User jon - MathOverflow

Exact solutions to nonlinear Klein-Gordon equation

2013-05-22T08:13:52Z

The nonlinear pde $$ \partial_t^2\phi-\partial_x^2\phi+\lambda\phi^3=0 $$ has the exact solution $$ \phi(x,t)=\mu\left(\frac{2}{\lambda}\right)^\frac{1}{4}{\rm sn}(p_0t-p\cdot x+\varphi,i) $$ with $\mu$ and $\varphi$ two integration constants and sn the snoidal Jacobi function, provided the dispersion relation holds $$ p^2_0=p^2+\mu^2\sqrt{\frac{\lambda}{2}}. $$ If I interpret $p_0$ as the energy, it seems that is finite. Computing the integral $$ E=\int d^Dx\left[\frac{1}{2}(\partial_t\phi)^2+\frac{1}{2}(\partial_x\phi)^2+\frac{\lambda}{4}\phi^4\right] $$ and extending the volume to infinity, this is divergent. Can one find a sound mathematical explanation for this? Is there a way to "regularize" this integral?

Thanks.

Answer by Jon for Solving systems of integral equations using Volterra series

2013-05-21T12:22:30Z

In order to iterate, you have to substitute the second equation into the first one. So, $$ n_{11}(x,z)=1+\int_{-\infty}^xdye^{-izy}u(y)\int_{-\infty}^ydy_1e^{izy_1}{\bar u}(y_1)n_{11}(y_1,z). $$ This equation is generally the starting pointing for an iterative procedure, the main tool of perturbation techniques. E.g., you can choose for the first iterate $n_{11}(x,z)=1$ and you will get $$ n_{11}(x,z)=1+\int_{-\infty}^xdye^{-izy}u(y)\int_{-\infty}^ydy_1e^{izy_1}{\bar u}(y_1) $$ $$ +\int_{-\infty}^xdye^{-izy}u(y)\int_{-\infty}^ydy_1e^{izy_1}{\bar u}(y_1)\int_{-\infty}^{y_1}dy_2e^{-izy_2}u(y_2)\int_{-\infty}^{y_2}dy_3e^{izy_3}{\bar u}(y_3)+\ldots. $$ You can stop the procedure at any desired order to get an approximation to the solution of the integral equations. Then, you put this approximation to $n_{11}$ back into the equation for $n_{12}$ and you will get an approximate solution for it at the given order.

Answer by Jon for A heat kernel for Schrödinger operator with low-order terms

2013-05-16T07:33:52Z

I think your formula is not correct. The right kernel is invariant by interchanging $x$ with $y$. This symmetry must be preserved.

Then, note that $$ L=-\Delta+ax^2+bx=-\Delta+a\left(x+\frac{b}{2\sqrt{a}}\right)^2-\frac{b^2}{4a} $$ and this operator is invariant for translations. This means that the kernel for $b\ne 0$ can be obtained from the kernel for $b=0$ by translation of a shift $\frac{b}{2\sqrt{a}}$ and multiplying for the overall factor $e^{\frac{b^2}{4a}t}$.

Once this is observed, we can consider the kernel for $b=0$, just the argument of the exponential, to be $$ S_0(x,y;t)=\sqrt{a}\left[(x^2+y^2)\coth 2\sqrt{a}t-\frac{xy}{2\sinh 2\sqrt{a}t}\right] $$ and apply the above translation to get $$ S(x,y;t)=\sqrt{a}\left[\left(x^2+y^2+(x+y)\frac{b}{\sqrt{a}}+\frac{b^2}{4a}\right)\coth 2\sqrt{a}t\right. $$ $$ \left.-\frac{1}{2\sinh 2\sqrt{a}t}\left(xy+\frac{b}{2\sqrt{a}}(x+y)+\frac{b^2}{4a}\right)\right] $$ and this gives the right kernel $$ K(x,y;t)=\left(\frac{\sqrt{a}}{2\pi\sinh 2\sqrt{a}t}\right)^\frac{1}{2}e^{\frac{b^2}{4a}t}e^{-S(x,y;t)}. $$

Answer by Jon for Heat integro - differential equation

2013-05-15T11:55:52Z

There is a simple way to manage this equation using a Fourier series. We assume a boundary at $0$ and $L$ and that exists the Fourier series for the solution $$ u(x,t)=\sum_{n=-\infty}^{\infty}u_n(t)e^{i\frac{2\pi n}{L}x} $$ then you note that $$ D(t)=\int_{-L}^L u(s,t)ds=\int_{-L}^L\sum_{n=-\infty}^{\infty}u_n(t)e^{i\frac{2\pi n}{L}x}=2L u_0(t) $$ and you are left with the following set of ordinary equations $$ \partial_t u_n(t)=-4\pi^2n^2u_0(t)u_n(t). $$ This yields $\partial u_0(t)=0$ and so, $u_0=constant=D_0$ and so for $n\ne 0$, $$ u_n(t)=e^{-4\pi^2n^2D_0t}u_n(0). $$

Answer by Jon for First moment of a function of a normally distributed random variable

2012-12-03T08:27:25Z

The best way to proceed is to use Legendre polynomials through the formula $$ \frac{1}{\sqrt{1-2yx+x^2}} = \sum_{n=0}^\infty P_n(y) x^n. $$ In this way you will get closed form integrals with the error function.

Answer by Jon for Finding Kuramoto Model coupling strength with limits?

2012-12-02T15:07:53Z

This problem can be approached by a series in $KR$ and assuming for $g$ a Gaussian distribution. So, we have to manage $$ 1=K\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2\theta\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{K^2R^2\sin^2\theta}{2\sigma^2}}d\theta. $$ The limit $R\rightarrow 0^+$ can be taken under the integrale but we prefer a series in $KR$ that yields $$ 1=K\frac{1}{\sqrt{2\pi}\sigma}\left(\frac{\pi}{2}-\frac{\pi}{2^4 \sigma ^2} K^2 R^2+\frac{\pi}{2^7 \sigma ^4}K^4 R^4+O(K^6R^6)\right). $$ The required limit provides $$ K=\sqrt{\frac{8}{\pi}}\sigma $$

Answer by Jon for Solution of Helmholtz-Equation where Phase is restricted by additional PDE

2012-11-11T12:08:40Z

The condition on $g$ gives a definite pde for $A$. This can be seen in the following way. Let us insert the solution $f=A(x,y)e^{ig(x,y)}$ into the Helmholtz equation. We get $$ \Delta A+2i(\partial_xg\partial_xA+\partial_yg\partial_yA)+\Phi(x,y)A=0 $$ being $$ \Phi(x,y)=i\Delta g-(\partial_xg)^2-(\partial_yg)^2+a. $$ Now, assuiming $L$ is a linear operator with the Green function $LG=\delta$, one can write $$ g(x,y)=g_0(x,y)+\int_\Omega dx'dy'G(x,x';y,y')h(x',y') $$ being $Lg_0=0$. By substituting this into $\Phi$ and the equation for $A$ we get a partial differential equation to solve. For some operator $L$, the final equation could be simple to manage but, for the general case, maybe some approximation techniques could help.

Answer by Jon for 2D Ising model partition function expansion

2012-10-18T13:34:14Z

This problem is known in literature due to the Onsager's solution of the 2d Ising model. You can already find this on an old paper by Fisher and Ferdinand. Some more recent papers, available also from arxiv are the following:

http://arxiv.org/abs/cond-mat/0009054

http://arxiv.org/abs/cond-mat/0110287

A paper just available with subscription is Partition function of a finite Ising model on a torus T Morita 1986 J. Phys. A: Math. Gen. 19 L1191.

Answer by Jon for Solvability of a nonlinear elliptic equation

2012-10-12T12:17:21Z

Firstly, turning to the $\mathbb{R}^1$ case, we note that $$ -u_{xx}+u^2u_x=-\partial_x\left(u_x-\frac{1}{3}u^3\right)=f(x) $$ and so we get a Chini equation (see http://mathoverflow.net/questions/104006/inhomogeneous-bernoulli-equation): $$ -u_x+\frac{1}{3}u^3=\int_{x_0}^xf(x')dx'+C $$ and a general solution to this equation is not known. Turning to $\mathbb{R}^3$, you can see that your equation can be written down in the form $$ \nabla\cdot {\bf v}=-f $$ being $$ {\bf v}=\left(u_x-\frac{1}{3}u^3,u_y-\frac{1}{3}u^3,u_z-\frac{1}{3}u^3\right) $$ and you have to satisfy the Gauss identity $$ \int_{\partial\Omega}{\bf v}\cdot d{\bf S}=-\int_{\Omega}dV f. $$ This means that one has to find three functions $v_1,\ v_2,\ v_3$ that satisfy three different Chini equations. Again, a general solution misses in this case. The conclusion to draw is that you need to specialize the problem to see if this becomes manageable in some case or resort to some perturbation technique.

Answer by Jon for prove that flat shape maximizes a functional

2012-06-28T07:00:32Z

The easiest way to prove this is using variational calculus. You have to put $$ \delta I(G(\omega))=0. $$ The calculation is quite straigthforward and provides the condition $$ \delta G(\omega)=0 $$ and so the extremum is for $G(\omega)=G=constant$. Finally, from the condition you have to set $$ \int_{-k\pi}^{k\pi}G(\omega)=2k\pi G=1. $$ This gives the value of the extremum $G=\frac{1}{2k\pi}$.

Expanded on OP request: The idea behind functional calculus (calculus of variations) is to consider a class of functionals, as in your case, that can be amenable to a generalized differentiation. You can find all the rules and the definition of a functional derivative here but for a more serious approach some lectures as the ones I pointed out in the comment area are needed. Your case is particularly simple as one is left in each term with the variation of $G(\omega)$ and this must be zero to find an extremum.

Update on OP request: Let us introduce the following functional $$ Z_m[G]=\int_{-k\pi}^{k\pi}\frac{A}{G(\omega)+A}e^{-im\omega}d\omega $$ The functional we are considering takes the form $$ I[G]=Z_0[G]-\frac{Z_1^*[G]Z_1[G]}{Z_0[G]}. $$ Now we have $$ \delta Z_m[G]=-\int_{k\pi}^{-k\pi}\frac{A}{(G(\omega)+A)^2}\delta G(\omega)e^{-im\omega}d\omega. $$ Chain rule applies also to functionals and we can evaluate $\delta I[G]$ immediately to give $$ \delta I[G]=\delta Z_0[G]-\frac{Z_1^*[G]Z_1[G]\delta Z_0[G]-Z_0[G]\delta(Z_1^*[G]Z_1[G])}{Z_0^2[G]} $$ and we see that the condition $\delta G(\omega)=0$ sets the variation to zero. This solution is consistent with the given constraint provided $G=\frac{1}{2k\pi}$. The application of the constarint a posteriori fixes the value of the constant.

Further clarification for OP: I will show that a functional that does not depend from at least a first derivative is a constant in one dimension. Let us consider the functional $$ S=\int_a^bL(q(t),q'(t),t)dt. $$ The condition for the extremum just gives $\delta S=0$ yielding Euler-Lagrange equation $$ \frac{d}{dt}\frac{\partial L}{\partial q'(t)}=\frac{\partial L}{\partial q(t)}. $$ Then, if there is no dependence on derivative we are left with $\frac{\partial L}{\partial q(t)}=0$ that implies immediately $L=L(t)$ and $q(t)=constant$.

Answer by Jon for Wave equation v.s.Schrödinger equation

2012-08-03T09:47:15Z

There is no direct link in the way you write it down. In order to decompose the wave equation you will need a Clifford algebra of matrices $\gamma$, such that $\gamma_i\gamma_j + \gamma_j\gamma_i = 2\eta_{ij}$, and define the operator $D=i\gamma_k\partial^k$. Then you will have $D^2=\partial_{tt}-\Delta$. But note that this operator will not apply on ordinary scalar functions but on spinors. Finally, when you consider the "massive" operator $D-I$, you are able to recover in some limit the Schroedinger equation.

Answer by Jon for Who came up with the Euler-Lagrange equation?

2012-07-31T18:26:11Z

I think the main reference here is Euler's archive. You have to look at the two books on mechanics firstly presented on 1736. Euler often presented his works to the community before publication. Being these authors almost contemporary, it is possible that the name to these equations is indeed the right one.

Stochastic processes with random matrices

2012-07-18T09:46:16Z

I am currently working on complex networks. I consider a matrix $\cal N$ with random entries $\delta_{ik}$. These entries are varying randomly in time and so I have a sequence of random matrices that determines a stochastic process.

My question is quite simple: Does it exist a generalization of standard stochastic differential equations to random matrices like the case I am considering? Also relevant references are a good answer.

Thanks beforehand.

Tanaka stochastic differential equation and Kolmogorov equation

2012-01-19T09:42:17Z

Given Tanaka sde

$$dX_t=[a{\rm sign}(X_t)+b]dW_t$$

is there associated a diffusion process and so a Kolmogorov (Fokker-Planck) equation? What is this equation?

References answering the question are welcome.

Thanks.

Square root of a stochastic process

2011-11-30T13:54:34Z

I am currently working on the understanding of the stochastic nature of the Schroedinger equation. This has a notable history dating back to Nelson's works and relative criticisms. But one can take a different path and, starting from a random walk process with probability

\begin{equation} P(k;N) = \binom{N}{k}\left(\frac{1}{2}\right)^N, \end{equation}

one can assume the existence of a "square root" of this process with a complex amplitude

\begin{equation} A(k;N) = \binom{N}{k}^\frac{1}{2}\left(\frac{1}{2}\right)^{N/2}e^{i\phi(k,N)} \end{equation}

such that $P(k;N)=|A(k;N)|^2$ and $\phi(k,N)$ are some phases exactly determined. These are quantum amplitudes. Is this anything making sense? Does any literature exist about?

Answer by Jon for Classical Limit of Feynman Path Integral

2012-07-17T16:51:02Z

Things stay in this way. Consider the action of a given particle that appears in the path integral. We consider the simplest case $$ L=\frac{\dot x^2}{2}-V(x) $$ and so, a functional Taylor expansion around the extremum $x_c(t)$ will give $$ S[x(t)]=S[x_c(t)]+\int dt_1dt_2\frac{1}{2}\left.\frac{\delta^2 S}{\delta x(t_1)\delta x(t_2)}\right|_{x(t)=x_c(t)}(x(t_1)-x_c(t_1))(x(t_2)-x_c(t_2))+\ldots $$ and we have applied the fact that one has $\left.\frac{\delta S}{\delta x(t)}\right|_{x(t)=x_c(t)}=0$. So, considering that you are left with a Gaussian integral that can be computed, your are left with a leading order term given by $$ G(t_b-t_a,x_a,x_b)\approx N(t_a-t_b,x_a,x_b)e^{\frac{i}{\hbar}S[x_c]}. $$ Incidentally, this is exactly what gives Thomas-Fermi approximation through Weyl calculus at leading order (see my preceding answer and refs. therein). Now, if you look at the Schroedinger equation for this solution, you will notice that this is what one expects from it just solving Hamilton-Jacobi equation for the classical particle. This can be shown quite easily. Consider for the sake of simplicity the one-dimensional case $$ -\frac{\hbar^2}{2}\frac{\partial^2\psi}{\partial x^2}+V(x)\psi=i\hbar\frac{\partial\psi}{\partial t} $$ and write the solution exactly in the form given above. Substitute it into the equation and impose that Hamilton-Jacobi equation holds $$ \frac{1}{2}|\nabla S|^2+V(x)=\frac{\partial S}{\partial t}. $$ You can see that both solutions agree neglecting higher order derivatives and a possible Heaviside function. This means that, at this order, the description using waves or classical paths is perfectly identical. This situation is not different from the case of geometric optics and a full wave equation description. You can describe your light waves as rays exactly as in quantum mechanics your particles become classical ones and you can describe them with paths.

Answer by Jon for Classical Limit of Quantum Mechanics

2012-07-16T10:26:29Z

There are two different views about the semiclassical limit in quantum mechanics, the first is based on a somewhat shaky ground due to the fact that the existence of the Feynman integral is not proved yet. On the other side, Wiener integral, its imaginary time counterpart does exist and one could pretend to work things out from this and then move to the Feynman integral. The other approach relies on substantial mathematical theorems due to Elliott Lieb and Barry Simon in the '70 and is essentially valid for many-body physics. These latter results make the limit $\hbar\rightarrow 0$ and $N\rightarrow\infty$ equivalent while the former is not really a physical limit due to the fact that Planck constant is never zero.

Starting from Feynman path integral, the standard formulation applies to a mechanial problem described from a Lagrangian $L$, normally $L=\frac{\dot x^2}{2}-V(x)$ but one can extend this to more general cases, and then the postulate is that, given a path $x(t)$, this must contribute to the full quantum mechanical amplitude of a particle going from the point $x_a$ to $x_b$ with a term $e^{\frac{i}{\hbar}S}$ being $S=\int_{t_a}^{t_b}dtL(\dot x,x,t)$ the action. All the possible paths contribute and so, the full amplitude will be given by the formal writing $$ A(x_a,x_b)\sim\int[dx(t)]e^{\frac{i}{\hbar}\int_{t_a}^{t_b}dtL(\dot x,x,t)}. $$ Be warned that this integral is not proved to exist yet, but the Wiener counterpart, that can be obtained changing $t\rightarrow it$, exists and describes Brownian motion. Now, if you take the formal limit $\hbar\rightarrow 0$ to this integral you will immediately recognize the conditions to apply the stationary phase method to it. This implies that the functional must have an extremum and this can be obtained by pretending that $$ \delta S=\delta \int_{t_a}^{t_b}dtL(\dot x,x,t)=0 $$ that is, the paths that give the greatest contribution are the classical ones and one recover the classical limit as a variational principle as learned from standard textbooks.

While this is a quite common approach, to extend what really happens to a macroscopic system that we can see to respect all the laws of classical mechanics, we have to turn our attention to the limit of a large number of particles $N\rightarrow\infty$. In this case one has more rigorous results. These are due to Lieb and Simon as already said above. They published two papers about

Lieb E. H. and Simon B. 1973 Phys. Rev. Lett. 31, 681.

Lieb E. H. and Simon B. 1977 Adv. in Math. 23, 22.

In the first paper, their theorem 4 states

Theorem: For $\lambda < Z$, let $E_N^0$ and $\rho_N^0(x)$ denote the ground-state energy and one-electron distribution function for N spin-$\frac{1}{2}$ electrons obeying the Pauli principle and interacting with $k$ nuclei as described above. Then (a) $N^{-\frac{7}{3}}E_N^0\rightarrow E_1$, as $N\rightarrow\infty$; (b) $N^{-2}\rho_N^0(N^{-\frac{1}{3}}x)\rightarrow\rho_1(x)$ as $N\rightarrow\infty$, where convergence in (b) means that for any domain $D\subset R^3$, the expected fraction of electrons in $N^{-\frac{1}{3}}D$ approaches $\int_D\rho_1 (x)d^3x$.

Where $\rho_1(x)$ and $E_1$ refer to the Thomas-Fermi distribution and the corresponding energy. This theorem states that the limit $N\rightarrow\infty$ for a quantum system, under some mild conditions, is the Thomas-Fermi distribution. A system with this distribution is a classical system. The fact that a system with a Thomas-Fermi distribution is a classical one can be seen through the following two references:

W. Thirring(Ed.), The Stability of Matter: From Atoms to Stars - Selecta of E. Lieb, Springer-Verlag (1997).

L. Hörmander, Comm. Pure. Appl. Math. 32, 359 (1979).

The second paper just gives the mathematical support to derive Thomas-Fermi approximation as the leading order of a classical expansion for $\hbar\rightarrow 0$ that I will not present here.

A class of Ito integrals

2012-01-11T09:36:48Z

I am currently working on stochastic processes and I have met a stumbling block in the Ito integral

$$\int_{t_0}^tdt'G(t')[dW(t')]^\alpha$$

with $\alpha\in\mathbb{R}$ and $\alpha>0$. Textbooks result is given for integer $\alpha$ but not in the more general case that could not exist. Of course, also some good references are welcome.

Answer by Jon for Bidirectional ODE

2012-07-11T12:30:22Z

This is a typical problem for time-dependent Schroedinger equation. Firstly, let us consider the unperturbed problem. We can write the free solutions as $$ \tilde a=a_0e^{i\beta z} \qquad \tilde b=b_0e^{i\beta(L-z)}. $$ Now, redefine $a$ and $b$ in you system as $$ a(z)=\bar a(z)e^{i\beta z} \qquad b(z)=\bar b(z)e^{-i\beta z} $$ and, by direct substitution, you will get $$ \bar a(z)=e^{-i\beta z}f\left[\bar a(z)e^{i\beta z},\bar b(z)e^{-i\beta z}\right] $$

$$ \bar b(z)=e^{i\beta z}g\left[\bar a(z)e^{i\beta z},\bar b(z)e^{-i\beta z}\right] $$ and you can integrate this immediately to get $$ \bar a(z)=a_0+\int_{0}^zdz'e^{-i\beta z'}f\left[\bar a(z')e^{i\beta z'},\bar b(z')e^{-i\beta z'}\right] $$

$$ \bar b(z)=b_0e^{i\beta L}+\int_{0}^zdz'e^{i\beta z'}g\left[\bar a(z')e^{i\beta z'},\bar b(z')e^{-i\beta z'}\right]. $$

These equations can be solved iteratively and you will get at first order $$ \bar a(z)=a_0+\int_{0}^zdz'e^{-i\beta z'}f\left[a_0e^{i\beta z'},b_0e^{i\beta (L-z')}\right]+\ldots $$

$$ \bar b(z)=b_0e^{i\beta L}+\int_{0}^zdz'e^{i\beta z'}g\left[a_0e^{i\beta z'},b_0e^{i\beta (L-z')}\right]+\ldots $$ You will recognize that we are iterating using the free solutions and this is nothing else than standard perturbation theory. This gives your solution in a closed form and if $f$ and $g$ are not too involved, the integrals can be easily worked out.

Answer by Jon for References/Papers on analytic solutions to SDEs

2012-07-09T08:13:55Z

This book by Lawrence Evans at Berkeley should be enough.

Answer by Jon for change the sign of volatility

2012-06-15T14:05:40Z

There is no change into the probability distribution due to the absolute value. The reason can be traced back on the Ito's lemma applied to the derivation of the (Fokker-Planck) equation for the probability distribution: This will depend on $\sigma^2$.

Answer by Jon for Asymptotic behaviour of a mean

2012-06-12T12:31:03Z

Let us consider the sum $$ m_N(x)=\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{cN}{2}-i(2c-1)\right). $$ The first step to get an asymptotic approximation is to extract the leading term in $N$ to obtain $$ m_N(x)=[f(x)]\log N+\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-\frac{i}{N}(2c-1)\right). $$ When $N$ is finite, we recognize a Riemann series and apply the average theorem. So, there exists a value of argument of the logarithm such that $$ m_N(x)=[f(x)]\log N+[f(x)]\log[z(x)]. $$ We can take $z(x)=\frac{c}{2}-t[f(x)](2c-1)$ being $t\in (0,1)$.

Indeed, we can define a partition with $x_i=x_{i-1}+\frac{1}{[f(x)]N}$ and so $$ \frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-\frac{i}{N}(2c-1)\right)=[f(x)]\Delta x\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-i[f(x)](2c-1)\Delta x\right) $$ being $\Delta x=\frac{1}{[f(x)]N}$. But this, in the given limit, is nothing else than $$ \int_{\frac{c}{2}}^{\frac{c}{2}-[f(x)](2c-1)}\log(z)dz<\infty $$ as it should.

Turing machines and Ising model

2012-05-30T09:39:30Z

I have currently started a new research line aiming to prove a mapping between a 2-symbol Turing machine and the one dimensional Ising model. The connection is seen by recognizing that a set of symbols on the tape of the machine is indeed a configuration of a one-dimensional Ising model.

The relevance of this connection relies on the fact that a limit on the entropy for information has been conjectured by Rolf Landauer in the '60. A mapping between a Turing machine and the Ising model can make this conjecture a theorem.

My question here is what are the main references on Turing machines, probabilistic Turing machine and so on? Does it exist any standard, research level, literature to consider?

Especially for probabilistic Turing machines, does it exist some result making them equivalent to a deterministic Turing machine and, in any case, there exists a theorem stating they can compute anything a deterministic Turing machine can?

Thanks.

Answer by Jon for Integral Equation with "convolution"

2012-05-30T08:34:56Z

The simplest approach one can think is an iterative one. Let us consider the given equation $$ f(x)=\int_{-\infty}^xG(x,y)f(y)f(x-y)dy. $$ Now, we assume as a first iterate $f^{(0)}(x)=1$ and so $$ f^{(1)}(x)=\int_{-\infty}^xdyG(x,y) $$

$$ f^{(2)}(x)=\int_{-\infty}^xdyG(x,y)\int_{-\infty}^ydwG(y,w)\int_{-\infty}^{x-y}dzG(x-y,z) $$

and so on. One is granted the existence of the n-th iterate provided the integral of $G$ exists and is properly bounded.

Answer by Jon for First eigenvalue of Schrödinger operator is simple

2012-05-20T17:04:36Z

Barry Simon's book, "Functional Integration and Quantum Physics", should fit the bill.

Answer by Jon for Fourier and Bessel

2012-05-19T12:25:41Z

There is a fundamental reference using Bessel functions in Fourier's works. This is "Théorie analytique de la chaleur" firstly published on 1822. You will find this series firstly given in chapter VI pag. 370. This chapter is about the propagation of the heat in a cylinder ("of course" let me add).

The modern nomenclature was invented by Bessel himself on 1824, just two years after Fourier's work. This is proved in F. Bessel, "Untersuchung des Theils der planetarischen Störungen", Berlin Abhandlungen (1824). Here the functions I and J get their names.

Answer by Jon for The perturbed KdV Equation

2012-05-09T07:58:16Z

I will follow the original paper as much as I can. So, let us consider the solution of the equation

$$u_t-6uu_x+u_{xxx}=\epsilon u.$$

You will have a general solution $u=u(x,t;\epsilon)=\sum_{n=0}^\infty\epsilon^n u_n(x,t)$ after a power expansion on $\epsilon$. Now, turning your attention to the Schroedinger equation

$$\psi_{xx}-[u(x,t,;\epsilon)-\lambda]\psi=0$$

you will notice that the problem is now

$$\psi_{xx}-[u_0(x,t)+\epsilon u_1(x,t)+\ldots-\lambda]\psi=0$$

and this is amenable to standard Rayleigh-Schroedinger perturbation theory. This means that the first correction preserves the discrete nature of the spectrum and will be real yet. You can iterate this procedure going to higher orders in $\epsilon$ still preserving the inverse scattering method. This can be seen in the follwing way. From the given Schroedinger equation one has

$$u(x,t;\epsilon)=\lambda+\frac{\psi_{xx}}{\psi}.$$

The condition granting that this solves KdV equation is $\lambda_t=0$ when $\psi\rightarrow 0$ given $|x|\rightarrow\infty$ and $u$ evolves by KdV equation. This must be true also for the approximate solution. So, from this equation we can evaluate the next-to-leading order correction by setting

$$\lambda=\lambda_0+\epsilon\lambda_1+\ldots$$ $$\psi = \psi_0+\epsilon\psi_1+\ldots$$

and we get

$$u(x,t;\epsilon)=\lambda_0+\epsilon\lambda_1+\frac{\psi_{0xx}}{\psi_0}-\epsilon\frac{\psi_1\psi_{0xx}-\psi_0\psi_{1xx}}{\psi_0^2}+\ldots$$

and so

$$u_0(x,t)=\lambda_0+\frac{\psi_{0xx}}{\psi_0}$$

$$u_1(x,t)=\lambda_1-\frac{\psi_1\psi_{0xx}-\psi_0\psi_{1xx}}{\psi_0^2}$$

$$\vdots$$

Turning back to KdV equation for the leading order one has

$$u_{0t}-6u_0u_{0x}+u_{0xxx}=0$$

and this is exactly the problem solved by inverse scattering method that we recover as it should. The next-to-leading order correction gives:

$$u_{1t}+6(u_0u_{1x}+u_1u_{0x})+u_{1xxx}=u_0$$

that is a linear equation. We notice here that we are consistent with the condition $\psi_1\rightarrow 0$ when $|x|\rightarrow\infty$ provided this is true for $\psi_0$ and this grants also $u_1\rightarrow 0$ as it should. One can check at this point that this is equivalent on doing the same perturbation approach on the scattering problem for the inverse scattering method (see eq.(5) in the original paper by Miura et al.).

Answer by Jon for Constructing an example of Hamiltonian flow

2012-05-08T12:00:10Z

This is a typical problem of special relativity. The next best choice is a constant acceleration $a$ providing $V(x)=ax$. This has the well-known solution

$$x = d+ \frac{1}{a}\sqrt{1+ a^2 t^2}$$

being $d$ a constant, that is unbounded as required.

Answer by Jon for Monotonicity of a combination of Bessel functions

2012-05-07T10:58:05Z

This function is quite interesting as it reaches a maximum in 0. This can be seen without difficulty computing the first and second derivatives. It is

$$B(a,r)=\frac{K_2(ar)I_2(a)-I_2(ar)K_2(a)}{I_2(a)}I_2(ar)$$

that gives

$$B(0,r)=\frac{1}{4}(1-r^4)>0$$

$$B'(0,r)=0$$

$$B''(0,r)=-\frac{1}{12}r^2(1-r^2)^2<0$$

for the given interval. This means that this function is decreasing for $a>0$. The problem is if there are some other points where the first derivative can become zero changing the concavity of the curve. So, the first derivative has the following involved expression

$$B'(a,r)=\frac{1}{2 I^2_2(a)}\left[I_2(ar)^2 (I_1(a)+I_3(a)) K_2(a)+\right.$$ $$I_2(a) I_2(ar) (-2 r (I_1(ar)+I_3(ar)) K_2(a)+$$ $$I_2(ar) (K_1(a)+K_3(a))+r I_2^2(a) ((I_1(ar)+I_3(ar)) K_2(ar)$$ $$\left.-I_2(ar) (K_1(ar)+K_3(ar)))\right]$$

Now we notice that $I_2$ is a monotonic increasing function that never becomes zero and $K_2$ is a monotonic decreasing function that never becomes zero for the given intervals and similarly is true for $I_1,\ I_3$ and $K_1,\ K_3$. All these functions are positive. So, it is not difficult to realize that the first derivative is a monotonic function and never hits zero again being just a balance of functions never reaching zero unless $a=0$ and having monotonic behavior, $K_i$ are decreasing functions and $I_i$ increasing functions. We also note that

$$\lim_{a\rightarrow\infty}B'(a,r)=0$$

that can be proved using the asymptotic formula for these Bessel functions. Now, combining monotonicity and positivity of these Bessel functions, starting from 0 and reaching asymptotically 0 at increasing values of the argument, they can just reach an extremum and never cross zero again. This can also be seen with a simple plot

evaluated at $r=0.1,0.3,0.5,0.7,0.9$.

Answer by Jon for Asymptotic question about time ordered exponentials

2011-12-01T09:40:59Z

This question has a solution presented in this paper even if with the jargon and notation of theoretical physics. So, I will use a somewhat different notation and I will change

$${\bf A}(t)\rightarrow -i{\bf A}(t).$$

Then, I will compute the eigenvalues and eigenvectors of ${\bf A}(t)$ through

$${A}(t)|n;t\rangle=\lambda_n(t)|n;t\rangle.$$

Now, you get a series with a leading order term

$${\bf B}(r)=\sum_n e^{i\gamma_n}e^{-ir\int_{-1}^1 dt\lambda_n(t)}|n;1\rangle\langle n;-1| \qquad r\rightarrow\infty$$

being $\gamma_n=\int_{-1}^1dt\langle n;t|i\partial_t|n;t\rangle$ known as geometric phase. Then, an expansion in the inverse of $r$ can be obtained with the matrix

$$\tilde {\bf A}(t)=-\sum_{n,m,n\ne m}e^{i(\gamma_n(t)-\gamma_m(t))}e^{-ir\int_{t_0}^tdt[\lambda_m(t)-\lambda_n(t)]}\langle m;t|i\partial_t|n;t\rangle|m;t_0\rangle\langle m;t_0|$$

being in this case

$$\tilde {\bf B}(r)=\prod_{-1}^1e^{-i\tilde {\bf A}(t)dt}$$

so that

$$B(r)=\sum_n e^{i\gamma_n}e^{-ir\int_{-1}^1 dt\lambda_n(t)}|n;1\rangle\langle n;-1|\tilde {\bf B}(r).$$

This represents a solution of the Schroedinger equation

$$-ir{\bf A}(t)B(r;t,t_0)=\partial_tB(r;t,t_0)$$

in the interval $t\in [-1,1]$ and $r\rightarrow\infty$.

An example:

$$ A(t) = \frac{1}{1+t^2} \begin{pmatrix} 2 & t\\ -t & -2 \end{pmatrix} $$

and one has to solve the problem $$ \dot U(t)=rA(t)U(t) $$ with $r\gg 1$. We want to apply the technique outlined above. We note that $A(t)$ is not Hermitian and so, solving the eigenvalue problem, we get $\lambda_{\pm}=\pm r\frac{\sqrt{4-t^2}}{1+t^2}$ and $$ v_+=\frac{1}{2}\begin{pmatrix} \sqrt{2+\sqrt{4-t^2}}\\ -\frac{t}{\sqrt{2+\sqrt{4-t^2}}}\end{pmatrix} \qquad v_-=\frac{1}{2}\begin{pmatrix}-\frac{t}{\sqrt{2+\sqrt{4-t^2}}} \\ \sqrt{2+\sqrt{4-t^2}}\end{pmatrix}. $$ But $v_+^Tv_-\ne 0$ and so these vectors are not orthogonal. We need to solve also the eigenvalue problem $u^T(A-\lambda I)=0$ producing the following eigenvectors $$ u_+=\frac{1}{2}\begin{pmatrix} \sqrt{2+\sqrt{4-t^2}}\\ \frac{t}{\sqrt{2+\sqrt{4-t^2}}}\end{pmatrix} \qquad u_-=\frac{1}{2}\begin{pmatrix} \frac{t}{\sqrt{2+\sqrt{4-t^2}}} \\ \sqrt{2+\sqrt{4-t^2}}\end{pmatrix}. $$ It is easy to see that $u_+^Tv_-=u_-^Tv_+=0$. It is important to note that $\lambda(t)=\lambda(-t)$ and $u_+(-t)=v_-(t)$ and $u_-(-t)=v_+(t)$ and so, these eigenvectors are just representing a backward evolution in time. Now, we want to study the time evolution of a generic eigenvector $$ \phi(t)=\begin{pmatrix}\phi_+(t) \\ \phi_-(t)\end{pmatrix} $$ and this can be done by putting $$ \phi(t)=c_+(t)e^{r\int_0^tdt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}}v_+(t)+ c_-(t)e^{-r\int_0^tdt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}}v_-(t) $$ that will produce the set of equations $$ \dot c_+=\gamma_+c_++e^{-2r\int_0^tdt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}}\frac{u_+^T\frac{dv_-}{dt}}{u_+^Tv_+}c_- $$

$$ \dot c_-=\gamma_-c_-+e^{2r\int_0^tdt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}}\frac{u_-^T\frac{dv_+}{dt}}{u_-^Tv_-}c_+ $$ having set $\gamma_+=\frac{u_+^T\frac{dv_+}{dt}}{u_+^Tv_+}$ and $\gamma_-=\frac{u_-^T\frac{dv_-}{dt}}{u_-^Tv_-}$. These equations are interesting because they provide the way time evolution is formed in a non-hermitian case. But this is also saying to us that each component may evolve in time differently: One can be really smaller than the other for $r\gg 1$. But we can also understand the form of the higher order corrections:

$$ c_+(t)=c_+(0)+\int_0^tdt'e^{\int_0^{t'}dt''(\gamma_+(t'')-\gamma_-(t''))}e^{-2r\int_0^{t'}dt''\frac{\sqrt{4-t^{''2}}}{1+t^{''2}}}\frac{u_+^T\frac{dv_-}{dt''}}{u_+^Tv_+}c_-(0)+\ldots. $$

Using a saddle point technique, we can uncover here that the correction is exponentially small and cannot be stated that is something like $e^{r}/r^k$ in the general case.

Now, we consider the simple case $c_+(0)=1$ and $c_-(0)=0$. The approximate solution will be

$$ \phi_+(t)=\frac{1}{2}\sqrt{2+\sqrt{4-t^2}}e^{r\int_0^{t}dt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}} \qquad \phi_-(t)=-\frac{1}{2}\frac{t}{\sqrt{2+\sqrt{4-t^2}}}e^{r\int_0^{t}dt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}} $$

and solving numerically the set of differential equations for $r=50$ we get the following

The agreement is strikingly good.

Comment by Jon

2013-05-22T13:53:44Z

Carlo, thanks a lot. It appeared like this idea does not apply to the nonlinear case by simply redefining the solution with the trick of the $\frac{1}{\sqrt{V}}$ normalization for plane waves. This is my main concern.

Comment by Jon

2013-05-22T08:48:01Z

@IgorKhavkine: Thanks for the comment. Yes, I am aware of the similarity with the plane waves but there one can circumvent the problem (at least looking at standard textbooks). Here the situation seems more awkward. Could you expand on your last sentence?

Comment by Jon

2012-12-18T10:20:49Z

Also su math.stackexchange math.stackexchange.com/questions/261289/…

Comment by Jon

2012-12-05T09:03:45Z

It is possible to freely download this paper (both pdf and djvu at numdam.org/item?id=MSM_1953__123__1_0) but does not seem to exist a translated version.

Comment by Jon

2012-10-19T07:53:33Z

You will need a more physical oriented book like Giorgio Parisi "Statistical Field Theory" amazon.com/…. If you need some more rigorous results but not specifically for Ising model, Ruelle's book is the right one books.google.it/….

Comment by Jon

2012-08-15T15:56:42Z

Any linear function going like $\theta(x)=x+b$, wiht $b$ a constant, has the property $\theta(x+2\pi)=\theta(x)+2\pi$. Is this anything different?

Comment by Jon

2012-08-11T08:01:59Z

I am not fooling you at all. In your example you need a Lagrange multiplier. You do not know calculus of variation. I have provided a proof with all equations in place, just check my answer. I fear that you are fooling me so, since now, I will stop further posting on this matter. Please, take some time to study.

Comment by Jon

2012-08-10T12:15:28Z

Please, check my answer and I will show you that this is a mathematical truth. I just invite you to study a meaningful textbook on calculus of variations.

Comment by Jon

2012-08-09T20:29:30Z

In your case you need to have also derivatives with respect to $\omega$: $G'(\omega)\ G''(\omega)\ldots$. This is the only way to get a non-trivial solution in your case. Indeed, when an increment is zero you are coping with a constant.

Comment by Jon

2012-08-09T12:17:02Z

Pierre, I am doing calculus of variations en.wikipedia.org/wiki/Calculus_of_variations and my conclusion is correct. In your case there are a couple of points to be fixed: 1) In your example you need a Lagrange multiplier that provides the right answer. 2) Your derivative is written incorrectly. It should eventually be $\frac{\delta Z_m[G]}{\delta G[\omega]}$.

Comment by Jon

2012-08-08T21:24:16Z

Yes,it is. For a given functional $\int_a^bL(q,q',t)dt$ the minimum is achieved through Euler-Lagrange equation en.wikipedia.org/wiki/…. What makes your case easier is that there are no derivatives with respect to $\omega$.

Comment by Jon

2012-08-08T15:55:52Z

Pierre, give me a few time to arrange an update to the answer.

Comment by Jon

2012-08-06T16:52:48Z

First three chapters of Tao's book are at math.ucla.edu/~tao/preprints/chapter.pdf (thanks to author). Of course, I agree with Tao's conclusion but you are completely off the track. Read this chapter 2 more carefully.

Comment by Jon

2012-08-05T14:27:43Z

I fear you should move your question to math.stackexchange.com.

Comment by Jon

2012-08-05T10:21:22Z

I should look at Tao's book, please cite. Anyhow, you cannot factorize wave equation as you did. The only way, devised by Dirac in the '20s, to get the square root of a Laplacian or a wave operator is using Clifford algebra of matrices. This can also be seen by the fact that wave equation is invariant under Lorentz group SO(1,3) and Schroedinger equation is not. Dirac approach preserves this symmetry of the wave equation as it must.