# Asymptotic Stability for an Axis-Symmetric Ohmic Heating Model in Thermal Electricity

Authors: Anyin Xia; Mingshu Fan; Shan Li
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

The asymptotic behavior of the solution for the Dirichlet problem of the parabolic equation with nonlocal termut=urr+ur/r+f(u)/(a+2πb∫01‍f(u)rdr)2, for0<r<1,t>0,u1,t=u′(0,t)=0, fort>0,ur,0=u0r,for0≤r≤1. The model prescribes the dimensionless temperature when the electric current flows through two conductors, subject to a fixed potential difference. One of the electrical resistivity of the axis-symmetric conductor depends on the temperature and the other one remains constant. The main results show that the temperature remains uniformly bounded for the generally decreasing function f(s), and the global solution of the problem converges asymptotically to the unique equilibrium.


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## Body

## 1. Introduction

In this paper, we consider the asymptotic behavior for the following initial boundary value problem:(1)ut=urr+urr+f(u)(a+2πb∫01‍f(u)rdr)2,for0<r<1,t>0,u(1,t)=u′(0,t)=0,fort>0,u(r,0)=u0(r),for0≤r≤1,
wherea and bare some positive parameters.fis a continuous, positive, and decreasing function, and the initial datau0(r)is a decreasing smooth positive function.The original motivation for studying such problems comes from the plasma Ohmic heating process. The plasma is an electrical axis-symmetric conductor and so it could be heated by passing a current through it. This is called Ohmic heating and it is the same kind of heating that occurs in thermistors. We consider that the axis-symmetric conductorAis a part of simple circuit in series with another constant one, Band a constant voltageEis applied (see Figure 1). Letu(x,t)andρ(u)be the temperature and the electrical resistivity of the conductorA, respectively.Figure 1
Electric current flows through two conductors.Here, the conductorA⊂ℝ3is a prismatic one with the lengthLand the cross-sectional area S, and the length of conductorBisL′. Assume that the diameter of the cross-sectionDis much less thanLand the temperatureu(x,t)is independent of the variablex1. Suppose that the curved surface of the conductorA,Γ0is well thermal, and we can specify
(2)u=0onΓ0.Based on the derivation in [1] (see also [2, 3]), we get the temperatureuof the material which satisfies the following:
(3)ut-Δu=f(u)(a+b∫D‍f(u)dx)2,x∈D,t>0,u(x,t)=0,x∈∂D,t>0,u(x,0)=u0(x),x∈D,
wheref(u)=ρ-1(u),  a=L/E,  and b=(ρ0/E)L′. The initial datau0(x)is a positive, smooth function and satisfiesu0(x)=0on∂D.In this paper, we focus on the problem (3) in radially symmetric case. So, we assume additionally that cross-section D is a unit disk and the initial data be radially and decreasing, that is,
(4)u0(x)=u0(r),u0′(r)<0,
wherer=|x|∈(0,1). Thus, the problem in axis-symmetric case can be formulated into the problem (1). Furthermore, it is easy to see that the axis-symmetric solution to the problem (1) is radially decreasing (see [4]).Here, we would like to address the works on the Ohmic heating model with one conductorA. The problem with only one conductor can be formulated into the following problem with different boundary conditions:
(5)ut=Δu+λf(u)(∫Ω‍f(u)dx)2,forx∈Ω,t>0,
whereΩ⊂ℝ2is an open, bounded domain,f(u)is the electrical conductivityσ(u)=1/ρ(u), and the parameterλis a positive constant, which is depending upon the electric current or potential difference and also upon the “size” of the conductor (see [1, 5–9]).For problem (5), Lacey et al. have proved that iffis an increasing function, then the blow up cannot take place (see [1, 10]). Iffis a decreasing function, Lacey proved that comparison techniques was valid, by which he studied the asymptotic behavior of the solutions to (5) for special f (see [1]). Taking the advantage of this fact, Lacey [1, 6] and Tzanetis [7] proved the occurrence of blow up for one-dimensional model (5) and for the two-dimensional radially symmetric model (5), respectively. On the other hand, they proved that the global solution of (5) for some specialf(u), such asf(u)=e-u, asymptotically converges to its unique steady state.In [2], Du and Fan considered the nonlinear diffusion model for two conductors with one of the conductors remains constant. Whenfis decreasing, they proved that comparison principle is valid and the solution of the model was always global in time. Furthermore, iffis a decreasing exponential function, they proved that the solution of the problem converges asymptotically to the unique steady state. See also [3] for some results on asymptotic behavior of the global solution in one-dimensional case.Inspired by these works, modified by the methods in [2], one can easily prove that the comparison principle for model (1) is valid and the solution of (1) is global in time. Finally, the main purpose of this paper is to give the asymptotic behavior and to show the asymptotic stability of the problem (1) with generally deceasing functionf(u).Theorem 1.
Assume thatfsatisfies
(6)-sf′(s)<f(s),fors>0,
the solutionu(r,t)of the problem (1) converges asymptotically to the unique steady stateω(r), namely,
(7)u(r,t)⟶ω(r),ast⟶+∞,
for0<r<1.Remark 2.
The equations in models (1), (3), and (5) are semilinear parabolic equations with nonlocal sources. For the works on the global existence and blow up of nonlocal parabolic equations, the authors would like to refer to [11–14] and the references therein.The analysis and techniques in this paper is based on the analysis for the ordinary differential equations and comparison arguments.


## 2. The Asymptotic Stability for Problem (1)

In this section, we will consider the asymptotic stability for the problem (1), and give the proof of Theorem 1.Proof of Theorem1. Firstly, we deal with the local steady solutionω(r;μ)corresponding to the problem (1), consider
(8)ω′′(r;μ)+1rω′(r;μ)+μf(ω(r;μ))=0,0<r<1,ω(1;μ)=ω′(0;μ)=0,
with the positive parameter
(9)μ=1(a+2πb∫01‍f(ω(r;μ))rdr)2.Moreover, multiplyingron both sides of the equation in (8), and integrating over[0,1]yield
(10)∫01‍f(ω(r;μ))rdr=-ω′(1;μ)μ.
Multiplyingω′(r;μ)on both sides of the equation in (8), and integrating over[0,1]yields
(11)(ω′(1;μ))2=2μ∫0M(μ)‍f(s)ds-2∫01‍(ω′(r;μ))2rdr,
whereM(μ)=ω(0;μ)=maxr∈[0,1]ω(r;μ).In view ofω′(r;μ)<0, for anyr∈(0,1), it follows from (11) that
(12)ω′(1;μ)=-2μ∫0M(μ)‍f(s)ds-2∫01‍(ω′(r;μ))2rdr.Combining this with (9) and (10) yields
(13)μ(a+2πb∫01‍f(ω(r;μ))rdr)2=a2μ+4π2b2(ω′(1;μ))2μ-4πabω′(1;μ)=1.It is easily seen that the problem (8) does not possess nontrivial solution with the parameterμ=0. Namely,ω≡0andM=0forμ=0. Therefore, it follows from (11) that
(14)limμ→0∫01‍(ω′(r;μ))2rdr⟶0,
since(ω′(1;μ))2=2μ∫0M(μ)‍f(s)ds-2∫01‍((ω′(r;μ))2/r)dr>0and∫01‍((ω′(r;μ))2/r)dr>0.Define the following:(15)F(μ):=a2μ+4π2b2(ω′(1;μ))2μ-4πabω′(1;μ)-1,forμ>0.
Then, it follows from a direct computation thatF(0+)<0andF(1/a2)>0. Then, there exists at least one rootμ∈(0,1/a2)to (13).Making odd extension to the problem (8), we get
(16)ω′′(r;μ)+1rω′(r;μ)+μf(ω(r;μ))=0,-1<r<1,ω(±1;μ)=0.We claim that the differentiation ofω(r;μ)with respect to the parameterμ is always positive, namely,ωμ(r;μ)>0, for any0≤r≤1. In fact, by differentiating on both sides of (16) respect to the parameterμ, one gets
(17)-ωμ′′(r;μ)-1rωμ′(r;μ)-μf′(ω(r;μ))ωμ(r;μ)=f(ω(r;μ))>0.In view ofωμ(±1;μ)=0, by comparison principle, we haveωμ(r;μ)>0, for anyr∈[-1,1]. Then,Mμ(μ)=ωμ(0;μ)>0.Set(18)φ(μ)∶=(ω′(1;μ))2μ=2μ∫0M(μ)‍f(s)ds-2∫01‍((ω′(r;μ))2/r)drμ.
Letz(r;μ)=ω(r;μ)/α,  α=μ, we can rewrite
(19)φ(μ)=(z′(1;μ))2.
Then, it follows from (8) thatzα(1;μ)=zα′(0;μ)=z(1;μ)=z′(0;μ)and
(20)z′′(r;μ)+z′(r;μ)r+αf(ω(r;μ))=0.Differentiating on both sides of (20) with respect to the parameterα, we have
(21)zα′′(r;μ)+zα′(r;μ)r+α2f′(ω(r;μ))zα(r;μ)=-f(ω(r;μ))-f′(ω(r;μ))ω(r;μ).If the condition (6) holds, then it follows from maximum principle and Hopf’s boundary lemma that
(22)zα(r;μ)>0,zα′(1;μ)<0.
Thus,
(23)φ′(μ)=12αddα(z′(1;μ))2<0,
sincez′(1;μ)=ω′(1;μ)/α<0. Therefore, we can conclude thatF′(μ)>0, for anyμ∈(0,1/a2). Furthermore, (13) possesses a unique rootμ*in(0,1/a2), which shows that the problem possesses unique steady stateω(r,μ*).Next, we will show that the global solution of the problem (1) converges to its steady state. Inspired by the form of steady stateω(r;μ*), we seek for a decreasing in time, upper solution of a form similar to the steady statev-(r,t)=ω(r,μ-(t))=ω-, whereμ-(t)will be determined later.Then, one has(24)v-t-v-rr-1rv-r-f(v-)(a+2πb∫01‍rf(v-)dr)2=ω-μ-·μ-′(t)-ω-rr-1rω-r-f(ω-)(a+2πb∫01‍rf(ω-)dr)2=ω-μ-·μ-′(t)-f(ω-)(a+2πb∫01‍rf(ω-)dr)2×[1-μ-(t)(a+2πb∫01‍rf(ω-)dr)2]=ω-μ-{μ-′(t)-f(ω-)ω-μ-(a+2πb∫01‍rf(ω-)dr)2f(ω-)ω-μ-(a+2πb∫01‍rf(ω-)dr)2×[1-μ-(t)(a+2πb∫01‍rf(ω-)dr)2]}.Sinceu0(r)and u0′(r)are bounded in[0,1], we can chooseμ-(0), such that
(25)v-(r,0)=ω(r,μ-(0))≥u0(r).Set(26)g(μ-(t))=(infr∈[0,1]f(ω-)ω-μ-)1(a+2πb∫01‍rf(ω-)dr)2×[μ-(t)(a+2πb∫01‍rf(ω-)dr)2-1].
Note thatg(μ-(t))≥0, provided thatμ-(t)≥μ*, whereμ*is the unique root of (13), sinceF(μ)is increasing with respect toμ. Thus, we can choose a decreasing functionμ-(t)such that
(27)0≥μ-′(t)≥-g(μ-(t)),limt→∞μ-(t)=μ*.
Then we have established an upper solutionv¯(r,t)=ω(r;μ-(t))to the problem (1), which satisfies
(28)v-(r,t)⟶ω(r;μ*)=ω(r),ast⟶∞.Similarly, we can construct a lower solution asv_(r,t)=ω(r;μ_(t)), whereμ_(t)is increasing in time and tends toμ*. Finally, by the comparison principle, we obtain a pair of upper-lower solutions(v-(r,t),v_(r,t)), such that
(29)v_(r,t)≤u(r,t)≤v-(r,t),forr∈[0,1],t>0,
and it completes the proof of Theorem 1.



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*Source: 387565-2013-08-12.xml*