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Hello everyone,

i'm trying to simulate a diffusion problem. It contains two connected regions in which a species is diffusing at different speeds. In one region (zeta) one boundary is set to be constant whereas in the other region (c) there is some oscillation at the boundary.The code i try to use is as follows:

sys1 := [diff(c(x, t), t) = gDiffusion*10^5*diff(c(x, t), x$2), diff(zeta(x, t), t) = KDiffusion*10^6*diff(zeta(x, t), x$2)]

pds := pdsolve(sys1, IBC, numeric, time = t, range = 0 .. 3000, spacestep = 3)

However the main problem are my boundary conditions:

IBC := {c(0, t) = 0, c(x > 0, 0) = 0, zeta(0, t) = .4, zeta(x > 0, 0) = .4, (D[1](c))(3000, t) = sin((1/100)*t), (D[1](zeta))(0, t) = 0}

Like this it principally works (however it is apparently ill-posed).

Now what i do like is that the two equations are coupled at x=2000 with the condition that c(2000,t)=zeta(2000,t). This however i dont seem to be able to implement.

I appreciate your comments

Goon

 

hi.i encountered this erroe  [Error, (in dsolve/numeric/bvp/convertsys) unable to convert to an explicit first-order system] with solving set of differential equation.please help me.thanks a lot  

dsys3 := {`1`*h1(theta)+`1`*(diff(h1(theta), theta, theta))+`1`*(diff(h2(theta), theta))+`1`*(diff(h2(theta), theta, theta, theta))+`1`*h3(theta)+`1`*(diff(h3(theta), theta, theta))+`1`*(diff(h1(theta), theta, theta, theta, theta)) = 0, `1`*h2(theta)+`1`*(diff(h2(theta), theta, theta, theta, theta))+`1`*(diff(h2(theta), theta, theta))+`1`*(diff(h1(theta), theta))+`1`*(diff(h1(theta), theta, theta, theta))+`1`*(diff(h3(theta), theta))+`1`*(diff(h3(theta), theta, theta, theta)) = 0, h3(theta)^5*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h3(theta), theta, theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h3(theta), theta, theta, theta, theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+h1(theta)*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h1(theta), theta, theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h2(theta), theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+(diff(h2(theta), theta, theta, theta))*h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+h3(theta)^4*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)+h3(theta)^4*(diff(h2(theta), theta, theta, theta, theta, theta, theta))*(`1`+ln(h3(theta))^2*`1`+2*ln(h3(theta))*`1`)-beta*h3(theta)^3*`1`-chi*ln(h3(theta))^2*`1`/kappa-chi*`1`/kappa-2*chi*ln(h3(theta))*`1`/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}; dsol5 := dsolve(dsys3, 'maxmesh' = 600, numeric, output = listprocedure);
%;
Error, (in dsolve/numeric/bvp/convertsys) unable to convert to an explicit first-order system

 

Hi,

I have a system of diff equations (see below). I am trying to obtain analytical solution. when I assume that z=wN, I receive such solution. Do anybody have idea if I know that z>wN, does this system has an analytical solution?

diff(K(t), t) = -(1/2)*(Q(t)^2*alpha^2*eta*upsilon-2*eta*alpha*(N*upsilon*w*C[max]-z*alpha*K(t))*Q(t)+N*w*(-2*C[max]*z*eta*alpha*K(t)+upsilon*((-N*w+z)*alpha+N*C[max]^2*w*eta)))*K(t)/((C[max]*w*N-alpha*Q(t))*upsilon*N*w)

diff(Q(t), t) = (1/2)*(-z*(Q(t)^2*alpha^2*eta-2*N*Q(t)*alpha*eta*w*C[max]+w*(w*(eta*C[max]^2-alpha)*N+z*alpha)*N)*K(t)-2*N*upsilon*w*(N*w-z)*(C[max]*w*N-alpha*Q(t)))/((C[max]*w*N-alpha*Q(t))*upsilon*N*w)

K(0) = K0, Q(0) = Q0

Thanks,

Dmitry

Hello fellow maple users,im new to the software,im trying to solve a differential system but it dosent work

 

This is the system :

DE1 := diff(Y(t), t) = 5*Y(t)*ln(b(t)/Y(t))-5*Y(t)

DE2 := diff(b(t), t) = 5*b(t)*Y(t)^(3/2)-5*Y(t)

 

Thank you for your help !

Hello,

I wanted to ask whether it's possible to use the taylor command together with vectors from the physics package, maybe I am just doing something wrong here. I tried the following:

and get an error message that diff cannot handle vectors. Of course I could do the expansion by hand an enter the result in Maple but I think it would be a very nice feature because an expansion of vector fields which vary in space and time is such a common problem e.g. in classical electrodynamics. I think of an expansion with non-projected vectors such as

+ higher order terms.

Thanks a lot!

Peter

Hi. I want to differentiate the following expression using "Diff", not "diff". but I want to apply "Diff" to differentiate each separate term based on the chain rule. How can I do that? Does "Diff" apply chain rule for differentiation?

 

 

  

Hello,
I have a system of first order diff. equations which I would like to solve symbolically. Unfortunately, Maple does not solve the system. Do anybody have suggestions how can I solve this system (please see below):

diff(S(t), t) = -eta*(C[max]*w*N-alpha*Q(t))*K(t)*S(t)/(w*N*(S(t)+K(t))),

diff(K(t), t) = S(t)*((z*eta*alpha*(C[max]*w*N-alpha*Q(t))*S(t)-upsilon*(eta*alpha^2*Q(t)^2-2*C[max]*w*N*eta*alpha*Q(t)+((-N*w+z)*alpha+N*C[max]^2*w*eta)*N*w))*K(t)^2+(2*((1/2)*z*eta*(C[max]*w*N-alpha*Q(t))*S(t)+N*w*upsilon*(N*w-z)))*S(t)*alpha*K(t)+N*S(t)^2*w*alpha*upsilon*(N*w-z))/((K(t)^2*alpha*z+3*S(t)*K(t)*alpha*z+S(t)*(2*S(t)*z*alpha+upsilon*(C[max]*w*N-alpha*Q(t))))*(S(t)+K(t))*N*w),

diff(Q(t), t) = (-alpha*z*(z*eta*(C[max]*w*N-alpha*Q(t))*K(t)+N*w*upsilon*(N*w-z))*S(t)^2+(-z^2*eta*alpha*(C[max]*w*N-alpha*Q(t))*K(t)^2-(eta*alpha^2*Q(t)^2-2*C[max]*w*N*eta*alpha*Q(t)+N*w*((2*N*w-2*z)*alpha+N*C[max]^2*w*eta))*z*upsilon*K(t)-N*w*upsilon^2*(N*w-z)*(C[max]*w*N-alpha*Q(t)))*S(t)-N*w*z*alpha*upsilon*K(t)^2*(N*w-z))/((2*S(t)^2*alpha*z+(3*z*alpha*K(t)+upsilon*(C[max]*w*N-alpha*Q(t)))*S(t)+K(t)^2*alpha*z)*N*w*upsilon)

where initials conditions are:

S(0) = S0, K(0) = K0, Q(0) = Q0

Thanks,

Dmitry

 

 

 

Hello Hello everybody 
   I have to solve the following differential equation numerically 


``

 

restart:with(plots):

mb:=765 : mp:=587 : Ib:=76.3*10^3 : Ip:=7.3*10^3 : l:=0.92 : d:=10: F:=490: omega:=0.43 :

eq1:=(mp+mb)*diff(x(t),t$2)+mp*(d*cos(theta(t))+l*cos(alpha(t)+theta(t)))*diff(theta(t),t$2)+mp*l*cos(alpha(t)+theta(t))*diff(alpha(t),t$2)+mp*(d*diff(theta(t),t)^2*sin(theta(t))+l*(diff(theta(t),t)+diff(alpha(t),t))^2*sin(alpha(t)+theta(t)))-F*sin(omega*t)=0;

1352*(diff(diff(x(t), t), t))+587*(10*cos(theta(t))+.92*cos(alpha(t)+theta(t)))*(diff(diff(theta(t), t), t))+540.04*cos(alpha(t)+theta(t))*(diff(diff(alpha(t), t), t))+5870*(diff(theta(t), t))^2*sin(theta(t))+540.04*(diff(theta(t), t)+diff(alpha(t), t))^2*sin(alpha(t)+theta(t))-490*sin(.43*t) = 0

(1)

eq2:=(mp+mb)*diff(z(t),t$2)-mp*d*(sin(theta(t)+alpha(t))+sin(theta(t)))*diff(theta(t),t$2)-mp*l*sin(alpha(t)+theta(t))*diff(alpha(t),t$2)+mp*(d*diff(theta(t),t)^2*cos(theta(t))+l*(diff(theta(t),t)+diff(alpha(t),t))^2*cos(alpha(t)+theta(t)))+9.81*(mp+mb)-F*sin(omega*t)=0;

1352*(diff(diff(z(t), t), t))-5870*(sin(alpha(t)+theta(t))+sin(theta(t)))*(diff(diff(theta(t), t), t))-540.04*sin(alpha(t)+theta(t))*(diff(diff(alpha(t), t), t))+5870*(diff(theta(t), t))^2*cos(theta(t))+540.04*(diff(theta(t), t)+diff(alpha(t), t))^2*cos(alpha(t)+theta(t))+13263.12-490*sin(.43*t) = 0

(2)

eq3:=mp*(d*cos(theta(t))+l*cos(alpha(t)+theta(t)))*diff(x(t),t$2)-mp*(l*sin(theta(t)+alpha(t))+d*sin(theta(t)))*diff(z(t),t$2)+(Ip+Ib+mp*(d^2+l^2)+2*mp*d*l*cos(alpha(t)))*diff(theta(t),t$2)+[Ip+mp*l^2+mp*d*l*cos(alpha(t))]*diff(alpha(t),t$2)-mp*sin(alpha(t))*(l*d*diff(alpha(t),t)^2-l*d*(diff(alpha(t),t)+diff(theta(t),t))^2)+mp*9.81*l*sin(alpha(t)+theta(t))+mp*9.81*d*sin(theta(t))=0;

587*(10*cos(theta(t))+.92*cos(alpha(t)+theta(t)))*(diff(diff(x(t), t), t))-587*(.92*sin(alpha(t)+theta(t))+10*sin(theta(t)))*(diff(diff(z(t), t), t))+(142796.8368+10800.80*cos(alpha(t)))*(diff(diff(theta(t), t), t))+[7796.8368+5400.40*cos(alpha(t))]*(diff(diff(alpha(t), t), t))-587*sin(alpha(t))*(9.20*(diff(alpha(t), t))^2-9.20*(diff(theta(t), t)+diff(alpha(t), t))^2)+5297.7924*sin(alpha(t)+theta(t))+57584.70*sin(theta(t)) = 0

(3)

eq4:=mp*l*cos(alpha(t)+theta(t))*diff(x(t),t$2)-mp*l*sin(alpha(t)+theta(t))*diff(z(t),t$2)+(Ip+mp*l^2+mp*d*l*cos(alpha(t)))*diff(theta(t),t$2)+(Ip+mp*l^2)*diff(alpha(t),t$2)-mp*9.81*l*sin(alpha(t)+theta(t))+l*d*mp*diff(theta(t),t$1)^2*sin(alpha(t))=0;

540.04*cos(alpha(t)+theta(t))*(diff(diff(x(t), t), t))-540.04*sin(alpha(t)+theta(t))*(diff(diff(z(t), t), t))+(7796.8368+5400.40*cos(alpha(t)))*(diff(diff(theta(t), t), t))+7796.8368*(diff(diff(alpha(t), t), t))-5297.7924*sin(alpha(t)+theta(t))+5400.40*(diff(theta(t), t))^2*sin(alpha(t)) = 0

(4)

CI:= x(0)=0,z(0)=0,theta(0)=0,alpha(0)=0,D(x)(0)=0,D(alpha)(0)=0,D(z)(0)=0,D(theta)(0)=0;

x(0) = 0, z(0) = 0, theta(0) = 0, alpha(0) = 0, (D(x))(0) = 0, (D(alpha))(0) = 0, (D(z))(0) = 0, (D(theta))(0) = 0

(5)

solution:=dsolve([eq1,eq2,eq3,eq4, CI],numeric);

Error, (in f) unable to store '[0.]/(0.17571268341557e16+[-0.25659510610770e15])' when datatype=float[8]

 

 

 

I don't know why it says : Error, (in f) unable to store '[0.]/(0.17571268341557e16+[-0.25659510610770e15])' when datatype=float[8]

 

Help pleaase!

thank you !!!

Download systéme_complet.mw

 

Diff_EQ_sample_questions.pdf

 

This is a link to two sample questions I am trying to learn how to solve using maple. I am using maple student edition of maple. Any help would be great. Thank you.

 

 

can we load part of a package not the package whole !? for example from physics package i only need diff tool not all of the package tools,can i do sth for that !? 

hi,i want to take differential with respect to another differential using physics package,but using D instead of diff,could anyone help me do that ? for example :

restart; with(Physics):
A1 := -(1/24)*1*rho*((diff(phi[1](x, t), t))^2)*(h^3)-(1/2)*1*rho*((diff(u[ref](x, t), t))^2)*h-(1/2)*rho*((diff(w(x, t), t))^2)*h+(1/24)*1*1*((diff(phi[1](x, t), x))^2)*(h^3)+(1/2)*1*(1*((diff(u[ref](x, t), x)+(1/2)*(diff(w(x, t), x))^2)^2)+K*1*((diff(w(x, t), x)+phi[1](x, t))^2))*h-1*q*w(x, t):

A2:=-diff(diff(A1,diff(u[ref](x,t),x)),x);

here i want to compute A2 using D command,not diff and i do not want use convert command ! i just need to calculate A2 directly using D command. tnx for your help.

 

Hello there,

i got a question regarding derivatives in Matlab.

 

I got a function for a example:

f:=f1(y)*f2(x)

 

from this function i need the partial derivatives with respect to x and y, i can easily get them with

diff(f,x)

diff(f,y)

 

now i want to compute the derivative of f with respect to time t, assuming that y and x both depend on t - how can i tell maple that y and x depend on t?

 

thanks

ben

I want to find the solution in a special form.
How can I do it?
Here is what I tried:

(Maple)

(Maple)


In the left hand side u_1 is not changed in  D(u_1).
I want to substitute and evalute (differentiate) it.

Thanks,  Sandor

 

 

 

 

Hell

I write this code and didn't work , I have some erorrs as

Warning, The use of global variables in numerical ODE problems is deprecated, and will be removed in a future release. Use the 'parameters' argument instead (see ?dsolve,numeric,parameters)
Warning, cannot evaluate the solution past the initial point, problem may be complex, initially singular or improperly set up

also I have question " How I can change the scale of plot"

parameters := [z = 0, Omega = 2.2758, tau = 13.8, T2 = 200, omega0 = 1, r = .7071, s = 2.2758, H = 1.05457173*10^(-34), omega = .5, k = 1666666.667, Delta = 1.7758]

 

sys1 := {diff(u(t), t) = s*v(t)-u(t)/T2, diff(v(t), t) = -s*u(t)-2*Omega*exp(-r^2/omega0^2-t^2*1.177^2/tau^2)*cos(kz-`ωt`)*w(t)-v(t)/T2, diff(w(t), t) = 2*Omega*exp(-r^2/omega0^2-t^2*1.177^2/tau^2)*cos(kz-`ωt`)*v(t)}; ICs1 := {u(-20) = 0, v(-20) = 0, w(-20) = -1}

 

ans1 := dsolve(`union`(eval(sys1, parameters), ICs1), numeric, output = listprocedure); plots:-odeplot(ans1, [[t, u(t)], [t, v(t)], [t, w(t)]], t = -20 .. 20, legend = [w, v, u])

 

U := eval(u(t), ans1); F := eval(((-2*10^33*Omega*H*r*U(t))*(1/omega0^2))*exp(-r^2/omega0^2-t^2*1.177^2/tau^2)*cos(kz-`ωt`), parameters)

 

plot(F, t = -20 .. 20)

Hello everyone,

I need help to type in the following type of initial condition.

diff(1/x*diff(F(x),x),x)=0 at x =0.

Thanks

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