Example of differential equation with periodic boundary conditions that has at least two simple eigenvalues

I need an example of a periodic function. $ q: mathbb {R} to mathbb {R} $ with period $ pi $ such that if we consider the differential equation
begin {ecation} tag {1}
and & # 39; & # 39; (x) + ( lambda -q (x)) y (x) = 0
end {equation}

and boundary conditions
begin {ecation} tag {2}
y (0) = y ( pi), & # 39; (0) = y & # 39; ( pi)
end {equation}

begin {ecation} tag {3}
y (0) = – y ( pi), & # 39; (0) = – y & # 39; ( pi),
end {equation}

then there are at least two simple eigenvalues ​​of the problem (1), (2) or there are at least two simple eigenvalues ​​of the problem (1), (3).

I know that yes $ q (x) = 0 $ for all $ x in mathbb {R} $, then (1), (2) has only one simple eigenvalue and all eigenvalues ​​of (1), (3) are multiple.

dnd 5e: is the command spell linked to conditions?

The spell is generally left to the GM, but some commands, such as "Drop!" have specific default effects

Much of the spell is left to the GM's interpretation:

You can issue a different command than the one described here. If it does, the GM determines how the target behaves.

That said, the "Drop" command in particular is an example that provides the spell:

release: The target releases whatever he is holding and then ends his turn.

So, for this example, an objective would simply drop what is participation and use standard English, which means that the target would not drop things he wears as a backpack, and he would not be paralyzed or anything like that.

The "Grovel!" command states:

Crawl: The target falls prone and then ends its turn.

If you fall prone They are prone, this means that it is affected by the prone condition. This is shown because everything that makes a target prone uses that wording or similar (such as hitting a prone target).

The "Stop!" He says:

Stop: The target does not move and does not perform actions …

All it does is make a target not move and cannot perform actions, nothing more. No conditions apply because spells only do what they say.

plugins: short code PHP file for two past conditions when the short code is made

I have created a shortcode for the PHP file add-in template, now the template has two HTML designs, one with icons that are displayed and another that shows only the Name.

Is it possible to activate a bool when do_shotecode for icons in functions.php and then in the verification template file for bool and say if do_shortcode has the parameter icon set to true and then echo the content with icons?

Paths for files I am working on

  • functions.php: child-theme /
  • plugin template: child-theme / folderName / plugin_template.php

Basic code, for example

Print Items with Icons

  • Coffee

Print Names only without icons

  1. Coffee

java – Conditions in cycle for

In my function, what I do is fill a matrix with random numbers:

 public int()() llenarMatriz(int matriz()()) {

    for (int i = 0; i < matriz.length; i++) {

        for (int j = 0; j < matriz(j).length; j++) {

            matriz(i)(j) = (int) (Math.random() * 10);

    return matriz;


The matrix that passed by parameter is the following:

int matrizP()() = new int(2)(3);
matrizP = matriz.llenarMatriz(matrizP);

But what I don't understand is why he sends me the following error:

java.lang.ArrayIndexOutOfBoundsException: 2

What returns me matriz(j).length It will always be 3 in this case, I mean this is the condition that I use as a reference to not move to another position that does not exist.

Note that matriz(j).length where j increase to a maximum of 1, but when a cycle passes and the matriz( "aqui j seria 1" ).length but send the error of java.lang.ArrayIndexOutOfBoundsException: 2.

So I don't send this error I can put matriz(i).length I mean I change the iterator and everything is fixed, but I don't understand why you send that error with the iterator j Yes always

matriz("sea j o i ").length; //sera 3 en este caso.

In short, the problem is in the for nested bone the second for, and its condition as matriz(j).length, because with the iterator i it works but with the iterator j no.

I was thinking that: it must be that a cycle for You can only work with a condition that is immutable until you finish your cycle, that is, a cycle for It is not dynamic if you want to change the value of your condition parameter. I don't know if this is correct.

How to avoid the race conditions while handling the UNIX process in C?

I am trying to build a simple UNIX shell for educational purposes. My intention is to correctly manage the processes in the background, showing a message every time you finish your work through the SIGCHLD controller in the parent process. But my code seems to repeat infinitely, generating secondary processes every time I add & to my order, for example: ls &

So I guess there are some alarming errors regarding race conditions, since I get my expected start when I use gdb and go step by step in debug mode, but I can't detect where I'm really going wrong. Are they using best practices for process management? (In my original code, I made sure to manage jobs through appropriate data structures and perform memory management through free, then that is not a problem)

Fragment of code for reproduction: (alternative mirror here)


int argLength = -1;
char** argVector;
int isBackground = 0;
int childPID = -1;
int parentPID = -1;

void printPrompt()
    printf("Shell~: $ ");

char* str_concat(char* s1, char* s2)
    int l1 = strlen(s1);
    int l2 = strlen(s2);
    char* op = (char*) calloc (l1 + l2, sizeof(char));
    for(int i=0; i 0)
                        printf("(bg)  %d finishedn", pid);

                    int status;
                    waitpid(pid, NULL, WUNTRACED);
                isBackground = 0;
    return 0;

Denial of Schengen visa: no justification was provided for the purpose and conditions of the planned stay

When you see "No justification was provided for the purpose and conditions of the planned stay."In a rejection from Schengen, it generally means that they decided that the request was & # 39; incoherent & # 39; or not credible, or both.


Consistency has a special meaning in the Schengen vocabulary; It means clear, sensitive, consistent and, most importantly, understandable. An application & # 39; incoherent & # 39; It lacks these qualities. Consistent applications will generally succeed; Incoherent applications will not. Some examples of inconsistency are given in the manual …

  • an applicant claims to travel to an industrial zone, stay in a cheap hotel, for tourist purposes;
  • An applicant claims to visit a professional event on dates that do not correspond to the actual dates of the event;
  • an applicant states that the purpose of the trip is to visit a friend, but it turns out that the person in question is absent during that period;
  • A jewelry merchant claims to have been invited to attend a medical conference.

While these examples may not correspond to your situation, they have a common thread in which the application was not consistent with the job, status or apparent lifestyle of the applicant.


The inconsistency also includes applications that did not successfully establish a premise for the visit. Commonly, there is no apparent connection between the applicant's itinerary and his career, apparent lifestyle or financial capacity. Some examples of this are …

  • an applicant expresses interest in an event or geographical area for which there are no signs of prior interest;
  • An applicant submits a letter of invitation from a sponsor whose relationship with the applicant is not clear or distant;
  • An applicant is continuously sponsored by the same person where there is no apparent justification for visiting that person;
  • An applicant cannot establish why a visit is contemplated at this particular time.


The credibility of an applicant can also result in a reason for rejection of "No justification was provided for the purpose and conditions of the planned stay.".

Some examples of credibility problems are …

  • The applicant has an existing support network of friends or family that would easily allow them to break the rules;
  • A disproportionate amount of income is being spent for the visit;
  • The applicant has changed the stories in the course of several interactions with Schengen officials and his credibility has been damaged;
  • The decision maker cannot determine whether the funds really belong to the applicant;
  • The decision maker cannot determine whether the funds were obtained legally;
  • The applicant has a history of prior rejections and there are no visible changes in the circumstances;
  • The applicant's presentation contains gaps or evidentiary deficiencies that cannot be explained;
  • The applicant goes to a place where previous visa holders have incurred non-compliance;
  • The sponsor has a dubious state or has sponsored people who have incurred in default;
  • The sponsor tries to attest to things that are beyond their control (such as ensuring that the person does not stay longer).


All Schengen members offer a judicial appeal for a refusal, namely to appeal the decision. Each member has their own procedure and the details are too heavy to cover here, but a refusal is accompanied by an explanation of how to proceed if the applicant wishes to appeal. There is a great discussion about the decision to file an appeal here.

Cool application

The other option is to correct the deficiencies in your original application and apply again. In this way, it is invariably faster and cheaper, and there is no required interval or "cooling period" between successive applications. However, it is important to recognize that a new request made immediately after a rejection is a recipe for disaster when one of the reasons for rejection is "No justification was provided for the purpose and conditions of the planned stay."Before submitting a new application, you must understand precisely what caused the rejection (especially if your credibility has been negatively evaluated). To obtain the best results, a new application must contain one or more of these …

  • An explanation of why you think the rejection occurred;
  • An explanation of how their circumstances have changed since the refusal;
  • If the decision maker relied on an assumption that was adverse to you, then a brief clarification may be helpful;
  • If the decision maker did not (apparently) examine the evidence that was favorable to you, then a brief statement highlighting this may be useful;
  • A detailed list of the evidence you are presenting to correct the problems along with why you think each item will be useful;

Legal counsel

Understanding the various formulas of Schengen rejection can be daunting. If you don't understand how "No justification was provided for the purpose and conditions of the planned stay."It applies to your case, or if you are not sure what evidence you can use to get a better result, then it is advisable to arrange a consultation with a lawyer who operates a practice area on Schengen visas.

The disadvantage is that it can be scammed, especially if it is in Africa or South Asia. For this reason, a safe strategy is to instruct a licensed lawyer to practice in the member state you are applying for. Each member state operates a professional body that authorizes lawyers and regulates their practice in accordance with EU standards.


In Belgium, one of those organizations is the Ordre des barreaux francophones

For others, you can use the search portal offered by the Council of Bars and Legal Societies of Europe (CCBE).

Keep in mind that lawyers of this caliber will attract a fee for their services.


In some of the most heinous cases, a refusal will also have this reason: "It was not possible to determine his intention to leave the territory of the member state before the visa expires ". In these cases, the decision maker concluded that you were not a good faith applicant.

General advice

  1. Tell the truth;
  2. Make sure your financial evidence contains movements that are well explained;
  3. Include all trips abroad in a detailed list;
  4. Make sure your visit is explained with a plausible itinerary;
  5. Do not make empty and useless promises about your intentions;
  6. Don't let your hosts secure your intentions that they can't keep;

simplifying expressions – Assumptions with conditions

How do you specify the assumption that a function is real for real arguments?


Simplify[Re[I f[x]],Assumptions->f[x_]∈ Reals]
Out[1]= 0


Simplify[Re[I f[x]],Assumptions->x∈ Reals&&f[x_/;x∈ Reals]∈ Reals]
Out[2]= -Im[f[x]]

How do I get zero in the second case?

polynomials: find conditions in $ a $ and $ b $ since two cubics are equal.

Polynomial $ x ^ 3 + x-3 = 0 $ it has roots $ alpha $, $ beta $ Y $ gamma $. Yes $ frac {a alpha + 1} { alpha-b} $, $ frac {a beta + 1} { beta-b} $ Y $ frac {a gamma + 1} { gamma-b} $ are the roots of another cubic, what are the conditions in $ to $ Y $ b $ given that the two cubics are equal?

Where should this start? Using Vieta's formulas, this is tedious and I got stuck with too many unknowns. Please help

symbolic – Obtain conditions for the variable to be real

I have a complex definition of some variables (which I obtained using s = Solve(1 - a x^2 - x^3 == 0, x)) and now I want to know under what conditions of "a" each of the solutions will be real.

For example, one of the variables I have is r3=-(a/3) + ((1 - I Sqrt(3)) a^2)/(
3 2^(2/3) (-27 + 2 a^3 + 3 Sqrt(3) Sqrt(27 - 4 a^3))^(
1/3)) + ((1 + I Sqrt(3)) (-27 + 2 a^3 +
3 Sqrt(3) Sqrt(27 - 4 a^3))^(1/3))/(6 2^(1/3))
, I want to know in what conditions r3 will be a real number.

I have tried to simplify this to the most basic example where I want to know what are the conditions in a + b * that I need to be real (where the answer I expect is b = 0), so I tried to use Resolve(a + b I, Reals) But this does not give me the answer I expect.

real analysis: how does convergence in the $ ell ^ 2 $ norm imply convergence in the form $ ell ^ infty $ -norm with Lipschitz conditions?

Suppose I have $ | f | 2 = int_ {S} f (x) ^ 2dx rightarrow 0 $ on a compact set $ S subset R ^ d $Y $ f $ It is Lipschitz and soft. What kind of additional condition can I add? $ S $ so that $ | f | _ { infty} rightarrow 0 $?

Yes $ S $ is $ R ^ d $, no additional conditions are needed, since for any $ | f (x) |> 0 $ we can always build a neighborhood $ B (x, r) subset S $ for some $ r = Theta (| f (x) |) $ and the integral over $ B (x, r) $ is $ Theta (| f (x) | ^ {d + 1}) $. However, for a compact set, it does not necessarily hold that $ B (x, r) subset S $, so I suppose that some regularization condition can be added so that the convergence is still maintained.