mehdi jafari

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11 years, 87 days

MaplePrimes Activity


These are replies submitted by mehdi jafari

@Preben Alsholm tnx for your help and answer .


actually Z and R are different directions and thus indendepnt parameters,

Xstar is a paramatere dependet to Boundary Y 
from the last BC we conclude :
Y(T,1)=(Xstar)^n -> Xstar=Y(T,1)^(1/n) .

can not solve this problem using maple ?! what is th idea for solving this problem with and witouth maple !?

@Carl Love for example ( not surely correct example ) i say if -0.12a<a<0.08  and -0.44<c<-0.02 and b in sth like them, how many inervals can be found for my three variables that my answers are real ? the varibale are allowed to vary between -1 and 1 , but what are the intervals for each varibale that in those intervals,the answers are real ?
how many intervals exist in which the result is real ? can we find group of intervals for each varibale that in those intervals the answers are real ? for example
group one ={ <a< , <b< , <c< } -> the answers are real 

group two = { .... } sth like this, -> the answers are real  

and ....
what are the groups of intervals in which the answers are real .

@acer @Carl Love can i find the exact intervals of each variable for which it has real answer ?
the varibale can change between -1 and 1 , but what are intervals for each varibale for which the answer is real ? how many intervals are there for the vrabiles for which the answer in them are surely real ?

please change e to exp() and specify the varibales, are your variables theta[1](t) or theta[1] ?
please make correction about the varibales . and then speciify the equation which are going to be solved simultaneously. after all . u can use the command like this : 
dsolve( {eq[1],eq[2],eq[3]});
or after correction , please upload your new sheet again.

@Kitonum it is exactly what i needed

what is your problem exactly ? post your mw code so that we can help

@Axel Vogt as u can see,the integration has not been solved , ! i want the integration with respect to x, to be solved. in your answers,there is no x, but in mine there are . whay i can not do that , what is your maple version ?

@Axel Vogt this does not answere in my question ! :( 

 

NULL

restart:

l := 16; m := .75; EIy := 0.2e5; EIz := 0.400e7; GJ := 0.1e5; mj := .1; Pz := 5000;

16

 

.75

 

0.2e5

 

0.400e7

 

0.1e5

 

.1

 

5000

(1)

eq1 := m*(diff(diff(w(x, t), t), t))+diff(diff(EIy*(diff(diff(w(x, t), x), x))+(EIz-EIy)*(diff(diff(w(x, t), x), x))*(theta(x)+phi(x, t))^2, x), x)-Pz = 0;

.75*(diff(diff(w(x, t), t), t))+0.2e5*(diff(diff(diff(diff(w(x, t), x), x), x), x))+0.398e7*(diff(diff(diff(diff(w(x, t), x), x), x), x))*(theta(x)+phi(x, t))^2+0.1592e8*(diff(diff(diff(w(x, t), x), x), x))*(theta(x)+phi(x, t))*(diff(theta(x), x)+diff(phi(x, t), x))+0.796e7*(diff(diff(w(x, t), x), x))*(diff(theta(x), x)+diff(phi(x, t), x))^2+0.796e7*(diff(diff(w(x, t), x), x))*(theta(x)+phi(x, t))*(diff(diff(theta(x), x), x)+diff(diff(phi(x, t), x), x))-5000 = 0

(2)

w := unapply(q[1](t)*cosh(1.8751*x/l),x, t);

proc (x, t) options operator, arrow; q[1](t)*cosh(.1171937500*x) end proc

(3)

theta := proc (x) options operator, arrow; 0 end proc;

proc (x) options operator, arrow; 0 end proc

(4)

#theta(x);eq1;

phi := unapply(q[4](t)*sin(1.5708*x/l),x, t);

proc (x, t) options operator, arrow; q[4](t)*sin(0.9817500000e-1*x) end proc

(5)

eq2:=unapply(lhs(eq1),x,t);

proc (x, t) options operator, arrow; .75*(diff(diff(q[1](t), t), t))*cosh(.1171937500*x)+3.772661154*q[1](t)*cosh(.1171937500*x)-302.9569534*q[1](t)*cosh(.1171937500*x)*q[4](t)^2*sin(0.9817500000e-1*x)^2+2515.691179*q[1](t)*sinh(.1171937500*x)*q[4](t)^2*sin(0.9817500000e-1*x)*cos(0.9817500000e-1*x)+1053.716523*q[1](t)*cosh(.1171937500*x)*q[4](t)^2*cos(0.9817500000e-1*x)^2-5000 end proc

(6)

f[1] := value(expand(int(expand(eq2(x,t)*cosh(1.8751*x/l)), x = 0 .. l)));

int(.75*(diff(diff(q[1](t), t), t))*cosh(.1171937500*x)^2+3.772661154*q[1](t)*cosh(.1171937500*x)^2-302.9569534*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*sin(0.9817500000e-1*x)^2+2515.691179*cosh(.1171937500*x)*q[1](t)*sinh(.1171937500*x)*q[4](t)^2*sin(0.9817500000e-1*x)*cos(0.9817500000e-1*x)+1053.716523*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*cos(0.9817500000e-1*x)^2-5000*cosh(.1171937500*x), x = 0 .. 16)

(7)

expand(eq2(x,t)*cosh(1.8751*x/l)):

Int(%, x = 0 .. l);

expand(%);

value(%);

 

Int(.75*(diff(diff(q[1](t), t), t))*cosh(.1171937500*x)^2+3.772661154*q[1](t)*cosh(.1171937500*x)^2-302.9569534*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*sin(0.9817500000e-1*x)^2+2515.691179*cosh(.1171937500*x)*q[1](t)*sinh(.1171937500*x)*q[4](t)^2*sin(0.9817500000e-1*x)*cos(0.9817500000e-1*x)+1053.716523*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*cos(0.9817500000e-1*x)^2-5000*cosh(.1171937500*x), x = 0 .. 16)

 

Int(.75*(diff(diff(q[1](t), t), t))*cosh(.1171937500*x)^2+3.772661154*q[1](t)*cosh(.1171937500*x)^2-302.9569534*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*sin(0.9817500000e-1*x)^2+2515.691179*cosh(.1171937500*x)*q[1](t)*sinh(.1171937500*x)*q[4](t)^2*sin(0.9817500000e-1*x)*cos(0.9817500000e-1*x)+1053.716523*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*cos(0.9817500000e-1*x)^2-5000*cosh(.1171937500*x), x = 0 .. 16)

 

int(.75*(diff(diff(q[1](t), t), t))*cosh(.1171937500*x)^2+3.772661154*q[1](t)*cosh(.1171937500*x)^2-302.9569534*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*sin(0.9817500000e-1*x)^2+2515.691179*cosh(.1171937500*x)*q[1](t)*sinh(.1171937500*x)*q[4](t)^2*sin(0.9817500000e-1*x)*cos(0.9817500000e-1*x)+1053.716523*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*cos(0.9817500000e-1*x)^2-5000*cosh(.1171937500*x), x = 0 .. 16)

(8)

 

 

NULL

 

Download 1.mw

f(x)=x^2= 1/(1/x)^2 , thus what does this plot will have that the other one won't ?

@Preben Alsholm but now it is optional, and u should subscribe for receiving email notifications.

i have tried sth like this, but it has some problems, for example for an unkown number of rows and columns,i should update Matrix dimensions ! but i do not know them. i want the code be general .


restart:

M:=Matrix(2,9):N:=Matrix(3,4):

file1:="ELIST.txt":fclose(file1);

file2:="NLIST.txt":fclose(file2);

for i from 1 to 2 do

M[i]:=(fscanf(file1,"%{9}fr")[1]);

od:

for i from 1 to 3 do

N[i]:=(fscanf(file2,"%{4}fr")[1]);

od:

N;M;

Matrix(3, 4, {(1, 1) = 1., (1, 2) = 0., (1, 3) = 0., (1, 4) = 0., (2, 1) = 2., (2, 2) = 50., (2, 3) = 0., (2, 4) = 0., (3, 1) = 3., (3, 2) = 25., (3, 3) = 0., (3, 4) = 0.})

 

Matrix(2, 9, {(1, 1) = 1., (1, 2) = 1., (1, 3) = 1., (1, 4) = 1., (1, 5) = 0., (1, 6) = 1., (1, 7) = 1., (1, 8) = 3., (1, 9) = 0., (2, 1) = 2., (2, 2) = 1., (2, 3) = 1., (2, 4) = 1., (2, 5) = 0., (2, 6) = 1., (2, 7) = 3., (2, 8) = 2., (2, 9) = 0.})

(1)

``


Download matrix.mw

@Mac Dude it only imports one row ! not more !

restart:

FileTools[Text][ReadFile]("NLIST.txt");

"       1     0.00000000000       0.00000000000       0.00000000000    
       2     50.0000000000       0.00000000000       0.00000000000    
       3     25.0000000000       0.00000000000       0.00000000000
        

"

(1)

op(FileTools[Text][ReadFile]("ELIST.txt"));

"       1   1   1   1   0   1      1     3     0
       2   1   1   1   0   1      3     2     0

      
      "

(2)

M:=ImportMatrix("NLIST.txt", source=delimited, datatype=float, delimiter=" ");

Error, (in ImportMatrix) end of input encountered while reading Matrix

 

M:=ImportMatrix("NLIST.txt", source=delimited, datatype=float);

M := Matrix(5, 1, {(1, 1) = 1., (2, 1) = 0., (3, 1) = 0., (4, 1) = 0., (5, 1) = 2.})

(3)

 

``


Download convert_to_array.mw

@schloemilch please insert your original question here, and maybe i can help u and upload a working version !
actually i do not know u can not use these. until u insert your worksheet here. tnx


restart:

df1:=a-y(x)=diff(y(x),x)*bc;

a-y(x) = (diff(y(x), x))*bc

(1)

dsolve(df1);

y(x) = a+exp(-x/bc)*_C1

(2)

df2:=diff(y(x),x)*bc-a+y(x);

(diff(y(x), x))*bc-a+y(x)

(3)

dsolve(df2);

y(x) = a+exp(-x/bc)*_C1

(4)

# do u think there is a difference between a=c and a-c=0 ? and according to one theorem in differential equation,if y1 and y2 are two answer to an ODE,c1*y1+c2*y2 is also an answer ! so if y(x) is an answer,-y(x) is also an answer,please note that _C1 is a constant number and u can assume _C2= - (_C1) .

 


Download dsolve.mw

@wo0olf maybe u want to notice Mr.Kitonum's suggestion about your input correctness.

khahesh mikonam aga, deggat kon bebin oun 1/1 boode ya 1/() kolle ebarat bode? good luck.

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