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Michael Hardy
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Out of curiosity, I was wondering why in the analysis of elliptic PDEs with "source" in $H^{-1}(\Omega)$ everyone writes the right-hand side in the form $$g_0- \mathrm{div} g$$$$g_0- \operatorname{div} g$$ with $g_0,g\in L^2(\Omega)$ instead of the (equivalent but apparently) more precise $$f-\mathrm{div} (\nabla f)$$$$f-\operatorname{div} (\nabla f)$$ with $f$ in $H_0^1(\Omega)$.

Any real advantage behind the first choice?

Out of curiosity, I was wondering why in the analysis of elliptic PDEs with "source" in $H^{-1}(\Omega)$ everyone writes the right-hand side in the form $$g_0- \mathrm{div} g$$ with $g_0,g\in L^2(\Omega)$ instead of the (equivalent but apparently) more precise $$f-\mathrm{div} (\nabla f)$$ with $f$ in $H_0^1(\Omega)$.

Any real advantage behind the first choice?

Out of curiosity, I was wondering why in the analysis of elliptic PDEs with "source" in $H^{-1}(\Omega)$ everyone writes the right-hand side in the form $$g_0- \operatorname{div} g$$ with $g_0,g\in L^2(\Omega)$ instead of the (equivalent but apparently) more precise $$f-\operatorname{div} (\nabla f)$$ with $f$ in $H_0^1(\Omega)$.

Any real advantage behind the first choice?

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Kosh
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Elementary curiosity on the dual of $H^{-1}(\Omega)$

Out of curiosity, I was wondering why in the analysis of elliptic PDEs with "source" in $H^{-1}(\Omega)$ everyone writes the right-hand side in the form $$g_0- \mathrm{div} g$$ with $g_0,g\in L^2(\Omega)$ instead of the (equivalent but apparently) more precise $$f-\mathrm{div} (\nabla f)$$ with $f$ in $H_0^1(\Omega)$.

Any real advantage behind the first choice?