The one-dimensional nonlocal Dominative p-Laplace equation
aa r X i v : . [ m a t h . A P ] D ec The one-dimensional nonlocalDominative p -Laplace equation Karl K. Brustad
Norwegian University of Science and Technology
December 23, 2020
Abstract
The explicit solution to the Dirichlet problem for a class of meanvalue equations on the real line is derived. It shed some light on thebehavior of solutions to general nonlocal elliptic equations.
Imagine you are standing at a point x on a ledge [ − , − ≤ r ≤ r metres to theright, but if it is negative you have to a take a step | r | metres towards, andperhaps over, the edge at x = −
1. The process is repeated until you eitherhave reached solid ground at x >
1, or until you fall into the abyss x < − y of surviving when starting from x ∈ [ − , (cid:18) cos(1 / − sin(1 /
2) sin( x/
2) + 1 − sgn( x ) h − cos( x/ i(cid:19) . It is the solution to the mean value equation u ( x ) = 12 ˆ x +1 x − u ( y ) d y, x ∈ [ − , , with boundary conditions u ( x ) = ( , − ≤ x < − , , < x ≤ . The equation above is a special case of the nonlinear problem u ( x ) = N +2 N + p B ǫ ( x ) u ( y ) d y + p − N + p sup | ξ | =1 u ( x − ǫξ ) + u ( x + ǫξ )2 , x ∈ Ω , (1.1) u ( x ) = f ( x ) , x ∈ Γ ǫ , (1.2)which is investigated in the paper [BLM20]. Here, p ∈ [2 , ∞ ) and ǫ > ⊆ R N is open and bounded, and Γ ǫ := { x / ∈ Ω | dist( x, Ω) ≤ ǫ } is the outer strip of width ǫ . Furthermore, f : Γ ǫ ∪ ∂ Ω → R is a given bounded and integrable function. (1.1) approximates theDominative p -Laplacian equation0 = D p u := ∆ u + ( p − λ max ( D u )in the sense that, if we denote the mean value operator by M ǫp , then M ǫp φ ( x ) − φ ( x ) = ǫ N + p ) D p φ ( x ) + o ( ǫ )as ǫ → C functions φ . The uniformly elliptic operator D p was intro-duced in [Bru20] in order to explain a superposition principle in the p -Laplaceequation.By Lemma 2.2 in [BLM20] we know that there is a unique solution to(1.1) - (1.2). We are going to derive the solution in the one-dimensional case, u ( x ) = p +1 · ǫ ˆ x + ǫx − ǫ u ( y ) d y + p − p +1 · (cid:16) u ( x − ǫ ) + u ( x + ǫ ) (cid:17) , x ∈ [ a, b ] ,u ( x ) = f ( x ) , x ∈ [ a − ǫ, a ) ∪ ( b, b + ǫ ] . (1.3)2hat is, when N = 1 and when Ω = ( a, b ) is an interval in R . We shallassume that ǫ = ( b − a ) /n for some even number n = 2 m , and that f iscontinuous. The main result of [BLM20] states that the solutions of (1.1) -(1.2), as ǫ →
0, converge uniformly to the solution of the corresponding localDirichlet problem. At least for well-behaved domains and boundary values.In our case, this amounts to the simple equation u ′′ ( x ) = 0 and the nonlocalsolutions will therefore converge to the affine function with endpoint values f ( a ) and f ( b ). Our explicit formulas give some insight to the nature of thisconvergence. In Section 2 we solve the problem for p = 2. The case p > u ( x ) = 12 (cid:16) u ( x − ǫ ) + u ( x + ǫ ) (cid:17) , x ∈ [ a, b ] ,u ( x ) = f ( x ) , x ∈ [ a − ǫ, a ) ∪ ( b, b + ǫ ] (1.4)which is obtained by sending p → ∞ , is considered in Section 3.The stochastic interpretation of the Dirichlet problem (1.3) is as follows.Suppose you start a random walk from x ∈ [ a, b ] where each step is chosenfrom the ǫ -neighbourhood of the previous one according to the rule with probability p +1 , the point x k +1 ∈ [ x k − ǫ, x k + ǫ ] is picked at random.with probability · p − p +1 , we set x k +1 = x k − ǫ .with probability · p − p +1 , we set x k +1 = x k + ǫ . (1.5)You stop the walk once you have left [ a, b ] at, say, step k = τ . Then u ( x ) isthe expected value of the random variable f ( x τ ). In particular, if f = 0 on[ a − ǫ, a ] and f = 1 on [ b, b + ǫ ], then u ( x ) is the probability of exiting at theright when starting the walk from x .Note that the sup disappears in (1.3). Thus, in contrast to the higherdimensional situation, there is no control over the stochastic process in onedimension and the equation remains linear for p > With N = 1 and p = 2 the Dirichlet problem reads u ( x ) = 12 ǫ ˆ x + ǫx − ǫ u ( y ) d y, for x ∈ [ a, b ] ,u ( x ) = f ( x ) , for x ∈ [ a − ǫ, a ) ∪ ( b, b + ǫ ] . (2.1)3f f is constant on each of the two boundary parts, we shall show that u isa piecewise trigonometric function . Specifically, on each of the n intervals[ a + ( k − ǫ, a + kǫ ], k = 1 , . . . , n , of length ǫ , u is on the form u ( x ) = a k + m X j =1 b k,j sin (cid:18) λ j ǫ x (cid:19) + c k,j cos (cid:18) λ j ǫ x (cid:19) (2.2)for computable coefficients a k , b k,j , c k,j , and where λ j = cos (cid:18) jπn + 1 (cid:19) . When f is not constant, the above formula is supplemented with some addi-tional terms involving trigonometric convolutions of the data. These termsare written out in (2.10) for the case [ a, b ] = [ − , n = 4.Although the solution has oscillations for all ǫ = ( b − a ) /n , our plots showthat the graph is almost indistinguishable from a straight line on the innerpart of the interval already from n ≥
4. However, near the endpoints thegraph is unmistakably curved and the numerics indicate that the convergence u ( a ) → f ( a ) and u ( b ) → f ( b ) is no better than linear in ǫ . See Figure 3. Onthe other hand, the comparison principle (Lemma 2.5 [BLM20]) ensures thatthe convergence is linear, since the solution has to lie between the two linespassing through the points ( a − ǫ, f ( a )) , ( b, f ( b )) and ( a, f ( a )) , ( b + ǫ, f ( b )),respectively. Unfortunately, this argument is purely one-dimensional anddoes not work for N ≥
2. 4igure 3: Convergence of the boundary values.
One may easily show that the solution u of (2.1) is C/ (2 ǫ )-Lipschitz on [ a, b ],where C = max f − min f . It follows that u is differentiable in ( a, b ) \ { a + ǫ, b − ǫ } with u ′ ( x ) = 12 ǫ (cid:16) u ( x + ǫ ) − u ( x − ǫ ) (cid:17) . (2.3) u is generally not differentiable at a + ǫ or b − ǫ since it is no reason u shouldbe continuous at a or b . A close inspection of Figure 2 reveals kinks in thegraph of u at these points. The formula (2.3) provides, however, a weakderivative for u on [ a, b ]. Let u be the solution of u ( x ) = 12 ǫ ˆ x + ǫx − ǫ u ( y ) d y, for x ∈ [ a, b ] ,u ( x ) = c l , for x ∈ [ a − ǫ, a ) ,u ( x ) = c r , for x ∈ ( b, b + ǫ ] . (2.4)where a < b , ǫ = ( b − a ) /n , for some even number n = 2 m , and where c l = c r are two constants. If c l = c r the solution is identically equal to this commonconstant.To exploit the symmetry in the problem, we assume that u is scaled,shifted, and translated so that [ a, b ] = [ − , c l = − c r = 1. Thesolution will then be odd. This is possible because the equation is linear andtranslation invariant. Also, constants are solutions.Divide the domain into n parts of length ǫ . For k = 1 , . . . , n define the5unctions v k : [0 , → R as v k ( t ) := 2 c r − c l (cid:18) u (cid:0) ǫt + a + ( k − ǫ (cid:1) − c l + c r (cid:19) = u (cid:0) ǫt − k − ǫ (cid:1) . (2.5)Now, each v k is differentiable in (0 , k = 1 we have by (2.3) v ′ ( t ) = ǫu ′ ( ǫt − ǫ ǫ (cid:16) u ( ǫt − ǫ ) − u ( ǫt − − ǫ ) (cid:17) = 12 (cid:16) u ( ǫt − ǫ ) − c l (cid:17) = 12 (cid:16) v ( t ) + 1 (cid:17) . Similarly, v ′ n ( t ) = (cid:16) − v n − ( t ) (cid:17) , and for k = 2 , . . . , n − v ′ k ( t ) = 12 (cid:16) v k +1 ( t ) − v k − ( t ) (cid:17) . This defines a non-homogeneous linear system of ODEs, v ′ v ′ v ′ ... v ′ n − v ′ n = 12 · · · − · · · − · · · · · · − · · · − v v v ... v n − v n + 12 , or v ′ ( t ) = A v ( t ) + c (2.6)in vector notation. The general solution is v ( t ) = e tA ( v + A − c ) − A − c .where e A := ∞ X k =0 k ! A k is the matrix exponential.Defining ˜ e := n X k =1 ( − k e k ,
6e note that A ˜ e = 12 · · · − · · · − · · · · · · · · · − − − = 12 = c and thus v ( t ) = e tA ( v + ˜ e ) − ˜ e . (2.7)The initial value v = v (0) contains the values of u at the nodes − k − ǫ and is of course unknown. By definition, we have the n − v k (0) = v k − (1), k = 2 , . . . , n . An n ’th equation can be obtained by usingthe fact that u is an odd function. For example, u ( −
1) = − u (1) whichcorresponds to v (0) = − v n (1). Thus, v (0) = B v (1) (2.8)where B is the orthogonal matrix B := · · · −
11 0 0 · · · · · · · · · = (cid:20) ⊤ − I n − (cid:21) . Inserting (2.7) into (2.8) with t = 1, the continuity of u in [ − ,
1] gives alinear equation for v , ( I − Be A ) v = B ( e A − I )˜ e , and the unique solution of (2.6), (2.8) is, after some simplifications, v ( t ) = e tA ( I − Be A ) − ( I − B )˜ e − ˜ e , ≤ t ≤ . The invertability of I − Be A is discussed later. Observe that v ( t ) does notdepend on anything but n .The solution u of the original problem (2.4) can now be assembled byinverting the relations in (2.5): 7 ( x ) = c l , x ∈ [ a − ǫ, a ) , c r − c l v k (cid:0) x − aǫ + 1 − k (cid:1) + c r + c l , x ∈ [ a + ( k − ǫ, a + kǫ ] , k = 1 , . . . , n,c r , x ∈ ( b, b + ǫ ] . = − , x ∈ [ − − ǫ, − ,v k (cid:0) x +1 ǫ + 1 − k (cid:1) , x ∈ [ − k − ǫ, − kǫ ] , k = 1 , . . . , n, , x ∈ (1 , ǫ ] . The coefficient matrix of the linear system is the skew-symmetric and tridi-agonal n × n matrix A := 12 · · · − · · · − · · · · · · · · · − where n = 2 m is even.In order to diagonalize A , we define the vectors ξ j ∈ C n with components e ⊤ k ξ j = c n i k +2 j sin (cid:18) kj πn + 1 (cid:19) , c n := r n + 1 , i := √− . They have unit length since n X k =1 sin (cid:18) kj πn + 1 (cid:19) = n + 12by Lagrange’s identity. Using the rulesin θ + sin φ = 2 cos (cid:18) θ − φ (cid:19) sin (cid:18) θ + φ (cid:19) , e ⊤ k Aξ j = c n (cid:18) i k +2 j +1 sin (cid:18) ( k + 1) j πn + 1 (cid:19) − i k − j sin (cid:18) ( k − j πn + 1 (cid:19)(cid:19) = i k +1+2 j c n (cid:18) sin (cid:18) ( k + 1) j πn + 1 (cid:19) + sin (cid:18) ( k − j πn + 1 (cid:19)(cid:19) = i k +1+2 j c n cos (cid:18) j πn + 1 (cid:19) sin (cid:18) kj πn + 1 (cid:19) = i cos (cid:18) j πn + 1 (cid:19) e ⊤ k ξ j . That is, ξ j is an eigenvector to A with eigenvalue i cos (cid:0) j πn +1 (cid:1) . Since theeigenvalues are distinct, this also implies that the eigenvectors are orthogonal.Moreover, for j = 1 , . . . , m one can show that ξ n +1 − j = ξ j , and A is thusdiagonalized by the unitary matrix U := [ ξ , ξ , . . . , ξ m , ξ m ] ∈ C n × n , producing Λ := U ⊤ AU = diag( iλ , − iλ , . . . , iλ m , − iλ m )where 0 < λ j := cos (cid:18) jn + 1 π (cid:19) < , j = 1 , . . . , m. The real and imaginary parts of the eigenvectors ξ j = a j + i b j are a j = c n m X k =1 ( − k + j sin (cid:18) kj πn + 1 (cid:19) e k , b j = c n m X k =1 ( − k + j +1 sin (cid:18) (2 k − j πn + 1 (cid:19) e k − . Here, a ⊤ j b k = 0 and a ⊤ j a k = b ⊤ j b k = δ j,k ξ ⊤ j ξ k = 0 and ξ j ⊤ ξ k = δ j,k . If we define the real skew-symmetricmatrices A j := − ξ j ξ ⊤ j = 2( a j b ⊤ j − b j a ⊤ j ) , j = 1 , . . . , m, A = U Λ U ⊤ = m X j =1 iλ j (cid:16) ξ j ξ ⊤ j − ξ j ξ ⊤ j (cid:17) = m X j =1 λ j A j . Furthermore, A j = − a j a ⊤ j + b j b ⊤ j ) = − ξ j ξ ⊤ j is the negative of a two-rank symmetric projection and e tA = U e t Λ U ⊤ = m X j =1 e iλ j t ξ j ξ ⊤ j + e − iλ j t ξ j ξ ⊤ j = 2 Re m X j =1 e iλ j t ξ j ξ ⊤ j = m X j =1 sin( λ j t ) A j − cos( λ j t ) A j . The solution on the interval [ − k − ǫ, − kǫ ] is therefore u ( x ) = v k (cid:18) x + 1 ǫ + 1 − k (cid:19) = e ⊤ k v (cid:18) x + 1 ǫ + 1 − k (cid:19) = e ⊤ k e ( x +1 ǫ +1 − k ) A ( v + ˜ e ) − e ⊤ k ˜ e = ( − k +1 + e ⊤ k e ( m +1 − k ) A e mxA ( v + ˜ e ) , ǫ = m, = ( − k +1 + e ⊤ k e ( m +1 − k ) A m X j =1 sin( λ j mx ) A j − cos( λ j mx ) A j ! ( v + ˜ e ) , which is on the form (2.2). n = 2 When n = 2, A = 12 (cid:20) − (cid:21) ± i/
2. We have λ = 1 / · π/ ( n + 1)) and A = (cid:20) − (cid:21) so that A = λ A . Next, A = − I and it follows that e tA = sin( λ t ) A + cos( λ t ) I = (cid:20) cos( t/
2) sin( t/ − sin( t/
2) cos( t/ (cid:21) . Also, B = − A = A ⊤ and I − Be A = (1 − sin(1 / I + cos(1 / A = (cid:20) − sin(1 /
2) cos(1 / − cos(1 /
2) 1 − sin(1 / (cid:21) with inverse( I − Be A ) − = 12(1 − s ) (cid:20) − s − cc − s (cid:21) = ( I − Be A ) ⊤ − s ) = I − e − A A − s ) . This makes v = ( I − Be A ) − ( Be A − B )˜ e = 12(1 − s ) ( I − A e − A )( I − e A ) A ˜ e = 12(1 − s ) ( I − sA − cI − A ( − sA + cI ) + A ) A ˜ e = 1 − sin(1 / − cos(1 / − sin(1 / (cid:20) (cid:21) , and v ( t ) = e tA ( v + ˜ e ) − ˜ e = (cid:20) cos( t/
2) sin( t/ − sin( t/
2) cos( t/ (cid:21) " − cos(1 / − sin(1 / − (cid:20) − (cid:21) . For k = 1 ,
2, the solution in [ − k, − k ] is therefore u ( x ) = v k ( x + 2 − k ).That is, u ( x ) = − , x ∈ [ − , − , − C cos( x/ /
2) + sin( x/ /
2) + 1 , x ∈ [ − , ,C sin( x/
2) + cos( x/ − , x ∈ [0 , , x ∈ (1 , , = − , x ∈ [ − , − ,C sin( x/ − sgn( x ) h − cos( x/ i , x ∈ [ − , , , x ∈ (1 , , C := cos(1 / − sin(1 / . It is the graph of ( u + 1) / f Let u be the solution of u ( x ) = 12 ǫ ˆ x + ǫx − ǫ u ( y ) d y, for x ∈ [ − , ,u ( x ) = f ( x ) , for x ∈ [ − − ǫ, − ∪ (1 , ǫ ] , where ǫ = 2 /n . Translate the data to the interval [0 ,
1] by writing f l ( t ) := f ( ǫt − − ǫ ) , f r ( t ) := f ( ǫt + 1) , and, as usual, define the functions v k : [0 , → R as v k ( t ) = u ( x k + ǫt ) , x k := − k − ǫ. As before, v ′ k ( t ) = 12 (cid:16) v k +1 ( t ) − v k − ( t ) (cid:17) for k = 2 , . . . , n −
1, and now v ′ ( t ) = 12 (cid:16) v ( t ) − f l ( t ) (cid:17) ,v ′ n ( t ) = 12 (cid:16) f r ( t ) − v n − ( t ) (cid:17) . The system then reads v ′ v ′ v ′ ... v ′ n − v ′ n = 12 · · · − · · · − · · · · · · − · · · − v v v ... v n − v n + 12 − f l f r , or v ′ ( t ) = A v ( t ) + f ( t )12n obvious notation. The general solution is v ( t ) = e tA (cid:18) v + ˆ t e − sA f ( s ) d s (cid:19) , ≤ t ≤ , (2.9)and the solution u on the k ’th interval [ x k , x k +1 ] is u ( x ) = e ⊤ k v (cid:18) x + 1 ǫ + 1 − k (cid:19) = e ⊤ k e ( x +1 ǫ +1 − k ) A v + ˆ x +1 ǫ +1 − k e − sA f ( s ) d s ! = e ⊤ k e ( x +1 ǫ +1 − k ) A v + 1 ǫ ˆ x − k − ǫ e ⊤ k e x − yǫ A f (cid:18) y + 1 ǫ + 1 − k (cid:19) d y. The first term is on the form (2.2) while the integral may be expanded as12 ǫ m X j =1 ˆ x − k − ǫ (cid:20) sin ( λ j ( x − y ) /ǫ ) (cid:16) a k,j f (cid:0) y + 2 + (1 − k ) ǫ (cid:1) − b k,j f (cid:0) y − kǫ (cid:1)(cid:17) − cos ( λ j ( x − y ) /ǫ ) (cid:16) c k,j f (cid:0) y + 2 + (1 − k ) ǫ (cid:1) − d k,j f (cid:0) y − kǫ (cid:1)(cid:17)(cid:21) d y (2.10)In order to find an expression for the initial condition v = v (0) we stillhave the n − v k (0) = v k − (1) , k = 2 , . . . , n, but now we cannot assume u to be odd and v (0) = − v n (1) is therefore notvalid. Thus we are one equation short of determining v .Set F := Γ ǫ f d x to be the average of the data and let U := ´ − u d x be the total integral ofthe solution. Then n X k =1 v k (0) + v n (1) = n X k =0 u ( − kǫ )= 12 ǫ n X k =0 ˆ − k +1) ǫ − k − ǫ u ( y ) d y = 12 ǫ (cid:18) ˆ − − − ǫ f ( y ) d y + 2 ˆ − u ( y ) d y + ˆ ǫ f ( y ) d y (cid:19) = F + 1 ǫ U , m X k =1 v k − (0) + v n (1) = m X k =0 u ( − kǫ )= 12 ǫ m X k =0 ˆ − k +1) ǫ − k − ǫ u ( y ) d y = F + 12 ǫ U . It follows that F = 2 m X k =1 v k − (0) + v n (1) ! − n X k =1 v k (0) + v n (1) ! = n X k =1 ( − k − v k (0) + v n (1)= v (0) + n X k =1 ( − k v k (1) . We now have n equations for v , namely v (0) v (0) v (0)... v n (0) = − − · · · −
11 0 0 0 · · ·
00 1 0 0 · · · · · · v (1) v (1) v (1)... v n (1) + F , or v = ˜ B v (1) + F e . By the general solution (2.9) we may write v = ˜ B v (1) + F e = ˜ Be A (cid:18) v + ˆ e − sA f ( s ) d s (cid:19) + F e , which means that v = ( I − ˜ Be A ) − (cid:18) ˜ Be A ˆ e − sA f ( s ) d s + F e (cid:19) . It seems hard to prove that I − Be A and I − ˜ Be A are invertible or,equivalently, that Be A and ˜ Be A does not have an eigenvalue equal to 1.14igure 4: The graph of det /n ( I − Be A ) as a function of n . The correspondingpicture with ˜ B instead of B is rather similar.The matrix Be tA is orthogonal for all t ∈ R and thus have eigenvalues onthe unit circle in C . The numerics indicate that det( I − Be tA ) > t ∈ [0 ,
1] and in all even dimensions n . However, the value t = 1 seems toplay a special role: If we define t n to be the smallest positive number suchthat det( I − Be t n A ) = 0, we conjecture that the sequence ( t n ) is strictlydecreasing with lim n →∞ t n = 1. ≤ p ≤ ∞ Recall that the Dirichlet problem for p > u ( x ) = p +1 · ǫ ˆ x + ǫx − ǫ u ( y ) d y + p − p +1 · (cid:16) u ( x − ǫ ) + u ( x + ǫ ) (cid:17) , x ∈ [ − , ,u ( x ) = f ( x ) , x ∈ [ − − ǫ, − ,u ( x ) = f ( x ) , x ∈ (1 , ǫ ] , where ǫ = 2 /n .As usual, we let x k := − k − ǫ denote the nodes, and we define thefunctions v k : [0 , → R as v k ( t ) := u ( x k + ǫt ) , k = 0 , . . . , n + 1 . Also, f l , f r : [0 , → R are given by f l ( t ) := f ( x + ǫt ) , f r ( t ) := f ( x n +1 + ǫt ) . In order to derive the solution, it is instructive to first look at the case p = ∞ . The equation is then u ( x ) = 12 (cid:16) u ( x − ǫ ) + u ( x + ǫ ) (cid:17) , k = 1 , . . . , n we have v k ( t ) = 12 (cid:16) v k − ( t ) + v k +1 ( t ) (cid:17) . Now, v ( t ) = f l ( t ) for 0 ≤ t < v n +1 ( t ) = f r ( t ) for 0 < t ≤
1. Thus, v ( t ) = 12 ( L + L ⊤ ) v ( t ) + 12 (cid:0) f l ( t ) e + f r ( t ) e n (cid:1) for 0 < t <
1, (3.1)where v ( t ) := [ v ( t ) , . . . , v n ( t )] ⊤ ∈ R n , L := (cid:20) ⊤ I n − (cid:21) ∈ R n × n . The well-known matrix 2 I − L − L ⊤ is invertible and a direct calculationconfirms that (2 I − L − L ⊤ ) w l = e and (2 I − L − L ⊤ ) w r = e n where w l := n X k =1 (cid:18) − kn + 1 (cid:19) e k , w r := n X k =1 kn + 1 e k . The solution of the linear equation (3.1) is therefore v ( t ) = n X k =1 (cid:20)(cid:18) − kn + 1 (cid:19) f l ( t ) + kn + 1 f r ( t ) (cid:21) e k , < t < . Figure 5: The solution of (1.3) with p = ∞ and n = 4. The dotted lineindicates how u ( x ) can be constructed from the boundary values.16or t = 0 we have v k (0) = 12 f l (0) + v (0) , for k = 1, v k − (0) + v k +1 (0) , for k = 2 , . . . , n , v n (0) + f r (1) , for k = n + 1.If we defineˆ v := [ v (0) , . . . , v n (0) , v n +1 (0)] ⊤ ∈ R n +1 , ˆ L := (cid:20) ⊤ I n (cid:21) ∈ R ( n +1) × ( n +1) , we get the same equation for ˆ v as we did for v ( t ), but in one dimensionhigher. Thus, ˆ v = n +1 X k =1 (cid:20)(cid:18) − kn + 2 (cid:19) f l (0) + kn + 2 f r (1) (cid:21) ˆ e k . Figure 5 shows the graph of the solution to the infinity-equation u ( x ) = 12 (cid:0) u ( x − /
2) + u ( x + 1 / (cid:1) , x ∈ [ − , , with boundary values u ( x ) = ( sin (cid:0) π ( x + 3 / (cid:1) − , x ∈ [ − / , − , − x − , x ∈ (1 , / . Notice the jumps lim x → x − k u ( x ) < u ( x k ) < lim x → x + k u ( x ).We now turn to the case 2 ≤ p < ∞ . For k = 0 , . . . , n + 1 define theintegrals V k := V k (1) where V k ( t ) := ˆ t v k ( s ) d s. Then V ( t ) = ˆ t f l ( s ) d s =: F l ( t ) and V n +1 ( t ) = ˆ t f r ( s ) d s =: F r ( t ) . The equation can be written as v k ( t ) = u ( x k + ǫt )= 3 p + 1 · ǫ ˆ x k +1 + ǫtx k − + ǫt u ( y ) d y + p − p + 1 · (cid:16) u ( x k − + ǫt ) + u ( x k +1 + ǫt ) (cid:17) = 3 p + 1 · (cid:16) V k − − V k − ( t ) + V k + V k +1 ( t ) (cid:17) + p − p + 1 · (cid:16) v k − ( t ) + v k +1 ( t ) (cid:17) k = 1 , . . . , n . We have v ( t ) = f l ( t ) when 0 ≤ t < v n +1 ( t ) = f r ( t )when 0 < t ≤
1. Thus, V ′ ( t ) = v ( t )= 3 p + 1 · (cid:0) ( L ⊤ − L ) V ( t ) + F r ( t ) e n − F l ( t ) e + ( L + I ) V + F l e (cid:1) + p − p + 1 · (cid:0) ( L + L ⊤ ) V ′ ( t ) + f l ( t ) e + f r ( t ) e n (cid:1) . That is, E p V ′ ( t ) = A V ( t ) + F ( t ) + 12 ( L + I ) V , < t < , (3.2)where E p := 13 (cid:18) ( p + 1) I − p −
22 ( L + L ⊤ ) (cid:19) , and where F ( t ) := 12 (cid:0)(cid:0) F l (1) − F l ( t ) (cid:1) e + F r ( t ) e n (cid:1) + p −
26 ( f l ( t ) e + f r ( t ) e n ) . The matrix E p is invertible since it is diagonal dominant. Since V ( t ) iscontinuous with V (0) = 0, the solution of (3.2) is V ( t ) = e tE − p A (cid:18) ˆ t e − sE − p A E − p (cid:20) F ( s ) + 12 ( L + I ) V (cid:21) d s (cid:19) = e tE − p A ˆ t e − sE − p A E − p F ( s ) d s + 12 (cid:16) e tE − p A − I (cid:17) A − ( L + I ) V . (3.3)This gives the linear equation (cid:20) I − (cid:16) e E − p A − I (cid:17) A − ( L + I ) (cid:21) V = e E − p A ˆ e − sE − p A E − p F ( s ) d s (3.4)for V = V (1), which, numerically, seems to be nondegenerate. In fact, onecan show that2 A (cid:20) I − (cid:16) e E − p A − I (cid:17) A − ( L + I ) (cid:21) ( L ⊤ + I ) − = I − e AE − p ˜ B and the question of solvability of (3.4) is, at least for p close to 2, equivalentto the solvability for v in the previous Section.18he formula for u on the interior of the intervals ( x k , x k +1 ) follows nowfrom (3.2): v ( t ) = E − p (cid:18) A V ( t ) + F ( t ) + 12 ( L + I ) V (cid:19) , < t < . When t = 0 we have v k (0) = 3 p + 1 · (cid:16) V k − + V k (cid:17) + p − p + 1 · (cid:16) v k − (0) + v k +1 (0) (cid:17) for k = 1 , . . . , n . Now, v (0) = 3 p + 1 · (cid:16) F l (1) + V (cid:17) + p − p + 1 · (cid:16) f l (0) + v (0) (cid:17) , and v n +1 (0) = 3 p + 1 · (cid:16) V n + F r (1) (cid:17) + p − p + 1 · (cid:16) v n (0) + f r (1) (cid:17) . We set Q := (cid:20) I ⊤ (cid:21) ∈ R ( n +1) × n , and get the solvable equationˆ E p ˆ v = 12 (cid:16) ˆ L + ˆ I (cid:17) Q V + 12 (cid:0) F l (1)ˆ e + F r (1)ˆ e n +1 (cid:1) + p − (cid:0) f l (0)ˆ e + f r (1)ˆ e n +1 (cid:1) for ˆ v := [ v (0) , . . . , v n +1 (0)] ⊤ in R n +1 .Some simplifications can be made when the boundary values are constant.If f l ( t ) = − f r ( t ) = 1, then F ( t ) = 12 (cid:0)(cid:0) F l − F l ( t ) (cid:1) e + F r ( t ) e n (cid:1) + p −
26 ( f l ( t ) e + f r ( t ) e n )= t e + e n ) + p − e n − p + 16 e = t e + e n ) + p −
26 ( e n − e ) + 12 ( L + I )˜ e , and an integration by parts, ˆ t se − sE − p A E − p d s = − (cid:12)(cid:12)(cid:12)(cid:12) t se − sE − p A ( E − p A ) − E − p + ˆ t e − sE − p A ( E − p A ) − E − p d s = − te − tE − p A A − − (cid:16) e − tE − p A − I (cid:17) A − E p A − , p = 5, n = 4, and boundary data ± V ( t ) = e tE − p A ˆ t e − sE − p A E − p h s ( e + e n ) + p − ( e n − e ) + ( L + I )( V + ˜ e ) i d s = e tE − p A (cid:16) − te − tE − p A A − − (cid:16) e − tE − p A − I (cid:17) A − E p A − (cid:17)
12 ( e + e n )+ e tE − p A ˆ t e − sE − p A E − p (cid:20) p −
26 ( e n − e ) + 12 ( L + I ) (cid:0) V + ˜ e (cid:1)(cid:21) d s = − (cid:16) tI + (cid:16) I − e tE − p A (cid:17) A − E p (cid:17) ˜ e − (cid:16) I − e tE − p A (cid:17) A − (cid:18) p −
26 ( e n − e ) + 12 ( L + I ) (cid:0) V + ˜ e (cid:1)(cid:19) = (cid:16) e tE − p A − I (cid:17) A − (cid:18) p −
26 ( e n − e ) + 12 ( L + I ) (cid:0) V + ˜ e (cid:1) + E p ˜ e (cid:19) − t ˜ e . The equation for V + ˜ e is then h I − (cid:16) e E − p A − I (cid:17) A − ( L + I ) i (cid:0) V + ˜ e (cid:1) = (cid:16) e E − p A − I (cid:17) A − (cid:0) p − ( e n − e ) + E p ˜ e (cid:1) , and a differentiation gives v ( t ) = V ′ ( t ) = e tE − p A (cid:18) p − E − p ( e n − e ) + 12 E − p ( L + I ) (cid:0) V + ˜ e (cid:1) + ˜ e (cid:19) − ˜ e . Finally, the equation for the values ˆ v := [ v (0) , . . . , v n +1 (0)] ⊤ at the20igure 7: p = 100 and n = 6. Constant boundary ± E p ˆ v = 12 (cid:16) ˆ L + ˆ I (cid:17) Q V + 12 (cid:0) F l (1)ˆ e + F r (1)ˆ e n +1 (cid:1) + p − (cid:0) f l (0)ˆ e + f r (1)ˆ e n +1 (cid:1) = 12 (cid:16) ˆ L + ˆ I (cid:17) Q V + p + 16 (cid:0) ˆ e n +1 − ˆ e (cid:1) . The Figures 6 - 8 show the graph of the solution to (1.3) for various n and p , and when the boundary values are constant. As p → ∞ the solutionconverges (slowly) to the solution of the infinity-equation.Figure 8: p = 25 and n = 10. Constant boundary ± p >