APMA 2811Q. Homework #1. Due: 9/25/13. 1 exp ( f (x) 2) dx, I[f] =

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1 APMA 8Q Homework # Due: 9/5/3. Ill-posed problems a) Consider I : W,, ) R defined by exp f x) ) dx, where W,, ) = f W,, ) : f) = f) = }. Show that I has no minimizer in A. This problem is not coercive or convex). Proof. Clearly, from strict positivity of the function gx) = exp x ) it follows that f A that I[f] >. Now consider the sequence of functions defined by nx, < x < nx + n, < x <. Calculating it follows that I[f n ] = e n dx = e n and hence lim n I[f n] =. Therefore, f A there exists N N such that I[f] > I[f N ] > proving there is no minimum in A. Remark: Notice that the minimizing sequence need not converge to anything. In fact, this is to be expected since the problem is neither convex nor coercive. b) Consider I : A R defined by xf x) dx, where A = f W,, ) : f) =, f) = }. Show that I has no minimizer in A. This problem shows that lack of coercivity at one point is enough to guarantee non-existence of a minimum). Proof. Clearly f A, I[f]. Let f n be defined by < x < n lnx) lnn) n < x <. Therefore, I[f n ] = n x lnn) dx = lnn) and consequently lim n I[f n] =. Now, suppose f A such that. Then f x) = a.e. which is not compatible with the boundary conditions.

2 Remark: The construction of the minimizing sequence is not trivial. The reason is if you try to confine the derivative to a small region then from dimensional analysis the value of I in this region will scale like an O) quantity. Instead, what I did was concentrate fx) = into a small region and then spread out the derivative over the interval, ). c) Consider I : A R defined by f x) dx, where A = f W,, ) : f) =, f) = }. Prove that minimizers of I are not unique. You first need to find a potential minimizer and prove that it is indeed a minimizer). Proof. From the Fundamental Theorem of Calculus it follows that f A, I[f]. The lower bound is obtained by any smooth monotone increasing function and hence the minimizer is not unique. d) Consider I : A R defined by x f x)) fx) dx, where A = f C, ) : f ) =, f) = }. Show that I has no minimum in A. What is the correct admissible set we should have considered this problem in? Proof. Clearly, I[f]. Moreover, this lower bound is obtained by the non-smooth function f defined by < x < fx) = x. < x < To obtain a minimizing sequence take any function f n A satisfying f n f strongly in W,, ). Now, suppose g A such that I[g] = a.e.. Then for every x in, ) we have that g x) = x or gx) = which for our boundary conditions cannot be satisfied by a smooth function. The correct space we should have considered is W,, ).. Euler-Lagrange Equations a) Consider I : A R defined by f x) ) + ɛ f x) dx, where A = W,, ) = f W,, ) : f) = f) = f ) = f ) = }. Determine the Euler-Lagrange equations for this functional. Find at least one solution to this equation and show that it cannot be a minimum for all values of ɛ. This is an example of a bifurcation). Proof. The formal calculation yields δ = 4 f x) )f x)δfx)) dx + ɛ f x)δfx)) dx 4 f x) ) ) f x) + ɛ f iv) x) δfx) dx.

3 Consequently, the Euler-Lagrange equations are ɛ f iv) x) + d [ f x) ) f x) ] =. dx One obvious solution is the function f x) =. To show that this cannot be a minimizer for all values of ɛ we will rewrite the functional as I [f] + ɛ I [f]. f x) minimizes I [f] alone. For large values of ɛ where I is dominate over I we expect f to be a minimizer. As ɛ decreases I dominates over I and we expect the minimizer to look something like f x) = x +. However, f does not satisfy our boundary conditions and is too rough. We need to smooth out the corners of the function. For simplicity we will only smooth out near x = and argue from symmetry. Define, x g w x) = w < x < w w < x < Then I[g w ] w + ɛ w. Minimizing over the choice of w we find that w = ɛ. Now, there are three corners so we get the upper bound that inf I[f] 3 + ) ɛ. f A Consequently, for ɛ small enough it follows that f cannot be a minimizer since I[f ] =. b) Consider I : A R defined as above with A = W,, ). Determine the natural boundary conditions that must be satisfied by a smooth minimizer of this functional. Proof. The formal calculation yields δ = 4 f x) )f x)δfx)) dx + ɛ f x)δfx)) dx 4 d dx [ f x) ) f x) ] ) + ɛ f iv) x) δfx) dx + 4 f x) ) f x)δfx) + ɛ f x) δfx)) ɛ f iii) x)δfx) Consequently, the natural boundary conditions are: 4 f ) ) f ) = f iii) ) 4 f ) ) f ) = f iii) ) f ) = f ) =. 3

4 .3 Weak-Convergence a) Prove that if p < and u n u in L p [, ]), v n v in L q [, ]) with p + q = then u n v n uv in L [, ]). Proof. Let g L [, ]). Then, u n v n uv)g dx = gu n v n v) dx + gvu n u) dx g u n v n v) dx + gvu n u) dx g L M v n v L q + gvu n u) dx, where M = sup n u n L q < by boundedness of weakly convergent sequences. Since gv L [, ]) the result follows from taking the limit. b) Prove that if u n u in L [, ]) and u n u in L [, ]) then u n u inl [, ]). Proof. u n u L = u n L u n u dx + u L. Since L [, ]) it follows that u n L u L. The results thus follows from taking the limit. c) Prove that for p the unit ball in L p [, ]) is not strongly compact. Proof. Let sinπnx). Clearly for all n, f n L p. Moreover, f n L = π and since the L p norms are monotone increasing in p it follows for all p that f n L p π. Now, for q satisfying q + p = it follows for all g Lq L that f n x)gx) dx = a n, where a n are the coefficients in the sine Fourier series of g. Hence, a n and consequently if f n has a strongly convergent subsequence it must converge to zero. However, from the bounds above this is a contradiction. d) Give an example of a bounded sequence in L [, ]) that does not have a weakly convergent subsequence. Proof. The delta sequence f n defined by n n < x < + n o.w. does not weakly converge to an L function. e) Find a sequence of functions f n with the property that f n in L [, ]), f n in L 3 [, ]) but f n does not converge strongly in L [, ]). 4

5 Proof. The sequence of functions defined by n n < x < + n o.w. works. 5