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I have a PDE from my model's equation of motion(step.3) and I was suggested to use NDSolve to solve it, I was wondering how many boundary conditions(in step.4 and I currently had used 3 boundary conditions) I would need in order to successfully solve this PDE and produce a time evolution simulation(step.5) using the result produced with 'ndsolve'? And why? (And if I made a mistake somewhere since my code seems to produce an invariant straight line?). Thank you for reading my question!

(*Step1.Define and plot the field*)
lhs = 24
rhs = 40
beta = 0
field = -2/(1 + Exp[-x + rhs]) + 2/(1 + Exp[-x + lhs]) - 1 + 
  beta (*our field value v.s.spatial dimension*)
Plot[field, {x, 0, 
  64}, PlotRange -> {{0, 64}, {-1, 2}}]

(*Step.2 Define the field potential derivative dv/dphi*)
b = 0.1
dv = (y[t, x]^2 - 1)*(y[t, x] + 
    b)       (*dv actually denotes dv/dphi*)

(*Step3.Define PDE from our E.O.M*)
pde = -D[y[t, x], {t, 2}] + 
   D[y[t, x], {x, 2}] + dv == 0

(*Step4.Use NDSolve to solve PDE with the help of known initial \
condition-field value v.s.spatial coordinate at t0*)
nsol = 
 NDSolve[{pde, y[0, x] == 0.9, D[y[t, 0], x] == 0, 
   D[y[t, 64], x] == 0}, y[t, x], {t, 0, 1}, {x, 0, 64}]

(*Step5.plot the field in spatial dimension at different times*)
\
nsol2 = nsol[[1, 1, 2]]
nsol3[t_, x_] = nsol2;
ListAnimate[
 Table[Plot[nsol3[t, x], {x, 0, 64}, PlotLabel -> t, 
   PlotRange -> {-1.5, 1.5}], {t, 0, 1, 0.01}]]
```
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  • $\begingroup$ You're getting an invariant straight line because NDSolve is throwing an error and returning the input unevaluated, and you're then picking out part [[1,1,2]] of the original input (which happens to be 0.) As to how to fix it, I'm writing up something now. $\endgroup$ Feb 28, 2022 at 15:28

1 Answer 1

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There are two problems in your code, one conceptual and one syntactical.

  • An initial value problem that is second-order in time derivatives needs two initial conditions (just as a second-order ODE in classical mechanics needs both an initial position and an initial velocity to uniquely specify a solution.) So you need to provide both $y(0,x)$ and $\dot{y}(0,x)$ to NDSolve.

  • The expression D[y[t, 0], x] returns 0, since y[t,0] does not literally depend on x. To get the quantity $\partial y/\partial x|_{x=0}$, use (D[y[t, x], x] /. x -> 0); this takes the derivative of y[t,x] with respect to x and then sets x to 0.

So replace your Step 4 with:

nsol = NDSolve[{pde, 
                y[0, x] == 0.9, 
                (D[y[t, x], t] /. t -> 0) == 0, (* or whatever IC you want *)
                (D[y[t, x], x] /. x -> 0) == 0, 
                (D[y[t, x], x] /. x -> 64) == 0}, 
               y[t, x], {t, 0, 1}, {x, 0, 64}]

and it works fine, returning a solution for y that doesn't depend on x but does depend on t. This makes sense, since if you impose homogeneous Neumann boundary conditions on your equation, then solutions of the form $y(x,t) = f(t)$ are allowed.

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  • $\begingroup$ Also, note that your potential is not a "double well" but is instead a "double hump" given the signs you've put in your code. This may or may not have been your intention. $\endgroup$ Feb 28, 2022 at 15:45

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