2
$\begingroup$

I have a functional $S$,

$$S = \left( \int_{x_0}^{x_f} dx \frac{1}{z^d} \sqrt{-f(z,u) u'^2 - 2 u' z' +1} \right) + \frac{1}{z(x_0)^{d-1}}$$

where it is composed of two terms, i.e. an integral plus a constant term.

Varying $S$,

$$\delta S = \left( \int_{x_0}^{x_f} dx \rm{\;EOM} \right) + \frac{-(z' + f(z,u) u') \delta u - u' \delta z}{ z^d \sqrt{-f(z,u) u'^2 - 2 u' z' +1}}\Biggr|^{x_f}_{x_0} - \frac{d-1}{z(x_0)^d} \delta z(x_0)$$

The boundary term at $x_f$ vanishes since the points will be fixed. The only term left is at $x_0$ which is written as,

$$\frac{(z'(x_0) + f(z(x_0),u(x_0)) u'(x_0))}{ z(x_0)^d \sqrt{-f(z(x_0),u(x_0)) u'(x_0)^2 - 2 u'(x_0) z'(x_0) +1}} \delta u(x_0) + \left[\frac{u'(x_0)}{z(x_0)^d \sqrt{-f(z(x_0),u(x_0)) u'(x_0)^2 - 2 u'(x_0) z'(x_0) +1}} - \frac{d-1}{z(x_0)^d}\right]\delta z(x_0)$$

If we let $u(x_0)$ and $z(x_0)$ be movable boundary points then $\delta u(x_0) \neq 0$ and $\delta z(x_0) \neq 0$, so we have,

$$z'(x_0) + f(z(x_0),u(x_0)) u'(x_0) = 0$$

$$u'(x_0) - (d-1)\sqrt{-f(z(x_0),u(x_0)) u'(x_0)^2 - 2 u'(x_0) z'(x_0) +1} = 0$$

which actually could be combined to form,

$$u'(x_0) = \frac{-(d-1)^2 z'(x_0) \pm \sqrt{(d-1)^4 z'(x_0)^2 + 4(d-1)^2}}{2}$$

The boundary conditions of the problem are,

$$z(x_0) = z_s, z(x_f) = \epsilon = 10^{-5}$$

$$u'(x_0) = \frac{-(d-1)^2 z'(x_0) \pm \sqrt{(d-1)^4 z'(x_0)^2 + 4(d-1)^2}}{2}, u(x_f) = 1$$

Finding a solution to this problem will extremize $S$ for some chosen $z(x_0) = z_s$. The goal is to find the $z_s$ such that $S$ is minimized for all the set of solution that extremizes $S$.

The plot of $z(x)$ should look approximately as shown (ignore the values of the axes) where the blue dot is the location of $z(x_0) = z_s$,

Image

I have tried several ways to write a code, however, not only I had issues with the boundary value ODE but also I'm confused how and where to insert the minimization process involving $z(x_0)$. Any advice would be greatly appreciated.

ClearAll["Global`*"]
Needs["VariationalMethods`"]
d = 3;
u1 = 2;
m = (1/u[x] + 1/u1)^(d + 1);
f = 1 - m z[x]^(d + 1);
L = (Sqrt[-f u'[x]^2 - 2 u'[x] z'[x] + 1]/z[x]^d);
eulerlageq1 = EulerEquations[L, u[x], x];
eulerlageq2 = EulerEquations[L, z[x], x];
s = Solve[{eulerlageq1, eulerlageq2}, {u''[x], z''[x]}][[1]]//Simplify;
eq01 = {w'[x] == s[[1, 2]] /. u'[x] -> w[x], w[x] == u'[x]};
eq02 = {v'[x] == s[[2, 2]] /. z'[x] -> v[x], v[x] == z'[x]};

x0 = 10^-8;
xf = 10^-1;
eps = 10^-5; 
ztest[a_?NumericQ, b_?NumericQ, zs_?NumericQ] := First[z[xf] /. NDSolve[{eq01, eq02, z[x0] == zs, v[x0] == a, u[x0] == b, w[x0] == (-(d - 1)^2 a - Sqrt[(d - 1)^4 a^2 + 4 (d - 1)^2])/2}, {z, u}, {x, x0, xf}, Method -> "ExplicitEuler"]]
utest[a_?NumericQ, b_?NumericQ, zs_?NumericQ] := First[u[xf] /. NDSolve[{eq01, eq02, z[x0] == zs, v[x0] == a, u[x0] == b, w[x0] == (-(d - 1)^2 a - Sqrt[(d - 1)^4 a^2 + 4 (d - 1)^2])/2}, {z, u}, {x, x0, xf}, Method -> "ExplicitEuler"]]

sol = FindRoot[{ztest[a, b, zs] == eps, utest[a, b, zs] == 1}, {{a, -5}, {b, 10}}, WorkingPrecision -> 20]
S[zs_?NumericQ] := NIntegrate[Rationalize[L, 0] /. sol, {x, x0, xf}] + 1/zs^(d - 1)
FindMinimum[S[zs], {zs, 10}]
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6
  • $\begingroup$ In your Mathematica code you write m = (1/u + 1/u1)^(d + 1), maybe you mean m = (1/u[x] + 1/u1)^(d + 1)? Since you say that you do not know how to deal this kind of problem, maybe it would be good to first consider a simpler problem that can be solved by hand, such as minimizing $S = u(1) + \int_0^1 ((u'(x))^2 + u(x)) dx$ with $u(0) = 0$. In fact, your question seems to be about math more than about Mathematica. Btw, if you can share the context where this problem shows up, that would be interesting. $\endgroup$
    – user293787
    Commented Nov 24, 2022 at 17:10
  • $\begingroup$ @user293787 Thanks for the heads-up on the typo. Actually, I have considered simpler problems, i.e. has some conserved quantities so that the equation of motion can be written analytically. In that case, I could just take the derivative of $S$ with respect to $z_s$ and then find at which $z_s$ will $S$ be a minimum. In this case, however, there is no analytic form and I'd rather rely on FindMinimum to automate (simplify?) things. With regard to the context, I'm computing entanglement entropies where $S$ represents the classical area and the constant term is a quantum correction. $\endgroup$
    – mathemania
    Commented Nov 24, 2022 at 18:15
  • $\begingroup$ @mathemania Could you map your region on (0,1) so that we can use wavelets as well? $\endgroup$ Commented Nov 25, 2022 at 4:58
  • $\begingroup$ Do you know any solution that satisfies the 3 boundary conditions for $z(x_f)$, $u(x_0)$, $u(x_f)$, no matter what $z(x_0)$? Btw, your eq01 still contains z' and your eq02 still contains u' (that is not necessarily a problem for NDSolve but I assume you intended to write the equation in standard 1st order form). Since $u,z>0$ I thought substituting $u = e^U$ and $z = e^Z$ could help, but maybe not. $\endgroup$
    – user293787
    Commented Nov 25, 2022 at 8:42
  • $\begingroup$ @user293787 As of the moment I haven't solved any solution yet but if it can be solved I guess it should work for any $z(x_0)$ as long as it does allow the equations to diverge (refer to the equation above, you can see a $1/z^d$ factor). Regarding the two equations being solved using NDSolve, I'm having issues using the ShootingMethod as suggested in the MM post $(72725) (57262)$. $\endgroup$
    – mathemania
    Commented Nov 25, 2022 at 15:03

1 Answer 1

4
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First, we transform S to remove boundary value from optimization as follows $S = \left( \int_{x_0}^{x_f} dx \frac{1}{z^d} \sqrt{-f(z,u) u'^2 - 2 u' z' +1} \right) + \frac{1}{z(x_0)^{d-1}}$. We note, that $\frac{1}{z(x_0)^{d-1}}=\frac{1}{z(x_f)^{d-1}}+(d-1)\int_{x_0}^{x_f}\frac{z'dx}{z^d}$. Therefore, we should change Lagrange function as $L = \frac {1} {z^d}\sqrt {-f (z, u) u'^2 - 2 u' z' + 1} + \frac {(d - 1) z'} {z^d} $, then action is given by

$S = \int_ {x_ 0}^{x_f} Ldx + \frac {1} {\epsilon^{d - 1}} $

Last term is a constant, hence we can solve Euler equations to optimize action. We use the Euler wavelets collocation method

ClearAll["Global`*"]
Needs["VariationalMethods`"]
d = 3; zs = 10;
uc = 2; xf = 1/10; x0 = 1/xf 10^-8;
eps = 10^-5;
m = (1/u[x] + 1/uc)^(d + 1);
f = 1 - m ( z[x])^(d + 1);
L = (Sqrt[-(f u'[x]^2 + 2  u'[x] z'[x]) + 1]/( z[x])^
      d + (d - 1) z'[x]/z[x]^d);
eulerlageq1 = EulerEquations[L, u[x], x];
eulerlageq2 = EulerEquations[L, z[x], x];
s = Solve[{eulerlageq1, eulerlageq2}, {u''[x], z''[x]}][[1]] // 
   Simplify;
eq01 = u''[x] - s[[1, 2]] == 0;
eq02 = z''[x] - s[[2, 2]] == 0;
UE[m_, t_] := EulerE[m, t];
psi[k_, n_, m_, t_] := 
  Piecewise[{{2^(k/2) UE[m, 2^k t - 2 n + 1], (n - 1)/2^(k - 1) <= t <
       n/2^(k - 1)}, {0, True}}];
PsiE[k_, M_, t_] := 
 Flatten[Table[psi[k, n, m, t], {n, 1, 2^(k - 1)}, {m, 0, M - 1}]]
k0 = 3; M0 = 4; With[{k = k0, M = M0}, 
 nn = Length[Flatten[Table[1, {n, 1, 2^(k - 1)}, {m, 0, M - 1}]]]];
dx = 1/(nn); xl = Table[l*dx, {l, 0, nn}]; zcol = 
 xcol = Table[(xl[[l - 1]] + xl[[l]])/2, {l, 2, nn + 1}]; Psijk = 
 With[{k = k0, M = M0}, PsiE[k, M, t1]]; Int1 = 
 With[{k = k0, M = M0}, Integrate[PsiE[k, M, t1], t1]];
Int2 = Integrate[Int1, t1];
Psi[y_] := Psijk /. t1 -> y; int1[y_] := Int1 /. t1 -> y;
int2[y_] := Int2 /. t1 -> y;
M = nn;
A = Array[a, {M}]; B = Array[b, M];
z2[x_] := A . Psi[x]; z1[x_] := A . int1[x] + a0; 
z0[x_] := A . int2[x] + a0 x + a1; u2[x_] := B . Psi[x]; 
u1[x_] := B . int1[x] + b0; u0[x_] := B . int2[x] + b0 x + b1;
var = Join[A, 
  B, {a0, a1, b0, b1}]; eqs = {u''[x] - s[[1, 2]], 
   z''[x] - s[[2, 2]]} /. {u''[x] -> u2[x]/xf^2, u'[x] -> u1[x]/xf, 
   u[x] -> u0[x], z''[x] -> z2[x]/xf^2, z'[x] -> z1[x]/xf, 
   z[x] -> z0[x]};
eq = Flatten[Table[eqs, {x, xcol}]];

bc = {z0[x0] == zs, z0[xs] == eps, u0[x0] == 15, 
    u0[xs] == 1} /. {xs -> 1};
action = 
 Table[L /. {u'[x] -> u1[x]/xf, u[x] -> u0[x], z'[x] -> z1[x]/xf, 
     z[x] -> z0[x]}, {x, xcol}] // Total; con = 
 Table[(-(f u'[x]^2 + 2 u'[x] z'[x]) + 1 >= 0) /. {u'[x] -> u1[x]/xf, 
    u[x] -> u0[x], z'[x] -> z1[x]/xf, z[x] -> z0[x]}, {x, xcol}];

sol2 = FindRoot[Join[Table[eq[[i]] == 0, {i, Length[eq]}], bc], 
  Table[{var[[i]], 1/10}, {i, Length[var]}], 
  MaxIterations -> Infinity, 
  Method -> {"Newton", "StepControl" -> "TrustRegion"}, 
  WorkingPrecision -> 30];

Visualization

{Plot[Evaluate[u0[x/xf] /. sol2], {x, x0, xf}, 
  AxesLabel -> {"x", "u"}, PlotRange -> All, PlotPoints -> 200], 
 Plot[Evaluate[z0[x/xf] /. sol2], {x, x0, xf}, 
  AxesLabel -> {"x", "z"}, PlotRange -> All, PlotPoints -> 200]}
 

Figure 1

Finally, we can evaluate action

action /. sol2

Out[]= 267.893419311596720291006268 

To find minimum action with respect to zs we use Do loop as follows

Do[sol[j] = 
   FindRoot[
    Join[Table[eq[[i]] == 0, {i, Length[eq]}], 
     bc /. {zs -> 10 + (j - 5)/10}], 
    Table[{var[[i]], 1/10}, {i, Length[var]}], 
    MaxIterations -> Infinity, 
    Method -> {"Newton", "StepControl" -> "TrustRegion"}, 
    WorkingPrecision -> 30];, {j, 0, 10}] 

Note that S is real in some region around zs=10, here we have table to look at 11 data with six real and 5 complex S (here we use unscaled S without constant part $1/\epsilon^2$)

S = 
 Table[{10 + (j - 5)/10, dx xf action /. sol[j]}, {j, 0, 10}]

Out[]= {{19/2, 
  1.57089633087946341494365634 + 
   0.000665399108477383332080644974 I}, {48/5, 
  1.57046931931274310853709700 + 
   0.000658066927631603946876114786 I}, {97/10, 
  1.57004300965468526183575503 + 
   0.000650226806636050815817267348 I}, {49/5, 
  1.56961754084009057627105968 + 
   0.000641969667063151355287558226 I}, {99/10, 
  1.67313437375896650260462628}, {10, 
  1.67433387069747950181878917}, {101/10, 
  1.67556219665128071840760756}, {51/5, 
  1.67682955598650942018621185}, {103/10, 
  1.67815118607052005825248523}, {52/5, 
  1.67955111804270498084313785}, {21/2, 
  1.52660724801448396934450389 + 0.0000609672885395141311715951931 I}}

Plot S shows that minimum is on the border between real and complex S at $9.8\le zs \le 9.9$

ListPlot[S]

Figure 2

Note, there is no big difference in 6 solutions as it shown in the picture below

{Plot[Evaluate[Table[u0[x/xf] /. sol[i], {i, 4, 9}]], {x, x0, xf}, 
  AxesLabel -> {"x", "u"}, PlotRange -> All], 
 Plot[Evaluate[Table[z0[x/xf] /. sol[i], {i, 4, 9}]], {x, x0, xf}, 
  AxesLabel -> {"x", "z"}, PlotRange -> All], 
 Plot[Evaluate[Table[z0[x/xf] /. sol[i], {i, 4, 9}]], {x, 0.035, .04},
   AxesLabel -> {"x", "z"}, PlotRange -> All, 
  PlotLegends -> Automatic]}

Figure 3

We can reproduce optimal solution with Haar wavelets as well. But in this case, there are no solutions with real action around zs=10. Nevertheless, plot for u looks like theoretical one, for example,

ClearAll["Global`*"]
Needs["VariationalMethods`"]
d = 3;
uc = 2; xf = 10^-1; x0 = 1/xf 10^-8; zs = 985/100;
eps = 10^-5;
m = (1/u[x] + 1/uc)^(d + 1);
f = 1 - m ( z[x])^(d + 1);
L = (Sqrt[-(f u'[x]^2 + 2  u'[x] z'[x]) + 1]/( z[x])^
      d + (d - 1) z'[x]/z[x]^d);
eulerlageq1 = EulerEquations[L, u[x], x];
eulerlageq2 = EulerEquations[L, z[x], x];
s = Solve[{eulerlageq1, eulerlageq2}, {u''[x], z''[x]}][[1]] // 
   Simplify;
eq01 = u''[x] - s[[1, 2]] == 0;
eq02 = z''[x] - s[[2, 2]] == 0;
J = 3; M = 2^J; dx = 1/(2*M); xl = Table[l*dx, {l, 0, 2*M}];
xcol = Table[(xl[[l - 1]] + xl[[l]])/2, {l, 2, 2*M + 1}];
h1[x_] := Piecewise[{{1, 0 <= x <= 1}, {0, True}}];
p1[x_, n_] := (1/n!)*x^n; 
h[x_, k_, m_] := 
 Piecewise[{{1, 
    Inequality[k/m, LessEqual, x, Less, (1 + 2*k)/(2*m)]}, {-1, 
    Inequality[(1 + 2*k)/(2*m), LessEqual, x, Less, (1 + k)/m]}}, 0]
p[x_, k_, m_, n_] := 
 Piecewise[{{0, x < k/m}, {(-(k/m) + x)^n/n!, 
    Inequality[k/m, LessEqual, x, 
     Less, (1 + 2*k)/(2*m)]}, {((-(k/m) + x)^n - 
       2*(-((1 + 2*k)/(2*m)) + x)^n)/n!, (1 + 2*k)/(2*m) <= 
     x <= (1 + k)/
      m}, {((-(k/m) + x)^n + (-((1 + k)/m) + x)^n - 
       2*(-((1 + 2*k)/(2*m)) + x)^n)/n!, x > (1 + k)/m}}, 0]
var1 = Flatten[Table[a[i, j], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}]];
var2 = Flatten[Table[b[i, j], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}]];
z2[x_] := 
  Sum[a[i, j]*h[x, i, 2^j], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + 
   a0*h1[x];
z1[x_] := 
  Sum[a[i, j]*p[x, i, 2^j, 1], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + 
   a0*p1[x, 1] + a1;
z0[x_] := 
  Sum[a[i, j]*p[x, i, 2^j, 2], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + 
   a0*p1[x, 2] + a1*x + a2;
u2[x_] := 
  Sum[b[i, j]*h[x, i, 2^j], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + 
   b0*h1[x];
u1[x_] := 
  Sum[b[i, j]*p[x, i, 2^j, 1], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + 
   b0*p1[x, 1] + b1;
u0[x_] := 
  Sum[b[i, j]*p[x, i, 2^j, 2], {j, 0, J, 1}, {i, 0, 2^j - 1, 1}] + 
   b0*p1[x, 2] + b1*x + b2;

var = Join[var1, var2, {a0, a1, a2, b0, b1, b2}];
eqs = {u''[x] == s[[1, 2]], 
    z''[x] == s[[2, 2]]} /. {u''[x] -> u2[x]/xf^2, u'[x] -> u1[x]/xf, 
    u[x] -> u0[x], z''[x] -> z2[x]/xf^2, z'[x] -> z1[x]/xf, 
    z[x] -> z0[x]};
eq = Flatten[Table[eqs, {x, xcol}]];

bc = {z0[x0] == zs, z0[xs] == eps, u0[x0] == 15, 
    u0[xs] == 1} /. {xs -> 1};
action = 
 Table[L /. {u'[x] -> u1[x]/xf, u[x] -> u0[x], z'[x] -> z1[x]/xf, 
     z[x] -> z0[x]}, {x, xcol}] // Total; con = 
 Table[(-(f u'[x]^2 + 2 u'[x] z'[x] zs)/xf^2 + 1 >= 0) /. {u'[x] -> 
     u1[x], u[x] -> u0[x], z'[x] -> z1[x], z[x] -> z0[x]}, {x, xcol}];

sol2 = FindRoot[Join[eq, bc], 
  Table[{var[[i]], 1/10}, {i, Length[var]}], 
  Method -> {"Newton", "StepControl" -> "TrustRegion"}, 
  MaxIterations -> Infinity, WorkingPrecision -> 30];

Visualization

{Plot[Evaluate[u0[x/xf] /. sol2], {x, x0, xf}, 
  AxesLabel -> {"x", "u"}, PlotRange -> All, PlotPoints -> 200], 
 Plot[Evaluate[z0[x/xf] /. sol2], {x, x0, xf}, 
  AxesLabel -> {"x", "z"}, PlotRange -> All, PlotPoints -> 200]} 

Figure 4

Action has imaginary part

action /. sol2

Out[]= 340.6006157974879079324493482 + 0.109047774482325545497193142 I
$\endgroup$
10
  • $\begingroup$ We know that the solution of the Euler-Lagrange equation extremizes the action $S$, my goal is to find which $z(x_0)$ will give a minimum out of all possible extremal solutions. You transformed the constant term $1/z(x_0)^{d-1}$ to $1/z(x_f)^{d-1}$ plus an integral is good since $1/z(x_f)^{d-1}$ now does not contribute to the variation, but how do you determine which $z(x_0)$ gives the minimum? In your code, you wrote zs=10. I think your code partly solves the problem where I don't have to worry about the term $1/z(x_0)^{d-1}$ since it's transformed into an integral. $\endgroup$
    – mathemania
    Commented Nov 30, 2022 at 7:19
  • $\begingroup$ @mathemania I run last part of code, it takes a time to finish computation. The good news is that action at $z\le 9$, and $11\le z$ is complex, therefore the interval to find minimum is not so large. $\endgroup$ Commented Nov 30, 2022 at 7:36
  • $\begingroup$ The x0 you used is a bit different from mine. Also, in the line bc you used xs->1 instead of xf->10^-1. Why? I'm running the code but it's taking quite some time. $\endgroup$
    – mathemania
    Commented Nov 30, 2022 at 8:42
  • $\begingroup$ @mathemania Since we use wavelets defined on $0\le x\le 1$ we map region $x0\le x\le xf$ to (0,1) using xf as a scale - see eqs definition. $\endgroup$ Commented Nov 30, 2022 at 9:00
  • $\begingroup$ I have found something a bit strange, when you evaluate for example, Evaluate[u0[x0/xf]/.sol2] you do not get u[x0]=15, what you get is u[x0] =14.980569299. I know this might have something to do with precision but I'm asking this since when I change my setup a little bit, using Haar wavelet produces a solution where it doesn't match the boundary condition by a considerable value. In this case, it just differs by an order of $10^{-2}$, but changing the setup can produce a difference of order $10^0$ which is bad. $\endgroup$
    – mathemania
    Commented Dec 4, 2022 at 11:10

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