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Solitons/Kinks/Conservation laws help



This is a sheet I'm working through to help me learn about solitons. Honestly have no idea how to tackle exercise 4
Original post by 0range

This is a sheet I'm working through to help me learn about solitons. Honestly have no idea how to tackle exercise 4


Isn't this a bit of cavalier differentiation under the integral sign? Differentiating E wrt t you get ututt+uxuxt+utV′u_t u_{tt} + u_x u_{xt} + u_t V' integrated over the real line. Substituting from the K-G equation then simplifies the integrand to utuxx+uxuxtu_t u_{xx} + u_x u_{xt} which is the differential (∂/∂x)(utux)(\partial / \partial x) (u_t u_x) and the integral of this is zero as the total momentum is finite and therefore utux u_t u_x must tend to zero as x→∞x \rightarrow \infty

Don't immediately see the momentum one, though...
Original post by Gregorius
Isn't this a bit of cavalier differentiation under the integral sign? Differentiating E wrt t you get ututt+uxuxt+utV′u_t u_{tt} + u_x u_{xt} + u_t V' integrated over the real line. Substituting from the K-G equation then simplifies the integrand to utuxx+uxuxtu_t u_{xx} + u_x u_{xt} which is the differential (∂/∂x)(utux)(\partial / \partial x) (u_t u_x) and the integral of this is zero as the total momentum is finite and therefore utux u_t u_x must tend to zero as x→∞x \rightarrow \infty

Don't immediately see the momentum one, though...


Maybe this works for the momentum...apologies if it doesn't...I'm tired after trying to get a statistical model to converge all day...

Differentiating under the integral sign, we get uxutt+utuxtu_x u_{tt} + u_t u_{xt} to integrate. The second term is (1/2)(∂/∂x)ut2(1/2) (\partial / \partial x) u_t^{2} and this contributes nothing to the integral because the total energy is finite. (The energy is a sum of positive terms and so each term must tend to zero as x→∞x \rightarrow \infty)

Substituting from the K-G equation, the first term becomes (1/2)(∂/∂x)ux2−uxV′(1/2) (\partial / \partial x) u_x^{2} - u_x V'. In this pair, the same argument applies to the first term and the second term uxV′=dV/dxu_x V' = dV/dx and yet another appeal to finite energy finishes it.

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