Solving $0=-\lambda \phi (t)^3+\mu ^2 \phi (t)+\phi ''(t)$

Question

I happen to know that the equation $$0=-\lambda \phi (t)^3+\mu ^2 \phi (t)+\phi ''(t)$$ has a simple solution: $$\phi(t) = \frac{\mu \tanh \left(\frac{\mu \left(t-t_0\right)}{\sqrt{2}}\right)}{\sqrt{\lambda }}$$ which is easily verified:

Block[
    {eqn, \[Phi]},
    \[Phi][t_] := \[Mu]/Sqrt[\[Lambda]] Tanh[(\[Mu] (t - t0))/Sqrt[2]];
    eqn = \[Phi]''[t] + \[Mu]^2 \[Phi][t] - \[Lambda] \[Phi][t]^3 == 0;
    FullSimplify[eqn, 
  Assumptions -> And[\[Mu] > 0, \[Lambda] > 0, t > t0]]
 ]
Out[137]=True

But Mathematica has a hard time providing a useful solution. It gives some very strange functions which I cannot reduce in any way to the known solution. In fact if you plot them they look like the given solution, rotated by 90 degrees. Is there a way to get it to simplify? Do not suppose the form of the solution given above in your answer to the question. How could you use Mathematica to find it? Here is the present solution using DSolve:

$$\left\{\left\{\phi (t)\to -\frac{2 i c_1 \sqrt{\lambda } \sqrt{-\frac{1}{\mu ^2-\sqrt{\mu ^4-2 c_1 \lambda }}} \text{sn}\left(\frac{\sqrt{\mu ^2 t^2+\sqrt{\mu ^4-2 \lambda c_1} t^2+2 \mu ^2 c_2 t+2 \sqrt{\mu ^4-2 \lambda c_1} c_2 t+\mu ^2 c_2{}^2+\sqrt{\mu ^4-2 \lambda c_1} c_2{}^2}}{\sqrt{2}}|\frac{\mu ^2-\sqrt{\mu ^4-2 \lambda c_1}}{\mu ^2+\sqrt{\mu ^4-2 \lambda c_1}}\right)}{\mu ^2+\sqrt{\mu ^4-2 c_1 \lambda }}\right\},\left\{\phi (t)\to \frac{2 i c_1 \sqrt{\lambda } \sqrt{-\frac{1}{\mu ^2-\sqrt{\mu ^4-2 c_1 \lambda }}} \text{sn}\left(\frac{\sqrt{\mu ^2 t^2+\sqrt{\mu ^4-2 \lambda c_1} t^2+2 \mu ^2 c_2 t+2 \sqrt{\mu ^4-2 \lambda c_1} c_2 t+\mu ^2 c_2{}^2+\sqrt{\mu ^4-2 \lambda c_1} c_2{}^2}}{\sqrt{2}}|\frac{\mu ^2-\sqrt{\mu ^4-2 \lambda c_1}}{\mu ^2+\sqrt{\mu ^4-2 \lambda c_1}}\right)}{\mu ^2+\sqrt{\mu ^4-2 c_1 \lambda }}\right\}\right\}$$

Answer 1

Applying the chain rule \[Phi]''[t]==\[Phi]s[\[Phi]] \[Phi]s'[\[Phi]]the ode can be reduced to first order for \[Phi]s[\[Phi]]==\[Phi]'[t] .

From the initial ode we know \[Phi]''[0]==0( because \[Phi][0]==0 ) which implies \[Phi]s[\[0]]==0

sol = DSolve[{\[Phi]s[\[Phi]] \[Phi]s'[\[Phi]] + \[Mu]^2 \[Phi] - \[Lambda] \[Phi] ^3 == 0 ,\[Phi]s[0]\[Equal]0  }, \[Phi]s, \[Phi]]

    (*{{\[Phi]s -> Function[{\[Phi]}, -(Sqrt[-\[Phi]^2 (2 \[Mu]^2 - \[Lambda] \[Phi]^2)]/Sqrt[2])]}, 
    {\[Phi]s ->Function[{\[Phi]},Sqrt[-\[Phi]^2 (2 \[Mu]^2 - \[Lambda] \[Phi]^2)]/Sqrt[2]]}}*)

Transforming back into time space we get for t[\[Phi]]

Integrate[1/\[Phi]s[\[Phi]] /. sol, \[Phi] ]// Simplify[#, \[Phi] > 0] &
(*{-(ArcTan[Sqrt[-\[Mu]^2 + (\[Lambda] \[Phi]^2)/2]/\[Mu]]/\[Mu]), ArcTan[Sqrt[-\[Mu]^2 + (\[Lambda] \[Phi]^2)/2]/\[Mu]]/\[Mu]}*)

Dissolve after \[Phi] gives the expected solution!

addendum

Perhaps phasespace gives some more insight

Module[{}, 
 Manipulate[
  Show[{StreamPlot[{\[Phi]p, -\[Mu] \[Phi] + \[Lambda] \[Phi]^3}, {\
\[Phi], -Pi/2, Pi/2}, {\[Phi]p, 0, 1}, 
     FrameLabel -> {"\[Phi]", "\[Phi]'[\[Phi]]"}], 
    ParametricPlot[
     Evaluate[{(\[Mu] Tanh[(t \[Mu])/Sqrt[2]])/Sqrt[\[Lambda]], 
       D[(\[Mu] Tanh[(t \[Mu])/Sqrt[2]])/Sqrt[\[Lambda]], 
        t]}], {t, -Pi , Pi }, PlotStyle -> Red]}], {{\[Lambda], 1}, 0,
    3, Appearance -> "Labeled"}, {{\[Mu], 1}, 0, 3, 
   Appearance -> "Labeled"} , ControlPlacement -> {Top }] 
 ]

Red curve is the special solution (OP's simple solution) which separates stable and unstable regions (separatrix)

Answer 2

In general our equation can be rewritten as a first order ODE if we multiply it by $\;\phi'(t)$ and integrate once: $$0=-\lambda \phi (t)^3 \phi'(t)+\mu ^2 \phi (t) \phi'()t+\phi ''(t) \phi'(t)$$ $$0=-\frac{c_1}{4}-\frac{\lambda}{4} \phi (t)^4 +\frac{\mu ^2}{2} \phi(t)^2 +\frac{\phi'(t)^2}{2}$$ This is a standard technique of recasting certain autonomous nonlinear second order ODEs to first order ones, usually used in case of elliptic functions underlying behind.
Now we can rewrite it in the following way: $$\int\frac{ dt}{\sqrt{2}}=\int \frac{d w}{\sqrt{w^4 -2\mu ^2 w^2 +c}}$$ where $w =\sqrt{\lambda}\; \phi\;$ and $c_1 \lambda =c$. The rhs in general is an elliptic integral an cannot be calculated in terms of elementary functions (this Integrate[1/Sqrt[w^4 -2 \[Mu]^2 w^2 + c], w] // TraditionalForm yields

and an inverse function is an elliptic function expressed in terms of JacobiSN which we get when solving this ODE without initial conditions),

however for a special case when $c=\mu^4$ it might be reduced to $$\int \frac{d w}{w^2 -\mu ^2}=-\frac{arcth(\frac{w}{\mu})}{\mu}$$ and this is the case of the solution in the question.

Answer 3

Clear["Global`*"]

eqn = ϕ''[t] + μ^2 ϕ[t] - λ ϕ[t]^3 == 0;

rule = ϕ -> (μ/Sqrt[λ] Tanh[(μ (# - t0))/Sqrt[2]] &);

eqn /. rule // Simplify

(* True *)

To get the desired form you need consistent initial conditions.

{ϕ0, ϕp0} = {ϕ[0], ϕ'[0]} /. rule

(* {-((μ Tanh[(t0 μ)/Sqrt[2]])/Sqrt[λ]), 
    (μ^2 Sech[(t0 μ)/Sqrt[2]]^2)/(Sqrt[2] Sqrt[λ])} *)

Using the initial conditions

sol = DSolve[{eqn, ϕ[0] == ϕ0, ϕ'[0] == ϕp0}, ϕ, t][[1]]

(* {ϕ -> Function[{t}, (μ Tanh[1/2 (Sqrt[2] t μ - Sqrt[2] t0 μ)])/
   Sqrt[λ]]}8)

(ϕ[t] /. sol // Simplify) === (ϕ[t] /. rule)

(* True *)

Answer 4

If we look up the Jacobi theta functions in the DLMF, we'll find limiting values not included in the expressions returned by DSolve. They lie on the closure of the solution manifold, and it is typical for the general solution returned by DSolve to be missing such special solutions.

Following the implicit advice of the DLMF, we can recover the complete solution.

eqn = ϕ''[t] + μ^2 ϕ[t] - λ ϕ[t]^3 == 0;
dsol = DSolve[{eqn}, ϕ, t];

Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.

Length@dsol            (* two solutions *)
ϕ[t] /. dsol // Total  (* of opposite signs *)

(*
  2
  0
*)

phisol = ϕ[t] /. First@dsol // FullSimplify (* show just one *)

$$\frac{\displaystyle i\, \text{sn}\left(\frac{\sqrt{\left(\mu ^2+\sqrt{\mu ^4-2 \lambda c_1}\right) (t+c_2){}^2}}{\displaystyle \sqrt{2}}\,\Bigg|\,\frac{\displaystyle \mu ^4-\sqrt{\mu ^4-2 \lambda c_1} \mu ^2-\lambda c_1}{\lambda c_1}\right)}{\sqrt{\frac{\displaystyle \lambda }{\displaystyle -\mu ^2+\sqrt{\mu ^4-2 c_1 \lambda }}}}$$

One limiting value is the limit of $\text{sn}(z,k)$ as $k\rightarrow1$. This yields the solution the desired by the OP, if we clean it up with some assumptions about the parameters:

lim = Solve[(μ^4 - λ C[1] - μ^2 Sqrt[μ^4 - 2 λ C[1]])/(λ C[1]) == 1, 
   C[1]][[1, 1]]

(*  C[1] -> μ^4/(2 λ)  *)

Limit[phisol, lim] // Simplify[#, {λ, μ} ∈ PositiveReals] &

$$\frac{\mu \tanh \left(\frac{\mu \sqrt{(t+c_2){}^2}}{\sqrt{2}}\right)}{\sqrt{\lambda }}$$

The other limiting value is the limit of $\text{sn}(z,k)$ as $k\rightarrow0$. This yields a solution even simpler than the one desired by the OP, namely the trivial solution. This is impossible to solve for algebraically directly from the expression for $k$. Instead we solve for an ansatz and verify it.

(* numerator of k *)
Solve[μ^4 - λ C[1] - μ^2 Sqrt[μ^4 - 2 λ C[1]] == 0, C[1]]

Solve::nongen: There may be values of the parameters for which some or all solutions are not valid.

(*  {{C[1] -> 0}}  *)

(* limit of k *)
Limit[(μ^4 - λ C[1] - μ^2 Sqrt[μ^4 - 2 λ C[1]])/(λ C[1]),
  C[1] -> 0,
  Assumptions -> μ ∈ Reals]

(*  0  *)