Manipulate[
tMax = 4 Pi;
tTicks = {0, Pi, 2Pi, 3Pi, 4Pi};
u0min = 0;
Which[
σ == -1 && τ == -1, minV = -.5; maxV = .1; um =1; u0m = .2; minE = 0; maxE = 1; tMax = 3; tTicks = Automatic,
σ == -1 && τ == 1, minV = -0.26; maxV = 1; um = 2;u0min = 1;u0m = 1.78;minE = 0; maxE = 1,
σ == 1 && τ== -1,minV = -0.05; maxV = .26; um = 1.5; u0m = 1.0;minE =0; maxE = .25,
σ == 1 && τ == 1, minV = -0.1; maxV = 2; um = 1.5;u0m = 1.4;minE = 0; maxE = 2] ;
u0 = u0min + u0slider * (u0m- u0min);
energy = V[u0, σ, τ];
upMax = (Sqrt[2(maxE-minV)]); (* u prime max *)
pVE =Plot[{V[u,σ, τ], energy}, {u, -um,um}, PlotRange ->{minV, maxV}, PlotStyle ->{Blue, Red},AxesLabel -> {"u", "V, E"}, AspectRatio -> 1, Epilog -> Point[{u0, V[u0, σ, τ]}]];
puup = ContourPlot[(1/2)p^2 + V[u,σ,τ] == energy, {u, -um,um}, {p, -upMax,upMax}, Frame -> False, Axes -> True, AxesLabel ->{"u", "u'"}, AspectRatio -> 1];
soln = Part[NDSolve[{u''[t] + f[u[t],σ, τ] == 0, u[0] ==  u0, u'[0] ==  0}, u,{t, 0, tMax}], 1]; (* takes the first (and only) numerical solution to the ODE *)
uvst = Plot[u[t]/.soln, {t, 0, tMax}, PlotRange -> {-um, um}, AxesLabel ->{"t", "u"}, Ticks ->{tTicks, Automatic}, AspectRatio -> 1];
GraphicsRow[{pVE, puup, uvst}, ImageSize -> 600]
,{{σ, -1}, {-1,1}}, {{τ, 1},{-1,1}}, {{u0slider, .5311, "(Subscript[u, 0])"}, 0, 1}, ControlPlacement -> Top, Initialization ->{V[u_, s_, t_] := (1/2) s u^2 + (1/4) t u^4, f[u_, s_, t_]:= s u + t u^3; u0 = 0.25671}]

The left figure shows a graph of the potential, V(u) = 1/2 σ (u^2) + 1/4 τ (u^4), and the constant energy, E.  By scaling of position u and time t, we can set any nonzero σ and τ to ±1.
The middle figure shows the energy surface, 1/2 (((u') )^2)+ V(u) = E, 
The right figure shows the solution u as a function of t, to the IVP u’’= -V’(u), u(0) = (Subscript[u, 0]), u’(0) = 0