mehdi jafari

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7 years, 156 days

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These are replies submitted by mehdi jafari

steady state is a state in which variables are independent of time and actucally the "state" of the system is "steady". thus if differential equations are available, the differentiation with respect to the time becomes zero. ( see: https://en.wikipedia.org/wiki/Steady_state). 
then i think you are maximizing your function with respect to the varibale "x" , and your function is complex, is that right?

@Preben Alsholm @Kitonum 
tnx for the answer but as i already said, it is in the loop and the loop is updating the term f,
for example in the next loop f is as follows :


 

restart

f2:=-2*(sqrt(6*beta[11]^2-8*beta[11]*sigma[11]+12*beta[12]^2-24*beta[12]*sigma[12]+4*sigma[11]^2+12*sigma[12]^2)*(-sigma[11]^2-3*sigma[12]^2-3/2*(beta[11]^2)+2*beta[11]*sigma[11]+6*beta[12]*sigma[12]-3*beta[12]^2)+E*delta_gamma*omega*(beta[11]^2+6*beta[12]^2-12*beta[12]*sigma[12]+6*sigma[12]^2))*(1/sqrt(6*beta[11]^2-8*beta[11]*sigma[11]+12*beta[12]^2-24*beta[12]*sigma[12]+4*sigma[11]^2+12*sigma[12]^2))*(1/(3*beta[11]^2-4*beta[11]*sigma[11]+6*beta[12]^2-12*beta[12]*sigma[12]+2*sigma[11]^2+6*sigma[12]^2))

-2*((6*beta[11]^2-8*beta[11]*sigma[11]+12*beta[12]^2-24*beta[12]*sigma[12]+4*sigma[11]^2+12*sigma[12]^2)^(1/2)*(-sigma[11]^2-3*sigma[12]^2-(3/2)*beta[11]^2+2*beta[11]*sigma[11]+6*beta[12]*sigma[12]-3*beta[12]^2)+E*delta_gamma*omega*(beta[11]^2+6*beta[12]^2-12*beta[12]*sigma[12]+6*sigma[12]^2))/((6*beta[11]^2-8*beta[11]*sigma[11]+12*beta[12]^2-24*beta[12]*sigma[12]+4*sigma[11]^2+12*sigma[12]^2)^(1/2)*(3*beta[11]^2-4*beta[11]*sigma[11]+6*beta[12]^2-12*beta[12]*sigma[12]+2*sigma[11]^2+6*sigma[12]^2))

(1)

 

 

NULL


isn't there a direct substitution way?

Download problem2.mw

@minhhieuh2003 maybe u want to try this :

 

restart:

ans:=int((a*x+c)/(b*x+d), x);

a*x/b-ln(b*x+d)*a*d/b^2+ln(b*x+d)*c/b

(1)

eval(ans,{a=2,b=1,c=1,d=1,x=6})-eval(ans,{a=2,b=1,c=1,d=1,x=2})

8-ln(7)+ln(3)

(2)

 


now , if u change parameters, the answer will be change.
also maybe u want to try these commands : subs,eval,simplify...

Download parametric.mw

@tomleslie thank you, you are great.

@tomleslie in order to use the results in matlab, the command 'double' is used ( in order to convert symbolic result to floating point number used in matlab), but when there is more than one unkown in the system, it returns some errors, what should i do ? could pls help? tnx

@tomleslie yes, they all work, tnx for your comments, they were useful

@acer thnx for quick answer, it solved my problem, can i introduce more than one variable simultaneously?
for example sth like this :
setmaple('a,b',5,6) ? ( it returns errors)
or i should define them seprately ? 
with Regards, M.Jafari

@tomleslie tnx for your compelete answer

@tomleslie sorry for the inconvenience, you are right, but a1 is not constant and is a function of time, it is int(a1(t)*A(t),t)...
actually a1=(t)->(wl/gamma1)*cos(w*t);
can it be solved?

@tomleslie actually diff(x(t),t)=A(t) as u have defined, can i use directly A(t) ? even i used :
int(eval(A(t),ans),t=0..1); but it does not computed ...

odeProb.mw

@tomleslie i want to fit a function to these three plots...
i used polynomial interpolation (CurveFitting:-PolynomialInterpolation([seq]([t, (eval(B(t),ans))(t)],t=0.1..1,0.1),t)).
but it does not give the functions, i think it can be fitted using trigonomic functions, but actually i don't know how, could u help ?
and actually i want to calculate the integral :
int(a1*diff(x(t),t) +a2*diff(y(t),t)+a3*diff(z(t),t) );
can i calculate this without interpolating the functions? tnx for the help...
odeProb.mw

@tomleslie thanak you, and i made use of these deriavatives...

@tomleslie that was interesting idea ! tnx for the help ! Good Luck !

@Mariusz Iwaniuk tnx for your help,where in the help i can find answers for these questions? could please give me a link of the related halp page? tnx

@rlopez tnx for explanation, you are right sir

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