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I am programming a solution to Lamb's problem for a point source in 3-D that is outlined in Richards (1979), but I am confused by how the polynomial

$$ (A - 2P^2)^4 - 16X^2Y^2P^4 $$

is cubic in $P^2$. $\\$

The dimensionless quantities $A, T, X, \ \text{and} \ Y$ are defined as: $\\$

\begin{aligned} A &= \frac{\alpha^2}{\beta^2} (\alpha=\text{P-wave speed}, \ \beta=\text{S-wave speed})\\ T &= \frac{\alpha t}{r} (t=\text{time}, \ r=(x_1^2 + x_2^2)^{1/2} \ (x_1, x_2) \ \text{being the location of the receiver at the free surface.)}\\ X &= (1 - P^2)^{1/2} \ \text{or} \ -i(P^2 - 1)^{1/2} \\ Y &= (A - P^2)^{1/2} \ \text{or} \ -i(P^2 - A)^{1/2} \\ \end{aligned} $\\$

Apparently, $$ (A - 2P^2)^4 - 16X^2Y^2P^4 = 16(1 - A)(P^2 - R_1)(P^2 - R_2)(P^2 - R_3) $$ where $R_1, R_2, \ \text{and} \ R_3 $ are roots of the Rayleigh cubic in $P^2$. $\\$

Richards notes that an effective approach would be to find the largest root with $R_3^{1/2} = \frac{\alpha^2}{\gamma^2}$ ($\gamma = $ Rayleigh wave speed) and factorise $P^2 - R_3$ from the cubic and solve a quadratic for $R_1$ and $R_2$. $\\$

Ultimately I need to be able to solve for the roots for any given $A$, so my thought was to simply put it in standard polynomial form and use an algorithm such as numpy.roots([c1, c2, c3, c4]), where c1, c2, c3, and c4 are the coefficients of the standard form polynomial. This makes sense, but the polynomial is quartic in $P^2$ when I expand it. $\\$

I completely understand the ways of solving for the roots of a polynomial, but I just can't understand how this is a cubic polynomial as Richards describes. Clearly I am misunderstanding what Richards means by 'cubic in $P^2$'. $\\$

If anyone could kindly explain what this means that would be very helpful.

GeoJoe
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1 Answers1

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$X = (1 - P^2)^{1/2} \ \text{or} \ -i(P^2 - 1)^{1/2}$

$\\ Y = (A - P^2)^{1/2} \ \text{or} \ -i(P^2 - A)^{1/2}$

Assuming the "or" are paired between the two, this means either $\,X^2Y^2=(1-P^2)(A-P^2)\,$ or $\,X^2Y^2=(-i)^2(P^2-1)(-i)^2(P^2-A)=(-i)^4(P^2-1)(P^2-A)=(1-P^2)(A-P^2)\,$. Then:

$$ \require{cancel} \begin{align} (A - 2P^2)^4 - 16X^2Y^2P^4 &= \big(A^4 - 8 A^3 P^2 + 24 A^2 P^4 - 32 A P^6 + \bcancel{16 ​P^8}\big) \\ &\quad - 16 P^4 \big(A - (A+1)P^2 + \bcancel{P^4}\big) \\ &= 16\big(- A + 1\big) P^6 + 8A \big( 3 A - 2) P^4 - 8 A^3 P^2 + A^4 \end{align} $$

The latter is a polynomial in $\,P\,$ containing only the $\,0^{th}, 2^{nd}, 4^{th}, 6^{th}\,$ powers, so a cubic in $\,P^2\,$.

dxiv
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