The minimizer for the second order differential problem with one nonlocal condition

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1 Lietuvos matematikos rinkinys ISSN Proc. of the Lithuanian Mathematical Society, Ser. A Vol. 58, 17 DOI: /LMR.A.17.6 pages 8 33 The minimizer for the second order differential problem with one nonlocal condition Gailė Paukštaitė, Artūras Štikonas Institute of Applied Mathematics, Vilnius University Naugarduko str. 4, LT-35, Vilnius, Lithuania gaile.paukstaite@mif.vu.lt; arturas.stikonas@mif.vu.lt Abstract. In this paper, we investigate the minimizer of the residual for the second order differential problem with one initial and other nonlocal Bitsadze Samarskii condition. We obtain the representation of the minimizer and present an example. Keywords: nonlocal conditions, least squares, Moore Penrose inverse, minimizer. Introduction Second order differential problems with nonlocal conditions, various direct representations of the unique solution using Green s functions were obtained by Roman [4]. For example, the unique solution to the problem u = f(x), x [,1], (1) u() = g 1, u (1) = γu()+g, () where f L [,1], g 1,g,γ R and (,1), exists and is given by u = g 1 + γg 1 +g +γ G(x,y)f(y)dy x+ 1 γ if and only if γ 1. Here we used the Green s function G(x,y)f(y)dy G(x,y) = { y, y x, x, y x (3) of the problem (1) () with γ = [4]. Ifγ = 1, the problem does not have the unique solution: there may be infinitely many solutions (consistent problem) or no solutions (inconsistent problem). In this paper, we will look for the unique function, which minimizes the residual of the problem (1) () and is "smallest" among all minimizers of the residual. 1 Existence of the minimizer We rewrite the problem (1) () into the equivalent vectorial form Lu = f, (4)

2 The minimizer for the second order differential problem 9 where f = (f,g 1,g ) L [,1] R. For the Hilbert space L [,1] R, we use the inner product (f, f) = (f, f) L [,1] +g 1 g 1 +g g and the norm f = f L [,1] + g 1 + g. Theorem 1 The operator L : H [,1] L [,1] R is continuous and linear with the domain D(L) = H [,1] and the closed range R(L). Proof. Domain. The operator u is defined on the whole Sobolev space H [,1]. According to the Sobolev embedding theorem [], every function u H [,1] belongs to C 1 [,1]. Thus, conditions () are defined for every u H [,1]. It means that the operator L is defined on the whole H [,1]. Linearity. It is obvious that the operator L is linear, since the differential operator (1) and nonlocal conditions () are linear with respect to u. Continuity. From the triangle inequality, we have Lu = u L [,1] + u() + u (1) γu() C u H [,1], since u L [,1] u H [,1] and the Sobolev embedding theorem says that for all u H [,1] there exist such particular positive constants L 1 and L that u() max x [,1] u(x) u C 1 [,1] L 1 u H [,1], u (1) γu() u (1) + γ u() (1+ γ ) u C1 [,1] L u H [,1]. Precisely, we obtain the estimate Lu = ( 1+L 1 +L u H [,1] C u H [,1] for C = 1+L 1 +L and all u H [,1], which means that the linear operator L is continuous. Closeness. Finally, we consider the consistent problem (4). The equation u = f has the general solution u = c 1 +c x+ )1 G(x,y)f(y)dy, c 1,c R. Substituting this general solution to nonlocal conditions, we obtain c 1 = g 1 and (1 γ)c = γg 1 +γ yf(y)dy +γ f(y)dy +g. If γ 1, then we solve c uniquely and obtain the unique solution to the problem (4) with every right hand side (f,g 1,g ) L [,1] R. It means that the range R(L) = L [,1] R is coincident with the whole space and, so, is closed. If γ = 1, we solve g = γg 1 γ yf(y)dy f(y)dy and obtain the representation of the range { } R(L) = (f,g 1, γg 1 γ yf(y)dy f(y)dy) (5) for every f L [,1], g 1 R. Now we take the sequence f n = (f n,g 1n, γg 1n γ yf n(y)dy f n(y)dy) R(L), which converges in the space L [,1] R, say, tof = (f,g 1,g ) L [,1] R. Does this limit belong tor(l)? From the extended Liet. matem. rink. Proc. LMS, Ser. A, 58, 17, 8 33.

3 3 G. Paukštaitė and A. Štikonas form of the limit f n f L [,1] R, if n, we get f n f L [,1], g 1n g 1 and γg 1n γ yf n(y)dy f n(y)dy g. Moreover, we obtain γg 1 +γ yf(y)dy + γ g 1n g 1 +γ f(y)dy γg 1n γ y(f(y) f n (y)) + yf n (y)dy f(y) f n (y) dy γ g 1n g 1 +γ f f n L [,1] + f f n L [,1], f n (y)dy if n. Since the limit is unique, we obtain g = γg 1 γ yf(y)dy f(y)dy, what means that (f,g 1,g ) R(L). According to [1], the properties of the operator L, collected in Theorem 1, are sufficient to exist functions u g H [,1], those minimize the residual of the problem (4) Lu g f L [,1] R = min u H [,1] Lu f L [,1] R. Among all those minimizers, there also exists the unique function u o of the minimum norm, i.e. u o H [,1] < u g H [,1] for all u g u o. The minimizer u o is often called the minimum norm least squares solution. Theorem The problem (4) always has the minimizer u o = L f, where L : L [,1] R H [,1] is the Moore Penrose inverse of the operator L. Proof. It follows from Theorem 1. Precisely, L is the continuous linear operator with D(L) = H [,1] and the closed range. Then according to [1], it has the Moore- Penrose inverse L, which represents the minimizer by u o = L f. Since the residual of the consistent problem (1) () is equal to zero, then from the proof of Theorem 1, the general solution to the consistent problem u g = g 1 +cx+ G(x,y)f(y)dy, c R, represents all minimizers of the problem with the trivial residual. Minimizing its H [,1] norm or calculatingp N(L) u g [1], we obtain the minimum norm least squares solution u o = g 1 +c o x+ G(x, y)f(y) dy, (6) where c o = 3 8 g ( 4 x G(x,y)f(y)dy + x f(y)dy) dx. How to represent the minimizer to the inconsistent problem, which has no solutions? First, we need to discuss about the discretized problem and its minimizer.

4 The minimizer for the second order differential problem 31 The discrete minimizer Now we introduce the mesh ω h := {x i = ih, nh = 1, i =,1,...,n} for n N and suppose is coincident with the mesh point, i.e. = sh for s {1,,...,n }. Then we discretize the problem (1) () as follows u i+1 u i +u i 1 h = f i, i = 1,n 1, (7) u n u n 1 u = g 1, = γu s +g, (8) h which can be rewritten in the equivalent matrix form Au = b for the matrix A R (n+1) (n+1) and the right hand sideb = (f 1,f,...,f n 1,g 1,g ) R (n+1) 1, where we can take f i = f(x i ) if f C[,1]. For every u, b R (n+1) 1 introducing two discrete norms ( n n ( ) u H (ω h ) = u ui u i 1 ( ) ) 1/ ui+1 u i +u i 1 ih+ h+n 1 h h h, i= i=1 i=1 ( n ) 1/ b L (ω h ) R = b ih+ b n 1 + b n i= we consider the operatora : H (ω h ) L (ω h ) R. According to [1], there exists its Moore-Penrose inverse A : L (ω h ) R H (ω h ), which represents the minimizer u o = A b R (n+1) 1 of the discretized problem (7) (8). 3 Minimizer of the inconsistent problem Theorem 3 The minimizer u o of the differential problem (1) () is equal to the minimizer of the consistent problem Lu = f (v,f) v, (9) where v(x) = ({ γx, x, 1, x, γ,1) R(L). (1) Remark 1. If the problem (4) is consistent, i.e. f R(L), then (v,f) = and we obtain the same consistent problem (4). Proof. Let us consider the problem (1) () with γ = 1. The minimizer u o to the problem Lu = f is the minimizer to the consistent problem Lu = P R(L) f [1]. Thus, u o is the exact solution to this consistent problem. Similarly, the minimizer u o of the discretized problem Au = b is the minimizer to the consistent problem Au = P R(A) b. We obtain the general solution to the discrete consistent problem n u h,g i = g 1 +cx i + G ij b j h, c R, (11) j= Liet. matem. rink. Proc. LMS, Ser. A, 58, 17, 8 33.

5 3 G. Paukštaitė and A. Štikonas where { xj+1, j i, G ij = x i, j i is the discrete Green s function to the problem (7) (8) with γ = [4]. For simplicity, we suppose f C[,1], take f i = f(x i ) and f = (f(x ),f(x 1 ),...,f(x n )). Substituting (11) into (8), we get the representationg = γg 1 γ s i=1 x if i h n 1 i=s+1 f ih, which can be rewritten as follows g = γg 1 γ xf(x)dx f(x)dx+o(h). So, now the range of the discrete problem is given by R(A) = {( f,g 1, γg 1 γ s x i f(x i )h i=1 n 1 i=s+1 ) } f(x i )h, which, as we observe, represents the discretization of (5). Precisely, if b R(A), then b = f h + O, where f h = (f,g 1, γg 1 γ xf(x)dx f(x)dx) is the discretization of f and O i = O(h) for i =,n. In general for b R(A), we get b = P R(A) b = P N(A ) b = b P N(A )b. According to [3], dimn(a ) = 1 ant the nullspace of the adjoint matrix N(A ) is composed of the one vector v = ({ γxi+1, i+1 s, 1, i+1 s, γ,1) R (n+1) 1 (i =,n ), which converges to the function (1). The limit function v R(L), because taking f R(L), we get (v,f) =. Then P R(A) b = b v (v,b) = f h + O. We can directly verify that f = f v (v,f) R(L) for every f L [,1], since it is of the corresponding form (5). It means that the problem Lu = f is consistent. This problem has the minimizer ũ o = L f = L ( f v ) (v,f) = L f (v,f) L v = L f = u o, which is equal to the minimizer u o of the problem Lu = f, since v R(L) = N(L ) = N(L ) [1]. Since the problem (9) is consistent, its minimizer is of the corresponding form (6). Then expanding the right hand side of (9) and substituting g 1 γ (v,f) and f(x) (v,f) { γx, x, 1, x instead of g 1 and f(x), respectively, in (6), we obtain the representation of the minimizer u o for the (in)consistent problem (1) (). The obtained representation of u o simplifies to (6) if the problem (4) is consistent, because (v,f) =. The residual of the problem (1) () is equal to r(u) := Lu f. So, r(u o ) = Lu o f = f (v,f) v f = (v,f) v, r(uo ) L [,1] R = (v,f).

6 The minimizer for the second order differential problem 33 Example. Let us now consider the particular problem (1) () with f(x) =, x [,1], g 1 =, g = 1, γ = and = 1/. We can easily verify that this problem has no solutions and find the least minimizer u o of the residual. Let us note that the residual r(u o ) r(x) = 1. First, we calculate the right hand side of the consistent problem (9) f = f (v,f) v = 1 ({ 6x, x 1/, 17 3, x 1/,6, 14). Substituting the first two components instead of f(x) and g 1 into (6), we get the minimizer u o = x+ 1 { 8x 3 18x, x 1/, x 4x+1, x 1/. As we predicted the norm of the minimal residual r(u o ) = 3/17 is less than the residual with the function x, i.e. r(x) = 1. We can easily verify that the obtained u o H [,1]. References [1] A. Ben-Israel and T.N.E. Greville. Generalized Inverses. Theory and Applications. Springer Verlag, New York, 3. [] L.C. Evans. Partial Differential Equations, volume 19. Amer. Math. Soc., edition, [3] G. Paukštaitė and A. Štikonas. Nullspace of the m-th order discrete problem with nonlocal conditions. Liet. matem. rink., Ser A, 54:59 64, 16. [4] S. Roman. Green s functions for boundary-value problems with nonlocal boundary conditions. Doctoral dissertation, Vilnius University, 11. REZIUMĖ Antrosios eilės diferencialinio uždavinio su viena nelokaliąja sąlyga minimalusis sprendinys G. Paukštaitė ir A. Štikonas Šiame darbe nagrinėsime antrosios eilės diferencialinio uždavinio su viena pradine ir kita nelokaliąja sąlygomis netiktį minimizuojančią funkciją. Gausime šios minimizuojančios funkcijos išraišką ir pateiksime pavyzdį. Raktiniai žodžiai: nelokalios sąlygos, mažiausi kvadratai, Moore Penrose atvirkštinis atvaizdis, minimizuojanti funkcija Liet. matem. rink. Proc. LMS, Ser. A, 58, 17, 8 33.