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मान लीजिए B(0,1) = {(x,y) ∈ ℝ2|x2 + y2 < 1} ℝ2 में खुला इकाई डिस्क है, ∂B(0, 1) B(0,1) की सीमा को दर्शाता है, और v ∂B(0, 1) के लिए इकाई बाह्य अभिलम्ब है। मान लीजिए f : ℝ2 → ℝ एक दिया गया सतत फलन है। न्यूनीकरण समस्या

\(\rm min \left\{\frac{1}{2}\iint_{B(0,1)}|\nabla u|^2dxdy+\frac{1}{2}\iint_{B(0, 1)}e^{u^2}dxdy+∈t_{\partial B(0, 1)}fuds\right\}\)

जिसमें u ∈ C1 \(\rm \overline{B(0, 1)}\) है, का ऑयलर-लैग्रेंज समीकरण है

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CSIR-UGC (NET) Mathematical Science: Held on (2024 June)
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  1. \(\rm \left\{\begin{matrix}\Delta u=-ue^{u^2}&\rm in \ B(0, 1)\\\ \frac{\partial u}{\partial \nu}=f&\rm on\ \partial B(0, 1\end{matrix}\right.\)
  2. \(\rm \left\{\begin{matrix}\Delta u=ue^{u^2}+f&\rm in \ B(0, 1)\\\ u=0&\rm on\ \partial B(0, 1\end{matrix}\right.\)
  3. \(\rm \left\{\begin{matrix}\Delta u=ue^{u^2}&\rm in \ B(0, 1)\\\ \frac{\partial u}{\partial \nu}=-f&\rm on\ \partial B(0, 1\end{matrix}\right.\)
  4. \(\rm \left\{\begin{matrix}\Delta u=ue^{u^2}&\rm in \ B(0, 1)\\\ \frac{\partial u}{\partial \nu}+u=f&\rm on\ \partial B(0, 1\end{matrix}\right.\)

Answer (Detailed Solution Below)

Option 3 : \(\rm \left\{\begin{matrix}\Delta u=ue^{u^2}&\rm in \ B(0, 1)\\\ \frac{\partial u}{\partial \nu}=-f&\rm on\ \partial B(0, 1\end{matrix}\right.\)
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सही उत्तर (3) है।

हम बाद में हल अपडेट करेंगे।
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