14
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Mathematica 10 has a splendid new function, NDEigensystem, that makes it possible to solve Sturm-Liouville problems numerically in a single step. I have not however been able to find a way to get it to solve a problem with periodic boundary conditions. In particular, the u[0]==u[1] idiom that works in NDSolve causes NDEigensystem to return unevaluated. DirichletCondition and NeumannValue, by the nature of their calling sequences, don't seem to allow one to specify a relationship between function values or derivatives at distinct points (for easily understandable reasons having to do with the finite element implementation).

Some details: I'm trying to solve a fairly simple eigensystem on the unit circle. My operator is

$$L\phi(\theta)={\phi^{\prime\prime}(\theta)+(V^{\prime\prime}(\theta)-V^\prime(\theta)^2)\phi(\theta)}$$

$V(\theta)$ is a potential energy function to be specified numerically. (This, by the way, is not in S-L form, but it is equivalent to an S-L problem, after massaging to get rid of a weighting function that NDEigensystem doesn't know how to handle.) NDEigensystem has no problem with this on straightforward domains like $[0,1]$. It's only the periodic problem that I don't know how to solve.

Possibly I just need to wait for the next version of Mathematica, but any ideas for solving it now would be appreciated.

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  • $\begingroup$ Unfortunately, the FEM in V10.3 does not yet allow for periodic boundary conditions. If really needed one could use the low level functions to write that up though. $\endgroup$ – user21 Dec 9 '15 at 19:05
  • $\begingroup$ Yes, I saw your answer at mathematica.stackexchange.com/questions/71710/… . $\endgroup$ – Leon Avery Dec 9 '15 at 19:07
7
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In the apparent absence of a direct solution using NDEigensystem to the periodic problem posed in the question, ParametricNDSolveValue can be used.

V[θ_] := Cos[θ];
s = ParametricNDSolveValue[{ϕ''[θ] + (V''[θ] - V'[θ]^2 - a) ϕ[θ] == 0, 
    ϕ'[Pi/2] == 1, ϕ[0] == ϕ[2 Pi]}, ϕ, {θ, 0, 2 Pi}, {a}];

(The condition ϕ'[Pi/2] == 1 can, in principle, be applied for any value of θ, but Pi/2 works well in practice.) Periodicity then requires that s[a]'[0] - s[a]'[2 Pi] == 0. A plot of this function indicates the approximate locations of the eigenvalues a. (This Plot is slow.)

Plot[s[a]'[0] - s[a]'[2 Pi], {a, -30, 1}, AxesLabel -> {δϕ, a}, 
    PlotRange -> {Automatic, {-10, 10}}]

enter image description here

after which FindRoot can be used to obtain more precise roots, for instance.

a /. Quiet@FindRoot[s[a]'[0] - s[a]'[2 Pi] == 0, {a, -1.65, -1.64}]
(* -1.64737 *)

The first six eigenvalues, obtained in this way, and corresponding eigenfunctions are

ev = {0, -1.647365415724475`, -4.540588488149288`, -9.517577894696032`, 
      -16.509830069588382`, -25.506276514311562`};
norm[a_] := Sqrt[NIntegrate[(s[a][θ])^2, {θ, 0, 2 Pi}]]
Plot[Evaluate[Quiet@Table[s[a][θ]/norm[a], {a, Join[{0}, ev]}]], {θ, 0, 2 Pi},
    AxesLabel -> {θ, ϕ}]

enter image description here

A similar computation can be performed with the roles of ϕ[θ] and ϕ'[θ] reversed.

s = ParametricNDSolveValue[{ϕ''[θ] + (V''[θ] - V'[θ]^2 - a) ϕ[θ] == 0, 
    ϕ[Pi/2] == 1, ϕ'[0] == ϕ'[2 Pi]}, ϕ, {θ, 0, 2 Pi}, {a}];
Plot[s[a][0] - s[a][2 Pi], {a, -30, 1}, AxesLabel -> {δϕ, a},
    PlotRange -> {Automatic, {-10, 10}}]

enter image description here

The crossing are the same to about five significant figures, except that none occur at a = 0. With the 0 eigenvalue deleted from ev,

enter image description here

Comparing the two eigenfunction plots indicates that, with the exception of the a = 0 mode, there are two eigenfunctions for each eigenvalue, much like Sin[m θ] and Cos[m θ]. Depending on the details of how periodicity is imposed, one or another linear combination of each eigenfunction pair is obtained.

Asymmetric effective potentials

Note that the effective potential above, V''[θ] - V'[θ]^2, is symmetric in the domain {θ, 0, 2 Pi}. Effective potentials that are not symmetric would break the two-fold degeneracy. Below is an example.

V[θ_] := Sin[θ] + Sin[2 θ];
Plot[s[a][0] - s[a][2 Pi], {a, -15, 1}, AxesLabel -> {a, δϕ}, 
    PlotRange -> {Automatic, {-10, 10}}]

enter image description here

ev = {0, -0.23268071142546354`, -4.682799571386772`, -6.89410542471997`, 
      -7.6379026933633805`, -11.384296067999848`, -12.495331911000417}

enter image description here

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  • 1
    $\begingroup$ It's a painful solution, but perhaps the best simple (relatively) method for now. In principle much more efficient methods are available by hacking the FEM internals, something like in this answer: mathematica.stackexchange.com/questions/71710/…. Not sure it's worth it to me to work that hard right now. Thanks. $\endgroup$ – Leon Avery Dec 7 '15 at 11:35
  • $\begingroup$ @LeonAvery You also could convert your ODE into a finite difference equation with a the eigenvalue, and solve it with Eigensystem or, perhaps faster, using matrix routines. Alternatively, decompose the equation into Fourier components and solve the result as a matrix equation. Which is better depends on the form of V. $\endgroup$ – bbgodfrey Dec 7 '15 at 14:29
  • 1
    $\begingroup$ I've done the second -- that's basically the Ritz method. It works but is computationally heavy. It's also hard to get a good Fourier approximation when the potential varies widely or rapidly, which can happen. The finite difference method would work quite well, I think, maybe with some adjustment of grid density where the potential misbehaves. Of course, this is close to what happens in the FEM guts of NDEigensystem. Reinventing the wheel is annoying... Finally, I could just use an external FEM package and not try to get MMA to do everything. $\endgroup$ – Leon Avery Dec 7 '15 at 14:37
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I think here is another way to do it. For this we use the low level FEM functions. First we have a utility function that converts a PDE specification to a discrete version.

Needs["NDSolve`FEM`"]
PDEtoMatrix[{pde_, \[CapitalGamma]___}, u_, r__, 
  opts : OptionsPattern[NDSolve`ProcessEquations]] :=
 Module[{ndstate, feData, sd, vd, bcData, methodData, pdeData},
  {ndstate} =
   NDSolve`ProcessEquations[Flatten[{pde, \[CapitalGamma]}], u, 
    Sequence @@ {r}, opts];
  sd = ndstate["SolutionData"][[1]];
  (*vd = ndstate["VariableData"];*)
  feData = ndstate["FiniteElementData"];
  pdeData = feData["PDECoefficientData"];
  bcData = feData["BoundaryConditionData"];
  methodData = feData["FEMMethodData"];
  {pdeData, bcData, vd, sd, methodData}
  ]

We specify the problem as a time dependent problem to get a stiffness and a damping matrix. The initial conditions and time integration domain are arbitrary as we never do an actual time integration:

V[x_] := Cos[x];
eqn = D[u[t, x], t] + 
    Laplacian[u[t, x], {x}] + (V''[x] - V'[x]^2) u[t, x] == 0;
ic = u[0, x] == 0;

We then convert this setup to a discretized version and extract the stiffness and damping matrix.

{pdeData, bcData, vd, sd, md} = 
  PDEtoMatrix[{eqn, ic}, 
   u, {t, 0, 1}, {x} \[Element] Line[{{0}, {2 \[Pi]}}]];
dpde = DiscretizePDE[pdeData, md, sd];
{load, stiffness, damping, mass} = dpde["SystemMatrices"];

Now, for periodic boundary conditions we add a constraint such that the value of node on the left is equal to the value of the node at the right:

dof = md["DegreesOfFreedom"];
mesh = md["ElementMesh"];
{leftNode, rightNode} = 
  Flatten[ElementIncidents[mesh["PointElements"]]];
constraints = ConstantArray[0., {dof}];
constraints[[{leftNode, rightNode}]] = {-1., 1.};
(*Anti-periodic*)
(*constraints[[{leftNode,rightNode}]]={1.,1.};*)

If one wants anti-periodic boundary conditions that can be done as well.

The idea is now to compute the NullSpace of this constraint and project the system matrices to the subspace spanned.

projection = SparseArray[NullSpace[{constraints}]];
projectionT = Transpose[projection];
stiff = projection.stiffness.projectionT;
damp = projection.damping.projectionT;

(As a side note, if anyone has an efficient way/code to compute that sparse null space please let me know)

The eigensystem computation is now done on the subspace:

{vals, funs} = 
  Reverse /@ Eigensystem[{-stiff, damp}, -5, Method -> "Arnoldi"];
(*vals*)

The eigenvectors need to be projected back to create the interpolation functions:

ifuns = 
  ElementMeshInterpolation[{mesh}, #] & /@ (projectionT.# & /@ funs);

Visualize the result:

Plot[Evaluate[Through[ifuns[x]]], {x, 0, 2 \[Pi]}]

Exampe 1

V[x_] := Cos[x];
vals
{0.0000220456, 1.64739, 1.64742, 4.54151, 4.54152}
Plot[Evaluate[Through[ifuns[x]]], {x, 0, 2 \[Pi]}]

enter image description here

For a higher accuracy a finer mesh can be used:

{pdeData, bcData, vd, sd, md} = 
  PDEtoMatrix[{eqn, ic}, 
   u, {t, 0, 1}, {x} \[Element] Line[{{0}, {2 \[Pi]}}], 
   Method -> {"MethodOfLines", 
     "SpatialDiscretization" -> {"FiniteElement", 
       "MeshOptions" -> {"MaxCellMeasure" -> 0.01}}}];

With this I get eigenvalues

{2.71943*10^-11, 1.64737, 1.64737, 4.54059, 4.54059}

Using "Method"->Direct for `Eigensystem get them down to

{0.`, 1.6473653985486567`, 1.6473653985607`, 4.540588371453302`, \
4.540588371506683`}

Example 2 - Anti-symmetric potential

V[x_] := Cos[x] + x;
vals
{0.318589, 1.99807, 3.25784, 5.59323, 5.8009}
Plot[Evaluate[Through[ifuns[x]]], {x, 0, 2 \[Pi]}]

enter image description here

Example 3 - Anti-symmetric potential and anti-periodic BCS:

(*needs*)
constraints[[{leftNode, rightNode}]] = {1., 1.};
V[x_] := Cos[x] + x;
vals
{0.323106, 1.9253, 3.68747, 4.38815, 7.85327}
Plot[Evaluate[Through[ifuns[x]]], {x, 0, 2 \[Pi]}]

enter image description here

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  • 1
    $\begingroup$ Just a quick observation: Comparing to the ParametricNDSolveValue solution by @bbgodfrey, one difference is that this answer finds the degeneracies for the symmetric Cos[x]-derived potential automatically, whereas they are obtained in separate calculations when using ParametricNDSolveValue. But in this symmetric case it looks like the numerical accuracy of this answer is not as good as the previous approach, e.g., if I compare the ground state, which should come out to be zero and is 0.0000220456 in this answer. $\endgroup$ – Jens Mar 18 '16 at 3:40
  • $\begingroup$ @Jens Thanks, added a more accurate version. You would not know about a more efficient way/alternative to compute the NullSpace? $\endgroup$ – user21 Mar 18 '16 at 7:43
  • 1
    $\begingroup$ Sorry, can't think of anything better... anyway, the Q only wanted periodic bc, so the projectors aren't strictly needed here (although it's nice to know how it can be done, of course). But I noticed that you should probably delete the line vd = ndstate["VariableData"] in PDEtoMatrix because it throws an error in version 10.3 (no method for VariableData). Since those data aren't needed anywhere, cutting that line is OK, I think. For other bc, one might also think about using Bloch's theorem (but that's a separate answer). $\endgroup$ – Jens Mar 19 '16 at 4:30
  • 1
    $\begingroup$ @Jens, yes the projection is only needed if the Method->"Arnoldi" is used in Eigensystem, the Method->"Direct" should work. If I can find a solution for the slow NullSpace then this is a very general way to apply boundary conditions. Bloch's theorem is something I have never understood. If I could see a Mathematica code that would help me.... I'll look at this again after the week end. $\endgroup$ – user21 Mar 19 '16 at 7:53
  • $\begingroup$ I posted an answer using Bloch's theorem and finite difference. $\endgroup$ – Jens Mar 19 '16 at 18:17
7
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Periodic potentials and Bloch waves

This is a completely different approach, using functionality that was already present in Mathematica version 8. I'm posting this because it's quite robust and also allows me to demonstrate how Bloch's theorem can be used here to get a more general class of non-periodic solutions in a periodic potential. This is equivalent to imposing boundary conditions that continuously interpolate between between periodic and anti-periodic boundary conditions (as measured by the phase difference between the function values at the left and right boundary). The original question is a special case of the solutions I obtain here.

The starting point would be a periodic potential $U(x)$ with periodicity interval $\ell$, and look for solutions of the form

$$\psi_k(x)\equiv \exp(i k x) u_k(x)$$

where $u_k(x) = u_k(x+\ell)$ is the periodic part. The additional parameter $k$ is the Bloch wave number, and it gives us periodic solutions $\psi_k$ if $k=0$ (or an integer multiple of $2\pi/\ell$). On the other hand, if $k=\pi/\ell$, the solution is "antiperiodic." Any value $k\in[-\pi/\ell,\pi/\ell]$ leads to a unique solution (that's the "first Brillouin zone").

Inserting this Bloch form into the wave equation, we get an equation for $u_k(x)$ which is always solved with purely periodic boundary conditions. The answer to the original question would simply be to set $k=0$ in the following. But the generalization to $k\ne 0$ is of great practical importance for wave propagation in periodic media, where you think of $U(x)$ as unwrapped along the entire $x$ axis.

Implementation with finite-difference method

To exploit the periodicity of $U(x)$ further, one could now expand it in a Fourier series and find $u_k(x)$ in Fourier space as well. However, I will instead do the calculation directly in real space because this allows me to leverage the built-in FiniteDifferenceDerivative capabilities contained in NDSolve. The essential feature relevant for this question is that FiniteDifferenceDerivative supports the option PeriodicInterpolation -> True. This makes it possible to convert the entire problem into a matrix equation using only version-8 functionality to construct all the matrices.

The equation to be solved is written here in a more generic form than in the question. The derivatives of $V(\theta)$ are really just additional terms in the periodic potential that I call $U$. So I'll only use the latter:

$$-\frac{1}{2}\frac{d^2}{dx^2}\psi_k + U(x) \psi_k = E \psi_k$$

Inserting the Bloch form, the resulting equation is (after dividing out the $\exp(i k x)$ factor):

$$-\frac{1}{2}\left(\frac{d^2}{dx^2}u_k+2i k \frac{d}{dx}u_k\right)+ \frac{k^2}{2} u_k + U(x) u_k = E u_k$$

To solve this eigenvalue equation, use the function spectrum. It's actually very short, but I tried to add a lot of comment, so the code looks a bit long:

spectrum[n_, dim_: 10000][potential_, {var_, varMin_, varMax_},  kBloch_: 0] := 
Module[
  {e, v, vRange, dx, grid, potentialGrid, eKin, ePot, min, interpolate},
  vRange = varMax - varMin;
  interpolate = 
   ListInterpolation[Append[#, First[#]], {{varMin, varMax}}, 
     PeriodicInterpolation -> True] &;
  dx = N[vRange/dim];
  grid = Range[varMin, varMax, dx];
  eKin = -(1/2)
       NDSolve`FiniteDifferenceDerivative[2, grid, 
       PeriodicInterpolation -> True]["DifferentiationMatrix"] - 
    I kBloch NDSolve`FiniteDifferenceDerivative[1, grid, 
       PeriodicInterpolation -> True]["DifferentiationMatrix"];
  potentialGrid = Table[potential + kBloch^2/2, {var, Most[grid]}];
  (* eKin is periodically interpolated, 
  so its last element is internally dropped by  FiniteDifferenceDerivative, as redundant. Therefore, 
  I also have to drop the last grid element in potentialGrid. *)
  min = Min[potentialGrid];
  (* Matrix for the potential is shifted so its minimum entry is  zero, guaranteeing that eigenvalues will be sorted in descending order: *)
  ePot = DiagonalMatrix[SparseArray[potentialGrid - min]];
  {e, v} = Eigensystem[eKin + ePot, -n];
  (* Final step: turn vectors on spatial grid back into functions of x by interpolation: *)
  Append[
   Reverse /@ {e + min, Map[interpolate[#/Max[Abs[#]]] &, v]},
   (* In the eigenvalues, 
   potential offset min was added back to get original energy scale. *)
      interpolate[potentialGrid]
   ]]

In order to test this, let's define a potential identical to the one used in other answers (except that I defined my entire equation with a factor $1/2$ to be closer to usual quantum-mechanical choice of units):

potential[x_] = 1/2 (-D[Cos[x], x, x] + D[Cos[x], x]^2);
{eigenvals, eigenvecs, pot} = spectrum[7][potential[x], {x, -Pi, Pi}];
2 eigenvals
(*
==> {-1.80943*10^-10, 1.64737, 1.64737, 4.54059, 4.54059, 9.51758, 9.51758}
*)

In the function call, the first argument [7] is the number of desired eigenstates (starting from the lowest one). This can optionally be followed by an integer giving the grid size (default is 10000). As a second set of arguments, you provide the potential and periodicity range, in the same way you'd pass it to Plot.

My factor $1/2$ is un-done by multiplying the resulting eigenvalues by $2$ above. The results agree with the other answers for this example, also showing the degeneracies expected due to symmetry.

The above results were obtained by omitting the optional argument kBloch, which gives it the default value 0. Since the periodicity interval is $2\pi$, we'll get anti-periodic boundary conditions for $k=1/2$:

{eigenvals, eigenvecs, pot} = spectrum[7][potential[ x], {x, -Pi, Pi}, 1/2];
2 eigenvals
(*
==> {0.0204952, 1.18926, 2.73105, 2.90459, 6.77308, 6.77789,    12.7628}
*)

Plotting the results

With these results, you can make various types of plots. For each Bloch wave number $k$, you can plot the wave functions. Or you can plot the eigenvalues as a function of $k$ (that's the band structure of the periodic potential).

I'll just show the wave functions for the anti-periodic boundary condition here, but I'll wrap it into a plotting function that can be applied directly to the output of spectrum:

plotSpectrum[{ev_, efn_, potential_}, kBloch_: 0] := Module[
  {evReal, evRange, eLow, eHigh, amplitude, eTicks, n = Length[ev], 
   n1, n2, min, max},
  Check[{min, max} = First[efn[[1]]["Domain"]], 
   Print["plotSpectrum needs the output of spectrum as input"]; Abort[]];
  evReal = Re[ev];
  evRange = Last[ev] - First[ev];
  amplitude = evRange/(n + 2);
  eLow = First[ev] - amplitude;
  eHigh = Last[ev] + amplitude;
  eTicks = {{Automatic, Automatic}, {Automatic, None}};
  Show[Plot[Evaluate[
     Table[
      ev[[i]] + amplitude Re[Exp[I kBloch x] efn[[i]][x]], {i,n}]], {x, min, max},
    Epilog -> {Gray, Map[Tooltip[Line[{{min, #}, {max, #}}], #] &, Chop[ev]]},
    Frame -> True, Axes -> None, FrameTicks -> eTicks,
    PlotStyle -> Map[ColorData[1], Range[n]]],
   Plot[potential[x], {x, min, max},
    PlotPoints -> 250, ExclusionsStyle -> Automatic,
    PlotStyle -> Directive[Thick, Opacity[.5], Orange], 
    Filling -> -10000], PlotRange -> {eLow, eHigh},
   PlotRangePadding -> {{Automatic, Automatic}, {Scaled[.1], Scaled[.05]}},
   FrameLabel -> {"\!\(\*FormBox[\(x\),TraditionalForm]\)", 
     "\!\(\*FormBox[\(\*SubscriptBox[\(ℰ\), \(n\)], \*SubscriptBox[\(ψ\), \(n\)]\),TraditionalForm]\)"}, RotateLabel -> False
   ]
  ]

plotSpectrum[{eigenvals, eigenvecs, pot}, 1/2]

potential and efns

Shown in orange is the potential energy. The wave functions are offset by their respective energy eigenvalues (now in my units with a factor $1/2$ in the kinetic energy). The plot only shows the real part of the wave function. The anti-periodicity can be seen in those functions whose real part doesn't happen to vanish at the boundaries. The Bloch wave number 1/2 has to be added as an argument to plotSpectrum. This is useful because it allows you to also plot the purely periodic part $u_k$ of the Bloch wave by setting $k=0$ in the last line.

Bound states of non-periodic potentials

The periodic boundary conditions implemented in spectrum become irrelevant when the potential is much higher than the desired range of eigenvalues. In that case, we're essentially dealing with bound states and the potential can have arbitrary shape. This means you can use the above code to calculate bound states in arbitrary potentials, too. Since the computation is also quite fast, it can be done in real-time inside a manipulate.

For example, consider the quartic potential with asymmetry:

Manipulate[
  Show[plotSpectrum@
    spectrum[10, 2000][x^4 - α x^2 + β x^3, {x, -5, 4}],
   PlotLabel -> 
    Framed[Row[{"Potential \!\(\*FormBox[\(\[ScriptCapitalV](x)\),
TraditionalForm]\) = ", ("\!\(\*FormBox[\(x\),
TraditionalForm]\)")^4 - α ("\!\(\*FormBox[\(x\),
TraditionalForm]\)")^2 + β ("\!\(\*FormBox[\(x\),
TraditionalForm]\)")^3}], FrameStyle -> None, RoundingRadius -> 7, 
     Background -> Lighter[Orange]]],
  {{α, 2, "Quadratic coefficient α"}, 0, 
   10}, {{β, 0, "Cubic coefficient β"}, 0, 2}]

quartic potential, made with ManToGif, and convert -scale 66% -colors 64

So this code can do the same things that NDEigensystem does. But you have to choose the simulation interval such that the desired eigenfunctions decay to zero at the boundaries, which makes the periodic boundary conditions unnecessary. In practice, having periodic boundary conditions is useful even in this type of application, because it will continue to give reasonable results when the interval isn't small enough to make all functions decay completely to zero.

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  • $\begingroup$ Thanks Jens. I'll need some time to work through it. Very much appreciated. $\endgroup$ – user21 Mar 20 '16 at 21:40
  • $\begingroup$ Would you mind if I use your visualization in a documentation example? $\endgroup$ – user21 Mar 20 '16 at 21:42
  • 1
    $\begingroup$ @user21 Sure, you can use the visualization - I'd of course be curious to find out where exactly. If anything needs clarification, just let me know. $\endgroup$ – Jens Mar 20 '16 at 22:29
  • $\begingroup$ FYI, there is a solvable model for cosine potential, which leads to MathieuCharacteristicA. $\endgroup$ – xslittlegrass Jun 17 '16 at 16:59
  • 1
    $\begingroup$ @xslittlegrass Yes, the cosine potential is special because it has only two Fourier components. But of course for general V things are impossible to solve without doing numerics. Usually people go to momentum space (and I have a notebooks for that too), but here I wanted to focus on the "real-space* implications of lattice periodicity on the boundary conditions. $\endgroup$ – Jens Jun 17 '16 at 18:06
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Version 11.0 has an additional boundary condition PeriodicBoundaryCondition. With this you can then use with NDEigensystem:

Here is an example with an unsymmetrical potential finding 5 anti-periodic eigenvalues and vectors:

V[x_] := Cos[x] + x;
{vals, funs} = 
  NDEigensystem[{-u''[x] - (V''[x] - V'[x]^2) u[x], 
    PeriodicBoundaryCondition[-u[x], x == 2 \[Pi], 
     TranslationTransform[{-2 \[Pi]}]]}, u[x], {x, 0, 2 \[Pi]}, 5];

Plot[funs, {x, 0, 2 \[Pi]}]

enter image description here

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  • $\begingroup$ Excellent! This is just what the doctor ordered. Or it will be, when I get the 11.0 upgrade. $\endgroup$ – Leon Avery Aug 8 '16 at 18:54
  • $\begingroup$ I hope the drug will not disappoint.... ;-) Positive feedback to the support is welcome too ... $\endgroup$ – user21 Aug 8 '16 at 18:57
3
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Though not particularly satisfying, the following produces periodic eigenfunctions. For specificity, assume

V[θ_] := Cos[θ]

Then, use

s = NDEigensystem[{ϕ''[θ] + (V''[θ] - V'[θ]^2) ϕ[θ]}, ϕ[θ], {θ, 0, 2 Pi}, 12];
ls = Map[Chop[(# /. θ -> 0) - (# /. θ -> 2 Pi)] == 0 &, Last@s];
Plot[Evaluate[Pick[Last@s, ls]], {θ, 0, 2 Pi}, AxesLabel -> {θ, ϕ}]

enter image description here

to visualize the eigenfunctions and

Pick[First@s, ls]
(* {-0.0000220456, -1.64739, -4.54152, -9.52717, -16.5614, -25.6947} *)

to obtain the corresponding eigenvalues.

Caveat: The approach works only for potentials v that are symmetric or anti-symmetric about θ = Pi, or can be made so by a shift of the range of integration.

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  • $\begingroup$ Additional caveat: as described, this misses half the eigenfunctions, even for a potential that is symmetric around 0. The problem is that it requires the functions to satisfy Neumann boundary conditions at 0 and $2\pi$. In addition to those shown, which are even functions of $\theta$, there is a series that are eigenfunctions that are odd in $\theta$ (which can be found by imposing a homogeneous Dirichlet boundary condition at 0 and $2\pi$).... (cont) $\endgroup$ – Leon Avery Dec 6 '15 at 11:22
  • $\begingroup$ I'm pretty sure that, together, these are the complete set for symmetric potentials (and actually, I had gotten that far before I posted the question). For potentials that show no symmetry, I'm still stuck. In the general case, neither Neumann nor Dirichlet BCs are predictable (or at least I can't predict them). $\endgroup$ – Leon Avery Dec 6 '15 at 11:22
  • $\begingroup$ @bbgodfrey, if you could have a look at my answer below and let me know what you think that would be great. Thanks. $\endgroup$ – user21 Mar 18 '16 at 2:41
  • $\begingroup$ @user21 This seems like a good approach, although I would need to spend a bit of time (which I do not have just now) to be sure. As Jens remarked a few minutes ago, accuracy could be better. Perhaps, you could rerun the symmetric example with more finite elements to see whether anything changes. In any case, +1. $\endgroup$ – bbgodfrey Mar 18 '16 at 3:51
  • $\begingroup$ Thanks. Added a more accurate version. $\endgroup$ – user21 Mar 18 '16 at 7:44

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