guras

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These are replies submitted by guras

 

Thanks for your nice answer. What if we have a system of equations like

Eq1:=(a-4*b+1)*exp(4*x)+(c-2)=0

Eq2:=(a+b)*exp(4*x)+(c+d^2)=0

How can we solve this system? I tried many versions of the command solve('identity'(Eq, x)); both with Eq1 and Eq2, but 

it was useless. Thank you in advance.

Regards,

@vv Of course not. They are the solutions of a pde and I need to determine whether they are singular or not. Since they are really complicated it is hard to determine this.

@Markiyan Hirnyk  

I used your comments on another example. When I use

Search(U,strategy=globalsearch,maximize);

I have a warning "Warning, complex or non-numeric value encountered; trying to find a feasible point" with an answer. Here the question is that if we do not have HFloat(\infty) as a result to above command, can we say that the function is continuous? Can we rely on the answer?

Best Regards,
 

 

 

Download anotherexample.mw

 

 

@Markiyan Hirnyk 

Specifically,  I am dealing with the following function: 

V:=(x,t)->((5185/2592)*cos(x+(7/4)*t)+(5/72)*cos(3*x+(9/4)*t)+(1/18)*cos(2*x+(1/2)*t)+(1/72)*cos(4*x+4*t)+41489/20736)/((5/36)*cos(x+(7/4)*t)+(1/32)*cos(2*x+(7/2)*t)+(5185/11664)*cos(3*x+(9/4)*t)+(1/2592)*cos(6*x+(9/2)*t)+(20737/41472)*cos(2*x+(1/2)*t)+(1297/10368)*cos(4*x+4*t)+29985553/26873856);
 

The graph of the function seems that it is not discontinuous. I followed your commands. But I could not be sure about the result. 

Thank you.

Regards,

@tomleslie 

  Sorry that I am not so clear. I want to find conditions solving the equation: In the conditions a[1],b[1],c[1], and d[1] must be related to each other and a[2], b[2],c[2], and d[2] must be related to each other. For instance, conditions like  b[1]=alpha*a[1]^6, b[2]=alpha*a[2]^6, c[1]=beta*a[1]^4+gamma*d[1], c[2]=beta*a[2]^4+gamma*d[2]; alpha, beta, gamma will be determined. I am trying to find such conditions on the parameters a[i], b[i], c[i], d[i] with such restriction that I explained. I hope this explanation makes sense.

 

 

  More specifically,

I want to solve the following equation:

(a[1]-a[2])^8+(a[1]-a[2])^2*(b[1]-b[2])+(c[1]-c[2])^2+(a[1]-a[2])*(d[1]-d[2])=0,

with the relations between (a[1],b[1],c[1],d[1]) and (a[2],b[2],c[2],d[2]).

@Markiyan Hirnyk I need a portion of a huge output. So I just want to copy the part of the output that I need, then paste it as input. Maybe there is a problem about the program I installed. Thanks for the answer. 

@Rouben Rostamian  thank you so much for the answer. The functions that I am working are messy, so I did not give them in my question. One of the functions is the following;

restart;
f := rho -> 3/2-1/2*cosh(2*rho);
h := rho -> ((3/2-1/2*cosh(2*rho)-2)*(3/2-1/2*cosh(2*rho)-3)*(3/2-1/2*cosh(2*rho)-4))^(-1/2);
v := unapply(Int(h(s), s=0..rho), rho);
plot([v(rho), f(rho), rho=0..3]);

I tried many intervals for rho to see the graph. But I could not get a result.  What do you suggest?

Thanks.

 

I am just trying to calculate the following integral; 

int(exp(I*t*x)/((x+a[1]+I*b[1])*(x+a[2]+I*b[2])),x=0..infinity)

How wil I express that a[i] and b[i] are real?(t is also a real parameter)

Thanks for your quick answer. I see that I can use RootOf(something) in equations. Since it is important for my work, for the last time, I will ask again. Sorry for repetition. For instance, I found b=Rootof(_Z^6+...), which means there are 6 roots i.e. 6 values for b. I inserted b=Rootof(_Z^6+...) without calculating the roots explicitly(because I had huge expressions for b) into, for instance, ac+b=0. I saw that this equality satisfied automatically. So this means for each values of b, equation ac+b=0 is satisfied even I did not write the values of b explicitly. Am I correct?

Thanks.

Thanks for your quick answer. I see that I can use RootOf(something) in equations. Since it is important for my work, for the last time, I will ask again. Sorry for repetition. For instance, I found b=Rootof(_Z^6+...), which means there are 6 roots i.e. 6 values for b. I inserted b=Rootof(_Z^6+...) without calculating the roots explicitly(because I had huge expressions for b) into, for instance, ac+b=0. I saw that this equality satisfied automatically. So this means for each values of b, equation ac+b=0 is satisfied even I did not write the values of b explicitly. Am I correct?

Thanks.

 

The exact link is the following. By the way, I indeed tried to contact to authors but I did not get any message for now. And I thought maybe there is a unique way of installing a package in Maple 14.

http://cpc.cs.qub.ac.uk/summaries/AEHL_v1_0.html

Manuscript Title: [SADE] A Maple package for the Symmetry Analysis of Differential Equations
Authors: Tarcísio M. Rocha Filho, Annibal Figueiredo
Program title: SADE
Catalogue identifier: AEHL_v1_0
Distribution format: tar.gz
Journal reference: Comput. Phys. Commun. 182(2011)467
Programming language: MAPLE 13 and MAPLE 14.
Computer: PCs and workstations.
Operating system: UNIX/LINUX systems and WINDOWS.
Keywords: symmetry transformations, invariant solutions, first integrals, nöther theorem.
PACS: 02.70.Wz, 11.30.-j, 02.30.Jr.
Classification: 4.3.

Nature of problem:
Determination of analytical properties of systems of differential equations, including symmetry transformations, analytical solutions and conservation laws.

Solution method:
The package implements in MAPLE some algorithms (discussed in the text) for the study of systems of differential equations.

Restrictions:
Depends strongly on the system and on the algorithm required. Typical restrictions are related to the solution of a large over-determined system of linear or non-linear differential equations.

Running time:
Depends strongly on the order, the complexity of the differential system and the object computed. Ranges from seconds to hours.
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