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mehdi jafari

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These are questions asked by mehdi jafari

solving system of odes...

mehdi jafari 764 Maple
ode dsolve differential-equations
December 20 2019
0 0

I have 4 ode equations. i just want to know can i use any option or simplification to have a analytical solution or NOT? Thanks in Advance

 

``

> 

restart:
> 

ode1 := -2*diff(lambda(t),t)*y1(t) - lambda(t)*diff((y1)(t),t)-0*diff(eta(t),t) - diff((y1)(t),t$3) + diff((y1)(t),t)*(y1(t)^2 + y2(t)^2) +4*y1(t)*sqrt(y1(t)^2 + y2(t)^2)*diff(sqrt(y1(t)^2 + y2(t)^2),t)+diff((y1)(t),t)/r^2
+ y1(t)^2*diff(y1(t),t) + y1(t)*y2(t)*diff(y2(t),t) - 2*diff(y1(t),t)/r^2 ;
> 

 

-2*(diff(lambda(t), t))*y1(t)-lambda(t)*(diff(y1(t), t))-(diff(diff(diff(y1(t), t), t), t))+(diff(y1(t), t))*(y1(t)^2+y2(t)^2)+2*y1(t)*(2*y1(t)*(diff(y1(t), t))+2*y2(t)*(diff(y2(t), t)))-(diff(y1(t), t))/r^2+y1(t)^2*(diff(y1(t), t))+y1(t)*y2(t)*(diff(y2(t), t))

 (1)
> 

ode2 := diff((lambda)(t),t$2) + lambda(t)*(y1(t)^2 + y2(t)^2) - 2*y1(t)*diff((y1)(t),t$2) - y1(t)^2*(y1(t)^2 + y2(t)^2) - y1(t)^2/r^2 - diff((y1)(t),t)^2 - 2*diff(sqrt(y1(t)^2 + y2(t)^2),t)^2 - 2*sqrt(y1(t)^2 + y2(t)^2)*diff(sqrt(y1(t)^2 + y2(t)^2),t$2) - diff((y2)(t),t)^2 - 2*y2(t)*diff((y2)(t),t$2) - y2(t)^2*(y1(t)^2 + y2(t)^2)

diff(diff(lambda(t), t), t)+lambda(t)*(y1(t)^2+y2(t)^2)-2*y1(t)*(diff(diff(y1(t), t), t))-y1(t)^2*(y1(t)^2+y2(t)^2)-y1(t)^2/r^2-(diff(y1(t), t))^2-(1/2)*(2*y1(t)*(diff(y1(t), t))+2*y2(t)*(diff(y2(t), t)))^2/(y1(t)^2+y2(t)^2)-2*(y1(t)^2+y2(t)^2)^(1/2)*(-(1/4)*(2*y1(t)*(diff(y1(t), t))+2*y2(t)*(diff(y2(t), t)))^2/(y1(t)^2+y2(t)^2)^(3/2)+(1/2)*(2*(diff(y1(t), t))^2+2*y1(t)*(diff(diff(y1(t), t), t))+2*(diff(y2(t), t))^2+2*y2(t)*(diff(diff(y2(t), t), t)))/(y1(t)^2+y2(t)^2)^(1/2))-(diff(y2(t), t))^2-2*y2(t)*(diff(diff(y2(t), t), t))-y2(t)^2*(y1(t)^2+y2(t)^2)

 (2)
> 

ode3 := 2*diff((lambda)(t),t)*y2(t) + lambda(t)*diff((y2)(t),t) - y1(t)*y2(t)*diff((y1)(t),t) - 4*y2(t)*sqrt(y1(t)^2 + y2(t)^2)*diff((sqrt(y1(t)^2 + y2(t)^2)),t) - y2(t)^2*diff((y2)(t),t) - (y1(t)^2 + y2(t)^2)*diff((y2)(t),t) - diff((y2)(t),t$3) ;

2*(diff(lambda(t), t))*y2(t)+lambda(t)*(diff(y2(t), t))-y1(t)*y2(t)*(diff(y1(t), t))-2*y2(t)*(2*y1(t)*(diff(y1(t), t))+2*y2(t)*(diff(y2(t), t)))-y2(t)^2*(diff(y2(t), t))-(y1(t)^2+y2(t)^2)*(diff(y2(t), t))-(diff(diff(diff(y2(t), t), t), t))

 (3)
> 

ode4 := lambda(t)*y1(t)/r + mu(t)*r - diff((y1)(t),t$2)/r -1/r*y1(t)*(y1(t)^2 + y2(t)^2) - y1(t)/r^3-2/r*diff(y1(t),t$2)

lambda(t)*y1(t)/r+mu(t)*r-3*(diff(diff(y1(t), t), t))/r-y1(t)*(y1(t)^2+y2(t)^2)/r-y1(t)/r^3

 (4)
> 

sys := [ode1, ode2, ode3, ode4]:
> 

dsolve(sys,[y1(t),y2(t),lambda(t),mu(t)],'implicit')

``

``


 

Download 1.1.mw

a math question...

mehdi jafari 764 Maple
permutation combination
December 02 2019
2 3

There are 10 questions. each question has one negetive point and three positive points. what is the number of students so that there will not be any repeatitive grades so all the students have distinct grades. Thanks in advance

procedure with if statement...

mehdi jafari 764 Maple
proc heaviside
October 29 2019
2 4

I have a question with a proc containing "if" statement. 
i want maple to do the calculations without just like Heaviside function does. how can it be done?is it possible?

> 

restart
> 

f1:=proc(x)
if x>=0 then 1
else 0
fi:end proc:
> 

Matrix(2,[[f1(x)+2,2*f1(x)],[5,4]]);

Error, (in f1) cannot determine if this expression is true or false: 0 <= x

  
> 

P:=f1(x)*exp(x)+5*x;

Error, (in f1) cannot determine if this expression is true or false: 0 <= x

  
> 

diff(P,x)

0

 (1)
> 

 

> 

 


but i can do all of the above with Heaviside(x). can i do the same with my procedure?

Download if.mw

some problems with maple...

mehdi jafari 764 Maple
parameters linearalgebra eigenvectors
August 27 2019
2 4

i have a matrix named S. it has two variables with the names sigma1 and tau. i do some computations with these varibels to obtain a matrix named S_NEW. when i calculate S_NEW with parameters and then i substitute parameter tau with zero i got the answer a zero Matrix. but when i put tau=0 at the initiation of my code, i got another answer. what is the problem?

the second issue is related with the name of parametes in maple. when the varible has the name sigma1 (and tau=0), the matrix S_NEW is a matrix with the only component of (1,1). when i change the parameter name to sigma the matrix S_NEW is a matrix with the only component of (3,3). where is the problem?

 

> 

restart:with(LinearAlgebra):
> 

tau:=0;

0

 (1)
> 

S:=Matrix(3,[[sigma1,tau,0],[tau,0,0],[0,0,0]]):B1:=Matrix(3):
> 

A12,B12:=(Eigenvectors(S)):
> 

V12:= Matrix(Normalize~([Column(B12,1..3)],Euclidean)):
> 

S_NEW:=simplify(V12^+.S.V12) assuming real ;

Matrix(3, 3, {(1, 1) = sigma1, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0})

 (2)
> 

eval(S_NEW,tau=0)

Matrix(3, 3, {(1, 1) = sigma1, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0})

 (3)
> 

 

> 

 

> 

 

Download problem.mw



how to keep set members in order of addi...

mehdi jafari 764 Maple
ordering set
August 15 2019
0 3

i have a question about sets. how can i keep set members in order of addition not the defualt maple ordering.
in maple help, i saw the command setsort=0..8 but i do not know how to use it.
consider exmaple below:


 

> 

restart
> 

L:={}:
> 

for j in [3,5,6,1] do
birth:=j:
L:=`union`(L ,{j});
od;

3

  

{3}

  

5

  

{3, 5}

  

6

  

{3, 5, 6}

  

1

  

{1, 3, 5, 6}

 (1)
> 

 


how can i keep L as order of addition? L={3,5,6,1}. thanks in advance

Download setoder.mw

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