t} \right) - 2usec{h^2}\left( {x - t} \right)$ $ = 4sec{h^2}\left( {x - t} \right){\tanh ^2}\left( {x - t} \right) - 2sec{h^4}\left( {x - t} \right)$ Substituting these in ${u_{xxx}}$ ${u_{xxx}} = - 2{u_{xx}}\tanh \left( {x - t} \right) - 6u{u_x}$ gives ${u_{xxx}} = - 8sec{h^2}\left( {x - t} \right){\tanh ^3}\left( {x - t} \right) + 4sec{h^4}\left( {x - t} \right)\tanh \left( {x - t} \right)$ $ + 12sec{h^4}\left( {x - t} \right)\tanh \left( {x - t} \right)$ ${u_{xxx}} = - 8sec{h^2}\left( {x - t} \right){\tanh ^3}\left( {x - t} \right) + 16sec{h^4}\left( {x - t} \right)\tanh \left( {x - t} \right)$ In summary we have: $u\left( {x,t} \right) = sec{h^2}\left( {x - t} \right)$ ${u_t} = 2sec{h^2}\left( {x - t} \right)\tanh \left( {x - t} \right)$ ${u_x} = - 2sec{h^2}\left( {x - t} \right)\tanh \left( {x - t} \right)$ ${u_{xxx}} = - 8sec{h^2}\left( {x - t} \right){\tanh ^3}\left( {x - t} \right) + 16sec{h^4}\left( {x - t} \right)\tanh \left( {x - t} \right)$ Evaluate $4{u_t} + {u_{xxx}} + 12u{u_x}$ $4{u_t} + {u_{xxx}} + 12u{u_x}$ $ = 8sec{h^2}\left( {x - t} \right)\tanh \left( {x - t} \right)$ ${\ \ }$ $ - 8sec{h^2}\left( {x - t} \right){\tanh ^3}\left( {x - t} \right) + 16sec{h^4}\left( {x - t} \right)\tanh \left( {x - t} \right)$ ${\ \ }$ $ - 24sec{h^4}\left( {x - t} \right)\tanh \left( {x - t} \right)$ $ = 8sec{h^2}\left( {x - t} \right)\tanh \left( {x - t} \right)$ ${\ \ }$ $ - 8sec{h^2}\left( {x - t} \right){\tanh ^3}\left( {x - t} \right)$ ${\ \ }$ $ - 8sec{h^4}\left( {x - t} \right)\tanh \left( {x - t} \right)$ $ = \left( {8sec{h^2}\left( {x - t} \right)\tanh \left( {x - t} \right)} \right)\left( {1 - {{\tanh }^2}\left( {x - t} \right) - sec{h^2}\left( {x - t} \right)} \right)$ But the hyperbolic functions have the identity: $1 - {\tanh ^2}\left( {x - t} \right) = sec{h^2}\left( {x - t} \right)$ Therefore, $4{u_t} + {u_{xxx}} + 12u{u_x} = 0$. Hence, $u\left( {x,t} \right) = sec{h^2}\left( {x - t} \right)$ satisfies the Korteweg-deVries equation.