I'd say these belongs to the family of shape optimization problems in the Calculus of Variations (of whom the most famous example is the isoperimetric problem). As to the first one, you should make some assumption on $f$ to ensure existence (for e.g. $f(x):=\frac{1}{1+|x|^2}$ gives a non attained infimum, equal to 0). Natural assumptions are e.g.: $f$ is smooth, coercive, and its only critical point is its global minimum. This should give as minimizer a level set of $f$ -the last assumption is just to get a regular simple curve. (The latter problem seems less easy and should require at least some small computation. I guess the solution is a circle). PS: Say, what about u=1 any $\omega$ as minimizer, as $u=1$ constant in $\omega$ ? :-). -) The latter problem was a joke, wasn't it?
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I'd say these belongs to the family of shape optimization problems in the Calculus of Variations (of whom the most famous example is the isoperimetric problem). As to the first one, you should make some assumption on $f$ to ensure existence (for e.g. $f(x):=\frac{1}{1+|x|^2}$ gives a non attained infimum, equal to 0). Natural assumptions are e.g.: $f$ is smooth, coercive, and its only critical point is its global minimum. This should give as minimizer a level set of $f$ -the last assumption is just to get a regular simple curve. (The latter problem seems less easy and should require at least some small computation. I guess the solution is a circle). PS: Say, what about u=1 constant? :-). |
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I'd say these belongs to the family of shape optimization problems in the Calculus of Variations (of whom the most famous example is the isoperimetric problem). As to the first one, you should make some assumption on $f$ to ensure existence (for e.g. $f(x):=\frac{1}{1+|x|^2}$ gives a non attained infimum, equal to 0). Natural assumptions are e.g.: $f$ is smooth, coercive, and its only critical point is its global minimum. This should give as minimizer a level set of $f$ -the last assumption is just to get a regular simple curve. (The latter problem seems less easy and should require at least some small computation. I guess the solution is a circle). |
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