Problem 9 - ’ = 0 Integrate both sides of the equation...

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Problem 9 Long wavelength limit of the Rayleigh-Taylor Instability In this problem, we obtain the growth rate of the RT instability in the long wavelength limit, ie, for kL << 1. The opposite limit was done in class. Proceed as follows: All the equations governing the RT cells as derived in class are still valid. Just that now we will make the assumption that k << |n 0 ’/ n 0 |. For simplicity, assume that |p 0 ’/ p 0 | << |n 0 ’/ n 0 |, and thus S 0 ’ = - γ n 0 ’/ n 0 ; in fact, for simplicity, assume that p 0 ’ = 0. Primes denote d/dx. Since we want n 0 ’ to be sharply varying compared to k, we may go to the limit that the density variation in x is a step function, ie, n 0 (x) = n 1 + (n 2 – n 1 ) H(x), where H(x) is a step function going from 0 to 1. In this case, we solve the eigenvalue equation [the equation immediately following (7) in Topic 6] separately for x < 0 and x > 0 and then match across x=0. To do the matching, we will need “jump conditions” across x=0. Thus, 1. Start with Eq (7) for p
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Unformatted text preview: ’ = 0. Integrate both sides of the equation from x=-ε to x=+ ε , proceed to ε-> 0, and in so doing obtain the jump, if any, in u x . You may assume that u y is at worst piecewise discontinuous across x=0? Is u x continuous across x=0? 2. Now write down the eigenvalue equation, simplifying S ’ as suggested above. Integrate both sides of this equation from x=-ε to x=+ ε and in so doing obtain the jump in u x ’ across x=0. 3. Solve the eigenvalue equation for u x separately for x > 0 and for x < 0, assuming that u x (x) -> 0 as |x| -> infinity. There will be 2 arbitrary constants when you do this. 4. Apply the 2 jump conditions from 1 and 2 above. Thus, obtain the dispersion relation and so find ω 2 as a function of given k. 5. Compare with the short wavelength limit as done in class (for p ’ small). Make a sketch of | ω | vs k, for kL <<1 and kL >> 1, with a reasonable interpolation thru kL ~ 1....
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Problem 9 - ’ = 0 Integrate both sides of the equation...

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