Calculate the Hopf-Lax function explicitly
$begingroup$
The Cauchy problem I'm dealing with leads to the following Hopf-Lax function:
$$u(t,x) = inf_{yin mathbb{R}} left{ 2|y|^2 + frac{|x-y|^2}{2t} right},$$
but I don't know how to explicitly calculate this infimum. Could any solutions be step by step, please.
pde supremum-and-infimum cauchy-problem hamilton-jacobi-equation
edited Dec 11 '18 at 13:27
Harry49
6,21331132
$endgroup$
add a comment |
$begingroup$
The Cauchy problem I'm dealing with leads to the following Hopf-Lax function:
$$u(t,x) = inf_{yin mathbb{R}} left{ 2|y|^2 + frac{|x-y|^2}{2t} right},$$
but I don't know how to explicitly calculate this infimum. Could any solutions be step by step, please.
pde supremum-and-infimum cauchy-problem hamilton-jacobi-equation
edited Dec 11 '18 at 13:27
Harry49
6,21331132
$endgroup$
add a comment |
$begingroup$
The Cauchy problem I'm dealing with leads to the following Hopf-Lax function:
$$u(t,x) = inf_{yin mathbb{R}} left{ 2|y|^2 + frac{|x-y|^2}{2t} right},$$
but I don't know how to explicitly calculate this infimum. Could any solutions be step by step, please.
pde supremum-and-infimum cauchy-problem hamilton-jacobi-equation
edited Dec 11 '18 at 13:27
Harry49
6,21331132
$endgroup$
The Cauchy problem I'm dealing with leads to the following Hopf-Lax function:
$$u(t,x) = inf_{yin mathbb{R}} left{ 2|y|^2 + frac{|x-y|^2}{2t} right},$$
but I don't know how to explicitly calculate this infimum. Could any solutions be step by step, please.
pde supremum-and-infimum cauchy-problem hamilton-jacobi-equation
pde supremum-and-infimum cauchy-problem hamilton-jacobi-equation
edited Dec 11 '18 at 13:27
Harry49
6,21331132
edited Dec 11 '18 at 13:27
Harry49
6,21331132
edited Dec 11 '18 at 13:27
Harry49
6,21331132
edited Dec 11 '18 at 13:27
Harry49
6,21331132
edited Dec 11 '18 at 13:27
Harry49
6,21331132
6,21331132
asked May 4 '17 at 18:15
user383264
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1 Answer
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You don't really need the absolute values if everything is real, so you're looking for the minimum of a quadratic in $y$. As long as $2 + 1/(2t) > 0$, the $y^2$ term in this quadratic is positive, so the critical point is a minimum. You can find it by the usual calculus method: take the derivative, set it to $0$, and solve. But if $2 + 1/(2t) < 0$ (or $=0$ with $x ne 0$), the infimum will be $-infty$.
answered May 4 '17 at 18:21
Robert IsraelRobert Israel
322k23212465
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You don't really need the absolute values if everything is real, so you're looking for the minimum of a quadratic in $y$. As long as $2 + 1/(2t) > 0$, the $y^2$ term in this quadratic is positive, so the critical point is a minimum. You can find it by the usual calculus method: take the derivative, set it to $0$, and solve. But if $2 + 1/(2t) < 0$ (or $=0$ with $x ne 0$), the infimum will be $-infty$.
answered May 4 '17 at 18:21
Robert IsraelRobert Israel
322k23212465
$endgroup$
add a comment |
$begingroup$
You don't really need the absolute values if everything is real, so you're looking for the minimum of a quadratic in $y$. As long as $2 + 1/(2t) > 0$, the $y^2$ term in this quadratic is positive, so the critical point is a minimum. You can find it by the usual calculus method: take the derivative, set it to $0$, and solve. But if $2 + 1/(2t) < 0$ (or $=0$ with $x ne 0$), the infimum will be $-infty$.
answered May 4 '17 at 18:21
Robert IsraelRobert Israel
322k23212465
$endgroup$
add a comment |
$begingroup$
You don't really need the absolute values if everything is real, so you're looking for the minimum of a quadratic in $y$. As long as $2 + 1/(2t) > 0$, the $y^2$ term in this quadratic is positive, so the critical point is a minimum. You can find it by the usual calculus method: take the derivative, set it to $0$, and solve. But if $2 + 1/(2t) < 0$ (or $=0$ with $x ne 0$), the infimum will be $-infty$.
answered May 4 '17 at 18:21
Robert IsraelRobert Israel
322k23212465
$endgroup$
You don't really need the absolute values if everything is real, so you're looking for the minimum of a quadratic in $y$. As long as $2 + 1/(2t) > 0$, the $y^2$ term in this quadratic is positive, so the critical point is a minimum. You can find it by the usual calculus method: take the derivative, set it to $0$, and solve. But if $2 + 1/(2t) < 0$ (or $=0$ with $x ne 0$), the infimum will be $-infty$.
answered May 4 '17 at 18:21
Robert IsraelRobert Israel
322k23212465
answered May 4 '17 at 18:21
Robert IsraelRobert Israel
322k23212465
answered May 4 '17 at 18:21
Robert IsraelRobert Israel
322k23212465
answered May 4 '17 at 18:21
Robert IsraelRobert Israel
322k23212465
322k23212465
add a comment |
add a comment |
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