Mixing
solve an Isoperimetric problem in calculus of variation using Maxima open source.
Clash Royale CLAN TAG#URR8PPP
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Kind help to solve the below problem via maxima program
Find the extremal of the functional $int_0^1((y^')^2 + x^2)dx$ under the constraint $int_0^1 y^2 dx,=,2$ and having $y(0),=,0$ and
$y(1),=,1.$ Using Maxima programming.
I tried Like the below code kind help not coming
kill(all)$
ratprint: false$
depends(f,[x,y])$
depends(g,[x,y])$
depends(y,x)$
declare(P,real)$
assume(P>0)$ (As positive or negative or zero error was giving)
z: diff(y,x);
f:z^2+x^2;
g:y^2;
h:f+P*g;
h1:diff(h,y);
h2:diff(h,z);
h3:h1-diff(h2,x);
h4:ode2(h3,y,x);
h5: bc2(h4,x=0,y=0,x=1,y=1);
hr: trigrat(h5);
h6:integrate((rhs(hr))^2,x,0,1);
h7:solve(trigrat(h6)=2,P);
trigrat(rhs(h7[1]));
Need help
Answer expected from maxima
y= %k1*%e^(sqrt(l)*x)+%k2*%e^(-sqrt(l)*x)
calculus-of-variations maxima-software
edited Jul 20 at 15:02
Daniel Fischer♦
171k16154274
asked Jul 20 at 14:49
sriram
43
add a comment |Â
up vote
0
down vote
favorite
Kind help to solve the below problem via maxima program
Find the extremal of the functional $int_0^1((y^')^2 + x^2)dx$ under the constraint $int_0^1 y^2 dx,=,2$ and having $y(0),=,0$ and
$y(1),=,1.$ Using Maxima programming.
I tried Like the below code kind help not coming
kill(all)$
ratprint: false$
depends(f,[x,y])$
depends(g,[x,y])$
depends(y,x)$
declare(P,real)$
assume(P>0)$ (As positive or negative or zero error was giving)
z: diff(y,x);
f:z^2+x^2;
g:y^2;
h:f+P*g;
h1:diff(h,y);
h2:diff(h,z);
h3:h1-diff(h2,x);
h4:ode2(h3,y,x);
h5: bc2(h4,x=0,y=0,x=1,y=1);
hr: trigrat(h5);
h6:integrate((rhs(hr))^2,x,0,1);
h7:solve(trigrat(h6)=2,P);
trigrat(rhs(h7[1]));
Need help
Answer expected from maxima
y= %k1*%e^(sqrt(l)*x)+%k2*%e^(-sqrt(l)*x)
calculus-of-variations maxima-software
edited Jul 20 at 15:02
Daniel Fischer♦
171k16154274
asked Jul 20 at 14:49
sriram
43
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Kind help to solve the below problem via maxima program
Find the extremal of the functional $int_0^1((y^')^2 + x^2)dx$ under the constraint $int_0^1 y^2 dx,=,2$ and having $y(0),=,0$ and
$y(1),=,1.$ Using Maxima programming.
I tried Like the below code kind help not coming
kill(all)$
ratprint: false$
depends(f,[x,y])$
depends(g,[x,y])$
depends(y,x)$
declare(P,real)$
assume(P>0)$ (As positive or negative or zero error was giving)
z: diff(y,x);
f:z^2+x^2;
g:y^2;
h:f+P*g;
h1:diff(h,y);
h2:diff(h,z);
h3:h1-diff(h2,x);
h4:ode2(h3,y,x);
h5: bc2(h4,x=0,y=0,x=1,y=1);
hr: trigrat(h5);
h6:integrate((rhs(hr))^2,x,0,1);
h7:solve(trigrat(h6)=2,P);
trigrat(rhs(h7[1]));
Need help
Answer expected from maxima
y= %k1*%e^(sqrt(l)*x)+%k2*%e^(-sqrt(l)*x)
calculus-of-variations maxima-software
edited Jul 20 at 15:02
Daniel Fischer♦
171k16154274
asked Jul 20 at 14:49
sriram
43
Kind help to solve the below problem via maxima program
Find the extremal of the functional $int_0^1((y^')^2 + x^2)dx$ under the constraint $int_0^1 y^2 dx,=,2$ and having $y(0),=,0$ and
$y(1),=,1.$ Using Maxima programming.
I tried Like the below code kind help not coming
kill(all)$
ratprint: false$
depends(f,[x,y])$
depends(g,[x,y])$
depends(y,x)$
declare(P,real)$
assume(P>0)$ (As positive or negative or zero error was giving)
z: diff(y,x);
f:z^2+x^2;
g:y^2;
h:f+P*g;
h1:diff(h,y);
h2:diff(h,z);
h3:h1-diff(h2,x);
h4:ode2(h3,y,x);
h5: bc2(h4,x=0,y=0,x=1,y=1);
hr: trigrat(h5);
h6:integrate((rhs(hr))^2,x,0,1);
h7:solve(trigrat(h6)=2,P);
trigrat(rhs(h7[1]));
Need help
Answer expected from maxima
y= %k1*%e^(sqrt(l)*x)+%k2*%e^(-sqrt(l)*x)
calculus-of-variations maxima-software
edited Jul 20 at 15:02
Daniel Fischer♦
171k16154274
asked Jul 20 at 14:49
sriram
43
edited Jul 20 at 15:02
Daniel Fischer♦
171k16154274
edited Jul 20 at 15:02
Daniel Fischer♦
171k16154274
edited Jul 20 at 15:02
Daniel Fischer♦
171k16154274
171k16154274
asked Jul 20 at 14:49
sriram
43
asked Jul 20 at 14:49
sriram
43
asked Jul 20 at 14:49
sriram
43
43
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
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This problem can be handled easily without the help of a symbolic processor so calling
$$
f(x,y,y',lambda) = (y')^2+x^2+lambda y^2
$$
we have
$$
f_y-left(f_y'right)'=y''-lambda y = 0
$$
and after solving the ODE we have
$$
y = c_1 e^sqrt lambda x+c_2 e^-sqrtlambda x
$$
with the boundary and integral conditions we need to solve
$$
beginarrayrcl
c_1+c_2 & = & 0\
c_1 e^sqrt lambda+c_2 e^-sqrtlambda & = & 1\
frace^-sqrtlambda left(e^sqrtlambda -1right) left(c_1 e^sqrtlambda +c_2right)sqrtlambda & = & 2
endarray
$$
thus obtaining the constants $c_1, c_2, lambda$ Here making the substitution $c_2 =-c_1$ we have
$$
beginarrayrcl
c_1 & = & -c_2\2 c_1 cosh left(sqrtlambda right) & = & 1\
frac2 c_1 left(cosh left(sqrtlambda right)-1right)sqrtlambda & = & 2
endarray
$$
This last step should be concluded using an iterative process which is not explicit in the MAXIMA script.
I hope this helps.
answered Jul 20 at 16:13
Cesareo
5,7432412
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
This problem can be handled easily without the help of a symbolic processor so calling
$$
f(x,y,y',lambda) = (y')^2+x^2+lambda y^2
$$
we have
$$
f_y-left(f_y'right)'=y''-lambda y = 0
$$
and after solving the ODE we have
$$
y = c_1 e^sqrt lambda x+c_2 e^-sqrtlambda x
$$
with the boundary and integral conditions we need to solve
$$
beginarrayrcl
c_1+c_2 & = & 0\
c_1 e^sqrt lambda+c_2 e^-sqrtlambda & = & 1\
frace^-sqrtlambda left(e^sqrtlambda -1right) left(c_1 e^sqrtlambda +c_2right)sqrtlambda & = & 2
endarray
$$
thus obtaining the constants $c_1, c_2, lambda$ Here making the substitution $c_2 =-c_1$ we have
$$
beginarrayrcl
c_1 & = & -c_2\2 c_1 cosh left(sqrtlambda right) & = & 1\
frac2 c_1 left(cosh left(sqrtlambda right)-1right)sqrtlambda & = & 2
endarray
$$
This last step should be concluded using an iterative process which is not explicit in the MAXIMA script.
I hope this helps.
answered Jul 20 at 16:13
Cesareo
5,7432412
add a comment |Â
up vote
0
down vote
This problem can be handled easily without the help of a symbolic processor so calling
$$
f(x,y,y',lambda) = (y')^2+x^2+lambda y^2
$$
we have
$$
f_y-left(f_y'right)'=y''-lambda y = 0
$$
and after solving the ODE we have
$$
y = c_1 e^sqrt lambda x+c_2 e^-sqrtlambda x
$$
with the boundary and integral conditions we need to solve
$$
beginarrayrcl
c_1+c_2 & = & 0\
c_1 e^sqrt lambda+c_2 e^-sqrtlambda & = & 1\
frace^-sqrtlambda left(e^sqrtlambda -1right) left(c_1 e^sqrtlambda +c_2right)sqrtlambda & = & 2
endarray
$$
thus obtaining the constants $c_1, c_2, lambda$ Here making the substitution $c_2 =-c_1$ we have
$$
beginarrayrcl
c_1 & = & -c_2\2 c_1 cosh left(sqrtlambda right) & = & 1\
frac2 c_1 left(cosh left(sqrtlambda right)-1right)sqrtlambda & = & 2
endarray
$$
This last step should be concluded using an iterative process which is not explicit in the MAXIMA script.
I hope this helps.
answered Jul 20 at 16:13
Cesareo
5,7432412
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This problem can be handled easily without the help of a symbolic processor so calling
$$
f(x,y,y',lambda) = (y')^2+x^2+lambda y^2
$$
we have
$$
f_y-left(f_y'right)'=y''-lambda y = 0
$$
and after solving the ODE we have
$$
y = c_1 e^sqrt lambda x+c_2 e^-sqrtlambda x
$$
with the boundary and integral conditions we need to solve
$$
beginarrayrcl
c_1+c_2 & = & 0\
c_1 e^sqrt lambda+c_2 e^-sqrtlambda & = & 1\
frace^-sqrtlambda left(e^sqrtlambda -1right) left(c_1 e^sqrtlambda +c_2right)sqrtlambda & = & 2
endarray
$$
thus obtaining the constants $c_1, c_2, lambda$ Here making the substitution $c_2 =-c_1$ we have
$$
beginarrayrcl
c_1 & = & -c_2\2 c_1 cosh left(sqrtlambda right) & = & 1\
frac2 c_1 left(cosh left(sqrtlambda right)-1right)sqrtlambda & = & 2
endarray
$$
This last step should be concluded using an iterative process which is not explicit in the MAXIMA script.
I hope this helps.
answered Jul 20 at 16:13
Cesareo
5,7432412
This problem can be handled easily without the help of a symbolic processor so calling
$$
f(x,y,y',lambda) = (y')^2+x^2+lambda y^2
$$
we have
$$
f_y-left(f_y'right)'=y''-lambda y = 0
$$
and after solving the ODE we have
$$
y = c_1 e^sqrt lambda x+c_2 e^-sqrtlambda x
$$
with the boundary and integral conditions we need to solve
$$
beginarrayrcl
c_1+c_2 & = & 0\
c_1 e^sqrt lambda+c_2 e^-sqrtlambda & = & 1\
frace^-sqrtlambda left(e^sqrtlambda -1right) left(c_1 e^sqrtlambda +c_2right)sqrtlambda & = & 2
endarray
$$
thus obtaining the constants $c_1, c_2, lambda$ Here making the substitution $c_2 =-c_1$ we have
$$
beginarrayrcl
c_1 & = & -c_2\2 c_1 cosh left(sqrtlambda right) & = & 1\
frac2 c_1 left(cosh left(sqrtlambda right)-1right)sqrtlambda & = & 2
endarray
$$
This last step should be concluded using an iterative process which is not explicit in the MAXIMA script.
I hope this helps.
answered Jul 20 at 16:13
Cesareo
5,7432412
edited Jul 20 at 18:45
answered Jul 20 at 16:13
Cesareo
5,7432412
answered Jul 20 at 16:13
Cesareo
5,7432412
answered Jul 20 at 16:13
Cesareo
5,7432412
5,7432412
add a comment |Â
add a comment |Â
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