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hi all.
i have a system of ODE's including 9 set of coupled OED's . 

i have  converted second deravaties to dd2 , in other words : diff(a[i](t),t,t)=dd2[i](t) . i =1..9 :

and i have set these 9 equations in form of vibrational equations such :  (M.V22)[i]+(K(t).V(t))[i]+P(t)[i] = eq[i] , where M is coefficient Matrix of second  derivatives , V22 is Vector of second derivaties , for example V22[1] = diff(a[1](t),t,t) , and  P(t) is the numeric part of equations ( they are pure number and do not contain any symbolic function ) and K(t).V(t) is the remaining part of equations such that : (K(t).V(t))[i] = eq[i] - (M.V22)[i] - P(t)[i]  , and V(t) are vector of a[i](t)'s which V(t)[1] = a[1](t) ,

i have used step by step time integration method (of an ebook which i have attachted that part of ebook here), when i set time step of solving process to h=0.01 , i can solve this system up to time one second or more, but when i choose h=0.001 or smaller, the answer diverges after 350 steps . i do not know whether the problem is in my ODS system, or maple can not handle this ?the answer about the time t=0.3 are the same in both steps, but after that, the one with stpe time h=0.001 diverges. my friend has solved this in mathematica without any problem, could any body help me ?! it is urgent for me to solve this problem,thnx everybody.


ebook.pdf  step_=_0.001.mw  step_=_0.01.mw 

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

I am trying to illustrate the chain rule for multivariet functions

 

diff(f(u(x,y),v(x,y),x)

 

The Maple responce is D1(f)(u(x,y),v(x,y)*(partial of u(x,y) wrt x) +..etc

 

I would like to replace the D- notation with the standard notation for the "partial of f wrt u" for obvious reasons - this is what students are familar with. The convert cmnd Doe Not Work in this case.

 

Similarly the cmnd diff(u(x,y),v(x,y),x,x) gives rise to D1,D11, D12 symbols which I would likee to convert to standard partial notation.

 

All this is a BIG DEAL when trying to illstrate the chain rule in Cal III.

 

Joe Salacuse

Mathematics

Kettering University

analytical solution...

January 13 2014 goli 130

Dear guys. I want to solve this equation analytically:

diff(Q(t),t)^2 = ln(t)^(b) + b*ln(t)^(b-1)

I think it is impossible. So I assumed that diff(Q(t),t)^2 = diff(P(t),t) and now I can solve it for P(t) easily. I obtained:

P(t) = t*ln(t)^(b)

Now, I want to know that there is anyway to obtain Q(t) using the latter relation and  diff(Q(t),t)^2 = diff(P(t),t) ?

Also I have a function as V(P)=P^(-2). What can I say about the behaviour of V(Q)?

Thanks a lot.

i have got alot of mixed and high degree derivatives. For example:

u[x]*u[x,t]*eta[x,t]+u[]^2*u[x]*eta[x]+kis(x,y)u[x,t]^2*u[]+eta(x, y)*u[]*u[x]^2+ksi[x,t]*u[x]^2*u[x,t]+......

like this alot of terms

my question is how can i solve divided by the derivative of the u(x,t) partial differential equations system and so  how can i find eta(x,t,u) and ksi(x,t,u) 

How do I differentiate with respect to lnx? So for example df(x)/d lnx = xdf(x)/dx

 

What I am specifically after is a way to obtain d^n f(x)/d (lnx)^n (or equivalently (x*d/dx)^n f(x) ) so simply writing df(x)/d lnx = x*diff(f(x),x) is no good.

restart;

diffeq := diff(w(r), `$`(r, 1))+2*beta*(diff(w(r), `$`(r, 1)))^3-(1/2)*S*(r-m^2/r) = 0;

con := w(1) = 1;

ODE := {con, diffeq};

sol := dsolve(ODE, w(r), type = numeric);

 

How can i have numerical solution of the above differential equation with corresponding boundary condition?

 

Hi everyone,

I have been  trying to solve a coupled system of 2 differencial equations, 1 PDEs with 1 ODE.

The code is below.

> restart;

> with(linalg):with(plots):
> PDE:=[diff(x(z,t),t)=(a/Pe)*diff(x(z,t),z$2)-a*diff(x(z,t),z)-(1-epsilon)/epsilon*3*Bi*(x(z,t)-1)/(1-Bi*(1-1/ksi(z,t)))]:
> a=2:Pe=3:Bi=5:epsilon=0.85:

> ODE := [diff(ksi(z,t),t) = (b*Bi*x(z,t)-1)/ksi(z,t)^2/(1-Bi*(1-1/ksi(z,t)))]:
> IC1:=c(z,0)=0:
> IC2:=ksi(z,0)=1:
> bc2:=diff(x(z,t),z):
> bc1:=x(z,t)-1/pe*diff(x(z,t),z):
> N:=10:
> L:=1:
> dyduf:=1/2*(-x[2](t)-3*x[0](t)+4*x[1](t))/h:
> dydub:=1/2*(-x[N-1](t)+3*x[N+1](t)+4*x[N](t))/h:
> dydu:=1/2/h*(x[m+1](t)-x[m-1](t)):
> d2ydu2:=1/h^2*(x[m-1](t)-2*x[m](t)+x[m+1](t)):
> bc1:=subs(diff(x(z,t),z)=dyduf,x(z,t)=x[0](t),z=1,bc1):
> bc2:=subs(x(z,t)-1/pe*diff(x(z,t),z)=dydub,x(z,t)=x[N+1](t),t=0,bc2):
> eq[0]:=bc1:
> eq[N+1]:=bc2:
> eq[m]:=subs(diff(x(z,t),z$2)=d2ydu2,diff(x(z,t),z)=dydu,diff(x(z,t),t)=dydu,x(z,t)=x[m](t),z=m*h,PDE):
> for i from 1 to N do eq[i]:=subs (m=i,eq[m]);od:
> x[0](t):=(solve(eq[0],x[0](t)));

> x[N+1](t):=solve(eq[N+1],x[N+1](t));

> for i from 1 to N do eq[i]:=eval(eq[i]);od:
> eqs:=[seq((eq[i]),i=1..N)]:
> Y:=[seq(x[i](t),i=1..N)];

> A:=genmatrix(eqs,Y,'B1'):
Error, (in linalg:-genmatrix) equations are not linear

> evalm(B1);

[B1[1], B1[2], B1[3], B1[4], B1[5], B1[6], B1[7], B1[8], B1[9],

B1[10]]

> B:=matrix(N,1):for i to N do B[i,1]:=B1[i]:od:evalm(B);

> h:=eval(L/(N+1));

 

> A:=map(eval,A);

 

> if N > 10 then A:=map(evalf,A);end:
> evalm(A);

 

> mat:=exponential(A,t);
Error, (in linalg:-matfunc) input must be a matrix

> mat:=map(evalf,mat):
> Y0:=matrix(N,1):for i from 1 to N do
> Y0[i,1]:=evalf(subs(x=i*h,rhs(IC)));od:evalm(Y0);
Error, invalid input: rhs received IC, which is not valid for its 1st argument, expr

 

> s1:=evalm(Y0+inverse(A)&*b):
Error, (in linalg:-inverse) expecting a matrix

> Y:=evalm(mat&*s1-inverse(A)&*b):
Error, (in linalg:-inverse) expecting a matrix

> Y:=map(simplify,Y):
> Digits:=5;

Digits := 5

> for i from 1 to N do x[i](t):=evalf((Y[i,1]));od:
Error, invalid subscript selector

> for i from 0 to N+1 do x[i](t):=eval(x[i](t));od;

 

> setcolors(["Red", "Blue", "LimeGreen", "Goldenrod", "maroon",
> "DarkTurquoise", "coral", "aquamarine", "magenta", "khaki", "sienna",
> "orange", "yellow", "gray"]);

["Red", "LimeGreen", "Goldenrod", "Blue", "MediumOrchid",

"DarkTurquoise"]

> pp:=plot([seq(x[i](t),i=0..N+1)],t=0..10,thickness=4);pt:=textplot([[0.28,0.15,typeset("Follow the arrow: ",x[0],"(t), ..., ",
> x[N+1],"(t).")]]):
Warning, unable to evaluate the functions to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct


pp := PLOT(THICKNESS(4), AXESLABELS(t, ""),

VIEW(0. .. 10., DEFAULT))

> display([pp,pt,arw],title="Figure 5.13",axes=boxed,labels=[t,"x"]);
Error, (in plots:-display) expecting plot structures but received: [arw]

 

In advance, thanks for the time of reading it!

Regards

Ozlem

Hello everybody

I'm new at using Maple

so what I'm trying to do is " solve system of differential equations numerically " and plot the result 

I use the floweing code

 

PDEtools[declare]((u, v, w)(t), prime = t)

> params := z = 0;

Omega= 2.2758;

tau = 13.8;

T2 = 200; s = 1;

r = 0.7071;

\[CapitalDelta] = 1.7758;

s = 2.2758;

Eta= 1.05457173*10^-34;

omega = 0.5; k = 1666666.667;

> sys1 := {diff(u(t), t) = Omega*v(t)-u(t)/T2,

diff(v(t), t) = -Omega*u*{t}-2*s*exp(-r^2/omega0^2-t^2*1.177^2/tau^2)*cos(k*z-omega*t)*w(t)-v(t)/T2,

diff(w(t), t) = 2*s*exp(-r^2/omega0^2-t^2*1.177^2/tau^2)*cos(k*z-omega*t)*v(t)};

Cs1 := {u(-20) = 0, v(-20) = 0, w(-20) = -1}

> ans1 := dsolve*RealRange(Open({ICs1, sys1}), {u(t), v(t), w(t)});
%;
Error, (in RealRange) invalid arguments

plot([u(t),t=-20..20])
plot([v(t),t=-20..20])
plot([w(t),t=-20..20])

 

 

:::::::::

also I need to use the result of v(t) in another equation as,

x=2*v(t)*cos(k*z-omega*t)

How I can do that ?

 

hello 

please say me how can i make diff and implicit diff with Set Point answer?

for example how can i write these?

(x^2+y^2=6)' => 2*x+2*y*y'=0 => y'(-2,5)=2/5

or

(y=x^3+4*x)' => y'=3*x^2+4 => y'(3)=31

thanks very much

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