STEADY STATE TUMOURS AND MOMENTS

OF MONOTONIC FUNCTIONS

\authors{K A Landman and E R Love} \affiliation{Department of Mathematics and Statistics, University of Melbourne} \authors{R Vyborny} \affiliation{Department of Mathematics, University of Queensland} In a paper on tumour growth dynamics, Please et al [1] show that a steady state spherically symmetric tumour exists with a necrotic region if the first and second moments of a function S(C) are both identically zero. The S(C) represents the growth and death rate of cells, as a function of the oxygen concentration C, and C is a function of radial distance x. The problem reduces to the following: Problem 1\enspace Let $S(x)$ be an increasing function on $(0,\infty)$. Find $a$ and $b$ such that $0 < a < b$ and \begin{equation}\displaystyle \int_a^b x S(x) dx = \int_a^b x^2 S(x) dx = 0 .\label{eq1st} \end{equation} Here a and b are the necrotic region radius and total tumour region radius respectively. In the tumour scenario, the consequence is that no such equilibrium solution exists if equation (

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{eq1st}) is only true for S(x)\equiv 0 on (a,b). Vyborny [2] recently considered a related problem with the following formulation: \\ Problem 2\enspace {\it{Is every (not necessarily strictly) increasing function $f$ equal to zero if } \begin{equation} \int_a^b f(x) dx = \int_a^b x f(x) dx = 0 ? \label{eq2nd} \end{equation}} To establish an affirmative answer, Vyborny uses the substitution \begin{equation} x = a + (b-a)t \label{sub} {equation} yielding the same moment equations now reduced to integrations on [0,1], and proves a more powerful result namely: \\ Theorem 1\enspace \label{th1} {\it{If $f$ is bounded and increasing (not necessarily strictly) on $(0, 1)$ and if there exist $m$ and $n$ with $0 \le m < n$ such that}} \begin{equation} \int_0^1 x^mf(x) dx = \int_0^1 x^n f(x) dx = 0 , \label{eq3rd} {equation} {\it{then $f(x) = 0$ on $(0, 1)$.}} Although it may appear that Theorem 1 can be applied to Problem 1, this is not the case. Indeed, if we try the substitution (

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{sub}) in Problem 1, the moment equations (

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{eq1st}) become \begin{equation} \int_0^1 \left[a + (b-a)t \right] \bar{S}(t) dt = \int_0^1 \left[a + (b-a)t \right]^2 \bar{S}(t) dt = 0 , \label{eq4th} {equation} where \bar{S}(t) = S(a + (b-a)t). Together these give \[ \int_0^1 \left[a t + (b-a)t^2 \right] \bar{S}(t) dt = 0 \] so that \[ \int_0^1 t^2 \bar{S}(t) dt = -\frac{a}{b-a}\int_0^1 t\bar{S}(t) dt = \frac{a^2}{(b-a)^2} \int_0^1 \bar{S}(t) dt = \frac{a^2}{(b-a)^3} \int_a^b S(x) dx . \] This is zero if and only if either a = 0 or the average of S(x) over (a, b) is zero (or both). Supposing the average of S(x) is not so restricted, equations (

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{eq4th}) are equivalent to \[ \int_0^1 t \bar{S}(t) dt = 0 = \int_0^1 t^2 \bar{S}(t) dt \] if and only if a = 0. Thus Theorem 1 can be used to prove that \bar{S}(t) = 0 in (0, 1), and hence S(x) = 0 in (a, b), if and only if a = 0. But Problem 1 requires that a > 0 so it cannot be solved by appeal to Theorem 1. In fact, for Problem 1 the best one can do using a change of integration variable is to transform (a, b) into (c, 1) with 0 < c < 1. In this case, following the method in [2], we find \begin{equation} \int_c^1 (x^m - x^n) dS(x) = - S(c)(c^m - c^n) . \label{eq5th} {equation} Certainly if S(x) is strictly increasing, then the integrals in (

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{eq1st}) must imply that S(c) < 0. Then the right hand side of (

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{eq5th}) is positive, which does not yield the desired result that S(x) = 0. Another attempt to answer Problem 1 using Theorem 1 is to write f(x) = xS(x). However, if we choose the increasing function S(x) = x - (a+b)/2, then f(x) = x \left[x - (a+b)/2 \right] is not a monotonic function over the domain (a, b) if 3a < b, since it has a local minimum at x = (a+b)/4 inside (a, b). Hence choices of S(x) can be made such that S(x) is increasing but f(x) is not. We have now established that Theorem 1, or variations to it, cannot provide an answer to Problem 1 - two different approaches will now be presented. The problem posed at (

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{eq1st}) by Please et al [1] is answered in the first approach by Theorem 2 with p=2 and q =3, and answered in the second approach by Theorem 3 with m=1 and n=2. Lemma\enspace \it{ If $a, b, p, q$ and $x$ are real numbers such that $0\le a0 .$$ } Proof\enspace \rm Let A=a^p, X=x^p, B=b^p and r=q/p>1. By the convexity of the graph of Y = X^r in X > 0, (X^r-A^r)/(X-A) is strictly increasing with respect to X, and consequently
\frac{x^q-a^q}{x^p-a^p}<\frac{b^q-a^q}{b^p-a^p}
for a < x \le b, since the left side is strictly increasing with respect to x. The stated inequality follows from this. Theorem 2\enspace \it{If $a, b, p$ and $q$ are real numbers such that $0\le a Proof\enspace \rm Using Riemann-Stieltjes integration by parts, \begin{eqnarray*} p\int_a^b x^{p-1}S(x) dx &=& b^pS(b)-a^pS(a)-\int_a^b x^{p} dS(x)\\ &=& (b^p -a^p)S(b)+a^p\{S(b)-S(a)\}- \int_a^b x^{p} dS(x)\ {eqnarray*} so that \begin{equation} 0= (b^p -a^p)S(b)-\int_a^b (x^{p}-a^p) dS(x) . \label{eq1} {equation} Similarly
0= (b^q -a^q)S(b)-\int_a^b (x^{q}-a^q) dS(x) .
Thus \begin{equation} (b^q -a^q)\int_a^b (x^{p}-a^p) dS(x)=(b^q -a^q)(b^p -a^p)S(b) = (b^p-a^p)\int_a^b (x^{q}-a^q) dS(x) . \label{eq2} {equation} Let
f(x)=(b^q -a^q)(x^p -a^p)-(b^p -a^p)(x^q -a^q) ;
then by (

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{eq2}) \begin{equation} \int_a^b f(x) dS(x)=0. \label{eq3} {equation} Suppose that S(a+0). Since S has at most countably many discontinuities, there must exist two points of continuity, called a' and b', such that a and S(a') By the lemma there is an m>0 such that f(x)\ge m in a'\le x\le b'. By (

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{eq3}), since f and S are continuous at a' and b', \begin{eqnarray*} 0=(\int_a^{a'}+\int_{a'}^{b'}+\int_{b'}^{b})f(x) dS(x) &\ge& \int_{a'}^{b'}f(x) dS(x)\ &\ge& m\int_{a'}^{b'} dS(x)= m\{S(b')-S(a')\}>0 . {eqnarray*} This contradiction shows that S(a+0)=S(b-0). Then by the monotonicity of S \begin{equation} S(a+0)=S(x)=S(b-0) \mbox{for $a < x < b$} , \label{eq4} {equation} so that S(x) is constant in a By (

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{eq1}),
(b^p -a^p)S(b)=\int_a^b (x^{p}-a^p) dS(x)=(b^p -a^p)\{S(b)-S(b-0)\} ,
since the only contribution to the integral comes from the two end-points. This implies that S(b-0)=0; whence by (

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{eq4}) S(x)=0 for a as required. Remark 1\enspace \rm The values of S(a) and S(b) remain almost arbitrary. They only have to satisfy S(a)\le 0 \le S(b), a restriction imposed by the monotonicity of S. Theorem 3\enspace \it{ If $ a, b, m$ and $n$ are real numbers such that $0\le a Proof\enspace \rm Clearly S(b-0)\ge 0, since otherwise the integrals (

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{th3}) would be negative. It suffices to prove that S(b-0) = 0; for then S(a+0) cannot be negative because of (

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{th3}); hence S(a+0) = 0 and it follows that S(x)=0 for a. We will assume that S(b-0)>0, (not excluding the possibility of an infinite limit), and show that this leads to a contradiction. Let H(x)=\int_a^xt^mS(t) dt. For some C, a, we have \int_C^b t^mS(t) dt>0, consequently H(C)<0. The continuous function H assumes on [a,b] its negative minimum value. Let c_1 and c_2 be the leftmost and rightmost point where H attains this minimum value. Clearly a and
0\ge H'(c_1-0)=\lim_{x\to c_1-0}x^mS(x) .
It follows that S\le 0 on (a,c_1). Actually S must be negative on (a,c_1), otherwise c_1 would not be the leftmost minimum point. Similar arguments show that S=0 on (c_1, c_2) and that S>0 on (c_2, b). By the mean value theorem, there exists \xi with a<\xi such that
\int_a^{c_1} x^nS(x) dx=\xi^{n-m}\int_a^{c_1}x^m S(x) dx .
Similarly
\int_{c_2}^b x^nS(x) dx=\eta^{n-m}\int_{c_2}^bx^m S(x) dx ,
for some \eta with c_2<\eta. Consequently \begin{equation} \int_a^bx^nS(x) dx=\xi^{n-m}\int_a^{c_1}x^m S(x) dx+ \eta^{n-m}\int_{c_2}^bx^m S(x) dx .\label{eqn12} {equation} Combining (

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{th3}) and (

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{eqn12}) yields
0=\xi^{n-m}\int_a^{b}x^m S(x) dx+ (\eta^{n-m}-\xi^{n-m})\int_{c_2}^bx^mS(x) dx = (\eta^{n-m}-\xi^{n-m})\int_{c_2}^bx^mS(x) dx>0 .
This contradiction completes the indirect proof. Remark 2\enspace \rm Since S is {\em not} assumed to be bounded in Theorems 2 and 3, the integrals in (

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{th2}) and (

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{th3}) are understood either as improper integrals or as Lebesgue integrals. Remark 3\enspace \rm Theorems 2 and 3 also hold for S(x) non-increasing, since we can replace S(x) by -S(x). Hence both theorems hold for all monotonic functions.
\bf References \begin{description}
\it Avascular tumour dynamics and necrosis , Math Models and Methods in App Sciences, to appear.
\it Moments of an increasing function , Aust. Math. Soc. Gazette, Vol. 25, No. 3, (1998) p.142. {description}

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