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Solving ODE: DSolve gives no output

Posted 6 months ago

I have this ODE equation:

$\frac{1}{3\phi } \left( \frac{\dot{a}\dot{\phi} }{a} + \ddot{\phi} \right) + 2 \frac{\ddot{a}}{a} = \partial_i \partial_j \phi$

Where $ \phi(x,t) = \frac{1}{4} e^{k(t-x)} - \frac{1}{4} e^{-k(t-x)} $

Can this equation be solved analytically to get $a(t)$?

Here is my trial:

f[x_, t_, k_] := (1/4)*Exp[k*(t - x)] - (1/4)*Exp[(-k)*(t - x)]

eq[x_, t_, k_] = (1/(3*f[x, t, k]))*((D[a[t], t]*D[f[x, t, k], t])/a[t] + 
          D[f[x, t, k], {t, 2}]) + 2*(D[a[t], {t, 2}]/a[t]) - D[f[x, t, k], {x, 2}]

DSolve[{eq[x, t, k] == 0, a[0] == 0}, a[t], t]

Which returns the same DSolve input. So any help to get a[t]

I think seprations of variables is needed, cause a part of the equation depends on t and the other part depends on x.

POSTED BY: Safinaz Salem
2 Replies
Posted 6 months ago

Thank you for answer.

POSTED BY: Safinaz Salem

The function a[t] is at the denominator of the equation. This makes the initial datum a[0] == 0 singular. However, removing the datum does not solve the equation, except in the special case

DSolve[{eq[1, t, 1] == 0}, a[t], t]

Taking the inspiration from the solution formula I tried this change of independent variable:

DSolve[Numerator@Simplify[eq[x, t, k] /.
      {a -> Function[z, b[Exp[k (z - x)]]]} /.
     (Solve[E^(k (t - x)) == z, t][[1]] /. C[1] -> 0)] == 0,
 b[z], z]

which gives a solution formula in version 13.3.

POSTED BY: Gianluca Gorni
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