Finding functional extremals
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So my problem is as follows: find the extremals of the functional
$$I[x_1(t), x_2(t)]=int_{0.5}^1(dot{x}_1^{2}-2x_{1}dot{x}_2t)dt,$$
given:
$$x_1(0.5)=2, x_2(0.5)=15, x_1(1)=1, x_{2}(1)=1.$$
calculus ordinary-differential-equations derivatives optimization calculus-of-variations
asked Dec 5 '18 at 19:35
mar3gmar3g
33
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add a comment |
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So my problem is as follows: find the extremals of the functional
$$I[x_1(t), x_2(t)]=int_{0.5}^1(dot{x}_1^{2}-2x_{1}dot{x}_2t)dt,$$
given:
$$x_1(0.5)=2, x_2(0.5)=15, x_1(1)=1, x_{2}(1)=1.$$
calculus ordinary-differential-equations derivatives optimization calculus-of-variations
asked Dec 5 '18 at 19:35
mar3gmar3g
33
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1
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Welcome to Math.SE! What have you tried so far? What about using the Euler-Lagrange equation? The first term inside the integral is $dot{x}_1^2$ or $dot{x}_2^2$?
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– rafa11111
Dec 5 '18 at 20:02
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@rafa11111 Corrected the integral. I tried using Euler-Lagrange equation, but I go stuck and don't know how to interprete the "dt" at the end of integral.
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– mar3g
Dec 5 '18 at 20:35
add a comment |
$begingroup$
So my problem is as follows: find the extremals of the functional
$$I[x_1(t), x_2(t)]=int_{0.5}^1(dot{x}_1^{2}-2x_{1}dot{x}_2t)dt,$$
given:
$$x_1(0.5)=2, x_2(0.5)=15, x_1(1)=1, x_{2}(1)=1.$$
calculus ordinary-differential-equations derivatives optimization calculus-of-variations
asked Dec 5 '18 at 19:35
mar3gmar3g
33
$endgroup$
So my problem is as follows: find the extremals of the functional
$$I[x_1(t), x_2(t)]=int_{0.5}^1(dot{x}_1^{2}-2x_{1}dot{x}_2t)dt,$$
given:
$$x_1(0.5)=2, x_2(0.5)=15, x_1(1)=1, x_{2}(1)=1.$$
calculus ordinary-differential-equations derivatives optimization calculus-of-variations
calculus ordinary-differential-equations derivatives optimization calculus-of-variations
asked Dec 5 '18 at 19:35
mar3gmar3g
33
asked Dec 5 '18 at 19:35
mar3gmar3g
33
edited Dec 5 '18 at 21:48
mar3g
asked Dec 5 '18 at 19:35
mar3gmar3g
33
asked Dec 5 '18 at 19:35
mar3gmar3g
33
asked Dec 5 '18 at 19:35
mar3gmar3g
33
33
1
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Welcome to Math.SE! What have you tried so far? What about using the Euler-Lagrange equation? The first term inside the integral is $dot{x}_1^2$ or $dot{x}_2^2$?
$endgroup$
– rafa11111
Dec 5 '18 at 20:02
$begingroup$
@rafa11111 Corrected the integral. I tried using Euler-Lagrange equation, but I go stuck and don't know how to interprete the "dt" at the end of integral.
$endgroup$
– mar3g
Dec 5 '18 at 20:35
add a comment |
1
$begingroup$
Welcome to Math.SE! What have you tried so far? What about using the Euler-Lagrange equation? The first term inside the integral is $dot{x}_1^2$ or $dot{x}_2^2$?
$endgroup$
– rafa11111
Dec 5 '18 at 20:02
$begingroup$
@rafa11111 Corrected the integral. I tried using Euler-Lagrange equation, but I go stuck and don't know how to interprete the "dt" at the end of integral.
$endgroup$
– mar3g
Dec 5 '18 at 20:35
1
1
$begingroup$
Welcome to Math.SE! What have you tried so far? What about using the Euler-Lagrange equation? The first term inside the integral is $dot{x}_1^2$ or $dot{x}_2^2$?
$endgroup$
– rafa11111
Dec 5 '18 at 20:02
$begingroup$
Welcome to Math.SE! What have you tried so far? What about using the Euler-Lagrange equation? The first term inside the integral is $dot{x}_1^2$ or $dot{x}_2^2$?
$endgroup$
– rafa11111
Dec 5 '18 at 20:02
$begingroup$
@rafa11111 Corrected the integral. I tried using Euler-Lagrange equation, but I go stuck and don't know how to interprete the "dt" at the end of integral.
$endgroup$
– mar3g
Dec 5 '18 at 20:35
$begingroup$
@rafa11111 Corrected the integral. I tried using Euler-Lagrange equation, but I go stuck and don't know how to interprete the "dt" at the end of integral.
$endgroup$
– mar3g
Dec 5 '18 at 20:35
add a comment |
1 Answer
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Your Lagrange function is
$$L(x_1,x_2,dot{x_1}, dot{x_2},t)=dot{x_1}^2-2x_1dot{x_2}$$
And the Euler-Lagrange equations are
$$frac{partial L}{partial x_1}-frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_1}}right)=0$$
$$frac{partial L}{partial x_2}-frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_2}}right)=0$$
Which will give you a system of differential equations. And your second Euler-Lagrange equation will be a bit easier, because the Lagrange function is independent of $x_2$:
$$frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_2}}right)=0$$
$$frac{partial L}{partial dot{x_2}}=text{constant}$$
answered Dec 5 '18 at 20:37
BotondBotond
5,9252832
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Comments are not for extended discussion; this conversation has been moved to chat.
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– Aloizio Macedo♦
Dec 5 '18 at 22:57
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Your Lagrange function is
$$L(x_1,x_2,dot{x_1}, dot{x_2},t)=dot{x_1}^2-2x_1dot{x_2}$$
And the Euler-Lagrange equations are
$$frac{partial L}{partial x_1}-frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_1}}right)=0$$
$$frac{partial L}{partial x_2}-frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_2}}right)=0$$
Which will give you a system of differential equations. And your second Euler-Lagrange equation will be a bit easier, because the Lagrange function is independent of $x_2$:
$$frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_2}}right)=0$$
$$frac{partial L}{partial dot{x_2}}=text{constant}$$
answered Dec 5 '18 at 20:37
BotondBotond
5,9252832
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Comments are not for extended discussion; this conversation has been moved to chat.
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– Aloizio Macedo♦
Dec 5 '18 at 22:57
add a comment |
$begingroup$
Your Lagrange function is
$$L(x_1,x_2,dot{x_1}, dot{x_2},t)=dot{x_1}^2-2x_1dot{x_2}$$
And the Euler-Lagrange equations are
$$frac{partial L}{partial x_1}-frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_1}}right)=0$$
$$frac{partial L}{partial x_2}-frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_2}}right)=0$$
Which will give you a system of differential equations. And your second Euler-Lagrange equation will be a bit easier, because the Lagrange function is independent of $x_2$:
$$frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_2}}right)=0$$
$$frac{partial L}{partial dot{x_2}}=text{constant}$$
answered Dec 5 '18 at 20:37
BotondBotond
5,9252832
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 5 '18 at 22:57
add a comment |
$begingroup$
Your Lagrange function is
$$L(x_1,x_2,dot{x_1}, dot{x_2},t)=dot{x_1}^2-2x_1dot{x_2}$$
And the Euler-Lagrange equations are
$$frac{partial L}{partial x_1}-frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_1}}right)=0$$
$$frac{partial L}{partial x_2}-frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_2}}right)=0$$
Which will give you a system of differential equations. And your second Euler-Lagrange equation will be a bit easier, because the Lagrange function is independent of $x_2$:
$$frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_2}}right)=0$$
$$frac{partial L}{partial dot{x_2}}=text{constant}$$
answered Dec 5 '18 at 20:37
BotondBotond
5,9252832
$endgroup$
Your Lagrange function is
$$L(x_1,x_2,dot{x_1}, dot{x_2},t)=dot{x_1}^2-2x_1dot{x_2}$$
And the Euler-Lagrange equations are
$$frac{partial L}{partial x_1}-frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_1}}right)=0$$
$$frac{partial L}{partial x_2}-frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_2}}right)=0$$
Which will give you a system of differential equations. And your second Euler-Lagrange equation will be a bit easier, because the Lagrange function is independent of $x_2$:
$$frac{mathrm{d}}{mathrm{d}t}left(frac{partial L}{partial dot{x_2}}right)=0$$
$$frac{partial L}{partial dot{x_2}}=text{constant}$$
answered Dec 5 '18 at 20:37
BotondBotond
5,9252832
answered Dec 5 '18 at 20:37
BotondBotond
5,9252832
answered Dec 5 '18 at 20:37
BotondBotond
5,9252832
answered Dec 5 '18 at 20:37
BotondBotond
5,9252832
5,9252832
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Comments are not for extended discussion; this conversation has been moved to chat.
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– Aloizio Macedo♦
Dec 5 '18 at 22:57
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 5 '18 at 22:57
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 5 '18 at 22:57
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 5 '18 at 22:57
add a comment |
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Welcome to Math.SE! What have you tried so far? What about using the Euler-Lagrange equation? The first term inside the integral is $dot{x}_1^2$ or $dot{x}_2^2$?
$endgroup$
– rafa11111
Dec 5 '18 at 20:02
$begingroup$
@rafa11111 Corrected the integral. I tried using Euler-Lagrange equation, but I go stuck and don't know how to interprete the "dt" at the end of integral.
$endgroup$
– mar3g
Dec 5 '18 at 20:35