This paper provides an alternative method for determining the ultimate bearing capacity of pile foundations based on Static Load Test (SLT) results. A differential equation modelling the relation between loads and settlements of SLT is presented as well as its solution in a form of continuous function. The notion of Lipschitz continuity in mathematical analysis is adopted and is applied to the function for finding the most critical load, which is defined as the ultimate bearing capacity. For a validation, interpretations of an SLT data using Davisson’s method and the method proposed in this paper are compared. The result shows that the interpretation using the proposed method is slightly greater than that of Davisson’s and gives a valid result. Keywords: Static Load Test, Lipschitz Function, Differential Equation, Bearing Capacity 1 Introduction Static Load Test (SLT) is an in-situ loading test for pile foundations (Das, 2011). The primary reason for the test to be conducted is the unreliability of available prediction methods according to the pre-existing data (Das, 2011). The standard procedure of the SLT may be different for each country in accordance with its respective applicable codes. Figure 1: Load vs Settlement Graphs of SLT The resulting data of SLT can be plotted for the values of loads vs settlements, and will have a typical form as illustrated by figure 1 (Das, 2011). During the loading process, the settlement tends to increase further up to the point of the maximum test load. During the unloading process, it tends to come back to its initial state before settlements even though it never really reaches its initial state and stops at some settlement value. Through out this paper, it will be assumed 1 that the settlement values are expressed in milimeters (mm), while the load values are expressed in kilonewtons (kN). The ultimate bearing capacity of piles can naturally be determined from the curve of the loading process. A loading point quin the load interval is the ultimate bearing capacity of pile if the differential of the settlement at quis significantly increase than that of points before qu.This notion is made an axiom of the ultimate bearing capacity formulation in this paper. As a preliminary, the readers are encouraged to explore the foundational concepts in mathe- matics that are incorporated in this paper, namely formal logic (Bergmann et al., 2014), set theory (Stoll, 1963), mathematical analysis (Rudin, 1964), metric spaces (Kreyszig, 1978; Andr´ e, 2020) as well as Lipschitz functions (Searc´ oid, 2007) and Ordinary Differential Equation (ODE) (Kreyszig, 2011). 2 Lipschitz Continuity of Functions Lipschitz continuity is a stronger continuity form of uniform continuity concept of functions (Searc´ oid, 2007), named after German mathematician Rudolf Lipschitz. This concept is appli- cable to functions between metric spaces (Andr´ e, 2020; Kreyszig, 1978). The formal definition of Lipschitz continuity is presented as follows. Definition 1 (Lispchitz Continuity) .Let ( X, d X) and ( Y, dY) be metric spaces (Searc´ oid, 2007; Andr´ e, 2020). A function (Stoll, 1963; Judson, 2012) f:X→Yis a Lipschitz function (Searc´ oid, 2007) if there exists some K≥0 such that ∀s, t∈X , d Y(f(s), f(t))≤KdX(s, t) holds. The function fis also referred to as a K-Lipschitz function (Searc´ oid, 2007). And the constant Kis also referred to as a Lipschitz constant. An alternative way in defining Lipschitz continuity involving difference expressions is presented on the following remark. Remark 1.Let ( X, d X) and ( Y, dY) be metric spaces (Searc´ oid, 2007; Andr´ e, 2020). Suppose a function f:X→Y .For a pair of points u, v∈X ,the inequality dY(f(u), f(v))≤KdX(u, v) is trivially satisfied by an equality in case u=v, as shown by dY(f(u), f(v)) =dY(f(u), f(u)) = 0 = K·0 =KdX(u, u) =KdX(u, v). The expression above always hold as long as ( X, d X) and ( Y, dY) are metric spaces. And it can be shown that fisK-Lipschitz if dY(f(u), f(v)) dX(u, v)≤K is satisfied with u̸=v . A derivative notion from Lipschitz continuity is presented on the following definition. Note that this notion will be important to the formulation of the load vs settlement graph as a function on the following section. Definition 2 (Locally Lipschitz) .Let ( X, d X) and ( Y, dY) be metric spaces (Searc´ oid, 2007; Andr´ e, 2020). A function f:X→Yis called locally K-Lispchitz if for every point x∈X , there exists a neighbourhood U⊆Xofxsuch that the restriction map (Stoll, 1963; Judson, 2012; Rudin, 1964) fU:U→YisK-Lipschitz. 2 3 Settlement Function Suppose a data set of a SLT result. Suppose there are αincremental applications of loading, for some α∈N.LetQ:={q0, q1, . . . , q α}be the set of the load increments. Since Qis the set of load increments, then ∀k∈ {1, . . . , n } qk∈[0,∞) ∧ ∀k∈ {1, . . . , n } qk−qk−1>0 (1) is necessarily satisfied. Also, suppose q0:= 0,since at the initial state before the application of loading, settlements are not assumed to happen. Consequently, qαis the maximum load of the SLT. Suppose a map s:Q→R+is the settlement map of the loading process, where R+:={x∈R|x≥0}. We may denote sk:=s(qk),for the settlement due to the load qk,for every k∈ {1, . . . , α }. 3.1 Differential Equation and Solution of Settlement The map s:Q→R+is necessarily approximated with a continuous function S:R+→R+. The function Scan be determined from a differential equation describing the load and settlement relationship from the SLT. Empirical results as represented by figure 1 show that the settlement keeps increasing as the load increases, and the rate of change of the settlement is proportional to the settlement itself. It can be inferred that dS dq∝S . Thus, a differential equation describing Sis given by ∀q∈R+,d dqS(q) =τS(q) +γ , (2) where τ, γ∈Rare some constants. Note that equation 2 above is separable (Kreyszig, 2011), hence we obtain ∀q∈R+,dS(q) τS(q) +γ=dq . By integrating the expression above, we have ZdS τS+γ=Z dq . Suppose W:=τS+γ .Then, dW/dS =τand consequently, dS=dW/τW . And the expression above becomes1 τZdW W=q+c1∴ln|W|=τq+c2, which again implies ∀q∈R+, S(q) =ceτq−γ τ. For convenience, let r:=γ/τ , and we have ∀q∈R+, S(q) :=ceτq−r , (3) which is the general solution of S(Kreyszig, 2011). A set of initial conditions shall be presented (Kreyszig, 2011) in order to determine randτ . At the load of 0, the settlement has not occurred, thus S(0) = 0 holds. And it is assumed that there exists a point ˜ q∈R+such that there exist ν, ξ∈ {1, . . . , α }with qν= ˜qandqξ= 2˜q. In other words, ˜ q,2˜q∈QandS(˜q) =sνas well as S(2˜q) =sξ.Empirical results show that 2 sν< sξ. Follows from S(0) = 0 ,we have 0 =S(0) = ceτ·0−r=c−r 3 which implies that c=r .Thus, ∀q∈R+, S(q) =reτq−r . (4) Follows from S(˜q) =sνandS(2˜q) =sξ,we have reτ˜q−r=sν and reτ2˜q−r=sξ, which imply τ= ln˜qr sν+r r= ln2˜qr sξ+r r. (5) Follows from equation 5 above, rcan be solved as follows. By the equality on equation 5, we obtain 0 = ln˜qr sν+r r−ln2˜qr sξ+r r= ln˜qq sν+r r 2˜qq sξ+r r, which implies ˜qq sν+r r 2˜qq sξ+r r=e0= 1. By multiplying both sides with the denominator of the far left side, we obtain ˜qr sν+r r=2˜qr sξ+r r. And by raising both sides to the power of 2˜ q, we obtain (sν+r)2=r(sξ+r) s2 ν+ 2rsν+r2=rsξ+r2, which implies s2 ν+ 2rsν=rsξ. And consequently, we obtain r=s2 ν sξ−2sν. (6) Since sνandsξcan be obtained from the SLT, thus, the exact value of rcan be determined by applying equation 6. As rhas already been determined, τcan be determined by applying τ=lnsν r+ 1 ˜q, (7) which is implied from equation 5. Thus, Scan be determined by applying equation 4. A further analysis on the settlement function Swill be presented on the following subsection. 3.2 Analysis on Settlement Function Follows from the empirical result that 2 sν< sξ,then there exists some ε >0 such that sξ sν= 2 + ε . Consequently, r=s2 ν sξ−2sν=sν sξ sν−2=sν ε>0. 4 This result gives an implication ˜qr sν+r r=˜qrsν r+ 1 =˜q√ ε+ 1>1, which shows that τ= ln˜qr sν+r r= ln˜q√ ε+ 1>ln 1 = 0 . Hence we conclude that r, τ > 0.A direct consequence to these results is presented on the following lemma. Lemma 1. The settlement function S:R+→R+as defined in accordance with equation 3 is a strictly increasing function (Bartle, 1991). Proof. The settlement function S:R+→R+is said to be strictly increasing if and only if ∀u, v∈R+ u < v ⇐⇒ S(u)< S(v)] holds. Let p, q∈R+.(Forward part of the proof ). Assume that p < q . Then S(p) +r S(q) +r=reτp−r+r reτq−r+r=eτ(q−p). Since r, τhave been shown to be positive on section 3.2, then eτ(q−p)>1,which implies S(q) +r S(p) +r>1. It shows that S(p) +r < S (q) +r , which again implies S(p)< S(q).(Backward part of the proof ). For the backward part, let us assume that S(p)< S(q).Now we need to show that p < q , which can be done by reversing the forward part. We have 1<S(q) +r S(p) +r=eτq eτp=eτ(q−p) which implies q−p=τlneτ(q−p)> τln 1 = τ·0 = 0 , which shows that p < q . Hence, we conclude that Sis strictly increasing. Another property of Sis the differentiabiliy of S .It is in fact a natural consequence that S is differentiable since it is a solution of a differential equation (Kreyszig, 2011). Thus, we present this property on the following remark. Remark 2.The settlement function S:R+→R+is differentiable. Moreover, given any pair of points p, q∈R+such that p < q , then we have S′(p)< S′(q), as shown by S′(p) S′(q)=τreτp τreτq=eτp eτq=eτ(q−p)> eτ(p−p)=e0= 1. An important property of the settlement function in this context is that it is not Lipschitz. This property is presented on the following theorem. Theorem 1. The settlement function S:R+→R+defined in accordance with equation 3 is not Lipschitz. 5 Proof. Note that the underlying metric spaces in this context are the same for both the domain and the codomain of S ,given by ( R+, ds),which is a metric subspace (Andr´ e, 2020) of the usual metric space (Andr´ e, 2020) on R.Thus, the metric dsis defined by ∀u, v∈R+, ds(u, v) :=|u−v|. We will prove this theorem by contradiction. Let K > 0.And assume that SisK-Lipschitz. Then we have ∀u, v∈R+, u < v =⇒|S(u)−S(v)| |u−v|≤K . Leta, b∈R+such that a < b . By lemma 1, we have that Sis strictly increasing. Thus, S(b)−S(a) b−a=|S(a)−S(b)| |a−b| holds. And the Lipschitz condition for S–as assumed–can simply be expressed by S(b)−S(a) b−a≤K . Follows from remark 2, Sis differentiable with a strictly increasing derivative. Let c∈R+such thatc > b . Then by mean value theorem (Rudin, 1964) there exists a point ˜b∈(a, b) and a point ˜c∈(b, c) such that S(b)−S(a) b−a=S′(˜b)< S′(˜c), since ˜b <˜c .LetK:=S′(˜c).Given another point d∈R+with d > c , follows from mean value theorem there exists a point ˜d∈(c, d) and we have S(d)−S(c) d−c=S′(˜d)> S′(c) =K , which shows that Sis not K-Lispchitz. Given any larger ˜K > K , with the similar process we will obtain that Sis not ˜K-Lipschitz. This is a contradiction to the initial assumption. Thus, Sis not a Lipschitz function. 4 Lipschitz Condition for Ultimate Bearing Capacity In order to find the ultimate bearing capacity of piles, the function Swill be restricted (Stoll, 1963; Judson, 2012) to a connected bounded subset of R+(Andr´ e, 2020). Suppose Qu⊂R+is the subset. It is necessarily assumed that infQu= 0. And since Quis bounded, then supQu<∞ holds. Follows from theorem 1, we find that S:R+→R+is not Lipschitz. However, it will be shown that the restriction S Qu:Qu→R+is Lipschitz, and consequently, Sis locally Lipschitz. Theorem 2. The restriction ˜S:=S Qu:Qu→R+on the settlement function S:R+→R+as defined in accordance with equation 3 is a R-Lipschitz function. Proof. By the boundedness of Qu,then let qu:= sup Qu. LetR:=S′(qu).By remark 2, we have ∀u∈Qu,˜S′(u) =S′(u)≤S′(qu) =R . 6 Given a pair of point p, q∈Quwith p < q , then by mean value theorem (Rudin, 1964) there exists some point s∈(p, q) such that ˜S(q)−˜S(p) q−p=˜S′(s). Note that s≤qu,since every point in Quis less than or equal to qu.Follows from remark 2, we obtain |˜S(p)−˜S(q)| |p−q|=˜S(q)−˜S(p) q−p=˜S′(s)≤S′(qu) =R , which shows that ˜SisR-Lipschitz. Remark 3.By theorem 1, the restriction ˜S:=S Qu:Qu→R+ofS:R+→R+isR-Lipschitz. Thus, by definition 2, Sis locally Lipschitz. The set Quis interpreted in terms of SLT as the points of loading in which the pile is still safe. The pile is assmued to collapse at the least upper bound of Qu,that is, the supremum (Rudin, 1964) of Qs.Thus, the ultimate bearing capacity of piles based on SLT is defined by qu:= sup Qu. (8) In order to find qu,we need to determine the best constant R .We propose two methods for this case. The first method is by considering the limit of settlement as determined by the standard (or code) in charge or by an empirical justification. In this scenario, the notion of the Lipschitz continuity of Sis redundant in terms of finding qu,but is still necessary for describing the nature ofS .The formal description of this method is presented as follows. Formulation 1 (Limit Settlement Method) .Suppose the settlement of piles has been limited to some number according to the standard in charge. Let L > 0 be the limit of the settlement. Follows from equation 3, the ultimate bearing capacity of piles quis assumed to satisfy L=S(qu) =reτqu−r . (9) And quis given by qu=lnL r+ 1 τ. (10) The second method, we limit the speed of settlement relative to the initial settlement. We postulate that the speed of the settlement at every point in Qucannot exceed twice the speed of the initial settlement. And the formal description of this postulate is presented as follows. Postulate 1. The speed of the settlement is the first derivative of the settlement function S: R+→R+.The derivative of Sexists as shown by remark 2, and we obtain ∀q∈R+, S′(q) =τrcτq< S′(qu). Then it is postulated that S′(qu) S′(0)= 2, which implies S′(qu) = 2 S′(0) = 2 τr . (11) Consequently, the restriction ˜S:=S Qu:Qu→R+is 2τr-Lipschitz, i. e., the Lipschitz constant Ris given by R= 2τr .And we have Qu={q∈R+|S′(q)<2τr}. (12) 7 Formulation 2 (Limit Settlement Speed Method) .Follows from postulate 1, the derivative of S atquis given by S′(qu) = 2 τr , Then we have 2τr=τreτqu, which implies qu=ln 2 τ(13) For a more conservative approach, formulation 1 and formulation 2 can be combined to obtain a more complete consideration. This combination is formally presented as follows. Formulation 3. Follows from equation 9 of formulation 1 and equation 11 of formulation 2, it is designated that S′(qu) = min {τ(L+r),2τr}, (14) where L > 0 is the limit settlement determined from the code in charge. And the restriction ˜S:=S Qu:Qu→R+is an R-Lipschitz with R=S′(qu) = min {τ(L+r),2τr}. (15) Thus, Qu:={q∈R+|S′(q)< τ(L+r)∧S′(q)<2τr}. (16) And quis given by qu= sup Qu=ln minL r+ 1,2  τ(17) 5 Discussion This section discusses the test on the method that has been built on earlier sections. A comparison will be made for the method constructed in this paper with an existing method called the Davisson’s method (Davisson, 1973). In Davisson’s method, the ultimate bearing capacity of pile is determined as a point qu∈R+such that the curve of load-settlement and a linear function δ:R+→R+is intersecting at qu.The linear function δis given in accordance with (Davisson, 1973) by ∀q∈R+, δ(q) := 0 .012Dr+ 0.1D Dr +qℓ ApEp, where Dris the reference pile diameter (taken equal to 300 mm), Dis the pile diameter in use (in mm), ℓis the pile length (in mm), Apis the pile cross section (in mm2) and Epis the Young’s modulus of pile material (in kN/mm2) (Das, 2011). Given a data set of compressive SLT result from a project in Indonesia. The data set consists only the load and settlement values during loading, and is presented on table 1. The data set is then modelled with the settlement function S .Several values are taken as follows: ˜q= 801 .9 kN ,2˜q= 1603 .8 kN , s ν= 2.398 mm , s ξ= 5.318 mm . And we obtain r= 11.000969 and τ= 0.000246 .And the particular solution to Sfor this data set is given by ∀q∈R+, S(q) = 11 .000969 e0.000246 q−11.000969 . The error metric is presented to measure the accuracy of the model with the actual data, which is given in the form of mean absolute error (MAE) (Willmott and Matsuura, 2005) by MAE =1 α+ 1αX k=0|sk−S(qk)|. And in this case, MAE = 0 .051474 mm, which is considerably small. The plot of the restriction S [0,5000]: [0,5000]→R+is presented in figure 2. 8 Table 1: SLT Data from a Project in Indonesia Load Settlement Load Settlement (kN) (mm) (kN) (mm) 0.000 0.000 980.100 3.015 44.600 0.023 1024.700 3.205 88.100 0.115 1069.200 3.375 133.700 0.228 1113.800 3.568 178.200 0.398 1158.300 3.678 222.800 0.490 1207.900 3.843 267.300 0.630 1247.400 3.995 311.900 0.758 1292.000 4.153 356.400 0.913 1336.500 4.318 410.000 1.055 1381.100 4.480 445.500 1.185 1425.600 4.643 490.100 1.340 1470.200 4.818 534.600 1.465 1514.700 4.995 579.200 1.653 1559.300 5.173 623.700 1.780 1603.800 5.318 668.300 1.933 1648.400 5.490 712.800 2.080 1692.900 5.638 757.400 2.250 1737.500 5.843 801.900 2.398 1782.000 5.993 846.500 2.545 1825.500 6.195 891.000 2.685 1871.100 6.387 935.600 2.870 1915.700 6.640 Figure 2: Plot from settlement function Srestricted to [0 ,5000] 5.1 Interpretation with Davisson Method Follows from Davisson’s method (Davisson, 1973), we obtain S(qu) =δ(qu) For this SLT data, the length of pile is given by ℓ= 23000 mm, the cross sectional area for the pile is given by Ap= 282743 .339 mm2,and the concrete Young’s modulus is given by 33 .234 kN/mm2. 9 Thus, we obtain 11.000969 e0.000246 qu−11.000969 = 3 .8 +qu 0.002447 The plot of the equation above is presented on figure 3. And quis given by qu≈2700 kN. Figure 3: Davisson’s Method 5.2 Interpretation with Lipschitz Condition In the project in which the SLT data was produced, the limit for pile settlement is set to be 25 mm. Hence we designate L:= 25 mm. And follows from equation 17, we obtain qu=ln min25 11.000969,2  0.000246= 2819 .25 kN. 6 Conclusion and Future Work As presented in section 5, it is concluded that the pile ultimate bearing capacity under the inter- pretation of Lipschitz condition is slightly greater than that of Davisson, given a SLT data set of a project in Indonesia. This result shows that the theory is valid. In terms of calculation process, it is believed that the use of the method presented in this paper is simple and direct, since one can simply apply equation 6 to calculate r ,apply equation 7 to calculate τ, apply equation 4 to calculate Sand apply equation 17 to calculate the ultimate capacity of pile under Lipschitz condition. Possible further research regarding this topic that can still be conducted is figuring out whether there exists a more accurate criterion to determine the ultimate bearing capacity with the Lipschitz condition based on more data sets as experimental samples. 10