Timeline for Operator on a Sobolev space
Current License: CC BY-SA 3.0
4 events
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| Jul 7, 2015 at 20:49 | history | edited | Upin | CC BY-SA 3.0 |
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| Jul 7, 2015 at 20:33 | comment | added | Upin | Yes, that is one of the whole points of using weak formulation: the solution is defined in a weaker way. Consider $-\Delta u = f$. A classical solution needs $u$ to be twice differentiable. But a weak formulation of this problem may be $\int_\Omega \nabla u \nabla v = \int_\Omega fv$ (holding for all $v \in H^1_0$) and this only needs one space derivative. You may ask: why is such a notion of solution good enough? For some answers see this thread. | |
| Jul 7, 2015 at 18:35 | comment | added | user75795 | Thanks. So, every operator $L$ in this form, BY DEFINITION, act in a particular way that doesn't count second derivaties? | |
| Jul 7, 2015 at 17:51 | history | answered | Upin | CC BY-SA 3.0 |