Items tagged with equation equation Tagged Items Feed


I have been trying to compute the analytical solution of two dimensional diffusion equation with zero neumann boundary conditions (no-flux) in polar coordinates using the solution in Andrei Polyanin's book. When I use 2d Gaussian function as initial condition, i cannot get the result. If I use some nicer function like f(r,phi)=1-r; there is no problem.  

Any idea why this happens? or any suggestion to compute the analytical solution?



M := Matrix([[3.83170597020751, 7.01558666981561, 10.1734681350627, 13.3236919363142, 16.4706300508776], [1.84118378134065, 5.33144277352503, 8.53631636634628, 11.7060049025920, 14.8635886339090], [3.05423692822714, 6.70613319415845, 9.96946782308759, 13.1703708560161, 16.3475223183217], [4.20118894121052, 8.01523659837595, 11.3459243107430, 14.5858482861670, 17.7887478660664], [5.31755312608399, 9.28239628524161, 12.6819084426388, 15.9641070377315, 19.1960288000489], [6.41561637570024, 10.5198608737723, 13.9871886301403, 17.3128424878846, 20.5755145213868], [7.50126614468414, 11.7349359530427, 15.2681814610978, 18.6374430096662, 21.9317150178022], [8.57783648971407, 12.9323862370895, 16.5293658843669, 19.9418533665273, 23.2680529264575], [9.64742165199721, 14.1155189078946, 17.7740123669152, 21.2290626228531, 24.5871974863176], [10.7114339706999, 15.2867376673329, 19.0045935379460, 22.5013987267772, 25.8912772768391], [11.7708766749555, 16.4478527484865, 20.2230314126817, 23.7607158603274, 27.1820215271905]]):

c := 10:

A := 5:

w := proc (r, phi, t) options operator, arrow; int(int(f(xi, eta)*G(r, phi, xi, eta, t)*xi, xi = 0 .. 5), eta = 0 .. 2*Pi) end proc:


Warning,  computation interrupted




Hi there,


I'd like to solve 7th order implicit simultaneous equation such as below, so I tried to do it by solve command.

However the calculation wasn't over although three hours passed.


eq1 := f1(a,b,c,d,e,f,g) = 0;

eq2 := f2(a,b,c,d,e,f,g) = 0;




eq7 := f7(a,b,c,d,e,f,g) = 0;


Just for your information, the eq1 and eq6 are written as follows specifically.

eq1 := -a-b-c-d-e-f-g+0.501857 = 0

eq6 := a*b*c*d*e*f+a*b*c*d*e*g+a*b*c*d*f*g+a*b*c*e*f*g+a*b*d*e*f*g+a*c*d*e*f*g+b*c*d*e*f*g+a*b*c*d*e+a*b*c*d*f+a*b*c*d*g+a*b*c*e*f+a*b*c*e*g+a*b*c*f*g+a*b*d*e*f+a*b*d*e*g+a*b*d*f*g+a*b*e*f*g+a*c*d*e*f+a*c*d*e*g+a*c*d*f*g+a*c*e*f*g+a*d*e*f*g+b*c*d*e*f+b*c*d*e*g+b*c*d*f*g+b*c*e*f*g+b*d*e*f*g+c*d*e*f*g-0.5281141885e-3+1.01894577*10^(-12)*I = 0


And the program code I used is:



Here is the specification of my computer.

OS: Windows 7 Enterprise 64bit

CPU: Intel Core i7-3520M 2.90 GHz

Memory: 4.00 GB


How can I handle this problem? Is the specification not enough to solve the equation? Do I need to leave my computer more and more time?

Any help would be appriciated.

Hi there. 

I'm kind of new to Maple and i'm trying to solve a Linear Algebra problem for my class of Linear Algebra of the course of Physics. Also, my first language is portuguese so forgive for my not-so-perfect english.

I have a (solved) linear system of 7 equations and 12 variables (A, B, C, D, E, F, G, H, I, J, K, L) that is the following:

  • A = 33 - K - L
  • B = 1 + F - J
  • C = -15 - F + J + K + L
  • D = 15 + H - K
  • E = 16 - F - H + J + K
  • G = 34 - H - J - L
  • I = 18 - J - K

Note: I'm using letters (A, B, ..., L) instead of X1X2, ..., X12 because it's easier to write it like this here and because I don't know if the Xn notation is allowed on Maple (i don't think so).

So, the system is possible but undetermined (with 5 degrees of freedom), being F, H, J, K and L the free variables.

Until here, everything's fine. The problem arises when the professor asks us for every solution of the system that satisfies the condition that all the variables (form A to L) are positive integers (A, B, C, D, E, F, G, H, I, J, K, L ϵ IN → natural numbers).

From my understanding, that gives rise to a system of linear inequalities with 12 variables and the following inequalities:

  • A = 33 - K - L > 0
  • B = 1 + F - J > 0
  • C = -15 - F + J + K + L > 0
  • D = 15 + H - K > 0
  • E = 16 - F - H + J + K > 0
  • G = 34 - H - J - L > 0
  • I = 18 - J - K > 0
  • > 0
  • > 0
  • > 0
  • > 0
  • > 0                            (and A,B,C,D,E,F,G,H,I,J,K,L ϵ IN)

After some research, i found that a possible way to solve this type of system of linear inequalities is trough a method of elimination (analog to Gauss-Jordan's elimination method for systems of linear equations) named Fourier-Motzkin. But it's hardwork and i wanted to do it on the computer. After some research, i came across with the following Maple command:


So, I tried to use that command to solve my system, with the following result (or non-result):

LinearMultivariateSystem({F > 0, H > 0, J > 0, K > 0, L > 0, 1+F-J > 0, 15+H-K > 0, 18-J-K > 0, 33-K-L > 0, 34-H-J-L > 0, -15-F+J+K+L > 0, 16-F-H+J+K > 0}, [F, H, J, K, L]);

Error, (in SolveTools:-Inequality:-Piecewise) piecewise takes at least 2 parameters

So, i really need help solving this as the professor told us that the first one to solve would win a book, eheh. I don't know what I'm doing wrong. Maybe this Maple command is not made for 12 variables? Or maybe i'm just writing something on a wrong form. I've never used Maple before so i can be doing something really stupid without knowing it.

I would really apreciate an answer, as my only goal as a future physicist is to unveil the secrets of the Cosmos to us all.

Thank you again.

Miguel Jesus


After I have substitute all the variables by hand ,maple still continious  processing(27 hours and still evaluating)for a simple

summarisation by the time that live math2 needed only 2 mins.Have i done something wrong?I want to solve the following equation which is shown in the attached file with the experimental data

I actually had the same problem on Maple 16. When i go to plot an equation (usually on implicit plots) it just repeats the equation in blue text rather than showing any sort of plot. Im sure i am just missing a simple option somewhere, but for the life of me I cant find it. Any and all help would be appreciated. 

Negative result...

September 24 2014 PytEin 0

I'm trying to solve a differential equation in Maple given by:

 when y(3) = -7.


I should get the result

but for some reason I get 


I solved it by writing:


I have Maple 18 if it makes a difference.

Dear Maple enthusiasts,

I am unable to find a working method to solve a system of 8 equations, of which 4 are differential equations. The system contains 8 unknown variables and the goal is to find an expression for each of these variables as a function of the time t. I have attached the code of my project at the bottom of this message.

I have tried the following:

  1. Using solve/dsolve to solve all 8 equations at once. This results in Maple eating up all of my memory and never finishing its calculations.
  2. First using solve to solve the 4 non-differential equations so that I get 4 out of 8 variables as a function of the 4 remaining variables. This results in an expression containing RootOf() for each of the 4 veriables I'm solving for, which prevents me from using these expressions in the 4 remaining differential equations.
  3. First using dsolve to solve the differential equations, which gives once again an expression for 4 variables as a function of the 4 remaining variables. I then use solve to solve the 4 remaining equations with the new found expressions. This results in an extremely long solution for each of the variables.

The code below contains the 3rd option I tried.

Any help or suggestions would be greatly appreciated. I have been scratching my head so much that I'm getting bald and whatever I search for on google or in the Maple help, I can't find a good reference to a system of differential equations together with other equations.




PARK - Mixed control



Input parameters



Projected interface area (m²)



Temperature of the process (K)



Densities (kg/m³)

Rho_m:=7000: metal

Rho_s:=2850: slag


Masses (kg)

W_m:=0.5: metal

W_s:=0.075: slag


Mass transfer coefficients (m/s)






Weight percentages in bulk at t=0 (%)






Weight percentages in bulk at equilibrium (%)






Weight percentages at the interface (%)




Atomic weights (g/mol)







Molecular weights (g/mol)






Gas constant (m³*Pa/[K*mol])









4 rate equations











3 mass balance equations









1 local equilibrium equation



Gibbs free energy of the reaction when all of the reactants and products are in their standard states (J/mol). Al and Si activities are in 1 wt pct standard state in liquid Fe. SiO2 and Al2O3 activities are in respect to pure solid state.




Expression of mole fractions as a function of weight percentages (whereby MgO is not taken into account, but instead replaced by CaO ?)

x_Al2O3_i(t):=(Pct_Al2O3_i(t)/MW_Al2O3)/(Pct_Al2O3_i(t)/MW_Al2O3 + Pct_SiO2_i(t)/MW_SiO2 + (100-Pct_SiO2_i(t)-Pct_Al2O3_i(t))/MW_CaO);
x_SiO2_i(t):=(Pct_SiO2_i(t)/MW_SiO2)/(Pct_Al2O3_i(t)/MW_Al2O3 + Pct_SiO2_i(t)/MW_SiO2 + (100-Pct_SiO2_i(t)-Pct_Al2O3_i(t))/MW_CaO);


Activity coefficients

Gamma_Al_Hry:=1: because very low percentage present  during the process (~Henry's law)

Gamma_Si_Hry:=1: because very low percentage present  during the process (~Henry's law)

Gamma_Al2O3_Ra:=1: temporary value!

Gamma_SiO2_Ra:=10^(-4.85279678314968+0.457486603678622*Pct_SiO2_b(t)); very small activity coefficient?


Activities of components



Expressions for the equilibrium constant K













i will solve the three equations below with numerical method,how?

eq1 := -2.517407096*10^12*q[1](t)^2-5.292771429*10^12*q[1](t)-1.888055322*10^12*q[2](t) = 0
eq2 := 2.246321962*10^12*q[1](t)^2+1.684741471*10^12*q[2](t)+8.110113889*10^12*q[1](t)-7.480938859*10^10*q[3](t) = 0
eq3 := int((-3.826000000*10^11*q[2](t)*cos(Pi*x)*Pi^2-3.826000000*10^11*q[1](t)^2*cos(Pi*x)*Pi^3*sin(Pi*x)+3.414000000*10^11*q[1](t)^2*sin(Pi*x)^2*Pi^4-3.414000000*10^11*q[1](t)^2*cos(Pi*x)^2*Pi^4+7*(int(exp(10*tau), tau = -infinity .. t))+q(x, t))*sin(Pi*x), x = 0 .. 1) = 0

Dear people in Maple Primes,


I have a question about how to solve a system of equations.

In the following equation, I want to eliminate D(a).

x := D(a*b*c = 3*d); 
y := D(a^2*b^3*c = 3*a);


For this purpose,

a code of 

d_a := isolate(x, D(a));

eval(y, d_a);

works well. But, for me, this code is a little laborious.

Is there any better way than the above way?


Thanks in advance.


taro yamada




   here are  equations like this

 sol := [abs(r)^2+abs(t)^2 = 1, r*conjugate(t)+t*conjugate(r), abs(r) = abs(t)]

when i solve this equations using command solve,the result  is none. and i used r=x+I*y,t=u+I*v in the equations,

sol:=[u^2+v^2+x^2+y^2 = 1, 2*u*x+2*v*y, sqrt(x^2+y^2) = sqrt(u^2+v^2)]

i still can't get a result.why,can you help me.



I am a problem with solve differential equation, please help me: THANKS 

g := (y^2-1)^2; I4 := int(g^4, y = -1 .. 1); I5 := 2*(int(g^3*(diff(g, y, y)), y = -1 .. 1)); I6 := int(g^3*(diff(g, y, y, y, y)), y = -1 .. 1); with(Student[Calculus1]); I10 := ApproximateInt(6/(1-f(x)*g)^2, y = -1 .. 1, method = simpson);

dsys3 := {I4*f(x)^2*(diff(f(x), x, x, x, x))+I5*f(x)^2*(diff(f(x), x, x))+I6*f(x)^3 = I10, f(-1) = 0, f(1) = 0, ((D@@1)(f))(-1) = 0, ((D@@1)(f))(1) = 0};

dsol5 := dsolve(dsys3, numeric, output = array([0.]));

              Error, (in dsolve/numeric/bvp) system is singular at left endpoint, use midpoint method instead

****************FORMAT TWO ********************************************************

g := (y^2-1)^2; I4 := int(g^4, y = -1 .. 1); I5 := 2*(int(g^3*(diff(g, y, y)), y = -1 .. 1)); I6 := int(g^3*(diff(g, y, y, y, y)), y = -1 .. 1); with(Student[Calculus1]); I10 := ApproximateInt(6/(1-f(x)*g)^2, y = -1 .. 1, method = simpson);
dsys3 := {I4*f(x)^2*(diff(f(x), x, x, x, x))+I5*f(x)^2*(diff(f(x), x, x))+I6*f(x)^3 = I10, f(-1) = 0, f(1) = 0, ((D@@1)(f))(-1) = 0, ((D@@1)(f))(1) = 0};

dsol5 := dsolve(dsys3, method = bvp[midrich], output = array([0.]));
                                   Error, (in dsolve) too many levels of recursion





Hi all


I am having a very complicated equation that has the form of 



Of course, the actual equation is more complicate than above. It is just an example. I want to solve the equation in terms of x. And I know that both x and y are real, and they are positive (greater than 0). My question is, how should I specify this when solving the equation?



PS: I try to run the program to the solve the equation (without specifying that they are real and positive), and at the output, it gave me something like:

"RootOf(y^2+2y+.......)".   What is that "RootOf" means? Square-root or what?

******************************************where d1 to d45 -kappa and chi are constant**********

dsys4 := {d1*h1(theta)+d2*(diff(h1(theta), theta, theta))+d3*(diff(h2(theta), theta))+d4*(diff(h2(theta), theta, theta, theta))+d5*h3(theta)+d6*(diff(h3(theta), theta, theta))+d7*(diff(h1(theta), theta, theta, theta, theta)) = 0, d8*h2(theta)+d9*(diff(h2(theta), theta, theta, theta, theta))+d10*(diff(h2(theta), theta, theta))+d11*(diff(h1(theta), theta))+d12*(diff(h1(theta), theta, theta, theta))+d13*(diff(h3(theta), theta))+d14*(diff(h3(theta), theta, theta, theta)) = 0, h3(theta)^5*(d16+ln(h3(theta))^2*d15+2*ln(h3(theta))*d17)+(diff(h3(theta), theta, theta))*h3(theta)^4*(d19+ln(h3(theta))^2*d18+2*ln(h3(theta))*d20)+(diff(h3(theta), theta, theta, theta, theta))*h3(theta)^4*(d22+ln(h3(theta))^2*d21+2*ln(h3(theta))*d23)+h1(theta)*h3(theta)^4*(d25+ln(h3(theta))^2*d24+2*ln(h3(theta))*d26)+(diff(h1(theta), theta, theta))*h3(theta)^4*(d28+ln(h3(theta))^2*d27+2*ln(h3(theta))*d29)+(diff(h2(theta), theta))*h3(theta)^4*(d31+ln(h3(theta))^2*d30+2*ln(h3(theta))*d32)+(diff(h2(theta), theta, theta, theta))*h3(theta)^4*(d34+ln(h3(theta))^2*d33+2*ln(h3(theta))*d35)+h3(theta)^4*(d37+ln(h3(theta))^2*d36+2*ln(h3(theta))*d38)+h3(theta)^4*(diff(h2(theta), theta, theta, theta, theta, theta, theta))*(d40+ln(h3(theta))^2*d39+2*ln(h3(theta))*d41)-beta*h3(theta)^3*d42-chi*ln(h3(theta))^2*d43/kappa-chi*d45/kappa-2*chi*ln(h3(theta))*d44/kappa = 0, h1(0) = 0, h1(1) = 0, h2(0) = 0, h2(1) = 0, h3(0) = 1, h3(1) = 1, ((D@@1)(h1))(0) = 0, ((D@@1)(h1))(1) = 0, ((D@@1)(h2))(0) = 0, ((D@@1)(h2))(1) = 0, ((D@@1)(h3))(0) = 0, ((D@@1)(h3))(1) = 0, ((D@@2)(h3))(0) = 0, ((D@@2)(h3))(1) = 0}; dsol6 := dsolve(dsys4, 'maxmesh' = 600, numeric, output = listprocedure)

Hy all.

I want to solve this equation, with„dd” as numerical result. What do I do wrong? Thanks. Nico

TTot := 70;
TC := 17;
GM := .26;
QMax := 870;
V := 3600*GM*QMax*TTot;
eq := V = int(QMax*exp((-t+TC)/dd)*(1+(t-TC)/TC)^(TC/dd), t = 0 .. TTot);
fsolve(eq, dd);

1 2 3 4 5 6 7 Last Page 1 of 18