magento2 – Magento 2 How to save Conditions in CMS page form

In my custom module, I add field conditions like below code…

cms_page_from.xml

<fieldset name="conditions" sortOrder="70">
    <settings>
        <collapsible>true</collapsible>
        <label translate="true">Conditions</label>
    </settings>
    <container name="conditions_apply_to" sortOrder="10">
        <htmlContent name="html_content">
            <block name="conditions_apply_to" class="[Vendor][ModuleName]BlockAdminhtmlPromoQuoteEditTabConditions" />
        </htmlContent>
    </container>
</fieldset>

How to save selected options in field conditions after click save button?!
Please advice me!
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commerce – How to get the general conditions of sale accepted in the purchase tunnel?

How to get the general conditions of sale accepted in the purchase tunnel?

The only solution I have found is to use this module, which has not been recently updated, and is incompatible with Drupal 9.

I would prefer a method that uses the existing interface and doesn’t require me to write code.

Second order elliptic PDE problem with boundary conditions whose solutions depend continuously on the initial data

Consider the following problem
$$begin{cases}
-Delta u+cu=f,&xinOmega\
u=g,&xinpartialOmega
end{cases}$$

where $Omegasubseteqmathbb R^n$ is open with regular boundary, $cgeq0$ is a constant, $fin L^2(Omega)$ and $g$ is the trace of a function $Gin H^1(Omega)$. If we consider $u$ a weak solution to this problem, and define $U=u-Gin H_0^1(Omega)$, it is easy to see that $U$ is a weak solution to the following problem
$$begin{cases}
-Delta U+cU=f+Delta G-cG,&xinOmega\
U=0,&xinpartialOmega
end{cases}$$

It is also easy to see that we can apply Lax-Milgram theorem with the bilinear form
$$B(u,v)=int_Omegaleft(sum_{i=1}^nu_{x_i}v_{x_i}+cuvright)$$
and the bounded linear functional
$$L_f(v)=int_Omega(f-cG)v-int_Omegasum_{i=1}^n G_{x_i}v_{x_i}$$
to conclude there exists a unique weak solution $U$ to the auxiliary problem defined above. If we define $u=U+Gin H^1(Omega)$, it is clear then that this function will be a solution to the original problem.

Now to the question: I would like to prove that this solution $u$ depends continuously on the initial data, that is, that there exists a constant $C>0$ such that
$$lVert urVert_{H^1(Omega)}leq C(lVert frVert_{L^2(Omega)}+lVert GrVert_{H^1(Omega)})$$
I feel that the work I have done to prove that $L_f$ is bounded should be relevant for our purposes, because
$$lVert urVert_{H^1(Omega)}leqlVert UrVert_{H^1(Omega)}+lVert GrVert_{H^1(Omega)}$$
and
$$lVert UrVert_{H^1(Omega)}leq C B(U,U)^{1/2}= C|L_f(U)|^{1/2}$$
The problem is that I don’t know how to manipulate $L_f(U)$ to obtain the result. I have managed to prove a completely useless inequality, for it involves the norm of $U$.

I would appreciate any kind of suggestion. Thanks in advance for your answers.

P.S. The problem is that a priori $Delta G$ doesn’t have to be in $L^2(Omega)$, which makes it hard to use the $H^2$ regularity of $U$ (which would solve the problem instantly).

P.S.S. Also posted this question in SE.

dg.differential geometry – Integrability conditions imply existence of potential

I’m looking for a proof of the following well-known theorem:

If $f$ is a continuously differentiable vector field in a simply connected region $Gsubset mathbb{R}^n$ which satisfies the integrability conditions, then there is potential $phi$ such that $nabla phi = f$.

I know a proof for star-shaped regions. By googling I also found proofs for $n=2$ (resp. $n=3$) which make use of the theorem of Green (resp. Stokes).

Question 1: Does someone know a reference where a proof of the general case is written down?

Question 2: Can simply-connected be weakened in $H^1(G)=0$?

fa.functional analysis – Poincare Inequality for $H^2$ function satisfying homogeneous Robin boundary conditions

Let $Omegasubsetmathbb{R}^3$ be a bounded smooth domain. In general, for a Poincare inequality of the type
$$|u|_{L^2}le C |nabla u|_{L^2}$$
to hold for all $uin Xsubset H^1(Omega)$ and $C$ independent of $u$, then $X$ needs to be such that it doesn’t contain constant translates. That is, if we consider $u+M$ for large $M>0$, the left hand side of the inequality increases indefinitely while the right hand side is unchanged, so we need some extra constraint in the definition of $X$. So common choices are $X=H^1_0(Omega)$ or $X={uin H^1(Omega)| int_Omega u,dx=0}$.

Here’s my question. Suppose, we’d like to say that there exists $C$ such that for all $uin X={uin H^2(Omega)|(partial_n u+u)_{|partialOmega}=0}subset H^2(Omega)$ we have
$$|u|_{L^2}le C|nabla u|_{L^2}.$$
First, is this true? If so, how does one prove such a statement? Essentially the requirement that $u$ satisfies the homogeneous Robin condition $(partial_n u+u)_{|partialOmega}=0$ should at least formally rule out constant translates, since $(partial_n u+u)_{|partialOmega}=0$ is not invariant under translation of $u$ by constants.

My guess is that it IS true, however, the usual proof I know of such statements usually relies on some compactness argument. For example, if $X$ were simply $H_0^1(Omega)$, then for the sake of contradiction, if we assume that there exists a sequence $u_nin H_0^1$ such that
$$|u_n|_{L^2}ge n|nabla u_n|_{L^2}$$
then, defining $v_n=u_n/|u_n|_{L^2}$, we have
$$frac{1}{n}ge |nabla v_n|_{L^2}.$$
Thus we have a bounded sequence in $H^1$ and a subsequence that converges strongly in $L^2$ and weakly in $H^1$ to some $vin H^1$. Because $|nabla v_n|_{L^2}to 0$, $v$ is constant. And since the trace map is continuous (and weakly continuous) from $H^1(Omega)$ to $H^frac{1}{2}(partialOmega)$ we have that $v$ is in fact in $H^1_0(Omega)$ and therefore $v=0$. Then we have a contradiction because $|v_n|_{L^2}=1$ for each $n$ implies that $|v|_{L^2}=1$.

Now this argument doesn’t work for Robin boundary conditions because now the relevant (Robin) trace operator is continuous and weakly continuous from $H^2(Omega)$ to $H^frac{1}{2}(partialOmega)$. In particular, if $v$ is the weak $H^1$ limit of a sequence $v_nin H^2$, then $v$ could be in $H^1$ but not $H^2$ and thus the notion of the normal derivative $(partial_n v)_{|partialOmega}$ may not even make sense for $v$. And without being able to say $(partial_n v+v)_{|partialOmega}=0$, we can’t necessarily say that $v=0$ like we did in the previous paragraph. So this is where I’m stuck. Any help would be appreciated.

Are there any necessary conditions of the existence of a Hamiltonian cycle on directed graphs

I’m trying to prove that one concrete directed graph has no Hamiltonian cycle, but didn’t seem to find any relevant theorems

dual nationality – Travel to Russia under current conditions: Do I need a visa if I reside in country X as citizen of Y (but would not require one as citizen of Z)?

Russia has recently lifted travel bans for nationals and residents of some countries (see, e.g., here). I happen to reside in one of these countries (Germany) and as an EU citizen (Italy) automatically have a residence permit. In addition, I have dual citizenship and also hold a passport of a country (Brazil) for which a visa is not required to travel to Russia under normal circumstances, but which is currently still affected by the Russian travel ban.

Can I travel to Russia under the current circumstances? Do I need a visa as an Italian citizen? Or can I enter Russia without a visa with my Brazilian passport (along with adequate evidence of residence in Germany)?

accounts – Best way to ask an user to accept the terms and conditions

I’ve currently working on a registration page, and after looking across the internet I’ve seen two different types of way to get an user to agree with the Terms and Conditions.

The first way is to use a checkbox, labelled with “I agree to the Terms and Privacy Policy”

The second way I’ve seen is some text labelled near the continue button, with the “By clicking “Continue”, you agree to the Terms and Privacy Policy” on it. Most technology companies seem to use this option.

Is there any difference between both of them, and if not, which one would be more appealing for the visitors?

How likely is a SHA256 hash to be brute forced on this conditions?

If a string has a lenght of 16 characters, but it is also known that the first 2 characters are fixed, and that the rest of the string has random letters but letters only, no digits nor special chars, would it still take years to crack a sha256 on these conditions or is it unsafe?

mining pools – How does Bitcoin prevent race conditions?

Ok lets say there are two pools. Pool A and Pool B they are on opposite sides of the world. Lets say that pool A finds a hash that works for the current block and that pool B also finds a hash for the current block. They both start broadcasting to others on the network. How long would this take to resolve. What is to prevent two competing instances of Bitcoin? Lets say that half of the pools accepts A solution and half Accept B solution. They would have roughly equal mining power, how many blocks would it take to say one is invalid and what if the two blocks stayed equally matched? Even if one eventually wins out wouldn’t a lot of people lose money because of how many transactions were invalidated?