M545 Homework 8. Mark Blumstein. December 9, [f (x)] 2 dx = 1. Find such a function if you can. If it cannot be found, explain why not.

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1 M545 Homework 8 Mark Blumstein ecember 9, 5..) Let fx) be a function such that f) f3), 3 [fx)] dx, and 3 [f x)] dx. Find such a function if you can. If it cannot be found, explain why not. Proof. We will use Theorem, the minimum principle for the first eigenvalue. That is, given the eigenvalue problem u λu and homogeneous irichlet boundary condition, the first eigenvalue is given by the minimum of the Rayleigh quotient w w where w is any sufficiently smooth function satisfying the boundary condition. For this problem, we have the one dimensional case. Let s begin by searching among all functions such that 3 wx) dx. Notice that the Rayleigh quotient becomes, 3 w x) 3 dx 3 wx) dx w x) dx 3 w x) dx. So by the minimum principle for the first eigenvalue, the minimum of this quotient is given exactly by the first eigenvalue for the eigenvalue problem u x) λux). This is an ordinary differential equation we ve solved many times before, and we know its eigenvalues are given by ) nπ 3 with corresponding eigenfunctions u n x) sin nπ 3 x).. Then, the first eigenvalue is λ π 9. So, λ is the minimum value of 3 w x) dx for all w as above. Now simply observe that π 9 >, so there can be no function fx) with 3 f x) dx where f is an arbitrary C smooth function satisfying the homogeneous irichlet condition.)..3)construct a direct but unmotivated) proof of Example as foolows, without using any konwledge of eigenvalues. Let wx) be a C function such that w) w). a) Expand [w x) πwx) cotπx)] dx and integrate the cross-term by parts. b) Show that w x) cotπx) as x or. c) educe that [w x)] dx π [wx)] dx [w x) πwx) cotπx)] dx. d)show that if wx) sinπx), then part c) is an equality and therefore the minimum of Example is π. Proof. a) We have [w x) πwx) cotπx)] w x) + π wx) cotπx)) π w x)wx) cotπx).

2 Let s integrate the last term by parts w x)wx) cotπx) w x) cotπx) wx) wx) cotπx)) dx wx) w x) cotπx) πwx) csc πx) ) dx. Now notice that the first term in the integral on the right hand side equals the term on the left hand side, so group these two terms together. This yields, w x)wx) cotπx) π w x) csc πx). Now let s plug this expression back into the original equation. [w x) πwx) cotπx)] w x) + π wx) cotπx)) π w x)wx) cotπx) w x) + π wx) cotπx)) + π w x) sec πx) w x) + π w x) cot πx) + csc πx) ) w x) π w x). b) For wx) as in the prompt, the C hypothesis implies that the following limits exist w ) wx) lim x x, and w wx) ) lim x x. Now for all positive x, we have that tanπx) > x and also tanπx) > x. Thus, lim w w x) x) cotπx) lim x x tanπx) w x) lim x x wx) lim x x lim wx) x. Therefore, lim x w x) cotπx). The computation is virtually the same to show that lim x w x) cotπx). c) We showed this equality in part a). d) If wx) sinπx), then [w x) πwx) cotπx)]. [π cosπx) π sinπx) cotπx)] [π cosπx) π cosπx)] Then using part a, we have [w x)] π [wx)] π. Now recall that Example stipulated wx), and since sinπx), we get the minimum eigenvalue π by using the eigenfunction wx) sinπx).

3 ..5) a) Find an expression for the lowest eigenvalue λ of with the Robin boundary condition u n + ax)u. b) Show that λ increases as ax) increases. { Proof. Let m min w dx+ } aw ds, where w ranges over all all C functions which satisfy the w dx Robin condition. Call the function which produces the minimum u. Consider the particular function w. u + ɛv for any constant ɛ and any function v which is sufficiently smooth and satisfies the Robin condition. efine fɛ). u + ɛv) + au + ɛv) u + ɛv) u + ɛ u v + ɛ v + au + ɛv) u + ɛuv + ɛ v ) By definition of u, the function f attains a minimum when ɛ. Therefore f ). We compute f ) u v + auv) u ) u + au ) uv) u ). Clearing the denominator and rearranging, we get u + au ) uv) u + au u ) uv u v + u v + auv) u ) Now observe that the term all the way to the left on the last line is by definition m. Moreover, we can use Green s first identity and the Robin condition, to reduce u v. We have, m uv u v + auv auv v u + auv v u. Therefore, vum + u)dx. Since v was arbitrary, it must be true that um + u. This occurs if and only if u um i.e. that m is an eigenvalue for the eigenvector u. It remains to show that m is the smallest eigenvalue for any eigenvector. Suppose that f, λ are any arbitrary eigenvector/eigenvalue of, which satisfy the Robin boundary condition. By Green s first formula, we know that f dx f f n ds f f dx f dx f af) ds f λf) dx f dx + af ds λ f dx auv. 3

4 Then, f dx + af ds f dx Therefore, for every eigenfunction f the quantity on the left hand side defines its eigenvalue. However, we showed before that u is the function which minimizes this quantity, and so m is in fact the smallest eigenvalue. λ. b) As a gets large, the quantity au ds also gets large. In turn, the ratio that defines the smallest eigenvalue, u dx+ au ds u dx will get large..3.) Let fx) be a function in and gx) a function on. Consider the minimum of the functional w dx fw dx gw ds, among all C functions wx). Assume that f dx + g ds. Show that a solution of this minimum problem leads to a solution of the Neumann problem u f in, u g on bdy. n Proof. Suppose that u is the function which minimizes the functional given in the prompt. Let ɛ be any constant and v any C function. efine fɛ). u + ɛv) dx fu + ɛv) dx gu + ɛv) ds Observe that f) is by definition a minimum, and so f ). Then abbreviating notation), u + ɛv) dx fu + ɛv) dx gu + ɛv) ds u + ɛ u v + ɛ v ) fu + ɛfv gu + ɛgv Thus, f ɛ) f ). u v + ɛ v u v fv v u v u v u g) fv gv gv fv v u + f) gv Since v was chosen arbitrarily, it must be that each integrand equals. Therefore, u g and u f, satisfying the given conditions. 4

5 .4.7) Consider the operator v pv m) ) m), where the superscript denotes the mth derivative, in an interval with the boundary conditions v v v v m ) at both ends. show that its eigenvalues are real. Proof. We will show that the operator is self-adjoint also called symmetric). It is a fact that the eigenvalues of self-adjoint operators are real. Call the operator A, and let u and v be smooth, real valued functions. Fix an integer m >. We have, Av, w. l px)v m) x)) m) wx) dx w x)px)v m) x)) m ) l l l px)v m) x)) m ) w x) IBP) px)v m) x)) m ) w x) boundary condition on w ) Now, via repeated integration by parts, one computes that upon k iterations, we have In particular for k m, we have Av, w ) k l px)v m) x)) m k) w k) x). l Av, w ) m px)v m) x)w m) x). If we replace the roles of v and w, the same computation reveals that l v, Aw ) m px)w m) x)v m) x). Therefore, Av, w v, Aw, so that the operator A is symmetric and all eigenvalues are real. 5