How is it possible to minimize a definite integral, if there are three parameters that can be varied in the...
I have been attempting to minimize the following integral:
$$
T=int_{a}^{b}sqrt{frac{1+y'^2}{2gy}}dx,
$$
knowing that $y$ is of the form:
$$
y = a_0 +a_1 x+a_2x^2
$$
and that $y(0)=2$ and $y(pi)=0$. I don't know that it is possible to solve the integral analytically, and so I have been thinking that I should use the Euler-Lagrange differential equation (see this for details).
The trouble is, that I've found $a_2$ as a function of $x$, by recognizing $a_0=2$ and $a_1=-Big(frac{2}{pi}+pi a_2Big)$ and substituting everything into the Euler-Lagrange differential equation, which obviously doesn't make sense.
Is there any other way to solve such a problem? Am I just messing up a calculation somewhere? I'm working in Python, which has been making it difficult to keep track of variables, so it may be that the latter is in fact true.
I would appreciate very much any help in figuring this out.
functional-equations maxima-minima classical-mechanics euler-lagrange-equation
asked Nov 25 at 3:47
T. Zaborniak
11
|
show 1 more comment
I have been attempting to minimize the following integral:
$$
T=int_{a}^{b}sqrt{frac{1+y'^2}{2gy}}dx,
$$
knowing that $y$ is of the form:
$$
y = a_0 +a_1 x+a_2x^2
$$
and that $y(0)=2$ and $y(pi)=0$. I don't know that it is possible to solve the integral analytically, and so I have been thinking that I should use the Euler-Lagrange differential equation (see this for details).
The trouble is, that I've found $a_2$ as a function of $x$, by recognizing $a_0=2$ and $a_1=-Big(frac{2}{pi}+pi a_2Big)$ and substituting everything into the Euler-Lagrange differential equation, which obviously doesn't make sense.
Is there any other way to solve such a problem? Am I just messing up a calculation somewhere? I'm working in Python, which has been making it difficult to keep track of variables, so it may be that the latter is in fact true.
I would appreciate very much any help in figuring this out.
functional-equations maxima-minima classical-mechanics euler-lagrange-equation
asked Nov 25 at 3:47
T. Zaborniak
11
Could you tell what is $g$ ?
– Claude Leibovici
Nov 25 at 4:30
The Euler-Lagrange equation is useful if your unknown is a function (i.e. if you're doing variational calculus). Here your unknowns are three scalars. What you need to do instead is compute the derivatives $dT/da_i$ using differentiation under the integral sign and set them to zero.
– Rahul
Nov 25 at 4:31
1
This may also be too hard to find an analytical solution, and you may have to resort to numerical methods.
– Rahul
Nov 25 at 4:33
@ClaudeLeibovici: $g$ is in this case a constant.
– T. Zaborniak
Nov 25 at 4:45
@Rahul: so because there is a single unknown scalar after accounting for initial conditions, what you're saying is that I can differentiate with respect to $a_2$ (in this case), and set the resultant intergrand to zero, to solve for $a_2$?
– T. Zaborniak
Nov 25 at 4:48
|
show 1 more comment
I have been attempting to minimize the following integral:
$$
T=int_{a}^{b}sqrt{frac{1+y'^2}{2gy}}dx,
$$
knowing that $y$ is of the form:
$$
y = a_0 +a_1 x+a_2x^2
$$
and that $y(0)=2$ and $y(pi)=0$. I don't know that it is possible to solve the integral analytically, and so I have been thinking that I should use the Euler-Lagrange differential equation (see this for details).
The trouble is, that I've found $a_2$ as a function of $x$, by recognizing $a_0=2$ and $a_1=-Big(frac{2}{pi}+pi a_2Big)$ and substituting everything into the Euler-Lagrange differential equation, which obviously doesn't make sense.
Is there any other way to solve such a problem? Am I just messing up a calculation somewhere? I'm working in Python, which has been making it difficult to keep track of variables, so it may be that the latter is in fact true.
I would appreciate very much any help in figuring this out.
functional-equations maxima-minima classical-mechanics euler-lagrange-equation
asked Nov 25 at 3:47
T. Zaborniak
11
I have been attempting to minimize the following integral:
$$
T=int_{a}^{b}sqrt{frac{1+y'^2}{2gy}}dx,
$$
knowing that $y$ is of the form:
$$
y = a_0 +a_1 x+a_2x^2
$$
and that $y(0)=2$ and $y(pi)=0$. I don't know that it is possible to solve the integral analytically, and so I have been thinking that I should use the Euler-Lagrange differential equation (see this for details).
The trouble is, that I've found $a_2$ as a function of $x$, by recognizing $a_0=2$ and $a_1=-Big(frac{2}{pi}+pi a_2Big)$ and substituting everything into the Euler-Lagrange differential equation, which obviously doesn't make sense.
Is there any other way to solve such a problem? Am I just messing up a calculation somewhere? I'm working in Python, which has been making it difficult to keep track of variables, so it may be that the latter is in fact true.
I would appreciate very much any help in figuring this out.
functional-equations maxima-minima classical-mechanics euler-lagrange-equation
functional-equations maxima-minima classical-mechanics euler-lagrange-equation
asked Nov 25 at 3:47
T. Zaborniak
11
asked Nov 25 at 3:47
T. Zaborniak
11
edited Nov 25 at 4:10
asked Nov 25 at 3:47
T. Zaborniak
11
asked Nov 25 at 3:47
T. Zaborniak
11
asked Nov 25 at 3:47
T. Zaborniak
11
11
Could you tell what is $g$ ?
– Claude Leibovici
Nov 25 at 4:30
The Euler-Lagrange equation is useful if your unknown is a function (i.e. if you're doing variational calculus). Here your unknowns are three scalars. What you need to do instead is compute the derivatives $dT/da_i$ using differentiation under the integral sign and set them to zero.
– Rahul
Nov 25 at 4:31
1
This may also be too hard to find an analytical solution, and you may have to resort to numerical methods.
– Rahul
Nov 25 at 4:33
@ClaudeLeibovici: $g$ is in this case a constant.
– T. Zaborniak
Nov 25 at 4:45
@Rahul: so because there is a single unknown scalar after accounting for initial conditions, what you're saying is that I can differentiate with respect to $a_2$ (in this case), and set the resultant intergrand to zero, to solve for $a_2$?
– T. Zaborniak
Nov 25 at 4:48
|
show 1 more comment
Could you tell what is $g$ ?
– Claude Leibovici
Nov 25 at 4:30
The Euler-Lagrange equation is useful if your unknown is a function (i.e. if you're doing variational calculus). Here your unknowns are three scalars. What you need to do instead is compute the derivatives $dT/da_i$ using differentiation under the integral sign and set them to zero.
– Rahul
Nov 25 at 4:31
1
This may also be too hard to find an analytical solution, and you may have to resort to numerical methods.
– Rahul
Nov 25 at 4:33
@ClaudeLeibovici: $g$ is in this case a constant.
– T. Zaborniak
Nov 25 at 4:45
@Rahul: so because there is a single unknown scalar after accounting for initial conditions, what you're saying is that I can differentiate with respect to $a_2$ (in this case), and set the resultant intergrand to zero, to solve for $a_2$?
– T. Zaborniak
Nov 25 at 4:48
Could you tell what is $g$ ?
– Claude Leibovici
Nov 25 at 4:30
Could you tell what is $g$ ?
– Claude Leibovici
Nov 25 at 4:30
The Euler-Lagrange equation is useful if your unknown is a function (i.e. if you're doing variational calculus). Here your unknowns are three scalars. What you need to do instead is compute the derivatives $dT/da_i$ using differentiation under the integral sign and set them to zero.
– Rahul
Nov 25 at 4:31
The Euler-Lagrange equation is useful if your unknown is a function (i.e. if you're doing variational calculus). Here your unknowns are three scalars. What you need to do instead is compute the derivatives $dT/da_i$ using differentiation under the integral sign and set them to zero.
– Rahul
Nov 25 at 4:31
1
1
This may also be too hard to find an analytical solution, and you may have to resort to numerical methods.
– Rahul
Nov 25 at 4:33
This may also be too hard to find an analytical solution, and you may have to resort to numerical methods.
– Rahul
Nov 25 at 4:33
@ClaudeLeibovici: $g$ is in this case a constant.
– T. Zaborniak
Nov 25 at 4:45
@ClaudeLeibovici: $g$ is in this case a constant.
– T. Zaborniak
Nov 25 at 4:45
@Rahul: so because there is a single unknown scalar after accounting for initial conditions, what you're saying is that I can differentiate with respect to $a_2$ (in this case), and set the resultant intergrand to zero, to solve for $a_2$?
– T. Zaborniak
Nov 25 at 4:48
@Rahul: so because there is a single unknown scalar after accounting for initial conditions, what you're saying is that I can differentiate with respect to $a_2$ (in this case), and set the resultant intergrand to zero, to solve for $a_2$?
– T. Zaborniak
Nov 25 at 4:48
|
show 1 more comment
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Could you tell what is $g$ ?
– Claude Leibovici
Nov 25 at 4:30
The Euler-Lagrange equation is useful if your unknown is a function (i.e. if you're doing variational calculus). Here your unknowns are three scalars. What you need to do instead is compute the derivatives $dT/da_i$ using differentiation under the integral sign and set them to zero.
– Rahul
Nov 25 at 4:31
1
This may also be too hard to find an analytical solution, and you may have to resort to numerical methods.
– Rahul
Nov 25 at 4:33
@ClaudeLeibovici: $g$ is in this case a constant.
– T. Zaborniak
Nov 25 at 4:45
@Rahul: so because there is a single unknown scalar after accounting for initial conditions, what you're saying is that I can differentiate with respect to $a_2$ (in this case), and set the resultant intergrand to zero, to solve for $a_2$?
– T. Zaborniak
Nov 25 at 4:48