I am trying to compute the solution of the fourth-order ODE
$$ y^{\prime \prime} + V(y) - \beta y^{\text{ (IV)} } = 0$$ with $V(x) = -2x + 4 x^3$, on the real line, with boundary conditions $$ \begin{cases} \lim_{x \pm \infty} y(x) = 0 \\ \lim_{x \pm \infty} y^ \prime (x) = 0 \end{cases}$$.
Clearly the trivial solution $ y = 0$ is there, but another solution should exist, with $y(0) \neq 0$. Exponential decay at infinity is expected, so the conditions $$ \begin{cases} \lim_{x \pm \infty} y ^ {\prime \prime} (x) = 0 \\ \lim_{x \pm \infty} y^ {\prime \prime \prime} (x) = 0 \end{cases}$$ should apply. On symmetry grounds, $y^ \prime (0) = 0$ is also expected.
I tried to solve it on an interval $(0,L)$ with NDSolve
such as in
L = 10; s = NDSolve[{y''[x] -2*y[x] + 4*y[x]^3 - y''''[x] == 0, y[L] == 0,
y'[L] == 0, y'[0] == 0, y''[L] == 0}, y, {x, 0, 10}, WorkingPrecision -> 60]
also trying other b.c., such as the third derivative, but no success
FindRoot::cvmit: Failed to converge to the requested accuracy or precision within 100
iterations.
NDSolve::berr: The scaled boundary value residual error of 2.9040010780303314`*^7 indicates
that the boundary values are not satisfied to specified tolerances. Returning the best solution
I tried the Method -> {"Shooting", "StartingInitialConditions" -> {y`[0] == 0}}
, using the boundary conditions at the extremity $L$ for NDSolve
, to no avail
NDSolve::mxst: Maximum number of 10000 steps reached at the point x ==
6.2422167664683852553263167777483585328023492585909723902636`30..
I have searched the site and found similar "separatrix" problems addressed in the answers Computing separatrix of second order nonlinear autonomous ode, as well as Numerical solution of nonlinear boundary value problem, but I am unsure they are applicable. The former seems to me to rely on a particular structure for a system of second order ODEs, an the latter requires knowledge of the $y(0)$ value, if I am not missing the point.
Any hint would be most appreciated, thanks
EDIT
The differential equation has a physical interpretation as the Euler-Lagrange equation for the functional
$$ \int _{- \infty} ^{\infty } \frac{1}{2} y^{\prime 2} + U(y) + \frac{\beta}{2} y^{\prime \prime 2} \mathrm{d}x $$ representing the energy of a stretchable beam on an elastic foundation $U(x) = x^2 - x^4$. The first term represents the stretching energy, the second the potential energy, the third the bending energy. I thought about using this fact, by adding a "guess" boundary condition $ y(0) = y_0$ and then optimise $y_0$ by making the energy, as defined by the functional above, stationary. But even choosing an arbitrary $y_0$ I struggle to solve the ODE, for example choosing two b.c. per end, function value and first derivative,
s = NDSolve[{y''[x] - 2*y[x] + 4*y[x]^3 - y''''[x] == 0, y[L] == 0,
y'[L] == 0, y'[0] == 0, y[0] == 1}, y, {x, 0, 10},
WorkingPrecision -> 30, MaxSteps -> 10000]
returns
FindRoot::cvmit: Failed to converge to the requested accuracy or precision
within 100 iterations.
NDSolve::berr: The scaled boundary value residual error of
3.0598614681511925`*^14 indicates that the boundary values are not
satisfied to specified tolerances. Returning the best solution found.
which I am puzzled about, I am not so sure what makes this BVP so challenging.
EDIT 2
I am trying to use the approach of heat convection differential equations from 1952 - Mathematica fails to converge. If it could be of any utility, the (interesting) solution of the above BVP for $\beta = 0$ is $$ y(x) = \operatorname{sech} (\sqrt{2}x)$$ as Mathematica confirms
eq = u''[x] - 2 u[x] + 4 u[x]^3 == 0
FullSimplify[eq /. u -> Function[{x}, Sech[ Sqrt[2] x]]]
(*true*)
y''[0]==1
, and not asy''[L]==1
. Is it correct? $\endgroup$y'' [L] = 0
. I think nothing can be said for the second derivative at the origin, while the conditiony' [0] = 0
is expected due to symmetry, on top of all the b.c. at infinity. $\endgroup$