13 - Solving PDEs

# 13 - Solving PDEs - MATH 220: SOLVING PDES We now return to...

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MATH 220: SOLVING PDES We now return to solving PDE using duality arguments and energy estimates. Before getting into details, we note that the ideal kind of well-posedness result we would like is the following. We are given a PDE (including various additional conditions), and we would like to have spaces of functions X and Y such that there is a unique solution of the PDE in X which depends continuously on the data in Y . To be speciﬁc, let’s consider the linear PDE Pu = f , with any other condition we may want to have made homogeneous (e.g. homogeneous Dirichlet BC) – which we have seen can be done – and we want to ﬁnd function spaces X and Y , which should have a notion of convergence, e.g. be normed spaces, such that (i) (Existence) For any f Y there exists u X such that Pu = f . (ii) (Uniqueness) For any f Y there is at most one u X such that Pu = f . (iii) (Stability) If f j f in Y then the unique solutions, u j , resp. u , in X , of Pu j = f j , Pu = f , satisfy u j u in X . Of course, Pu needs to have some meaning, so a minimal requirement is that X and Y are both subspaces of D 0 (Ω) for some Ω – e.g. Ω = R n , or Ω a bounded domain in R n , or Ω = T n = ( S 1 ) n , C functions on which, recall, can be identiﬁed with 2 π -periodic C functions in every coordinate on R n . Also, to make this problem meaningful, Y should be relatively large – typically it should include C c (Ω) at least. (If P is linear and Y = { 0 } , it is trivial to ﬁnd an X : X = { 0 } will do.) Now, to get a feeling for what this means, note that if we can solve a PDE in a space X (let’s keep Y ﬁxed for now), then we can also solve it in a bigger space ˜ X X , since the solution u in X also lies in ˜ X . On the other hand, if there is at most one solution of the PDE in a space X , then there is at most one solution in a smaller space ˜ X X since if u 1 and u 2 are solutions in ˜ X then they are solutions in X , thus by uniqueness in X they are equal. Thus, there is some tension between existence and uniqueness: for the former, one could try to increase the space to make the problem easier, but then may lose uniqueness, for the latter one may try to make the space smaller, but then may lose existence. To be extreme, it may be easy to show uniqueness if X = C c (Ω) (e.g. using the maximum principle or energy estimates), but existence may not hold, while it may be relatively easy to show existence when X = D 0 (Ω), but uniqueness may not hold then. As a concrete example, if Y = C c ( R ), P = d dx , then with X = C c ( R ) one has at most one solution of Pu = f , f Y (since by the fundamental theorem of calculus one can write u as the indeﬁnite integral of f from -∞ ), but may not have any solutions (this indeﬁnite integral may not vanish for large x ), while if X = D 0 ( R ), or indeed if X = C ( R ), there is always a solution (the aforementioned indeﬁnite integral), but it is not unique (one can always add a constant). If however one lets

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## This note was uploaded on 03/16/2010 for the course CME 303 taught by Professor Vasy during the Fall '10 term at Stanford.

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13 - Solving PDEs - MATH 220: SOLVING PDES We now return to...

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