differential geometry – Interpretation of transport equation for a two-form

The vorticity equation arising from the Euler equation for a fluid with velocity $$bf u$$ reads:

$$partial_t {bf{w}} + nabla times( {bf{w}} times {bf{u}}) =0$$

where the vorticity is $${bf{w}} =nabla times {bf{u}}$$.

Now, assume that we have a 1-form $$u$$ and its associated “vorticity” 2-form $$w=du$$. Is it possible to interpret the above vorticity equation in terms of Lie derivative $$L_u$$ (or other kind of derivatives) and, possibly, Hodge duality? This would give a geometric meaning to the second term in the equation. Maybe something like

$$partial_t {w} + L_u w =0 , ?$$

differential equations – Error on DSolve

I’ve been trying to solve this initial value problem using ‘DSolve()’:

$$frac{dy}{dt}=1+tspace sin(tspace y),quad y(0)=0, quad t=(0,2)$$

``````ClearAll(y, t)
eq1 := {y'(t) == 1 + t *Sin(t y(t)), y(0) == 0};
DSolve(eq1, y(t), {t, 0, 2})
``````

All I get is the Inverse function error.

``````
Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>

``````

The documentation suggests it has to do with the sine function but I’m not sure how to by-pass it.

Any help would be appreciated.

geometry – Symplectic Hodge Star and Koszul differential

Let $$M = text{Spec}(R)$$ be a symplectic affine variety of dimension $$n$$ with the symplectic form $$omega$$. There is the symplectic Hodge star $$star: Omega^k_{M} to Omega^{2n-k}_{M}$$ given by the condition that for any $$alpha, beta in Omega^k$$ we should have $$beta wedge alpha = Lambda^k G(beta, alpha) cdot v_n$$ where $$G(-, -): TM^* times TM^* to R$$ is a bivector field corresponding to $$omega$$ and $$v_n = omega^n/n!$$. I want to understand the first nontrivial (for me) case: $$M = T^* mathbb{A}^2$$. In this case $$omega = dx_1 dx_2 + dx_3 dx_4$$ and $$v_2 = dx_1 dx_2 dx_3 dx_4$$. Also, $$G = frac{partial}{partial_{x_1}} frac{partial}{partial_{x_2}} + frac{partial}{partial_{x_3}} frac{partial}{partial_{x_4}}.$$ It seems to me that we have the following

$$star(f) = f dx_1 dx_2 dx_3 dx_4$$

$$star(alpha) = alpha wedge omega$$ where $$alpha in Omega^1_{M}$$

$$star(zeta) = -zeta$$ for $$zeta in Omega^{2}_{M}$$

$$star(eta) = i_{G} eta$$ where $$eta in Omega^{3}_{M}$$ and $$i_{G}$$ is the contraction with $$G$$.

OK, then let us see how the Koszul differential works and it is the place where I met a contradiction in math (of course, the problem is with me but I cannot figure it out).

$$textbf{Setup.}$$ The Koszul differential $$B: Omega^{k}_{M} to Omega^{k-1}_{M}$$ has two interpretations, the first one is $$B = (i_G, d)$$. The second is $$B = (-1)^{k+1} star d star.$$

$$textbf{My misunderstanding.}$$ Let $$x in Omega^3_{M}$$. The first definition gives $$Bx = i_G dx – d i_{G}x$$. The second definition gives $$Bx = star d star x = star d i_{G} x = -d i_{G} x$$. As we see, we should have $$i_{G} dx=0$$ which is not the case, for example, for $$x_1 dx_2 dx_3 dx_4.$$

$$textbf{My quesion.}$$ Where is the mistake? Or, can we write down the star isomorphism in the general case? I think it is related to the multiplication by $$omega^{n-k}$$, but mostly I’m interested how to write down the formula for $$star: Omega^{k}_{M} to Omega^{2n-k}_{M}$$ for $$k>n$$. Is there any reference?

differential equations – How to add a velocity boundary condition with specific time period

I have a wave equation for displacement and velocity, I want to add this boundary condition $$v(x=0,t>0)=1$$
My mathematica code is

``````sol = NDSolve({1/1000 D(u(x, t), {t, 2}) == D(u(x, t), {x, 2}),
D(u(x, t), t) == v(x, t), u(x, 0) == 0, v(x, 0) == 0,
v(0, t > 0) == 1, v(L, t) == 0}, {u(x, t), v(x, t)}, {x, 0, L}, {t,
0, 1})
``````

L is a constant.

differential equations – Infinity: Indeterminate expression 0.0389874+Complexinfinity+Complexinfinity encountered

I’d like to solve 2nd order differential equation:

Which I’d like to draw a graph of x_M and h. m, g, and M is a real constant. Cd is well defined function that has a parameter as x_M’
theta is piecewise defined function,
that appears as below:

for this equation, I used code :

``````(Theta)0 = 0.5236;
(Omega)1 = 5.2359877;
(Omega)2 = 1.396;
t1 = (Theta)0 / (Omega)1;
t2 = (Theta)0 / (Omega)2;
T=2*(t1+t2);
f(t_)=Piecewise({{ -Mod(t,T)*(Omega)1 ,0<=Mod(t,T)<t1},{ -(Theta)0+((Mod(t,T)-t1)*(Omega)2), t1<=Mod(t,T)<(t1+2t2)},{(Theta)0-((Mod(t,T)-t1-2t2)*(Omega)1),(t1+2t2)<=Mod(t,T)<T}})
``````

It is clear that derivative of the function is

``````thetaprime(t_):=Piecewise({{-5.2359877 ,0<=Mod(t,T)<t1},{ 1.396, t1<=Mod(t,T)<(t1+2t2)},{-5.2359877,(t1+2t2)<=Mod(t,T)<=T}});
``````

for differential equation, I used a code

``````initconds = {x(0) == 0, x'(0) == 0.00001, h(0) == (M + m)/((Rho)*A), h'(0) == 0}
eqns = {m Cos((Theta)(t)) Sin((Theta)(t)) h''(t) + (M + m (Sin((Theta)(t))^2)) x''(t) == - Cd(Derivative(1)(x)(t)) *h(t)* Derivative(1)(x)(t) + m Sin((Theta)(t)) (g Cos((Theta)(t)) + r thetaprime(t)^2),( -M-(m*Cos((Theta)(t))^2))*h''(t) - m Cos((Theta)(t)) Sin((Theta)(t)) x''(t) == -g M + A g (Rho) h(t) - m Cos((Theta)(t)) (g Cos((Theta)(t)) + r thetaprime(t)^2)}
``````

But when I try to solve the Equation,

``````sol = NDSolve(Append(eqns, initconds), {x, h}, {t, 0, tf}, Method -> {"DiscontinuityProcessing" -> False})
``````

It gives following three errors. I can’t understand why these errors occur.

Infinity::indet: Indeterminate expression 0.0389874 +ComplexInfinity+ComplexInfinity encountered.

Infinity::indet: Indeterminate expression -0.160636+ComplexInfinity+ComplexInfinity+ComplexInfinity+ComplexInfinity+ComplexInfinity encountered.

NDSolve::ndnum: Encountered non-numerical value for a derivative at t == 0..

differential equations – NDSolve evolution in a specific point

m = 0; n = 0; B = 0.7; W = 4;

T1(m_,n_):= NIntegrate((-Log(t))^(m – 1)*LaguerreL(n, m, -Log(t))*LaguerreL(n, m, -Log(t)), {t, 0, 1});

T2(m_,n_):= NIntegrate((-Log(t))^(m – 1)*LaguerreL(n – 1, m, -Log(t))*LaguerreL(n – 1, m, -Log(t)), {t, 0, 1});

T3(m_,n_):= NIntegrate((-Log(t))^(m – 1)*LaguerreL(n, m, -Log(t))*LaguerreL(n – 1, m, -Log(t)), {t, 0, 1});

T4(m_,n_):= NIntegrate((-Log(t))^(m + 1)LaguerreL(n, m, -Log(t))^2Exp(-B/(f0(z))^2*(-Log(t))^mtLaguerreL(n, m, -Log(t))^2)(-2n*(-Log(t))^(m – 1)tLaguerreL(n, m, -Log(t))^2 +2*(m + n)(-Log(t))^(m – 1)tLaguerreL(n, m, -Log(t))LaguerreL(n – 1, m, -Log(t)) – (-Log(t))^mtLaguerreL(n, m, -Log(t))^2 -m*(-Log(t))^(m – 1)tLaguerreL(n, m, -Log(t))^2), {t, 0, 1});

s(m_,n_,B_,W_,z):=NDSolve({f0”(z) + 1/f0(z)(f0′(z))^2 == n!/((n + m)!(2n + m + 1))1/(f0(z))^3(((n + m)!)/n!(-2n – m + 1) + T1(m, n)(2n + m)^2 + 4T2(m, n)(m + n)^2 – 4T3(m, n)(m + n)(m + 2n) + BW^2*T4(m, n, B, z)), f0(0) == 1, f0′(0) == 0},f0(z), {z, 0, 1});

f0(z_)/.s

f0(0.1)

My code is given above. I have some integrations T1,T2,T3,T4 and one differentiations. I want to find the result of NDSolve at a specific value z=0.1, but I am getting some error.
Can you suggest me to how to write the code.

differential equations – Neumann boundary condition ignored

Im trying to solve a system of two PDE (im(t,x) and ia(t,x)) that are dependent on time and distance. I’m using my own anisotropic mesh (thanks to @TimLaska) that is aimed to represent an interface between a membrane and a liquid. The interface is located at L/2 where L = thickness of the system. “im” takes positive values in the interval 0 <= x <= L/2 and is equal to 0 at x > L/2. “ia” behaves similarly.

The problem that I’m encountering is that when I specify a Neumann boundary value at L/2, Mathematica doesn’t find this value in the mesh and the Neumann value is effectively ignored.

These are the functions for creating anisotropic meshes (kindly provided by @TimLaska in one of my previous questions)

``````(*Import required FEM package*)
Needs("NDSolve`FEM`");
(*Define Some Helper Functions For Structured Meshes*)
pointsToMesh(data_) := MeshRegion(Transpose({data}), Line@Table({i, i + 1}, {i,Length(data)- 1}));
unitMeshGrowth(n_, r_) :=  Table((r^(j/(-1 + n)) - 1.)/(r - 1.), {j, 0, n - 1})
meshGrowth(x0_, xf_, n_, r_) := (xf - x0) unitMeshGrowth(n, r) + x0
firstElmHeight(x0_, xf_, n_, r_) := Abs@First@Differences@meshGrowth(x0, xf, n, r)
lastElmHeight(x0_, xf_, n_, r_) := Abs@Last@Differences@meshGrowth(x0, xf, n, r)
findGrowthRate(x0_, xf_, n_, fElm_) :=  Quiet@Abs@ FindRoot(firstElmHeight(x0, xf, n, r) -    fElm, {r, 1.0001, 100000},Method -> "Brent")((1, 2))
meshGrowthByElm(x0_, xf_, n_, fElm_) := N@Sort@Chop@meshGrowth(x0, xf, n, findGrowthRate(x0, xf, n, fElm))
meshGrowthByElm0(len_, n_, fElm_) := meshGrowthByElm(0, len, n, fElm)
flipSegment(l_) := (#1 - #2) & @@ {First(#), #} &@Reverse(l);
leftSegmentGrowth(len_, n_, fElm_) := meshGrowthByElm0(len, n, fElm)
rightSegmentGrowth(len_, n_, fElm_) := Module({seg}, seg = leftSegmentGrowth(len, n, fElm);
flipSegment(seg))
extendMesh(mesh_, newmesh_) := Union(mesh, Max@mesh + newmesh)
``````

Here I define a couple of constants and create my own mesh

``````(* Constants *)
L = 0.1; dim = dia = 1.*^-6; kf = kr = 1; kr =
1./kf;
(* Mesh generation *)
seg1 = rightSegmentGrowth(L/2., 100, L/1000);
seg2 = leftSegmentGrowth(L/2., 100, L/1000);
totalseg = extendMesh(seg1, seg2);
rh = pointsToMesh@totalseg ;
crd = MeshCoordinates(rh);
mesh = ToElementMesh(crd);
``````

And here is the simple code I use to find the solution alongside the error I get

``````(*PDE system*)

eqsim = {D(im(t, x), t) - dim D(im(t, x), x, x) == NeumannValue(kf*ia(t, x), x == L/2.) + NeumannValue(0, x == L/2.),im(0, x) == 0.1};
eqsia = {D(ia(t, x), t) - dia D(ia(t, x), x, x) ==  NeumannValue(kr*im(t, x), x == L/2.) + NeumannValue(0, x ==   L/2.), ia(0, x) == 0.1};
sol = NDSolve({eqsim, eqsia}, {im, ia},x (Element) mesh, {t, 0, 60}, Method ->{"MethodOfLines", "SpatialDiscretization" -> {"FiniteElement"}})
(*NDSolve::bcnop: No places were found on the boundary where x==0.05 was True, so NeumannValue(ia,x==0.05) will effectively be ignored.*)
(*NDSolve::bcnop: No places were found on the boundary where x==0.05 was True, so NeumannValue(1. im,x==0.05) will effectively be ignored.*)
(* NDSolve::bcnop: No places were found on the boundary where x==0.05 was True, so NeumannValue(0,x==0.05) will effectively be ignored.*)
(*General::stop: Further output of NDSolve::bcnop will be suppressed during this calculation.*)
``````

I first thought that the discretisation in the mesh didn’t include the point L/2 but after double checking I found out that it’s there

``````Position(crd, L/2.)
(*{{100, 1}}*)
``````

Similar problems to this have been reported when using Dirichlet conditions here DiscretizeRegion does not include the boundary specified in ImplicitRegion (10.1)
, here PeriodicBoundaryConditions: missing points (a simpler example)
and here Solving Laplace’s equation in 2D using region primitives
, but after reading the solutions provided I haven’t been able to find a solution to my issue.

Any help, feedback or advice is more than welcome (I’m using v 12.2.0.0).

differential equations – Solve PDE with contraints and unknown boundary conditions

I have a 2D PDE

``````mu = -0.1;
lambda = -1;
x10 = 0;
x20 = 0;
eq = {mu*x1 + D(h1(x1, x2), x1)*mu*x1 +
D(h1(x1, x2), x2)*lambda*(x2 - x1^2) == mu*(x1 + h1(x1, x2)),
lambda*(x2 - x1^2) + D(h2(x1, x2), x1)*mu*x1 +
D(h2(x1, x2), x2)*lambda*(x2 - x1^2) ==
lambda*(x2 + h2(x1, x2))};

``````

The problem is that I don’t know the boundary condition of the PDE, but I do have a constraint that the gradient of h1 and h2 at point (x10, x20) is zero.

``````D(h1(x1, x2), x1)/.{x1 -> x10, x2 -> x20} == 0
D(h1(x1, x2), x2)/.{x1 -> x10, x2 -> x20} == 0
D(h2(x1, x2), x1)/.{x1 -> x10, x2 -> x20} == 0
D(h2(x1, x2), x2)/.{x1 -> x10, x2 -> x20} == 0
``````

How can I approximate h1(x1, x2) and h2(x1, x2) numerically in Mathematica?

Differential Equation Solution By Power Series

Solve $$(1 + x)y’ = py; y(0) = 1$$, where $$p$$ is an arbitrary constant.

First I plugged in the guess $$y = sum_{n = 0}^infty a_n x^n$$:

$$(1 + x)(sum_{n = 0}^infty a_n x^n)’ = psum_{n = 0}^infty a_n x^n$$

Then I expanded the derivative and multiplication:

$$sum_{n = 0}^infty n a_n x^{n – 1} + sum_{n = 0}^infty n a_n x^n = psum_{n = 0}^infty a_n x^n$$

Then I shifted the left index (the first term yielding $$0$$ allows the lower bound to remain $$0$$) and algebraically combined the summations:

$$sum_{n = 0}^infty (n + 1)a_{n + 1} x^n + (n – p)a_n x^n = 0$$

This leads to the following recurrence relation:

$$a_{n + 1} = frac{p – n}{n + 1}a_n$$

Thus for various values of $$n$$:

$$a_1 = p a_0$$, $$a_2 = frac{p(p – 1)}{2}a_0$$, $$a_3 = frac{p(p – 1)(p – 2)}{6} a_0$$, etc.

So applying definitions for the exponential taylor series and falling factorial, the guessed solution would be:

$$y = sum_{n = 0}^infty frac{p! a_0 x^n}{n! (p – n)!} = sum_{n = 0}^infty a_0 e^x p^{underline n}$$

Solving the initial value problem:

$$1 = sum_{n = 0}^infty a_0 e^0 p^{underline n} implies a_0 = frac{1}{sum_{n = 0}^infty p^{underline n}}$$

My final solution is:

$$y = frac{sum_{n = 0}^infty e^x p^{underline n}}{sum_{n = 0}^infty p^{underline n}}$$

However, the answer is supposed to be $$y = (1 + x)^p$$. Are these identical, or did I make an error somewhere?

differential equations – Normalizing x-axis to some parameter

I have the following code

``````Clear("Global`*")
fp = 5.97;
Δ = 0;
f1 = fp + Δ;
f2 = fp - Δ;
f4 = 5.38413;
ωp = 2 π*fp;
ω1 = 2 π*f1;
ω2 = 2 π*f2;
ω4 = 2 π*f4;
ω3 = (ω1 + ω2 - ω4);
a = 10;
NN = 200*a;
LL = 1200*10^-3;
I0 = 0.1;
I1 = 0.48*I0;
I2 = 0.48*I0;
zchar = 50;
Cj = 39*10^-6;
CC0 = 480*10^-6;

k1 = Sqrt(1 - Cj LL ω1^2)/(a*Sqrt(CC0 LL) ω1);
k2 = Sqrt(1 - Cj LL ω2^2)/(a*Sqrt(CC0 LL) ω2);
k3 = Sqrt(1 - Cj LL ω3^2)/(a*Sqrt(CC0 LL) ω3);
k4 = Sqrt(1 - Cj LL ω4^2)/(a*Sqrt(CC0 LL) ω4);

κ1 = -(k1/(8 I0^2*LL^2*(1 - LL*Cj*ω1^2)));
κ2 = -(k2/(8 I0^2*LL^2*(1 - LL*Cj*ω2^2)));
κ3 = -(k3/(8 I0^2*LL^2*(1 - LL*Cj*ω3^2)));
κ4 = -(k4/(8 I0^2*LL^2*(1 - LL*Cj*ω4^2)));
α11 = -(k1/(16*I0^2*LL^2*(1 - LL*Cj*ω1^2)));
α12 = -(k1/(8*I0^2*LL^2*(1 - LL*Cj*ω1^2)));
α13 = -(k1/(8*I0^2*LL^2*(1 - LL*Cj*ω1^2)));
α14 = -(k1/(8*I0^2*LL^2*(1 - LL*Cj*ω1^2)));
α21 = -(k2/(8*I0^2*LL^2*(1 - LL*Cj*ω2^2)));
α22 = -(k2/(16*I0^2*LL^2*(1 - LL*Cj*ω2^2)));
α23 = -(k2/(8*I0^2*LL^2*(1 - LL*Cj*ω2^2)));
α24 = -(k2/(8*I0^2*LL^2*(1 - LL*Cj*ω2^2)));
α31 = -(k3/(8*I0^2*LL^2*(1 - LL*Cj*ω3^2)));
α32 = -(k3/(8*I0^2*LL^2*(1 - LL*Cj*ω3^2)));
α33 = -(k3/(16*I0^2*LL^2*(1 - LL*Cj*ω3^2)));
α34 = -(k3/(8*I0^2*LL^2*(1 - LL*Cj*ω3^2)));
α41 = -(k4/(8*I0^2*LL^2*(1 - LL*Cj*ω4^2)));
α42 = -(k4/(8*I0^2*LL^2*(1 - LL*Cj*ω4^2)));
α43 = -(k4/(8*I0^2*LL^2*(1 - LL*Cj*ω4^2)));
α44 = -(k4/(16*I0^2*LL^2*(1 - LL*Cj*ω4^2)));

A10 = (I1*zchar)/(ω1*Sqrt(2));
A20 = (I2*zchar)/(ω2*Sqrt(2));

α1 = α11*Abs(A10)^2 + α12*Abs(A20)^2;
α2 = α21*Abs(A10)^2 + α22*Abs(A20)^2;
α3 = α31*Abs(A10)^2 + α32*Abs(A20)^2;
α4 = α41*Abs(A10)^2 + α42*Abs(A20)^2;

ΔkL = (k1 + k2) - k3 - k4;

system(j_) := {A1'(x) ==
I*κ1*Conjugate(A2(x))*A3(x)*A4(x)*
E^(-I*ΔkL*x) +
I*A1(x)*(α11*Abs(A1(x))^2 + α12*
Abs(A2(x))^2 + α13*Abs(A3(x))^2 + α14*
Abs(A4(x))^2),
A2'(x) ==
I*κ2*Conjugate(A1(x))*A3(x)*A4(x)*
E^(-I*ΔkL*x) +
I*A2(x)*(α21*Abs(A1(x))^2 + α22*
Abs(A2(x))^2 + α23*Abs(A3(x))^2 + α24*
Abs(A4(x))^2),
A3'(x) ==
I*κ3*A1(x)*A2(x)*Conjugate(A4(x))*
E^(I*ΔkL*x) +
I*A3(x)*(α31*Abs(A1(x))^2 + α32*
Abs(A2(x))^2 + α33*Abs(A3(x))^2 + α34*
Abs(A4(x))^2),
A4'(x) ==
I*κ4*A1(x)*A2(x)*Conjugate(A3(x))*
E^(I*ΔkL*x) +
I*A4(x)*(α41*Abs(A1(x))^2 + α42*
Abs(A2(x))^2 + α43*Abs(A3(x))^2 + α44*
Abs(A4(x))^2), A1(0) == A10, A2(0) == A20,
A3(0) == j*(A10 + A20), A4(0) == 0};

sol(j_) := NDSolve(system(j), {A1, A2, A3, A4}, {x, 0, NN})

LogLinearPlot({10*Log10(Abs((A3(NN)/A3(0)) /. sol(j))^2), 10*Log10(Abs((A3(NN)/A3(0)) /. sol(10^-5))^2) - 1}, {j, 10^-5, 5}, PlotRange -> {Full, All}, AxesOrigin -> {0, 0}, PlotStyle ->{Directive(Black), Directive(Gray, Dashed)}, ImageSize -> Large)

(A10 + A20) // N
``````

Upon running, it should generate a plot like the following:

One can see that under the `system(j_)` and `sol(j_)` function, I am attempting to solve the (four) differential equations (`A1(x)`,`A2(x)`,`A3(x)`,`A4(x)`) numerically. In particular, I’ve defined my initial condition for `A3(0)` with respect to `j` where `j` is the variable to be plotted on the x-axis. As one can see, I am plotting `j` normalized to `A10+A20` as shown in the condition `A3(0)==j*(A10+A20)`

In particular, the x intersection between the black and the gray-dashed curve should be sensitive to the parameter `I0` (set to `0.1` in this case). However, changing `I0` to any other parameter doesn’t change the plot whatsoever (one can check this). I confirmed my suspicion by redefining my initial condition for `A3(0)` as simply `A3(0)==j` (rather than `A3(0)==j*(A10+A20)`) and the intersection between the black and gray curve indeed changes as `I0` changes. This seems to suggest that I am normalizing to the wrong parameter (`A10+A20`), but this shouldn’t be a problem. Because `A3(0)` represents some wave amplitude and `A10+A20` represents the initial pump amplitudes and it is conventional to normalize the wave with respect to the pump amplitudes.

What is the problem here? Is `I0` some how dependent on the x-axis so that something cancels when I define my `A3(0)` as `A3(0)==j*(A10+A20)`? Any suggestions for normalizing the x-axis?

Thank you.