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Jul 22, 2013 at 16:52 history edited user9072
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May 10, 2013 at 10:29 answer added JCM timeline score: 1
Jul 19, 2012 at 16:36 comment added timur Yes, then you move the term $(L_p)'u_p$ to the right hand side and treat is as a source term.
Jul 19, 2012 at 16:04 comment added CuriousUser Can you expand a little bit on that? In particular: when you differentiate wrt $p$, how does the term $(L_pu_p)'$ behave? It feels like it becomes $(L_p)'u_p+ L_p(u_p')$ (here by "u'" i mean differentiation wrt $P$ of course). But then i don't see how the resulting equation is still parabolic...
Jul 19, 2012 at 13:29 comment added timur You differentiate (wrt $p$) the equation and derive an equation for the $p$-derivatives of $u$. This will be a parabolic equation so you are fine.
Jul 19, 2012 at 11:16 history asked CuriousUser CC BY-SA 3.0