Equivalence of projections in smaller von Neumann algebra

I came across the following assertion and I can’t understand why it’s true.

We are given with two finite equivalent projection $esim f$ in some von Neumann algebra $A$ (with a unit of course). It’s known that the projection $q=evee f$ is also finite, so we infer the algebra $qAq$ is a finite algebra (that is, the unit element is a finite projection).

Now suppose that $q-esim q-f$ in $qAq$, show that $q-esim q-f$ also in $A$.

Thanks in advance.


Comment: Technically it is true that $q-esim q-f$ in $qAq$ so we don’t really need to assume that.

differential equations – Solving PDE with Neumann boundary conditions

I want to solve a pde on a triangular domain, such that you can’t leave the domain (which I think would just be Neumann boundary conditions).

My pde takes the following form:

$frac{partial u}{partial x}(A(x,y)u(x,y)) + frac{partial u}{partial y}(B(x,y)u(x,y)) + frac{partial^2 u}{partial x^2}(C(x,y)u(x,y)) + frac{partial^2 u}{partial x partial y}(E(x,y)u(x,y)) + frac{partial^2 u}{partial y^2}(F(x,y)u(x,y)) = 0.$

However, I’m struggling to solve this numerically in mathematica. When I use the code below, mathematica doesn’t do anything, just returning exactly what I wrote. My code is as follows, where A,B,C,E,F are functions of x and y I have already defined:

domain = Triangle({{0, 0}, {1, 0}, {0, 1}});
eqn = D(A(x,y)*u(x, y), x) + 
    D(B(x, y)*u(x, y), y) + 
    D(C(x,y)*u(x, y), x, x) + 
    D((E(x,y))*u(x, y), x, y) + 
     D((F(x, y)*u(x, y), y, y) == 0;
s = NDSolve(eqn, NeumannValue(0, ...), u, {x, y} (Element) domain);

I also require that the integral of $u(x,y)$ over the domain be equal to 1 – is there a way of including this condition in the pde solver?

Von Neumann Random Byte Debuting in Python

I need to create a method for you should be biased bytes of the method get_raw_bytes(length) using the von Neumann deburring method. Here is my current code:

def get_debiased_bytes(length):
    arr_debiased_bytes = ()
    debiased_byte = 0
    bit_counter = 0
    while len(arr_debiased_bytes) < length:
        raw_bytes = get_raw_bytes(length * 5)
        for byte in raw_bytes:
            for k in range(0, 8, 2):
                bit1 = byte >> k & 1
                bit2 = byte >> k + 1 & 1
                if bit1 != bit2:
                    debiased_byte = debiased_byte << 1 | bit1
                    bit_counter += 1
                    if bit_counter == 8:
                        arr_debiased_bytes.append(debiased_byte)
                        debiased_byte = 0
                        bit_counter = 0
    return bytes(arr_debiased_bytes(:length))

I think my code can still be improved and sorted. Can someone help me with this?

pde – Solving the Cauchy problem for the wave equation using Neumann boundary conditions

I've been trying to solve this problem for a while, but I can't seem to figure out what to do with the Neumann Limit Condition. The problem is solving the following:

begin {equation *}
u_ {tt} -u_ {xx} = 0 \
u (x, 0) = 0 \
u_t (x, 0) = begin {cases}
pi cos ( pi (x-1)), y 1 <x <2 \
0 and more
end {cases}
end {equation *}

For the next
1) $$ x in (- infty, infty) $$
two) $$ x in (0, infty) text {y} u_x (0, t) = 0 $$
3) $$ x in (- infty, 3) text {y} u (3, t) = 0 $$

I know how to do 1, just apply D & # 39; Alemebert's formula. But I am lost by 2 and 3.

differential equations – Numerical solution for ODE 1-D with Neumann conditions only

I tried to solve a simple ODE with only Neumann conditions like
enter the description of the image here
But obviously this does not work.
I must add a useless Dirichlet condition to make it work
enter the description of the image here
I have verified that the solution is correct, but how can I get it without
including the useless condition?

The code is

Sol = NDSolveValue({Piecewise({{1, x < 0}, {2.25, x > 0}}, 0)*u(x) + 
 Derivative(2)(u)(x) == 
     NeumannValue(I*(2*E^(I*x) - u(x)), x == -4*Pi) + 
 NeumannValue((0. - 1.5*I)*u(x), x == 4*Pi), 
   DirichletCondition(1, False)}, u, {x, -4*Pi, 4*Pi})
ReImPlot(Sol(x), {x, -4*Pi, 4*Pi})

reference request – Estimation of the Dirichlet map to Neumann for mixed limit value problems

To be specific, consider a model problem
in $ Omega subset mathbb {R} ^ 2 $ with Lipschitz limit,
begin {cases}
– Delta u = 0 & text {in} Omega, \
u = g & mbox {on} Gamma_D, \
partial_n u = g_N & text {on} Gamma_N,
end {cases}

where $ partial Omega = bar Gamma_D cup bar Gamma_N $, $ Gamma_N cap Gamma_D = emptyset $.

I'm curious to know if there is literature on the DtN map (existence, estimates, explicit construction as layer potentials) in the Dirichlet boundary: $$ mathcal {V}: H ^ {1/2} ( Gamma_D) a H ^ {1/2} ( Gamma_D), $$
such that
$$
| mathcal {V} g | _ {- 1/2, Gamma_D} ^ 2 leq langle mathcal {V} g, g rangle _ { Gamma_D} + ( text {terms in} Gamma_N)
$$

Laplacian's smallest non-zero eigenvalue with Neumann contour conditions of a ball in a simply connected pinched negative curvature space

In the document, Small values ​​of multiple geometric finites, http://www.math.uni-bonn.de/people/ursula/eigenvalue.pdf, page 11, in the last paragraph, the author stated that "Now the Value own smaller than zero with Neumann contour conditions of a radio ball $ rho $ in a variety of simply connected curvature contained in $ (- κ, −1) $ is not smaller than $ (n – 1) ^ 2/4 $"I struggled to find a reference to support this claim. If we use the Dirichlet boundary condition, it follows from the comparison of the monotonicity of the domain and the curvature of the balls. However, for the Neumann condition it depends on the geometry of the multiple and the limit.I consulted (BCD) cited on the line and I still did not see the idea.Maybe I am missing something or Neumann's own value for a ball in pinched curvature has been explicitly calculated.

I also have doubts about the following statement, "therefore, by domain monotonicity, the second proper value for each of the sets $ U_i $ with mixed boundary conditions (i.e., Neumann boundary conditions in $ ∂U_i ∩ Ω $ and border conditions of Dirichlet in $ ∂Ω) is not less than $ (n – 1) ^ 2/4 $ Also here $ U_i $ is a Dirichlet domain as polygons within the radio ball $ rho $ previously mentioned. $ Ω It is an open set with compact closure and lower limit in the radius of injectivity $ rho $. It is not clear if the author is using the monotonicity of Neumann or the monotonicity of Dirichelt. For the ball and $ U_i $, since there are domains that do not cross $ ∂Ω, the mixed boundary condition becomes the Neumann boundary condition. We cannot conclude Neumann's own value of the smaller sets. $ U_i $ is greater than Neumann's own value of the ball it contains $ U_i $ directly, as the usual Neumann monotonicity requires a partition and actually gives the inverse inequality instead of the desired one. Maybe I made a mistake in my argument or is there any reference to Neumann's own value / monotonicity that works this way. Thank you.

or operational algebras or – Norms for calculating operator polynomials in the Hilbert space and generalized inequality of von Neumann

Leave $ T $ be an operator $ l ^ 2 ({ mathbb {Z} _ { geq 0}}) a l ^ 2 ({ mathbb {Z} _ { geq 0}}) $, $ e_n mapsto sqrt {1 – q ^ {2 (n + 1)}} e_ {n + 1} $, where $ 0 <q <1 $. I want to compute $ | f (T, T *) | $ (operator standard) for any $ f in mathbb {C} (z, bar z) $. Operator $ T *, of course, is Hilbert's conjugate and $ T ^ * e_0 = 0 $, $ T ^ {*} e_n = sqrt {1-q ^ {2n}} e_ {n-1} $ for $ n> 0 $.

I'm not sure if that calculation is possible and would be happy to show that $ | f (T, T *) | to sup limits_ {| z | leq 1} f (z, bar z) $ how $ q $ It goes to 1 because it's really why I need to calculate the rules first.

I think there is a generalization of von Neumann's inequality as $ q $-analysis or something (there are many generalizations) because the usual inequality proves it for polynomials $ g in mathbb {C} (z) $ (without $ bar z $)

Question: Does anyone know any useful facts or inequalities related to my question? Something that could help.

1D transport equation with Neumann conditions

I am trying to solve the second order differential equation of the type of heat transfer in a finite system x<0,L>, in particular:

eq = D(T(x, t), t) - k*D(T(x, t), {x, 2}) == 0;

(For simplicity, I suppose that hereafter k = 1 and L = 1).
The system is connected to a semi-infinite source of power 1 for x (-infty, 0), while the & # 39; temperature & # 39; initial for x <0.1) is 0. This provides the initial condition for the equation in parts form:

T(x, 0) == Piecewise({{1, x < 0}, {0, x >= 0}})

I also assume that Neumann's boundary conditions: 1) the "heat" flow in the system is proportional to the temperature gradient between the source and the system:

Derivative(1, 0)(T)(0., t) == (T(0., T) - 1)

and 2) there is no loss of & # 39; heat & # 39; in the second system limit:

Derivative(1, 0)(T)(1, t) == 0}

Merging all the elements I obtained a code:

sol = NDSolve({eq,
T(x, 0) == Piecewise({{1, x < 0}, {0, x >= 0}}),
Derivative(1, 0)(T)(0., t) == (T(0., t) - 1),
Derivative(1, 0)(T)(1, t) == 0},
T, {x, 0, 1}, {t, 0, 1})
Plot3D(Evaluate(T(x, t) /. sol), {x, 0, 1}, {t, 0, 1})

When I ran the Mathematica code (v8.0) he sent me a warning about the inconsistency of the initial conditions / limit: I guess I confused something with the behavior at x = 0. However, I got a graph, not exactly what I want (T (x, t) = 0). Then I changed the initial conditions to

T(x, 0) == Piecewise({{1, x <= 0}, {0, x > 0}})

to impose the heat flux through x = 0. Once again, I received the warning and a graph: this time I can see some distributions of & # 39; temperature & # 39; in the system, but the result seems to be bad, since the inconsistency of the initial conditions / limit and also the non-physical results (T> 1, while it cannot be higher than the source temperature T = 1).

Can anyone help me overcome the problem with the initial limits / conditions? I am using Mathematica 8, so I cannot use the NeumannValue function.

o.operator algebras – Trace extension in von Neumann subalgebra

No Yes $ Gamma $ is a group without torsion and with the property of infinite conjugation class then the algebra $ L Gamma $ it's a factor of type $ II $ and has a single normal state of affairs.

Now leave $ x $ in $ Gamma $ Different from identity. The von Neumann algebra $ S $ generated by $ x $ is isomorphic to $ L ^ infty (S ^ 1) $, the algebra of measurable functions in the circle. It has many normal trative states.