Question

1. Consider the following PDE: \( \phi_{t}+c_{0} \phi_{x}+\delta \phi_{x x x}+\mu \phi_{x x x x x}=0, \quad \) the linear 5th order Korteweg-de Vries equation. Assuming \( \phi(x, t)=a \cos (k x-\omeg

A) a cos(kx−ωt)
B) a sin(kx−ωt)
C) a cos(kx+ωt)
D) a sin(kx+ωt)

Answer 1

The sine function in the solution
`\(\phi(x, t) = a \sin(kx - \omega t)\)`allows for the cancellation of terms in the Korteweg-de Vries equation due to the specific properties of sine derivatives. The negative sign in the expression for
`\(\phi_t\)` and the alternating nature of sine derivatives enable the equation to balance, confirming the validity of option B) as the correct solution.

Step-by-step explanation:

The given linear 5th order Korteweg-de Vries (KdV) equation is
`\(\phi_(t) + c_(0) \phi_(x) + \delta \phi_(xxx) + \mu \phi_(xxxxx) = 0\)`. We are asked to determine the form of the solution
`\(\phi(x, t)\)` among the provided options. Let's consider the proposed solution
`\(\phi(x, t) = a \sin(kx - \omega t)\).`

For clarity, let's compute the partial derivatives involved in the KdV equation. The first-order partial derivative with respect to
`\(t\) (\(\phi_(t)\)) results in \(-a \omega \cos(kx - \omega t)\)`, and the first-order partial derivative with respect to
`\(x\) (\(\phi_(x)\)) yields \(a k \cos(kx - \omega t)\).` The third-order partial derivative with respect to
`\(x\) (\(\phi_(xxx)\)) is \`(-a k^3 \sin(kx - \omega t)\), and the fifth-order partial derivative with respect to
`\(x\) (\(\phi_(xxxxx)\)) is \(a k^5 \sin(kx - \omega t)\).`

Substituting these derivatives into the KdV equation, we find that the terms involving
`\(\phi_(t)\), \(\phi_(x)\), \(\phi_(xxx)\)`, and
`\(\phi_(xxxxx)\)`cancel each other, satisfying the KdV equation. Therefore, the given solution
`\(\phi(x, t) = a \sin(kx - \omega t)\)` is a valid solution to the linear 5th order Korteweg-de Vries equation.

The correct option is B) a sin(kx−ωt).