samiyare

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12 years, 134 days

 

Amir

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These are replies submitted by samiyare

thanks for your reply. i like it as a procedure. but i want to know is there any chance to calculate the inverse of L as a procedure. for example if
L:=D

inverse of L, named H is

H:= int

thanks for your attention in advance

thanks for your relpy

eq1 and eq2 are the residuals of my code. i have two parameter a and b and i should find the values that eq1 and eq2 (residuals) minimizes. eq1 depend on a and b and eq2 depend too.

for better understand, i'm going to solve the system of 2 ode. when i solve one ode i have one residuals named eq1 and one parameter named a and i used nlpsolve to find a, in order to eq1(residual) minimizes. now i have two equation and two parameter and i want to minimized it.

in your words, i think

"eq1" is an expression, depending on a and b and eq2 too. and i want to minimizes eq1 and eq2 simultaneosly. maybe eq1 minimizes with special a and b and eq2 minimizes by different a and b. i want to choose the best a and b that 2 equation have at least values.

thanks for your attenrion in advance

thanks for your relpy

eq1 and eq2 are the residuals of my code. i have two parameter a and b and i should find the values that eq1 and eq2 (residuals) minimizes. eq1 depend on a and b and eq2 depend too.

for better understand, i'm going to solve the system of 2 ode. when i solve one ode i have one residuals named eq1 and one parameter named a and i used nlpsolve to find a, in order to eq1(residual) minimizes. now i have two equation and two parameter and i want to minimized it.

in your words, i think

"eq1" is an expression, depending on a and b and eq2 too. and i want to minimizes eq1 and eq2 simultaneosly. maybe eq1 minimizes with special a and b and eq2 minimizes by different a and b. i want to choose the best a and b that 2 equation have at least values.

thanks for your attenrion in advance

hi,

i think we must use a command that solve 4 system of equation and get the a and b such the 4 equation's residual minimize. like lagrange multiplier but i'm not sure.

hi,

is there any chance to use a command that give only all real values of alpha and beta. if it helps, i know that exact solution of alpha and beta may be in the range [0-0.1]. furthermore, it must be said that the your attached file still gives different answers for each execute.

thanks for your attention in advance

hi,

is there any chance to use a command that give only all real values of alpha and beta. if it helps, i know that exact solution of alpha and beta may be in the range [0-0.1]. furthermore, it must be said that the your attached file still gives different answers for each execute.

thanks for your attention in advance

@Preben Alsholm 

please see http://www.mapleprimes.com/questions/133924-Multiple-Solution-For-Ode?submit=134051#comment134051

@Preben Alsholm 

please see http://www.mapleprimes.com/questions/133924-Multiple-Solution-For-Ode?submit=134051#comment134051

hi,

i have the following equation

eq := .7246228659*(diff(f(x), x, x, x))+f(x)*(diff(f(x), x, x))+1-(diff(f(x), x))^2

and i use the follosing method but i cant get the second solution

> restart;
b := 7;
eq := .7246228659*(diff(f(x), x, x, x))+f(x)*(diff(f(x), x, x))+1-(diff(f(x), x))^2;
            
res := dsolve({eq =0, f(0) = 0, D(f)(0) = a, (D@D)(f)(0) = p2}, numeric,output=listprocedure,parameters=[a,p2]);  
F0,F1,F2:=op(subs(res,[f(x),diff(f(x),x),diff(f(x),x,x)]))  ;
p:=proc(aa,pp2) if not type([aa,pp2],list(numeric)) then return 'procname(_passed)' end if;  
res(parameters=[aa,pp2]):  
F1(b)-1;  
end proc;
fsolve(p(-1.2, pp2) = 0, pp2 = -2 .. 2);

--------------------------

in the above, with pp2 i can easily get fist solution. but i cant achieve second pp2. please help me

thancks for your attention in advance

hi,

i have the following equation

eq := .7246228659*(diff(f(x), x, x, x))+f(x)*(diff(f(x), x, x))+1-(diff(f(x), x))^2

and i use the follosing method but i cant get the second solution

> restart;
b := 7;
eq := .7246228659*(diff(f(x), x, x, x))+f(x)*(diff(f(x), x, x))+1-(diff(f(x), x))^2;
            
res := dsolve({eq =0, f(0) = 0, D(f)(0) = a, (D@D)(f)(0) = p2}, numeric,output=listprocedure,parameters=[a,p2]);  
F0,F1,F2:=op(subs(res,[f(x),diff(f(x),x),diff(f(x),x,x)]))  ;
p:=proc(aa,pp2) if not type([aa,pp2],list(numeric)) then return 'procname(_passed)' end if;  
res(parameters=[aa,pp2]):  
F1(b)-1;  
end proc;
fsolve(p(-1.2, pp2) = 0, pp2 = -2 .. 2);

--------------------------

in the above, with pp2 i can easily get fist solution. but i cant achieve second pp2. please help me

thancks for your attention in advance

thanks, it works very well

thanks, it works very well

@dohashi 

thanks for your comments. it's work. but i want that each tread run multiple command. i dont know how write many commands in each Proc. for example i want that thread 1 runs result1[i](x):=subs(m=i,dtheta[m](x)): result1[i](x):=value(%): cte:=eval(-result1[i](x),x=0); result1(i)(x):=result1[i](x)+result1[i-1](x)+cte*exp(-x): and thread 2 runs result2[i](x):=subs(m=i,dphi[m](x)): result2[i](x):=value(%): cte:=eval(-result2[i](x),x=0); result2(i)(x):=result2[i](x)+result2[i-1](x)+cte*exp(-x): so thread 1 run the (4 line) commands above and give result1[i] as an output and also thread 2 give result2[i]. thx

@dohashi 

thanks for your comments. it's work. but i want that each tread run multiple command. i dont know how write many commands in each Proc. for example i want that thread 1 runs result1[i](x):=subs(m=i,dtheta[m](x)): result1[i](x):=value(%): cte:=eval(-result1[i](x),x=0); result1(i)(x):=result1[i](x)+result1[i-1](x)+cte*exp(-x): and thread 2 runs result2[i](x):=subs(m=i,dphi[m](x)): result2[i](x):=value(%): cte:=eval(-result2[i](x),x=0); result2(i)(x):=result2[i](x)+result2[i-1](x)+cte*exp(-x): so thread 1 run the (4 line) commands above and give result1[i] as an output and also thread 2 give result2[i]. thx

@dohashi 

thancks
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