partial differential equations – Show that eigenvalues and eigenvectors of the correspponding Sturm-Liouville

$u_{xx}+u_{yy}=frac{xy^{2}}{2}$ $0<x<pi$ and $0<y<1$

$u_{x}(0,y)=0$ , $u_{x}(pi,y)=0$ , $0leq yleq 1$ ………………..(1)

$u(x,0)=sin(x)$ , $u(x,1)=2x+1$ , $0leq xleq pi$

Show that eigenvalues and eigenvectors of the correspponding Sturm-Liouville problem are;

$lambda _{n}=n^{2},X_{n}(x)=cos(nx),n=1,2,3…$

Find the solution of the problem (1)?

I can solution of he problem but i couldnt evaluate eigenvalues eigenvectors…

probability theory – Is the distribution of a random variable determined by its partial integrals?

Let $(Omega_i, mathcal{F}_i, mathbb{P}_i)$ be a probability space for $i in {1,2}$, and let $X_i:Omega_irightarrow(-infty, infty)$ be a $mathbb{P}_i$-integrable random variable, respectively. If $int_{{X_1 in A}} X_1 dmathbb{P}_1 = int_{{X_2 in A}} X_2 dmathbb{P}_2$ for every Borel set $A$, does it follow that $X_1 overset{d}{=} X_2$?

sequences and series – How to prove that these partial binomial sums are zero?

I am trying to prove that the following equation is equal to zero.
$$
0=
sum_{j=J+1}^N Big(j (1-q)+ (j-J) (q N-j) Big) cdot q^{j} (1-q)^{N -j} binom{N}{j} label{zero1}$$

Where
$J,N in mathbb{Z}^+$ and $J<N$ and
$0<q<1$ is a probability.

Numerical simulations (see link) suggest that this is true.
Tips and hints (or solutions) are very welcome!

computability – Is it valid to make an admission of a topological space by a “partial quotient map”?

It is well-known that the Sierpiński space, ${F,T}$ endowed with topology ${emptyset, {F},{F,T}}$, is admissible. I tried to implement it in Haskell.

First I implement $mathbb{N}$ (including zero; some topologists might prefer denoting it by $S_omega$) by Peano definition:

data Peano = Zero | Succ Peano deriving (Eq, Ord)

This encodes $mathbb{N}$, but there is one additional value lurking behind: fix Succ. I denote it by $omega$, and I denote $mathbb{N} cup {omega}$ by $overline{S_omega}$. We observe that $overline{S_omega}$ is in order topology.

I abuse this fact and implement the Sierpiński space by taking a quotient space. The quotient map $q$ is:

$$
q(o) =
begin{cases}
0 & text{if } o < omega \
1 & text{o.w.}
end{cases}
$$

In Haskell, This can be realized by:

newtype Sierpinski = Sierpinski Peano

instance Eq Sierpinski where
    Sierpinski m == Sierpinski n = let
        q m = case m of
            Zero   -> False
            Succ n -> q n
        in q m == q n

But to think about it, q doesn’t halt on $omega$. In other words, q is partial, and doesn’t match $q$. Is this really a valid implementation of the Sierpiński space?

ap.analysis of pdes – eliminating first derivative terms from second order elliptic partial differential equation

I have a partial differential equation
$-afrac{partial^2 G}{partial x_1 partial x_2}-b(cosh(x_1)+cosh(x_2))G(x_1,x_2)+sinh(x_1)frac{partial G(x_1,x_2)}{partial x_1}+sinh(x_2)frac{partial G(x_1,x_2)}{partial x_2}=epsilon ,,G(x_1,x_2)$

I have to cast it in a form like
$-frac{1}{m_1}frac{partial^2 psi}{partial x^2}-frac{1}{m_2}frac{partial^2 psi}{partial y^2}+U(x,y),psi=epsilon ,,psi$

for some variable transformation from {$x_1,x_2rightarrow x,y$} and also some transformation in $G(x,y)$. How to do this?

numerical integration – When I define a Piecewise function containing a solution to a partial differential equation, the function cannot be evaluated numerically

I need to find an integral, in which there is a function YO(x), YO (x_)= sum (e ^ (((I * h * PI) / ( (lambda) * f)) * x ^ 2) * subscript (Y, h) (x, z = 13.5), {h, – 5, – 1}) /. S (where subscript (Y, h) (x, z) is a solution of partial differential equation). I run YO (3) in the positive interval of {x, 0, 30} and find that I can get a numerical value. After that, I want to extend YO which only contains x positive direction {x, 0, 30} to {x, -30,30}. The value of negative interval upper function and positive interval are symmetric. So I manually defined a piecewise function, and tried it with Piecewise and If functions. When the function YO (x) is in the negative interval, the value is equal to YO (Abs (x)), but I found that after calculation, running YO (3) can not get the numerical value, so I can only return YO (3) itself.

Based on the above situation, it is OK for me to integrate with the non expanded YO(x), but if I integrate with the expanded YO(x) with Piecewise or If function, I will report an error: the sampling point is non numerical at a certain point. At the same time, It also report errors when I find the maximum value of the integral result by Findargmax: Whether it is the function of YO before expansion or the function of YO after integration, It will report an error The display function value is not a real number. But my function value is the square after taking the module, which must be a real number. The relevant knowledge that I have been looking for for a long time these days has not been solved, so I would like to know what’s wrong with my program.

In addition, I can get the specific value by directly integrating the function YO (x) which is not expanded, but I can’t get the numerical value by integrating the function YO (x) which is expanded by piecewise and if, so I can only get the function formula which contains subscript (Y, h)(x,z).

This is my program for solving partial differential equations

ClearAll("Global`*");
(Theta) = 0;
(Lambda) = 1.2398/(19.5 10^3);
f = 4.72 10^3;
Subscript((Delta), 1) = 1.274/10^6;
Subscript((Delta), 2) = 4.304/10^6;
Subscript(bt, 1) = 5.254/10^9;
Subscript(bt, 2) = 2.435/10^7;
(Chi)1 = -2 Subscript((Delta), 1) + 2 I Subscript(bt, 1);
(Chi)2 = -2 Subscript((Delta), 2) + 2 I Subscript(bt, 2);
(CapitalDelta)(Chi) = (Chi)1 - (Chi)2;
k = 2 ((Pi)/(Lambda));
Table((Beta)(b) = -((2 (b x) Sin((Theta)))/f) - ((b x)/f)^2, {b, -5,
    5});
Table((Chi)(
    a) = ((CapitalDelta)(Chi) (1 - (-1)^
        Abs(a)))/(2 I a (Pi)), {a, -10, -1});
Table((Chi)(
    a) = ((CapitalDelta)(Chi) (1 - (-1)^Abs(a)))/(2 I a (Pi)), {a, 
   1, 10});
Table((Chi)(a) = ((Chi)1 + (Chi)2)/2, {a, 0, 0});
eqns = Join(Table((Sin((Theta)) + (h x)/f) !(
*SubscriptBox(((PartialD)), (x))(
(*SubscriptBox((Y), (h)))(x, z))) + Cos((Theta)) !(
*SubscriptBox(((PartialD)), (z))(
(*SubscriptBox((Y), (h)))(x, 
         z))) == ((I (Pi)) ((Beta)(h) Subscript(Y, h)(x, z) + !(
*UnderoverscriptBox(((Sum)), (l = (-5)), (5))((Chi)(
             h - l) 
(*SubscriptBox((Y), (l)))(x, z)))))/(Lambda), {h, -5, 5}), 
   Table(Subscript(Y, h)(x, 0) == If(h == 0, 1, 0), {h, -5, 5}));
s = NDSolve(eqns, 
   Table(Subscript(Y, h)(x, z), {h, -5, 5}), {x, 0, 30}, {z, 0, 30},
    Method -> {"MethodOfLines", 
     "SpatialDiscretization" -> {"TensorProductGrid", 
       "MaxPoints" -> 100, "MinPoints" -> 100, 
       "DifferenceOrder" -> 4}, "TemporalVariable" -> z});

This is my program to define YO(x)

YO(x_?NumericQ) := 
 Piecewise({{Sum(
      E^((I h (Pi))/((Lambda) f) x^2)
        Subscript(Y, h)(x, z = 13.5), {h, -5, -1}), 
     20 >= x >= 0}, {Sum(
      E^((I h (Pi))/((Lambda) f) x^2)
        Subscript(Y, h)((Abs(x)), z = 13.5), {h, -5, -1}), 
     0 > x >= -20}}, {x, -20, 20}) /. s

And this is what I want to do: integrate, find the maximum and plot the density around zo at the maximum

f1(xo_?NumericQ, zo_?NumericQ) := 
  NIntegrate((E^(I k zo) YO(x) E^(((I k) x^2)/(2 zo))
        E^(((I k) xo^2)/(2 zo))
        E^(-(((I k) x xo)/zo)))/(I (Lambda) zo^0.5), {x, -20, 
     20}) /. s;
g(xo_?NumericQ, zo_?NumericQ) := Abs(f1(xo, zo)^2);
FindArgMax({g(xo, zo), -20 <= xo <= 20 && 3000 <= zo <= 5000}, {{xo, 
   0}, {zo, 4000}})
DensityPlot(g(xo, zo), {xo, -20, 20}, {zo, 3000, 5000}, 
 PlotLegends -> Automatic)

But in fact, both the integral and the maximum value will prompt me to report an error. The integral will prompt me that some places are not numerical, and the maximum value will prompt me that some places are not real. But it’s obvious that the absolute value followed by the square must be real and numerical.

In addition, when I integrate with the unexpanded YO(x), I can integrate and draw a graph, but I never succeed in finding the maximum value.

Is there a pair of tuples N, M of prime numbers which (a) have the same product, and (b) whose partial products have the same sum?

If N is a finite list of numbers, let $p(N)$ be the product of the numbers in N, that is, $$p(N)=Pi_{i=1}^{|N|}N_i$$ and let $s(N)$ be the sum of the partial products of the numbers in N, that is $$s(N)=Sigma_{k=1}^{|N|}left(Pi_{i=1}^kN_iright)$$

Is there a pair of lists $Nneq M$ for which $p(N)=p(M)$ and $s(N)=s(M)$?

What if they only contain odd numbers 3 or higher?

What if $N$ and $M$ only contain prime numbers? Only odd primes?

What if they only contain elements of ${3,7,19}$? (This is the case I’m most interested in)

It seems possible that there’s no such (distinct) $N$ and $M$ for the last case, but I’ve no idea how to prove that.

Doubt in a partial derivative question.

Let us say that I have an equation $G = xG_text{Au} + (1-x)G_text{Ni}$.Here $x$ and $(1-x)$ are the weights.Now I am supposed to calculate $G_text{Au}$
and $G_text{Ni}$.How do I do that with the help of derivatives?

Mathematica Partial Fraction Decomposition full steps

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Why is the partial differential equation $u_{t} + a u_{x} = 0$ hyperbolic?

I don’t quite understand why this advection equation, with some initial condition $u(0,x) = u_{0}(x)$, is considered hyperbolic (as for instance here). If I apply the test mentioned in Farlow, comparing it to
$$Au_{xx}+Bu_{xy} + Cu_{yy} + D u_{x} + Eu_{y} + Fu = G$$
then both $B^{2}$ and $AC$ are zero, and thus $B^{2} – 4AC = 0 Rightarrow text{parabolic}$. Or does a different criterion apply when there aren’t any second-order terms?