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Review

The potential of circulating autoantibodies in the early diagnosis of Alzheimer’s disease

  • Received: 19 July 2017 Accepted: 21 August 2017 Published: 24 August 2017
  • Alzheimer’s disease (AD) is a devastating neurodegenerative disorder frequently diagnosed among the aged suffering with cognitive loss. Managing the disease has considerable economic impact on society. AD is characterized by the presence of amyloid plaque and neurofibrillary tangles, which accompany neuronal loss. There is currently no routine blood test to help to diagnose the disease. Direct tracking of AD-related molecules is difficult and costly because they are confined to the central nervous system. However, in early stage AD patients, some autoantibodies can cross the blood brain barrier to build the bridge from internal brain molecules to blood by crossing the blood brain barrier. Recent studies showed that autoantibodies which target AD-related molecules change quantitatively in the periphery along with AD pathology. More importantly, autoantibodies with different targets show diverse features in different stages of AD and in other similar kinds of dementias. This review introduces four main AD-related autoantibodies recognizing separately, amyloid-β precursor proteins, τ protein, S100b and phospholipid. While there is limited specificity and sensitivity for a single autoantibody biomarker for AD diagnosis, a combination analysis using several autoantibodies and traditional clinical diagnostics at the same time can be a promising topic for prospective research into early stage AD diagnosis.

    Citation: Wen Yin, Cordula M. Stover. The potential of circulating autoantibodies in the early diagnosis of Alzheimer’s disease[J]. AIMS Allergy and Immunology, 2017, 1(2): 62-70. doi: 10.3934/Allergy.2017.2.62

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  • Alzheimer’s disease (AD) is a devastating neurodegenerative disorder frequently diagnosed among the aged suffering with cognitive loss. Managing the disease has considerable economic impact on society. AD is characterized by the presence of amyloid plaque and neurofibrillary tangles, which accompany neuronal loss. There is currently no routine blood test to help to diagnose the disease. Direct tracking of AD-related molecules is difficult and costly because they are confined to the central nervous system. However, in early stage AD patients, some autoantibodies can cross the blood brain barrier to build the bridge from internal brain molecules to blood by crossing the blood brain barrier. Recent studies showed that autoantibodies which target AD-related molecules change quantitatively in the periphery along with AD pathology. More importantly, autoantibodies with different targets show diverse features in different stages of AD and in other similar kinds of dementias. This review introduces four main AD-related autoantibodies recognizing separately, amyloid-β precursor proteins, τ protein, S100b and phospholipid. While there is limited specificity and sensitivity for a single autoantibody biomarker for AD diagnosis, a combination analysis using several autoantibodies and traditional clinical diagnostics at the same time can be a promising topic for prospective research into early stage AD diagnosis.


    We are interested in positive classical solutions of semilinear heat equations ut=Δu+up, in Rn or in a complete Riemannian manifold (M,g) without boundary, where p>1.

    In their celebrated paper [7], Li and Yau showed how a Harnack inequality for the classical heat equation on a manifold with nonnegative Ricci tensor can be derived from a differential inequality for the logarithm of a solution. Subsequently, in the case of a manifold with nonnegative sectional curvatures and parallel Ricci tensor, Hamilton in [5] proved that the Harnack estimate of Li and Yau can actually be obtained as the trace of a full matrix inequality, under some more restrictive geometric assumptions. In some cases these inequalities can be useful in proving triviality of eternal solutions, see [1], moreover, "geometric" versions of them appear naturally and play a key role in the analysis of mean curvature flow and of Ricci flow (which are described by much more complicated systems of parabolic PDEs), see [4,6].

    Our aim is to extend to the semilinear setting the matrix Harnack estimate of Li & Yau–type for the heat equation developed by Hamilton in [5].

    We set some definitions and notations. In all of the paper, the Riemannian manifolds (M,g) will be smooth, complete, connected and without boundary. We will denote with the Levi–Civita connection of (M,g) and Δ the associated Laplace–Beltrami operator and we assume that (M,g) has nonnegative sectional curvatures and parallel Ricci tensor, that is, Ric=0. Finally, all the solutions we will consider are classical (C2 in space and C1 in time, at least).

    Remark 1.1. The hypothesis that M has parallel Ricci tensor and nonnegative sectional curvatures is satisfied on a torus or a sphere or a complex projective space, or a product of such, or a quotient of a product by a finite group of isometries.

    Definition 1.2. A quintuple of real numbers (a,b,c,d,θ) is admissible, if the following inequalities

    {da>c>0θ>b0(ac)2θ2a(θb)[(2θ+na)(ac)+a(n1)(θb)]0 (1.1)

    are satisfied.

    The above system turns out to have actually solutions.

    Proposition 1.3. There exists a nonempty cone C of admissible quintuples of parameters.

    We can then state our main result.

    Theorem 1.4. Suppose (M,g) is a complete, n–dimensional Riemannian manifold without boundary, with nonnegative sectional curvatures and parallel Ricci tensor. Let f=logu, where u is a positive classical solution of

    tu=Δu+up (1.2)

    in M×(0,T). Then, for any (a,b,c,d,θ)C there exists a constant ε>0 such that

    t(θfij+aΔfgij+bfifj+c|f|2gij+de(p1)fgij)1εgij, (1.3)

    in the sense of bilinear forms, in M×(0,T), for all 1<p<1+G(a,b,c,d,θ), where

    G(a,b,c,d,θ)=min{G1(b,d,θ),G2(a,b,c,d,θ)}, (1.4)

    given

    G1(b,d,θ)=4d(θb)θ2

    and G2(a,b,c,d,θ) the positive solution of

    (da)θ2x2+(dc)θ2x4cd(θb)=0. (1.5)

    Taking the trace with the metric g in inequality (1.3), we get a scalar Li & Yau–type inequality (which is actually weaker than the analogous one proved in [1])

    t[(θ+na)Δf+(b+nc)|f|2+nde(p1)f]nε (1.6)

    and substituting u=ef,

    (θ/n+a)Δu+(b/n+cθ/na)|u|2u+dupuεt

    in M×(0,T).

    Remark 1.5. As we mentioned, inequalities like (1.3) and (1.6) are relevant for ancient (and eternal) solutions u, that is, solutions defined in M×(,T) for some TR{+}, since by a standard argument (see [1,Section 3], for instance) choosing suitable intervals for their application, they imply

    θuij+aΔugij+(bθ)uiuju+(ca)|u|2ugij+dupgij0

    in the sense of bilinear forms and

    (θ/n+a)Δu+(b/n+cθ/na)|u|2u+dup0

    in M×(,T).

    It is possible to have an explicit bound for the function G defined in Theorem 1.4 only in terms of the dimension n of the manifold, as shown in the next proposition, hence giving a lower bound for the range of exponents p for which the results hold.

    Proposition 1.6. There holds

    sup(a,b,c,d,θ)CG(a,b,c,d,θ)˜G(n)

    where

    ˜G(n)=4(k(n)+1)2(1+1z(n)),

    and

    k(n)=3ncos(13arccos(1/n)),z(n)=k2(n)3n+k4(n)6nk2(n)6nk(n)3n.

    In particular, for any c>0

    ((z(n)+1)c,k(n)c,c,(z(n)+1)c,(k(n)+1)c)C

    and

    ˜G(n)=G((z(n)+1)c,k(n)c,c,(z(n)+1)c,(k(n)+1)c).

    In the next sections we show these results, while in the last one we derive, along the lines of [7] (and [5]), some consequent Harnack–type local estimates for the solutions.

    We follow the line of Hamilton in [5] and, by simplicity and clarity, we show the proof when (M,g) is Rn with its canonical metric. In the general case of a complete n–dimensional Riemannian manifold, the extra curvature terms which appear in the computations because of the operations of interchanging covariant derivatives "have the right sign" in the final inequality, since we assumed nonnegative sectional curvatures and parallel Ricci tensor (see the end of this section). Furthermore, thanks to standard localization arguments (as explicitly shown in [1], see also [4]), in applying the maximum principle – on which the proof is based – we can argue as if we were in a compact case (all the maximum/minimum points there exist).

    Let u:M×[0,T)R be a positive solution of ut=Δu+up. Setting f=logu, we have

    |f|=|u|uΔf=Δuu|u|2u2=Δuu|f|2
    ft=utu=Δuu+up1=Δf+|f|2+ef(p1).

    Moreover, by Eq (1.2), we also have the following relations:

    (tΔ)f=|f|2+e(p1)f(tΔ)fi=2fikfk+(p1)e(p1)ffi(tΔ)(fifj)=2fikfkfj+2fjkfkfi+2(p1)e(p1)ffifj2fikfjk(tΔ)|f|2=4flkflfk+2(p1)e(p1)f|f|22|2f|2(tΔ)fij=2fikjfk+2fikfjk+(p1)2e(p1)ffifj+(p1)e(p1)ffij(tΔ)Δf=2(Δf)kfk+2|2f|2+(p1)2e(p1)f|f|2+(p1)e(p1)fΔf(tΔ)e(p1)f=2(p1)e(p1)f|f|2p(p1)e(p1)f|f|2+(p1)e2(p1)f, (2.1)

    where fi=if, fij=jif, fijk=kjif and so on. Let (a,b,c,d,θ)C and define the symmetric bilinear form

    Fij=t(θfij+aΔfgij+bfifj+c|f|2gij+de(p1)fgij).

    Observe that

    kFijfkt=θfijkfk+a(Δf)kfkgij+b(fikfkfj+fjkfkfi)+2cflkflfkgij+d(p1)e(p1)f|f|2gij

    and, by a simple computation,

    (tΔ)Fij=Fijt+2kFijfkt+tQij, (2.2)

    where

    Qij=(p1)e(p1)fFijt+(p1)[b+(p1)θ]e(p1)ffifj+(p1)[c+a(p1)dp]e(p1)f|f|2gij+2(θb)fikfjk+2(ac)|2f|2gij. (2.3)

    Define fij:=fij(Δf/n)gij to be the tracefree part of the Hessian of f. Since

    |fij|2=|2f|21n(Δf)2,

    we have

    2(θb)fikfjk+2(ac)|2f|2gij=2(θb)θ2(θfik+aΔfgik)(θfjk+aΔfgjk)+2(ac)|2f|2gij4a(θb)θΔffij2a2(θb)θ2(Δf)2gij=2(θb)θ2(θfik+aΔfgik)(θfjk+aΔfgjk)+2(ac)|fij|2gij4a(θb)θΔffij2nθ2[a(θb)(2θ+na)(ac)θ2](Δf)2gij=2(θb)θ2(Fiktbfifkc|f|2gikde(p1)fgik)(Fjktbfjfkc|f|2gjkde(p1)fgjk)+2(ac)|fij|2gij4a(θb)θΔffij+2nθ2[(ac)θ2a(θb)(2θ+na)](Δf)2gij. (2.4)

    In order to estimate the last three terms, we use the following algebraic inequality for traceless symmetric bilinear forms

    fijρgijn1n|fij|gij,

    where ρ=max{|λi|: λieigenvalueoffi,j} denotes the spectral radius of fij. By Young's inequality, for every K>0, one has

    2Δffij2n1nΔf|fij|gijn1n[K(Δf)2+1K|fij|2]gij. (2.5)

    We set

    K:=n1na(θb)θ(ac),

    where K>0, by the first two inequalities of system (1.1) since (a,b,c,d,θ)C. Using inequality (2.5) to estimate the last three terms of Eq (2.4) we achieve

    2(ac)|fij|2gij4a(θb)θΔffij+2nθ2[(ac)θ2a(θb)(2θ+na)](Δf)2gij2nθ2(ac){(ac)2θ2a(θb)[(2θ+na)(ac)+a(n1)(θb)]}(Δf)2gij0,

    where the last inequality follows from the third inequality in system (1.1) and since (a,b,c,d,θ)C. The computation above yields

    2(θb)fikfjk+2(ac)|2f|2gij2(θb)θ2(Fiktbfifkc|f|2gikde(p1)fgik)(Fjktbfjfkc|f|2gjkde(p1)fgjk)=2(θb)θ2[F2ijt2bt(Fijfjfk+Fjkfifk)2ct|f|2Fij2dte(p1)fFij+b(b+2c)|f|2fifj+2bde(p1)ffifj+c2|f|4gij+2cde(p1)f|f|2gij+d2e2(p1)fgij]. (2.6)

    Now we claim that there exists ε>0 such that if Fij0 then

    θ2QijF2ijεt2, (2.7)

    where F2ij=FikFkj. Thus, let us suppose that Fij0. Combining the estimate (2.6) with the definition of Qij given in (2.3) we get

    θ2Qij(p1)θ2e(p1)fFijt+(p1)θ2[b+(p1)θ]e(p1)ffifj+(p1)θ2[c+a(p1)dp]e(p1)f|f|2gij+2(θb)[F2ijt2bt(Fijfjfk+Fjkfifk)2ct|f|2Fij2dte(p1)fFij+b(b+2c)|f|2fifj+2bde(p1)ffifj+c2|f|4gij+2cde(p1)f|f|2gij+d2e2(p1)fgij]2(θb)F2ijt2+[(p1)θ24d(θb)]e(p1)fFijt+{(p1)θ2[c+a(p1)dp]+4cd(θb)}e(p1)f|f|2gij+2d2(θb)e2(p1)fgij.

    Recalling that for (a,b,c,d,θ)C there holds (θb)>0, we only have to show that

    {(p1)θ24d(θb)0(p1)θ2[c+a(p1)dp]+4cd(θb)0

    Setting x=p10, we can recast the previous inequalities as

    {xθ24d(θb)0(da)θ2x2+(dc)θ2x4cd(θb)0 (2.8)

    Being 4cd(θb)>0, (da)θ2>0 and (dc)θ2>0, Eq (1.5) admits two solutions, of which only one is positive. Hence, defining G1(b,d,θ) and G2(a,b,c,d,θ) as in the statement of the Theorem and G(a,b,c,d,θ) as in (1.4), the fact that 0xG(a,b,c,d,θ) implies that x satisfies both the inequalities in (2.8). Thus, Eq (2.7) is satisfied with ε=12(θb).

    The application of maximum principle (see for example Theorem C.1.3 in [8] or Lemma 8.2 in [3]) concludes the proof.

    We point out that in the case of a general Riemannian manifold extra curvature terms appear in the computations. For instance, the second and the fifth line in formulas (2.1) would become respectively

    (tΔ)fi=2fikfk+(p1)e(p1)ffiRikfk

    and

    (tΔ)fij=2fikjfk+2fikfjk+(p1)2e(p1)ffifj+(p1)e(p1)ffij+2RikjlfklRikfjkRjkfik+(lRijiRjljRil)fl=2fijkfk+2fikfjk+(p1)2e(p1)ffifj+(p1)e(p1)ffij+2RikjlfklRikfjkRjkfik+(lRijiRjljRil)fl+2Rikjlfkfl

    where Rij and Rikjl denote the components of the Ricci and the curvature tensor, respectively (we follow the convention in [2]). At this point it is straightforward to verify that the extra curvature terms which would appear in Eq (2.2) either disappear or have the right sign under our geometric assumptions, as in [5,Theorem 4.3].

    First of all, observe that if (a,b,c,d,θ) is admissible and λ>0 is a constant, then (λa,λb,λc,λd,λθ) is still admissible. Thus, if we show that an admissible quintuple of parameters exists, then we have a cone of admissible parameters. Let us consider a quintuple of the form (a,b,c,a,θ) where ac>0 and θb>0. We want to find a,b,c,θ such that the third inequality of system (1.1) is satisfied. To do this, let us set ac=δ for some δ>0. Then, the third inequality of system (1.1) becomes

    δ2θ2δ(θb)[(2θ+nδ+nc)δ+δ(n1)(θb)+c(n1)(θb)]c(θb)[(2θ+nδ+nc)δ+δ(n1)(θb)+c(n1)(θb)]0.

    Now let us assume that θb=c and b=kc for some k>0 to achieve

    nδ3c+(k23n)δ2c2(3n+2k)δc3c4(n1)0.

    Being c>0, we can set z=δc and recast the previous inequality as

    nz3+(k23n)z2(3n+2k)z(n1)0.

    Let us set

    H(z,k):=nz3+(k23n)z2(3n+2k)z(n1).

    We want to find some suitable values for z and k such that H(z,k)0. First of all, let us determine some necessary conditions on k in such a way that H(,k) admits positive roots.

    Proposition 3.1. There always exists a negative number z<0 such that H(z,k)=0. If there exist two positive numbers z1,z2>0 such that H(zi,k)=0, then k>3n.

    Proof. Let us observe that H(0,k)=(n1)<0 and limzH(z,k)=+, hence we obviously have z. Moreover, let us observe that, by Descartes' rule of the signs, the maximum number of positive solutions of the equation H(z,k)=0 for fixed k is 0 if k23n0 and 2 if k23n>0, thus, if two positive solutions z1,z2 exist, then k>3n.

    We can use the previous necessary condition to achieve a necessary and sufficient condition on the existence of two positive roots.

    Proposition 3.2. Fix n2. Then H(z,k) admits two (possibly equal) positive roots if and only if kk(n), where

    k(n)=3ncos(13arccos(1/n)).

    Proof. As we have shown before, a necessary condition to have two positive roots is that k>3n. On the other hand, let us observe that

    H(z,k)=nz3+(k23n)z2+(3n+2k)z(n1)

    hence, by Descartes' rule of the signs we know that H(,z) admits at most one negative root. Moreover, since H(0,k)=(n1)<0 and limzH(z,k)=+, then we know that H(z,k) admits exactly one negative root for any k>3n. Thus, if H(z,k) admits three real roots, two of them have to be positive.

    The discriminant of H(,k) is given by

    ΔH(k)=18n(n1)(k23n)(3n+2k)+4(n1)(k23n)3+(k23n)2(3n+2k)24n(3n+2k)327n2(n1)2=4(1+k)3n(k3274kn274n).

    We have that ΔH(k)0 if and only if P1(k):=k3274kn274n0. P1(k) is a depressed cubic polynomial with p1(n)=q1(n)=274n. By Descartes' rule of the signs and the fact that P1(0)=274n<0 we know that P1(k) always admits a unique positive root. The discriminant of P1(k) is given by

    Δ1(n)=(27342n3+27342n2)=27316n2(n1)>0,

    thus P1(k) admits three real roots. Since we are under the casus irreducibilis, we have to provide trigonometric solutions to recognize what is the real solution we are interested in. To do this, we will use Viéte's procedure. Consider the equation P1(k)=0 and set k=ucos(θ) to achieve

    u3cos3(θ)274nucos(θ)274n=0.

    Multiplying everything by 4u3 we get

    4cos3(θ)27u2ncos(θ)27u3n=0.

    Now set 27nu2=3, that is to say u=3n, to achieve

    4cos3(θ)3cos(θ)1n=0.

    Recalling that 4cos3(θ)3cos(θ)=cos(3θ), we get

    cos(3θ)=1/n

    and then

    θ=13arccos(1/n)+2πj3,

    where j=0,1,2. Finally, we get the three real roots of P1(k) as

    kj=3ncos(13arccos(1/n)+2πj3), j=0,1,2.

    However, we know that P1(k) admits only one positive root. Being

    013arccos(1/n)π3,

    we have that k0>0 is the solution we are searching for. Thus, let us relabel

    k(n):=3ncos(13arccos(1/n))

    to conclude that P1(k)0 if and only if kk(n).

    Now we can conclude the proof of Proposition 1.3. Indeed, let us consider k>k(n), in such a way that ΔH(k)>0. For such fixed k, H(,k) admits two positive roots 0<z1<z2. Consider any z[z1,z2]. Setting, without loss of generality, c=1, we then know that the quintuple (z+1,k,1,z+1,k+1) is admissible.

    To prove Proposition 1.6, we want to exhibit an admissible quintuple (a,b,c,d,θ) such that G(a,b,c,d,θ) can be explicitly calculated. To do this, let us consider again a quintuple of the form (z+1,k,1,z+1,k+1) where kk(n). Moreover, let us consider z(k,n) to be a positive local maximum point of H(z,k), for fixed kk(n), such that H(z(k,n),k)0. If k>k(n), then this maximum always exists, by a simple application of Rolle's theorem on the interval [z1,z2], together with the fact that both z1,z2 are simple roots. If k=k(n), then z1=z2 and it coincides with such local maximum of the polynomial H(z,k(n)). Let us evaluate it explicitly.

    Proposition 4.1. For fixed n2 and k>3n, H(z,k) admits two (eventually equal) critical points if and only if these critical points are positive and kk0(n) where

    k0(n)=22ncos[13arccos(322n)].

    If k>k0(n), the local maximum point is given by

    z(k,n)=k23n+k46nk26nk3n.

    In particular, k0(n)k(n) and, if kk(n), there holds H(z(k,n),k)0.

    Proof. Let us first observe that

    H(z,k):=zH(z,k)=3nz2+2(k23n)z(3n+2k). (4.1)

    Since k>3n, Descartes' rule of the signs tells us that, if the solutions exist, they must be positive. Hence we have only to show that the discriminant is nonnegative. Let us determine the discriminant of the polynomial (4.1):

    Δz(k)=4((k23n)23n(3n+2k))=4k(k36nk6n). (4.2)

    Being k>0, Δz(k)0 if and only if

    P2(k):=k36nk6n0.

    Let us first show that this polynomial admits a unique positive root. Observe that P2(0)=6n<0 and limk+P2(k)=+, thus P2(k) admits a positive root k0(n). Moreover, Descartes' rule of the signs tells us that P2(k) admits at most one positive root, hence k0(n) is the unique positive root.

    Now let us determine k0(n). First of all, let us observe that, since P2(k) is a depressed cubic polynomial, its discriminant is given by

    Δk=(463n3+2762n2)=108n2(8n9)>0

    since n2. Thus we know P2(k) admits three different roots. Since we are under the casus irreducibilis, we have to provide trigonometric solutions to recognize what is the real solution we are interested in. Arguing again by Viéte's procedure and selecting the unique positive root, we get

    k0(n)=22ncos[13arccos(322n)].

    Hence, as k>k0(n), we can find two solutions to equation H(z,k)=0. Being H(,k) a polynomial with H(0,k)=(n1) and limzH(z,k)=+, we know that H(z,k) is decreasing as z0. Thus, the first critical point has to be a local minimum and the second critical point a local maximum. Therefore, writing explicitly the second solution of H(z,k)=0, we get

    z(k,n)=k23n+k46nk26nk3n. (4.3)

    Finally, observe that, by Rolle's theorem, if k>k(n), then the interval [z1,z2] has to admit one of the two critical points. Moreover, since H(z,k)>0 as z(z1,z2) and H(z1,k)=H(z2,k)=0, then such critical point is the local maximum and H(z(k,n),k) is positive. If k=k(n), then z1=z2 is a double root and H(z1,k(n))=H(z2,k(n))=0. Being H(z,k(n))0 for any z0 (since the other simple root is negative), we have that z1=z2=z(k(n),n) and H(z(k(n),n),k(n))=0. This also obviously implies k0(n)<k(n).

    Next, we want to evaluate G on the quintuple (z(k,n)+1,k,1,z(k,n)+1,k+1) as kk(n). To do this, we first need to exploit a simple property of z(k,n).

    Proposition 4.2. For any kk(n) there holds z(k,n)2.

    Proof. By Eq (4.3), we have that z(k,n)2 if and only if

    k46nk26nk9nk2. (4.4)

    Being kk(n), we already know that the quantity under the square root is nonnegative. On the other hand, if k3n, then inequality (4.4) holds true. Let us consider k(n)k<3n. Then inequality (4.4) is equivalent to

    k46nk26nk81n218nk2+k4

    that is to say

    4k22k27n0.

    Since k(n)>0, the previous inequality is verified as

    k1+1+108n4=:k1(n)>12.

    To conclude the proof, we have to show that k1(n)k(n). Being k0(n)k(n), this is obvious if k1(n)k0(n), thus let us suppose k1(n)>k0(n).

    Let us consider the function g(k):=H(z(k,n),k) for kk0(n). Being z(k,n) a local maximum of H, there holds Hz(z(k,n),k)=0. Thus, we have

    g(k)=2kz2(k,n)2z(k,n)=2z(k,n)(kz(k,n)1).

    In particular g(k)0 if and only if kz(k,n)1. The latter holds if and only if

    kk46nk26nk3n+3nkk3=Δz(k)4k3n(1+k), (4.5)

    where Δz(k) is defined in Eq (4.2). Being kk0(n), we have that Δz(k)0 and then inequality (4.5) is verified. This implies, in particular, that g(k) is increasing as kk0(n). Moreover, by definition of k(n), there holds g(k(n))0. On the other hand, being z(k1(n),n)=2 and k1(n)>12, we get

    g(k1(n))=H(2,k1(n))=2k1(n)+1<0.

    Hence, we have k1(n)<k(n), concluding the proof.

    With this property in mind, we can evaluate

    G(z(k,n)+1,k,1,z(k,n)+1,k+1).

    To do this, let us first observe that

    G1(k,z(k,n)+1,k+1)=4(z(k,n)+1)(k+1)2.

    Concerning G2(z(k,n)+1,k,1,z(k,n)+1,k+1), it is the unique solution of

    z(k,n)(k+1)2x4(z(k,n)+1)=0

    that is to say

    G2(z(k,n)+1,k,1,z(k,n)+1,k+1)=4(z(k,n)+1)z(k,n)(k+1)2.

    Hence, we have

    G(z(k,n)+1,k,1,z(k,n)+1,k+1)=4(z(k,n)+1)(k+1)2min{1,1z(k,n)}.

    However, being z(k,n)2 by the previous proposition, there holds

    min{1,1z(k,n)}=1z(k,n)

    then

    G(z(k,n)+1,k,1,z(k,n)+1,k+1)=4(k+1)2(1+1z(k,n)).

    Now we want to optimize on kk(n). To do this, let us show the following Proposition.

    Proposition 4.3. The function kz(k,n) is increasing on [k(n),+).

    Proof. Just observe that

    z(k,n)=13n(2k+2k36nk3nk46nk26nk)=13n(2k+Δz(k)4kk46nk26nk+k3+3nk46nk26nk)0,

    being Δz(k)0 by the fact that kk(n)k0(n).

    The previous proposition implies that the function kG(z(k,n)+1,k,1,z(k,n)+1,k+1) is decreasing as kk(n), thus it achieve its maximum value as k=k(n). Setting

    ˜G(n):=G(z(k(n),n)+1,k(n),1,z(k(n),n)+1,k(n)+1),n0,

    we conclude the proof of Proposition 1.6.

    We begin the section with the following technical lemma.

    Lemma 5.1. There exists a cone C of an admissible quintuples (a,b,c,d,θ) such that a=d and ba+c<0.

    Proof. Let us consider kk(n) with k(n) defined in Proposition 1.6 and let (z(k,n)+1,k,1,z(k,n)+1,k+1)C with z(k,n) defined in (4.3). The condition ba+c<0 becomes k<z(k,n), i.e.,

    3nkk2+3n<k46nk26nk.

    This is obviously true if

    k>3n+9n2+12n2.

    With this choice of k we conclude the proof.

    Remark 5.2. It is easy to see that if (a,b,c,d,θ)C then also (a,b,c,a,θ)C.

    We will now derive some Harnack–type inequalities as a consequence of Theorem 1.4.

    Proposition 5.3. Let (M,g) be an n–dimensional complete Riemannian manifold with nonnegative sectional curvatures and parallel Ricci tensor. Consider (a,b,c,a,θ)C and 1<p<1+G(a.b,c,a,θ). Let u:M×[0,T)R be a classical positive solution of the equation ut=Δu+up, then there exists ε=ε(n,p,a,b,c,θ) such that, given any 0<t1<t2T and x1,x2M, the following inequality holds

    u(x1,t1)u(x2,t2)(t2t1)1/εexp(ψ(x1,x2,t1,t2)),

    where

    ψ(x1,x2,t1,t2):=infγΓ(x1,x2)10[a|˙γ(s)|24(abc)(t2t1)+θ(t2t1)aρ(γ(s),(1s)t2+st1)]ds,

    with Γ(x1,x2) given by all the paths in M parametrized by [0,1] joining x2 to x1, f=logu and ρ=max{|λi|:λieigenvalueoffij}.

    Proof. By Theorem 1.4 we know that there exists ε such that

    θfij+aΔfgij+bfifj+c|f|2gij+ae(p1)fgij+1εtgij0.

    Recalling that f satisfies

    ft=Δf+|f|2+e(p1)f,

    we get

    ftgijθafij+bafifjaca|f|2gij+1aεtgij. (5.1)

    Since ρ is the spectral radius of fij we get fijρgij. On the other hand the matrix fifj has rank one and the only nonzero eigenvalue is |f|2, then fifj|f|2gij. Plugging these inequalities into (5.1) we obtain

    ftgij(θaρacba|f|2+1aεt)gij. (5.2)

    Now let us consider any γΓ(x1,x2) as well as η:[0,1]M×[t1,t2] defined as η(s)=(γ(s),(1s)t2+st1). Evaluating (5.2) in η(s) and applying it to ˙γ(s) we see that

    ft(η(s))|˙γ(s)|2(θaρ(η(s))acba|f(η(s))|2+1aε((1s)t2+st1))|˙γ(s)|2.

    Noticing that η(0)=(x2,t2) and η(1)=(x1,t1), we have

    f(x1,t1)f(x2,t2)=10(ddsf(η(s)))ds=10[f(η(s)),˙γ(s)(t2t1)fs(η(s))]ds10[|f(η(s))||˙γ|+(t2t1)(1aε[(1s)t2+st1]acba|f(η(s))|2+θaρ(η(s)))]ds=10t2t1aε[(1s)t2+st1]ds+10[(acb)(t2t1)a|f(η(s))||˙γ(s)|(acb)(t2t1)a|f(η(s))|2+θ(t2t1)aρ(η(s))]ds1aεlog(t2t1)+10[a|˙γ(s)|24(abc)(t2t1)+(t2t1)θaρ(η(s))]ds.

    Being γΓ(x1,x2) arbitrary, we conclude the proof.

    With the same strategy we can also prove the following variant.

    Proposition 5.4. Let (M,g) be an n–dimensional complete Riemannian manifold with nonnegative sectional curvatures and parallel Ricci tensor. Consider (a,b,c,a,θ)C and 1<p<1+G(a.b,c,a,θ). Let u:M×[0,T)R a classical positive solution of the equation ut=Δu+up, then there exists ε=ε(n,p,a,b,c,θ) such that, given any 0<t1<t2T and x1,x2M, the following inequality holds

    u(x1,t1)u(x2,t2)(t2t1)1/εexp(ψ(x1,x2,t1,t2)),

    where

    ψ(x1,x2,t1,t2):=infγΓ(x1,x2)10[na|˙γ(s)|24(nanbnc+θ)(t2t1)+θ(t2t1)aρ(γ(s),(1s)t2+st1)]ds,

    with Γ(x1,x2) given by all the paths in M parametrized by [0,1] joining x2 to x1, f=logu, fij=fijΔfngij and ρ=max{|λi|:λieigenvalueoffij}.

    Finally, we also have another one coming from the scalar (trace) version of the Li & Yau inequality (1.6). We underline that it holds up to an exponent p lower than the analogous one for the Harnack inequality obtained in [1], moreover, this latter holds under the only hypothesis of nonnegative Ricci tensor.

    Proposition 5.5. Let (M,g) be an n–dimensional complete Riemannian manifold with nonnegative sectional curvatures and parallel Ricci tensor. Consider (a,b,c,a,θ)C and 1<p<1+G(a.b,c,a,θ). Let u:M×[0,T)R a classical positive solution of the equation ut=Δu+up, then there exists ε=ε(n,p,a,b,c,θ) such that, given any 0<t1<t2T and x1,x2M, the following inequality holds

    u(x1,t1)u(x2,t2)(t2t1)1/εexp(ψ(x1,x2,t1,t2)),

    where

    ψ(x1,x2,t1,t2):=infγΓ(x1,x2)10[(θ+na)|˙γ(s)|24(nanc+θb)(t2t1)θθ+naup1(γ(s),(1s)t2+st1)]ds,

    with Γ(x1,x2) given by all the paths in M parametrized by [0,1] joining x2 to x1.

    Proof. Let f=logu, by Theorem 1.4 we know that there exists ε>0 such that inequality (1.6) holds, then the proof proceeds as in Proposition 5.3.

    The authors declare no conflict of interest.

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