## Solving a Pair of Differential Equations with Asym...

Is it possible to determine an analytic solution to the following system of two differential equations for $A$ and $B$ using Maple.  My suspicion is that trial and error would find an analytic solution in theory and so that Maple could find the solution.  M is a constant and \sigma is some arbitrary function of t and the spatial coordinates.

$\Bigg( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} + \frac{1}{2} \Bigg( 1 + \frac{M}{2 \sqrt{x^2 + y^2 + z^2}} \Bigg) \Bigg( \frac{\partial \sigma}{\partial x }\frac{\partial}{\partial x} +\frac{\partial \sigma}{\partial y}\frac{\partial}{\partial y} +\frac{\partial \sigma}{\partial z}\frac{\partial}{\partial z} \Bigg) \Bigg)B=0,$

$\frac{d A}{dt} = AB.$

Furthermore, the boundary conditions are

$B \rightarrow -1 \: \text{as} \: \sqrt{x^2 + y^2 + z^2} \rightarrow \infty,$

$A \rightarrow e^{-t} \: \text{as} \: \sqrt{x^2 + y^2 + z^2} \rightarrow \infty$

System_of_Equations.pdf

## Maple difference between "text" and "2d input" in ...

I do not quite understand why prof asks this question. Or I am doing right? Where can I improve? Or I understand this question completely wrong. To be honest, I did not get the point

## modp1(('Multiply')(...))...

Hi there.

As we all know if we multiply two polynomials f(x) and g(x) of degrees m and n respectively we get polynomial h(x)= f(x)*g(x) of degree m+n and with m+n+1 coefficients in general. Function modp1(('Multiply')(...)) doing this very well. But sometimes we don't need full resulting h(x) - just subset of monomials and subset of coefficients of h(x) - so we don't need to calculate all m+n+1 coefficients of h(x) and waste time and resources for that.

I would request some additional rework of modp1 package: by adding to modp1(('Multiply')(...)) two optional parameters - degrees of first and last calculating coefficients of h(x).

For example:

h:=modp1(Multiply(f, g,n-1,n+1), p) could calculate only monomials with n-1, n and n+1 degrees and set other monomials to zero.

Or maybe it should be new function:

h:=modp1(Multiply_Truncate(f, g,n-1,n+1), p)

Is it possible?

It would be great and very efficient in many tasks.

Thank you.

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## why PDEtools:-Solve hangs, but solve does not on ...

These are 4 equations in 4 unknowns. the equations are kinda long. But the issue is that PDEtools:-Solve hangs, while solve finishes instantly.

I have though before that  PDEtools:-Solve is a higher level API which ends up using solve? So why does it hang on this?

restart;

eqs:=[2 = c1+c2+c4-1, 0 = ((c2+c4-2)*(108+12*59^(1/2)*3^(1/2))^(1/3)+(1/12*c3*3^(1/2)+1/12*c2-1/6*c4)*(108+12*59^(1/2)*3^(1/2))^(2/3)+2*c3*3^(1/2)-2*c2+4*c4)/(108+12*59^(1/2)*3^(1/2))^(1/3), -1 = -1/6/(108+12*59^(1/2)*3^(1/2))^(2/3)*((((c2-2*c4)*3^(1/2)-3*c3)*59^(1/2)-33*c3*3^(1/2)+33*c2-66*c4)*(108+12*59^(1/2)*3^(1/2))^(1/3)+(2*c2+2*c4+12)*(108+12*59^(1/2)*3^(1/2))^(2/3)+((-12*c2+24*c4)*3^(1/2)-36*c3)*59^(1/2)-60*c3*3^(1/2)-60*c2+120*c4), -5 = ((((2*c2-4*c4)*3^(1/2)+6*c3)*59^(1/2)-30*c3*3^(1/2)-30*c2+60*c4)*(108+12*59^(1/2)*3^(1/2))^(1/3)+(((-c2+2*c4)*3^(1/2)+3*c3)*59^(1/2)+13*c3*3^(1/2)-13*c2+26*c4)*(108+12*59^(1/2)*3^(1/2))^(2/3)-96*(59^(1/2)*3^(1/2)+9)*(c2+c4))/(24*59^(1/2)*3^(1/2)+216)];

unknowns:=[c1, c2, c3, c4];

#han to put a timelimit, else it will never finish. I waited 20 minutes before.
timelimit(30,PDEtools:-Solve(eqs,unknowns));

#this completes right away
solve(eqs,unknowns);


Is this a known issue and to be expected sometimes?

 > restart;
 > interface(version);

 > Physics:-Version();

 > eqs:=[2 = c1+c2+c4-1, 0 = ((c2+c4-2)*(108+12*59^(1/2)*3^(1/2))^(1/3)+(1/12*c3*3^(1/2)+1/12*c2-1/6*c4)*(108+12*59^(1/2)*3^(1/2))^(2/3)+2*c3*3^(1/2)-2*c2+4*c4)/(108+12*59^(1/2)*3^(1/2))^(1/3), -1 = -1/6/(108+12*59^(1/2)*3^(1/2))^(2/3)*((((c2-2*c4)*3^(1/2)-3*c3)*59^(1/2)-33*c3*3^(1/2)+33*c2-66*c4)*(108+12*59^(1/2)*3^(1/2))^(1/3)+(2*c2+2*c4+12)*(108+12*59^(1/2)*3^(1/2))^(2/3)+((-12*c2+24*c4)*3^(1/2)-36*c3)*59^(1/2)-60*c3*3^(1/2)-60*c2+120*c4), -5 = ((((2*c2-4*c4)*3^(1/2)+6*c3)*59^(1/2)-30*c3*3^(1/2)-30*c2+60*c4)*(108+12*59^(1/2)*3^(1/2))^(1/3)+(((-c2+2*c4)*3^(1/2)+3*c3)*59^(1/2)+13*c3*3^(1/2)-13*c2+26*c4)*(108+12*59^(1/2)*3^(1/2))^(2/3)-96*(59^(1/2)*3^(1/2)+9)*(c2+c4))/(24*59^(1/2)*3^(1/2)+216)]; unknowns:=[c1, c2, c3, c4];

 > timelimit(30,PDEtools:-Solve(eqs,unknowns))

Error, (in expand/bigprod) time expired

 > solve(eqs,unknowns):
 >

## What is the best curve fit for Primes of the form ...

Hi all,

We want to find a curve fit for an integer sequence.

We have n such that n^2+n+17 is a prime number.

Use the Maple CurveFitting package.

I tried with(CurveFitting).

We do not know if this is best represented by a polynomial or exponential curve fit.

n2_and_n_and_17_in_OEIS_007635.mw

n2_and_n_and_17_in_OEIS_007635.pdf

Regards,

Matt

## How Does Maple Actually Solve an Identity, Say, fo...

Since I am a mathematician, I am wondering how Maple goes about solving an identity for 3 functions.
Let's say we have af1(t)+bf_2(t)+cf_3(t) = 0 for all t. How does maple actually find a triplet a,b,c that works for all real t?
It does with solve(identity( ),[a,b,c]). But what is the theory behind it?
We know, of course, a priori, that such a triplet exists.

Thank you!

mapleatha

## Computational Nash equilibrium...

Dear Colleagues,

Apologies for the generic question below.

I am trying to obtain the Nash equilibrium solutions for a two-person game. I am not sure of any in-built packages that can help me in obtaining the solutions computationally. The algorithms that I created do not seem to give good solutions that are meaningful in my application. Any suggestion would be much appreciated.

Regards,

Omkar

I've been studying the  drawing  of graph lately .    One of the themes is  1-planar graph .

A 1-planar graph is a graph that can be drawn in the Euclidean plane in such a way that each edge has at most one crossing point,  where it crosses a single additional edge. If a 1-planar graph, one of the most natural generalizations of planar graphs, is drawn that way, the drawing is called a 1-plane graph or 1-planar embedding of the graph.

I know it is NP hard to determine whether a graph is a 1-planar . My idea is to take advantage of some mathematical software to provide some roughly and  intuitive understanding before determining .

Now,  the layout of vertices or edges becomes important.  The drawing of a plane graph is a good example.

DrawGraph(G1)
DrawGraph(G1,style=planar)

K5 := CompleteGraph(5);
DrawGraph(K5);
vp:=[[-1,0],[1,0],[-0.2,0.5],[0.2,0.5],[0,1]];
SetVertexPositions(K5,vp);  #modified the vertex position

DrawGraph(K5);

My problem is that I see that  Maple2020 has updated a lot of layouts about DrawGraph  graph theory backpack , and I don’t know which ones are working towards the least possible number of crossing of  each edges of graph .

Some links that may be useful:

https://de.maplesoft.com/products/maple/new_features/Maple2020/graphtheory.aspx

https://de.maplesoft.com/support/help/Maple/view.aspx?path=GraphTheory/SetVertexPositions

I think the software can improve some calculations related to topological graph theory, such as crossing number of graph, etc.

## How to find sgn on maple?...

How to find sgn on maple?

signum.mw

## Is sorting here backwards?...

When I select "oldest first", the first post shown is the newest.  Vice-versa when selecting "newest first".  Am I misunderstanding the meaning of the term?  I suppose it doesn't matter as both options are available but it's weird.

## Maple Apps, Venn Diagram...

Maple Apps-Venn Diagrams does not work.  In box on the right there is an error message.

## Kind help if anyone to convert the c++ code to m...

#include<iostream>
#include<vector>
#include<cmath>
#define NODE 8

using namespace std;
int graph[NODE][NODE] = {
{0,1,1,0,0,0,0,0},
{1,0,1,1,1,0,0,0},
{1,1,0,1,0,1,0,0},
{0,1,1,0,0,0,0,0},
{0,1,0,0,0,1,1,1},
{0,0,1,0,1,0,1,1},
{0,0,0,0,1,1,0,0},
{0,0,0,0,1,1,0,0}
};
int tempGraph[NODE][NODE];
int findStartVert() {
for(int i = 0; i<NODE; i++) {
int deg = 0;
for(int j = 0; j<NODE; j++) {
if(tempGraph[i][j])
deg++; //increase degree, when connected edge found
}
if(deg % 2 != 0) //when degree of vertices are odd
return i; //i is node with odd degree
}
return 0; //when all vertices have even degree, start from 0
}
int dfs(int prev, int start, bool visited[]){
int count = 1;
visited[start] = true;
for(int u = 0; u<NODE; u++){
if(prev != u){
if(!visited[u]){
if(tempGraph[start][u]){
count += dfs(start, u, visited);
}
}
}
}
return count;
}
bool isBridge(int u, int v) {
int deg = 0;
for(int i = 0; i<NODE; i++)
if(tempGraph[v][i])
deg++;
if(deg>1) {
return false; //the edge is not forming bridge
}
return true; //edge forming a bridge
}
int edgeCount() {
int count = 0;
for(int i = 0; i<NODE; i++)
for(int j = i; j<NODE; j++)
if(tempGraph[i][j])
count++;
return count;
}
void fleuryAlgorithm(int start) {
static int edge = edgeCount();
static int v_count = NODE;
for(int v = 0; v<NODE; v++) {
if(tempGraph[start][v]) {
bool visited[NODE] = {false};
if(isBridge(start, v)){
v_count--;
}
int cnt = dfs(start, v, visited);
if(abs(v_count-cnt) <= 2){
cout << start << "--" << v << " ";
if(isBridge(v, start)){
v_count--;
}
tempGraph[start][v] = tempGraph[v][start] = 0; //remove edge from graph
edge--;
fleuryAlgorithm(v);
}
}
}
}
int main() {
for(int i = 0; i<NODE; i++) //copy main graph to tempGraph
for(int j = 0; j<NODE; j++)
tempGraph[i][j] = graph[i][j];
cout << "Euler Path Or Circuit: ";
fleuryAlgorithm(findStartVert());
}

Kind help

## conversions of latex equation to MS word...

Hi, I generated latex formate of an equation by using a command of maple but when I paste it into MathType, could not get the required equation, can anyone help me

${\frac {1}{51200\, \left( {x}^{2}+2 \right) ^{6}} \left( -187110\, \left( {x}^{2}+2 \right) ^{6}\sqrt {2} \left( {Q}^{3}+ \left( {\frac {18\,k}{11}}-{\frac{18}{11}} \right) {Q}^{2}+ \left( {\frac {320\,{k}^ {2}}{297}}-{\frac {40\,k}{27}}+{\frac{320}{297}} \right) Q+{\frac {80 \,{k}^{3}}{297}}-{\frac {80\,{k}^{2}}{189}}+{\frac {80\,k}{189}}+{ \frac {640\,\lambda}{2079}}-{\frac{80}{297}} \right) \arctan \left( 1/ 2\,x\sqrt {2} \right) -93555\, \left( {x}^{2}+2 \right) ^{6}\pi\, \left( {Q}^{3}+ \left( {\frac {18\,k}{11}}-{\frac{18}{11}} \right) {Q }^{2}+ \left( {\frac {320\,{k}^{2}}{297}}-{\frac {40\,k}{27}}+{\frac{ 320}{297}} \right) Q+{\frac {80\,{k}^{3}}{297}}-{\frac {80\,{k}^{2}}{ 189}}+{\frac {80\,k}{189}}+{\frac {640\,\lambda}{2079}}-{\frac{80}{297 }} \right) \sqrt {2}-374220\, \left( \left( {Q}^{3}+ \left( {\frac { 18\,k}{11}}-{\frac{18}{11}} \right) {Q}^{2}+ \left( {\frac {320\,{k}^{ 2}}{297}}-{\frac {40\,k}{27}}+{\frac{320}{297}} \right) Q+{\frac {80\, {k}^{3}}{297}}-{\frac {80\,{k}^{2}}{189}}+{\frac {80\,k}{189}}+{\frac {640\,\lambda}{2079}}-{\frac{80}{297}} \right) {x}^{10}+ \left( { \frac {34\,{Q}^{3}}{3}}+ \left( {\frac {204\,k}{11}}-{\frac{204}{11}} \right) {Q}^{2}+ \left( {\frac {10880\,{k}^{2}}{891}}-{\frac {1360\,k }{81}}+{\frac{10880}{891}} \right) Q+{\frac {2720\,{k}^{3}}{891}}-{ \frac {2720\,{k}^{2}}{567}}+{\frac {2720\,k}{567}}+{\frac {21760\, \lambda}{6237}}-{\frac{2720}{891}} \right) {x}^{8}+ \left( {\frac {264 \,{Q}^{3}}{5}}+ \left( {\frac {432\,k}{5}}-{\frac{432}{5}} \right) {Q} ^{2}+ \left( {\frac {512\,{k}^{2}}{9}}-{\frac {704\,k}{9}}+{\frac{512} {9}} \right) Q+{\frac {128\,{k}^{3}}{9}}-{\frac {1408\,{k}^{2}}{63}}+{ \frac {1408\,k}{63}}+{\frac {97280\,\lambda}{6237}}-{\frac{128}{9}} \right) {x}^{6}+ \left( {\frac {4496\,{Q}^{3}}{35}}+ \left( {\frac { 80928\,k}{385}}-{\frac{80928}{385}} \right) {Q}^{2}+ \left( {\frac { 287744\,{k}^{2}}{2079}}-{\frac {35968\,k}{189}}+{\frac{287744}{2079}} \right) Q+{\frac {3328\,{k}^{3}}{99}}-{\frac {3328\,{k}^{2}}{63}}+{ \frac {3328\,k}{63}}+{\frac {10240\,\lambda}{297}}-{\frac{3328}{99}} \right) {x}^{4}+ \left( {\frac {10672\,{Q}^{3}}{63}}+ \left( {\frac { 21344\,k}{77}}-{\frac{21344}{77}} \right) {Q}^{2}+ \left( {\frac { 1094656\,{k}^{2}}{6237}}-{\frac {136832\,k}{567}}+{\frac{1094656}{6237 }} \right) Q+{\frac {35584\,{k}^{3}}{891}}-{\frac {35584\,{k}^{2}}{567 }}+{\frac {35584\,k}{567}}+{\frac {235520\,\lambda}{6237}}-{\frac{ 35584}{891}} \right) {x}^{2}+{\frac {25376\,{Q}^{3}}{231}}+ \left( { \frac {12352\,k}{77}}-{\frac{12352}{77}} \right) {Q}^{2}+ \left( -{ \frac {7936\,k}{63}}+{\frac {63488\,{k}^{2}}{693}}+{\frac{63488}{693}} \right) Q-{\frac{512}{27}}+{\frac {512\,{k}^{3}}{27}}-{\frac {5632\,{ k}^{2}}{189}}+{\frac {102400\,\lambda}{6237}}+{\frac {5632\,k}{189}} \right) x \right) }$