Implementing periodic boundary conditions in NDSolve [duplicate]

Question

This question already has an answer here:

• PDE list of equations, derivative for boundarry condition evaluates to true 2 answers

I am wanting to solve the KdV equation using NDSolve. The exercise asks that I impose periodic boundary conditions on u and its derivatives. My approach is as follows:

a = 30;

eq = {D[u[t, x], {t, 1}] + D[u[t, x], {x, 3}] == 
6 u[t, x] D[u[t, x], {x, 1}], u[0, x] == -2 Sech[x] ^2, 
u[t, -a] == u[t, a]};

sol = NDSolve[eq, u, {t, 0, 30}, {x, -a, a}, MaxStepSize -> 0.07];

data = Flatten[Table[{t, x, u[t, x]} /. sol, {t, 0, 30}, {x, -a, a}], 
2];

ListPlot3D[data, Mesh -> None, ColorFunction -> "Rainbow", 
PlotRange -> {{0, 30}, {-30, 30}, {-2, 2}}, Lighting -> "Automatic", 
AxesLabel -> {"t", "x"}]

I am not really sure what it means to impose periodic boundary conditions on the derivatives, but in the other posts I have seen on this site, periodic boundary conditions are what I have implemented above.

How would I implement periodic boundary conditions on the derivatives?

I have tried adding the code:

D[u[t, -a], {x, 1}] = D[u[t, a], {x, 1}],D[u[t, -a], {t, 1}] = D[u[t, a], {tx, 1}]

In the code above in the 'eq' part of the code, but there was an error.

Does the Boundary Condition:

u[t, -a] == u[t, a]

Also take care of the derivatives?

Answer 1

The code in the question is correct.

a = 30; 
eq = {D[u[t, x], {t, 1}] + D[u[t, x], {x, 3}] == 6 u[t, x] D[u[t, x], {x, 1}], 
    u[0, x] == -2 Sech[x]^2, u[t, -a] == u[t, a]}; 
sol = Flatten@NDSolve[eq, u, {t, 0, 30}, {x, -a, a}, MaxStepSize -> 0.07]

and yields

Plot3D[u[t, x] /. sol, {x, -a, a}, {t, 0, 30}, PlotRange -> All, PlotPoints -> 100,
    BoxRatios -> {2, 1, 1/2},, AxesLabel -> {x, t, u}], ImageSize -> Large, 
    LabelStyle -> Directive[12, Bold, Black]

Although the PDE certainly supports two-dimensional solutions, the result here is very nearly one-dimensional, which allows a highly accurate symbolic expression to be derived. Assume u[t, x] == v[y], where y == x - c t.

eq1d = Unevaluated[D[u[t, x], {t, 1}] + D[u[t, x], {x, 3}] == 
    6 u[t, x] D[u[t, x], {x, 1}]] /. u[t, x] -> v[x - c t] /. x -> c t + y
(* -(c*v'y]) + v'''y] == 6*v[y]*v''y] *)

Now, insert the initial condition from the question.

Simplify[eq1d /. v[y] -> -2 Sech[y]^2 /. v'[y] -> D[-2 Sech[y]^2, y] 
    /. v'''[y] -> D[-2 Sech[y]^2, {y, 3}]]
(* (-4 + c) Sinh[y] == 0 *)

So, c == 4, and the symbolic solution is given by

u[t, x] == -2 Sech[x - 4 t]^2

for t < a/4, after which the soliton propagates beyond x == a and does not reenter at x == -a, because this expression for u[t, x] is not periodic in x. However, an approximate periodic solution is obtained easily by summing such solutions:

u[t, x] == -2 Sum[Sech[x - 4 (t - 15 n)]^2, {n, 0, Infinity}]
(* (-120 + QPolyGamma[1, -(1/60) Log[-I E^(4 t - x)], E^60] + 
       QPolyGamma[1, -(1/60) Log[I E^(4 t - x)], E^60])/1800 *)

Although this expression does not precisely satisfy the initial condition or even the PDE itself, the relative error is of order Exp[-2 a]. Indeed, plotting this last expression gives a figure indistinguishable from that plotted above. We have, therefor, both symbolic and numerical solutions, and they are essentially identical, as one would hope.