**4. Positivity-Preserving Scattered Data Interpolation**

In this section, we apply the proposed scheme discussed in the previous section to preserve the positivity of scattered data sets. To do this, first we derive the sufficient condition for the positivity of the quartic triangular spline defined in (5). Finally, the rational corrected scheme defined by (19) will be used to construct a positive surface with C1 continuity.

To derive the sufficient condition for the positivity of the quartic spline triangular patch, we adopted a similar approach to Saaban et al. [35] and Piah et al. [36]. Assume that the quartic ordinates at the vertices are strictly positive such *b*300, *b*030, *b*<sup>003</sup> > 0. Let *A* = *b*300, *B* = *b*030,*C* = *b*003, therefore *A*, *B*, *C* > 0 (see Figure 11). Meanwhile, let the other ordinates have the same value, that is −*t* < 0 where *t* > 0. Thus, Equation (5) becomes

$$\begin{aligned} P(u,v,w) &= Au^4 \quad +Bv^4 + Cw^4 \\ &\quad -t\Big(u^2v \ (3+u) + (3+u)u^2w + (3+v)v^2u + (3+v)v^2w + (3+w)w^2u \\ &\quad + (3+w)w^2v + 6uvw \Big) \\ &= Au^4 + Bv^4 + Cw^4 - t\Big(1 - u^4 - v^4 - w^4\Big) \\ &= (A+t)u^4 + (B+t)v^4 + (C+t)w^4 - t \end{aligned} \tag{23}$$

**Figure 11.** Quartic triangular ordinates arrangement for positivity preservation.

From (23) we can observe that when *t* = 0 then *P*(*u*, *v*, *w*) > 0. Meanwhile as *t* increases, *P*(*u*, *v*, *w*) decreases. We want to find the value of *t* = *t*<sup>0</sup> when the minimum value of *P*(*u*, *v*, *w*) = 0. By taking first partial derivatives, we will obtain the following:

$$\begin{array}{l} \frac{\partial P(u,v,w)}{\partial u} = 4(A+t)w^3, \\ \frac{\partial P(u,v,w)}{\partial v} = 4(B+t)v^3, \\ \frac{\partial P(u,v,w)}{\partial w} = 4(C+t)w^3. \end{array} \tag{24}$$

The minimum value of *P*(*u*, *v*, *w*) occurs when

∂*P* <sup>∂</sup>*<sup>u</sup>* <sup>−</sup> <sup>∂</sup>*<sup>P</sup>* <sup>∂</sup>*<sup>v</sup>* <sup>=</sup> 0 and <sup>∂</sup>*<sup>P</sup>* <sup>∂</sup>*<sup>u</sup>* <sup>−</sup> <sup>∂</sup>*<sup>P</sup>* <sup>∂</sup>*<sup>w</sup>* = 0 or equivalently.

$$\frac{\partial P}{\partial u} = \frac{\partial P}{\partial v} = \frac{\partial P}{\partial w} \tag{25}$$

From Equation (25) we have *u*3 *<sup>v</sup>*<sup>3</sup> <sup>=</sup> *<sup>B</sup>*+*<sup>t</sup> <sup>A</sup>*+*<sup>t</sup>* and *<sup>u</sup>*<sup>3</sup> *<sup>w</sup>*<sup>3</sup> <sup>=</sup> *<sup>C</sup>*+*<sup>t</sup> A*+*t* Hence:

$$u^3: v^3: w^3 = \frac{1}{A+t}: \frac{1}{B+t}: \frac{1}{C+t}.$$

Since *u* + *v* + *w* = 1, we obtain the following relations:

$$\mu = \frac{\frac{1}{\left(A+t\right)^{1/3}}}{\frac{1}{\left(A+t\right)^{1/3}} + \frac{1}{\left(B+t\right)^{1/3}} + \frac{1}{\left(C+t\right)^{1/3}}},$$

$$w = \frac{\frac{1}{\left(B+t\right)^{1/3}}}{\frac{1}{\left(A+t\right)^{1/3}} + \frac{1}{\left(B+t\right)^{1/3}} + \frac{1}{\left(C+t\right)^{1/3}}}, \text{ and } \frac{1}{\left(C+t\right)^{1/3}}, \text{ and } \frac{1}{\left(C+t\right)^{1/3}}.$$

$$w = \frac{\frac{1}{\left(C+t\right)^{1/3}}}{\frac{1}{\left(A+t\right)^{1/3}} + \frac{1}{\left(B+t\right)^{1/3}} + \frac{1}{\left(C+t\right)^{1/3}}}.$$

Substituting this value into (23) we obtain the minimum value of *P*(*u*, *v*, *w*) i.e., *<sup>P</sup>*(*u*, *<sup>v</sup>*, *<sup>w</sup>*) <sup>1</sup> which can be simplified to

$$\frac{1}{\left(\mu+t\right)^{1/3}} + \frac{1}{\left(\frac{1}{(t+t)^{1/3}} + \frac{1}{\left(\sum t\right)^{1/3}}\right)^3} \Bigg|\_{\text{min}}^3 \tag{26}$$

$$P(\boldsymbol{\mu}, \boldsymbol{\upsilon}, \boldsymbol{w}) \frac{t}{\left[\frac{1}{\left(\boldsymbol{A}/t+1\right)^{1/3}} + \frac{1}{\left(\boldsymbol{B}/t+1\right)^{1/3}} + \frac{1}{\left(\boldsymbol{C}/t+1\right)^{1/3}}\right]^3} \tag{27}$$

Now, *P*(*u*, *v*, *w*)*min* when

$$\frac{1}{\left(A/t+1\right)^{1/3}} + \frac{1}{\left(B/t+1\right)^{1/3}} + \frac{1}{\left(C/t+1\right)^{1/3}} = 1\tag{27}$$

Let *s* = 1/*t*, then

$$G(s) = \frac{1}{\left(As + 1\right)^{1/3}} + \frac{1}{\left(Bs + 1\right)^{1/3}} + \frac{1}{\left(Cs + 1\right)^{1/3}}\tag{28}$$

3

Then Equation (27) can be written as *G*(*s*) = 1,*s* ≥ 0. This equation can be solved by using regula-falsi method with suitable choice of initial guess.

Since *A*, *B*, *C* > 0 and *s* ≥ 0 then *G* (*s*) < 0 and *G*(*s*) > 0. (see Figure 12). Thus, the curve is convex on that region. Let *X* = *max*(*A*, *B*,*C*) and *Y* = min(*A*, *B*,*C*), then the following holds

3

**Figure 12.** Function *G*(*s*) for *s* ≥ 0.

Such that

$$G\left(\frac{26}{X}\right) \ge 1 \text{ and } G\left(\frac{26}{Y}\right) \le 1\dots$$

Figure 12 shows the example of the relative locations of <sup>26</sup> *<sup>X</sup>* and <sup>26</sup> *<sup>Y</sup>* and *s*0. Now we establish the main theorem for positivity preservation using the proposed scheme.

**Theorem 3.** *Consider the quartic triangular patch P*(*u*, *v*, *w*) *with vertex A* = *b*300, *B* = *b*030,*C* = *b*003, *such that <sup>A</sup>*, *<sup>B</sup>*,*<sup>C</sup>* <sup>&</sup>gt; 0. *If the remaining quartic triangular ordinates are equal to* <sup>−</sup>*t*<sup>0</sup> *where <sup>t</sup>*<sup>0</sup> = <sup>1</sup> *<sup>s</sup>*<sup>0</sup> *is a unique solution to (28), then P*(*u*, *v*, *w*) ≥ 0 *for all u*, *v*, *w* ≥ 0 *and u* + *v* + *w* = 1.

Some observations from Theorem 3 can be made as follows: Let *A* = *B* > *C* = 1. Then, we have

$$G(s) = \frac{2}{\left(As + 1\right)^{1/3}} + \frac{1}{\left(s + 1\right)^{1/3}}$$

Therefore, as *<sup>A</sup>* → ∞, then *<sup>G</sup>*(*s*) <sup>→</sup> <sup>1</sup> (*s*+1) 1/3 . Hence, *s*<sup>0</sup> → 0 and therefore *t*<sup>0</sup> → ∞. Thus, the ordinate values are unbounded compared with the work of Chan and Ong [7] in which the Bézier ordinates are bounded by a lower bound −1/3.

**Remark 1.** *The su*ffi*cient condition for the positivity of the quartic triangular patch developed in this study is the same as the su*ffi*cient condition for the quartic Bézier triangular patch developed in Saaban et al.* [35]*. The main di*ff*erence is that the proposed quartic polynomial only requires ten control points (or ordinates) as compared to the quartic Bézier triangular which requires 15 control points and involves some optimization problems as shown in Saaban et al.* [35] *and Hussain et al.* [37,38]*. Therefore, the proposed positivity preservation using a quartic triangular patch requires less computation time than some established schemes for scattered data interpolation.*

The final construction of the positive scattered surface is described below:


3. Assign the first partial derivative at the respective data sites and adjust if necessary, to provide the positivity preservation;

4. The *C*<sup>1</sup> triangular surface is constructed via convex combination between three local schemes;

5. Repeat Steps 1 through 4 for other positive scattered data sets.
