Extremum of functional via functional derivative
$begingroup$
I have the following functional:
$$
H[lambda(T)]
= intfrac{1}{2}left | nabla m_{0}hleft(frac{x}{lambda(T)}right) right |^{2}+frac{1}{2}r_{0}m^{2}_{0}h^{2}left(frac{x}{lambda(T)}right)+frac{1}{4!}u_{0}m^{4}_{0}h^{4}left(frac{x}{lambda(T)}right)-B(x)m_{0}hleft(frac{x}{lambda(T)}right)dx,$$
with $r_{0},m_{0},u_{0}$ being real constants and x a real variable. B(x) is some real valued function. And I should find the $lambda(T)$ that minimizes that functional. To start I would compute the functional derivative with respect to $lambda(T)$ and set this to zero:
$$
int m_{0}nabla^{2} hleft(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+r_{0}m_{0}hleft(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}-frac{1}{6}u_{0}m^{3}_{0}h^{3}left(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+B(x)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}=0.
$$
In this equation the $h^{'}left(frac{x}{lambda(T)}right)$ denotes the first derivative of $hleft(frac{x}{lambda(T)}right)$ with respect to its argument. Now I am not completely sure if I did the functional derivatives correctly.And otherwise I would not know how to solve this equation for $lambda (T)$ since all the explicit $lambda$ would cancel.
functional-analysis derivatives optimization
asked Nov 26 '18 at 12:47
zodiaczodiac
617
$endgroup$
add a comment |
$begingroup$
I have the following functional:
$$
H[lambda(T)]
= intfrac{1}{2}left | nabla m_{0}hleft(frac{x}{lambda(T)}right) right |^{2}+frac{1}{2}r_{0}m^{2}_{0}h^{2}left(frac{x}{lambda(T)}right)+frac{1}{4!}u_{0}m^{4}_{0}h^{4}left(frac{x}{lambda(T)}right)-B(x)m_{0}hleft(frac{x}{lambda(T)}right)dx,$$
with $r_{0},m_{0},u_{0}$ being real constants and x a real variable. B(x) is some real valued function. And I should find the $lambda(T)$ that minimizes that functional. To start I would compute the functional derivative with respect to $lambda(T)$ and set this to zero:
$$
int m_{0}nabla^{2} hleft(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+r_{0}m_{0}hleft(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}-frac{1}{6}u_{0}m^{3}_{0}h^{3}left(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+B(x)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}=0.
$$
In this equation the $h^{'}left(frac{x}{lambda(T)}right)$ denotes the first derivative of $hleft(frac{x}{lambda(T)}right)$ with respect to its argument. Now I am not completely sure if I did the functional derivatives correctly.And otherwise I would not know how to solve this equation for $lambda (T)$ since all the explicit $lambda$ would cancel.
functional-analysis derivatives optimization
asked Nov 26 '18 at 12:47
zodiaczodiac
617
$endgroup$
add a comment |
$begingroup$
I have the following functional:
$$
H[lambda(T)]
= intfrac{1}{2}left | nabla m_{0}hleft(frac{x}{lambda(T)}right) right |^{2}+frac{1}{2}r_{0}m^{2}_{0}h^{2}left(frac{x}{lambda(T)}right)+frac{1}{4!}u_{0}m^{4}_{0}h^{4}left(frac{x}{lambda(T)}right)-B(x)m_{0}hleft(frac{x}{lambda(T)}right)dx,$$
with $r_{0},m_{0},u_{0}$ being real constants and x a real variable. B(x) is some real valued function. And I should find the $lambda(T)$ that minimizes that functional. To start I would compute the functional derivative with respect to $lambda(T)$ and set this to zero:
$$
int m_{0}nabla^{2} hleft(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+r_{0}m_{0}hleft(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}-frac{1}{6}u_{0}m^{3}_{0}h^{3}left(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+B(x)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}=0.
$$
In this equation the $h^{'}left(frac{x}{lambda(T)}right)$ denotes the first derivative of $hleft(frac{x}{lambda(T)}right)$ with respect to its argument. Now I am not completely sure if I did the functional derivatives correctly.And otherwise I would not know how to solve this equation for $lambda (T)$ since all the explicit $lambda$ would cancel.
functional-analysis derivatives optimization
asked Nov 26 '18 at 12:47
zodiaczodiac
617
$endgroup$
I have the following functional:
$$
H[lambda(T)]
= intfrac{1}{2}left | nabla m_{0}hleft(frac{x}{lambda(T)}right) right |^{2}+frac{1}{2}r_{0}m^{2}_{0}h^{2}left(frac{x}{lambda(T)}right)+frac{1}{4!}u_{0}m^{4}_{0}h^{4}left(frac{x}{lambda(T)}right)-B(x)m_{0}hleft(frac{x}{lambda(T)}right)dx,$$
with $r_{0},m_{0},u_{0}$ being real constants and x a real variable. B(x) is some real valued function. And I should find the $lambda(T)$ that minimizes that functional. To start I would compute the functional derivative with respect to $lambda(T)$ and set this to zero:
$$
int m_{0}nabla^{2} hleft(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+r_{0}m_{0}hleft(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}-frac{1}{6}u_{0}m^{3}_{0}h^{3}left(frac{x}{lambda(T)}right)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}+B(x)h^{'}left(frac{x}{lambda(T)}right)frac{x}{lambda^{2}(T)}=0.
$$
In this equation the $h^{'}left(frac{x}{lambda(T)}right)$ denotes the first derivative of $hleft(frac{x}{lambda(T)}right)$ with respect to its argument. Now I am not completely sure if I did the functional derivatives correctly.And otherwise I would not know how to solve this equation for $lambda (T)$ since all the explicit $lambda$ would cancel.
functional-analysis derivatives optimization
functional-analysis derivatives optimization
asked Nov 26 '18 at 12:47
zodiaczodiac
617
asked Nov 26 '18 at 12:47
zodiaczodiac
617
asked Nov 26 '18 at 12:47
zodiaczodiac
617
asked Nov 26 '18 at 12:47
zodiaczodiac
617
asked Nov 26 '18 at 12:47
zodiaczodiac
617
617
add a comment |
add a comment |
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