## 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 $$U$$.

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.

## Which iOS relative to the initial iOS of an iPhone gives the best/fastest performance?

For example, if an iPhone is released with iOS x as its initial iOS, and which out of iOS x, x+1, …, x+5 or x+6 (the last iOS that supports it) typically give the phone the best/fastest performance?

Is it true that the further the iOS from the initial iOS, the worse/slower the performance of the phone?

## users – How to target the initial revision of a node?

How to target the initial revision of a node?

This gives me a list of node IDs in which the latest revision of the node is of author `\$user->id()`:

``````\$query = Drupal::entityQuery('node');
\$query->condition('uid', \$user->id());
\$author_nids = \$query->execute();
``````

I would like the same but for nodes in which the initial revision was created by `\$user->id()`.

## ordinary differential equations – Given boundary conditions and initial condition, solve the PDE

Question, solve the given PDE :

$$frac{partial C}{partial t} = a frac{partial^2 C}{partial x^2} -kC$$
where a, k are constants.

boundary conditions : $$C= C_0$$ at $$x=0$$, $$C=0$$ at $$x =$$ infinitum

initial condition : $$C=0$$ at $$t=0$$

My attempt :

$$C(x,t) = X(x)~T(t)$$
$$C = X~T$$
$$boxed{ frac{partial C}{partial t} = T^{prime} X,~ space frac{partial C}{partial x}= X^{prime}T, ~ frac{partial^2 C}{partial x^2} = X^{prime prime } T}$$
replacing the partial derivatives into the Original equation :
$$T^{prime}X = a ~X^{prime prime}T-k(X~T)$$
$$X (T^{prime} +KT)=a ~X^{prime prime}T$$
$$frac{T^{prime}}{T} = a ~frac{ X^{prime prime}}{X} -k$$
$$boxed{frac{T^{prime}}{T} = J, ~frac{ X^{prime prime}}{X} -k = J}$$
equating each side to a constant $$J = -lambda^2$$
$$boxed{frac{d T}{d t} = -lambda^2 T ,~ frac{d^2 X}{d x^2} = left (frac{-lambda^2 + k}{a} right) X}$$
solving each differential equation :
$$T(t) = Ae^{- lambda t} , ~ X(x) = B cos{left (x sqrt{frac{lambda^2 -k}{a}} right)} + C sin{left (-x sqrt{frac{lambda^2 -k}{a}} right) }$$
Note : D = AB and E = AC
$$boxed{C(x,t) = left(D cos{left (x sqrt{frac{+lambda^2 -k}{a}} right)} + ~E sin{left (x sqrt{frac{lambda^2 – k}{a}} right) } right) e^{- lambda t}}$$
first boundary condition $$C= C_0$$ at $$x=0$$ :
$$D e^{- lambda t} = C_0$$
second boundary condition $$C=0$$ at $$x =$$ infinitum : DNE
$$C(x,t) = C_0cos{left (x sqrt{frac{+lambda^2 -k}{a}} right)} + ~E sin{left (x sqrt{frac{lambda^2 – k}{a}} right) } e^{- lambda t}$$
using initial condition $$C=0$$ at $$t=0$$ :
$$C_0 cos{left (x sqrt{frac{lambda^2 -k}{a}} right)} = -E sin{left (x sqrt{frac{lambda^2 – k}{a}} right) }$$
final equation :
$$boxed{ C(x,t) = -E sin{left (x sqrt{frac{lambda^2 – k}{a}} right) } + E sin{left (x sqrt{frac{lambda^2 – k}{a}} right) }e^{- lambda t} }$$

Could you guys please verify the answer also is “second boundary condition : DNE” true?

## physics – Initial bullet speed

A bullet of mass m=0,05(kg) is fired at two blocks, the First is attached to the ceiling by a rope,as shown in the diagrama. It crosses block 1, with mass m1=2.24(kg) and is lodged inside block 2, with mass m2=5.03(kg). After crossing block 1, due to the impulse, this block, which is suspended to the ceiling by a rope,It rises to a height h=0.345(m).Block 2, with the bullet inside, travels with speed v=1.6(m/s) after being hit.
What was the initial speed of the bullet? (Use g=9.81(m/s^2))

initially I discovered the speed with which the bullet leaves block 1, that is, 0.05v = (0.05 + 5.03) 1.6 , then v = 162.56 (m / s). After that, I tried to find the initial speed, but I didn’t find the speed of the block, so I assumed that the speed of the block was √ (2gh) so i got 0.05Vi = (0.05) (162.56) + 2.24√6.7689.
So, Vi = 2791.16 (m / s)

## postgresql – Filter index slow initial use

I have a filtered index like this:

``````CREATE INDEX idx1
ON public.tbl1 USING btree
(col1 COLLATE pg_catalog."default" ASC NULLS LAST, col2 ASC NULLS LAST)
WHERE col3 IS NULL;
``````

and a query like:

``````SELECT * FROM tb1 WHERE col3 is null and col4 = 12
``````

This query scan idx1 which is fine as this index is only a few rows (max 400 or so) the underlying table is millions of rows.

The first time this query runs, it runs painfully slow (almost a minute) but the seconds time it runs it does so in 20ms.

Can someone help me understand what is happening? The index gets rebuilt every evening but seems to bloat (gets to a few hundred megs) during the day. From the behaviour it feels like it has to load data into memory and the next time its faster because the data is cached. The database is quite busy 2k tupples in and 4k tuples out.

## mobile application – use of filters vs. initial questionnaire/survey

I’m trying to find any articles about the use of filters vs. initial questionnaire/survey in apps, in general.

I’m developing a new app for matching cancer patients and survivors and wanted to get more insights about the either the use of an initial survey to determine the user’s profile or the use of search filters throughout the app – or both.

## differential equations – Ndsolve of random (periodic) initial conditions via BSplineFunction

Suppose we want to solve the heat equation for a random initial temperature field T(t,x,y). Therefore I define the initial conditions with the BSplineFunction which nicely incorporates periodic boundary conditions and try to solve it with NDSolve:

``````Clear("Global`*")
Lx = 1; Ly = 1;
eq = D(T(t, x, y), t) == Laplacian(T(t, x, y), {x, y});
pbc1 = T(t, x, 0) == T(t, x, Ly);
pbc2 = T(t, 0, y) == T(t, Lx, y);
T0(x_, y_) =
BSplineFunction(RandomReal(1, {30, 30, 1}), SplineClosed -> True)(x,
y)
Plot3D(T0(x, y), {x, 0, Lx}, {y, 0, Ly}, PlotRange -> All,
AxesLabel -> {"x", "y", "T"}, ColorFunction -> "DarkRainbow",
Mesh -> All, MeshStyle -> Opacity(.2))
T0(0, 0)
ic = T(0, x, y) == T0(x, y)
ic2 = T(0, x, y) == 0.5*Exp(-4 ((x - Lx/2)^2 + (y - Ly/2)^2));
Monitor(AbsoluteTiming(
sol = NDSolve({eq, pbc1, pbc2, ic},
T, {t, 0, 1}, {x, 0, Lx}, {y, 0, Ly},
Method -> {"MethodOfLines",
"SpatialDiscretization" -> {"TensorProductGrid",
"MaxPoints" -> 25}},
EvaluationMonitor :> (currentTime =
Row({"t = ", CForm(t)})))), currentTime)
``````

The code unfortunately throws an error “Data … is not a rectangular tensor with dimensions …”. I assume this results from the fact that T0 evaluated e.g. at grid point (0,0) returns a list {} and NDSolve probably needs just the float value.

I tried T0(0,0)((1)) which yields the expected value. But this does not work for the definition of the initial condition: ic = T(0, x, y) == T0(x, y)((1)), where x,y are not set yet. This returns just x, in reality it should first put in the values of x,y and only after that take list element 1 (if I am not mistaken).

Apart from that the code works fine which can be verified if you simply use ic2 instead of ic as initial condition in NDSolve.

Can someone help to get this running?

## WordPress redirect to initial landing page

We have 3 landing pages, for 3 target customer types, with different deals.

The issue is that customer lands on a page, then click around the other pages in the site, then goes to a generic “contact-us” page (currently using wpforms).

We would prefer to redirect the customer back their the initial landing page, so they can see any special deals, from a campaign that they clicked on.

Is there a way to redirect them back to the initial landing page, when they go to the generic contact-Us page?