Solution to16g″=vg3+kg3−kg

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ophelialee4xn · Accepted Answer

Step 1It depends, of course, on the boundary conditions. As written the problem actually has no solution because&amp;nbsp;g→&amp;nbsp;as&amp;nbsp;x→∞&amp;nbsp;is obtainable only when&amp;nbsp;v=0. For nonzero v you find that assuming&amp;nbsp;g→1&amp;nbsp;gives the second derivative a nonzero limit which is contradictory. If you mean to force a constant value at infinity, you must allow this constant to vary with v and k so that it properly corresponds to a zero of g&#039;&#039;. Then, it turns out, you always have the hyperbolic tangent solution.To wit, render&amp;nbsp;g=ah&amp;nbsp;for some constant a, and then16ah&#039;&#039;=(v+k)a3h3-kahThen you eliminate the constant term from the first integral equation by setting a to match&amp;nbsp;(v+k)a3&amp;nbsp;with ka. So for a nonzero solution as you presumably want,a=±kv+kwhere the sign corresponds to an overall factor of&amp;nbsp;±1&amp;nbsp;in the solution for&amp;nbsp;g&amp;nbsp;itself (the differential equation has odd parity). For this value of a the constant-limit boundary condition turns out to beh=1,∴g=a=generally1When the substitution is carried through, the equation for h becomes identical to the equation for g already solved with&amp;nbsp;v=0, soh=tanh(3kx),g=atanh(3kx).

Solution to \frac{1}{6}g''=v g^3+k g^3-k g

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