AHybrid Solutions Method for Solving One Dimensional Parabolic Partial Differential Equations
Sunday Babuba*
Department of Mathematics, Federal University Dutse, Nigeria
Submission: March 19, 2018; Published: July 02, 2018
*Corresponding author: Sunday Babuba, Department of Mathematics, Federal University Dutse, Nigeria, Tel: ; Email: vishaaldeo@gmail.com
How to cite this article: Sunday B. AHybrid Solutions Method for Solving One Dimensional Parabolic Partial Differential Equations. Biostat Biometrics Open Acc J. 2018; 7(3): 555713. DOI: 10.19080/BBOAJ.2018.07.555713
Abstract
A new continuous numerical method based on polynomials approximation is here proposed for solving the equation arising from heat transfer along a copper rod and a hollow tube subject to initial and boundary conditions. The method results from discretization of the heat equation which leads to the production of a system of algebraic equations. By solving the system of algebraic equations we obtain the problem approximate solutions.
Keywords: Polynomials; Interpolation; Multistep collocation; Parabolic partial differential Equations; Numerical method; One dimensional; Hybrid solutions; Algebraic equations; Heat equation; Heat conduction equation; Interpolation; Collocation; Temperature; Heat flow; Polynomials; Numerical accuracy; Evaporates; Zero; Ethyl alcohol; Initial temperature; Interpolation point.
Introduction
The development of continuous numerical techniques for solving heat conduction equation in science and engineering subject to initial and boundary conditions is a subject of considerable interest. In this paper, we develop a new numerical method which is based on interpolation and collocation at some point along the coordinates [1-3]. To do this we let U(x,t) represents the temperature at any point in the rod and the tube. Heat is flowing from one end to another under the influence of temperature gradient .Ux∂∂ To make a balance of the rate of heat flow in and out of the media, we consider R for thermal conductivity of the steel, C the heat capacity which we assume constants, and ρ the density and D the thermal diffusivity of alcohol. Heat flow in the rod is given by
Where A and B are the cross sections of the rod and the tube respectively. Our new method strives to provide solutions to the heat flow eqns. (1.0) and (1.1).
The solution method
To set up the solution method we select an integer N such that .0>N We subdivide the interval Xx≤≤0 into N equal subintervals with mesh points along the space axis given by , where .Nh=X Similarly, we reverse the roles of x nd t, we select another integer Msuch that .0>M We also subdivide the interval Tt≤≤0 into M equal subintervals with mesh points along time coordinate given by , where ,TMk= and kh, are the mesh sizes along space and time coordinates [4-6]. We seek for the approximate solution , to U(x,t), of the form
over h>0,k>0 mesh sizes, such that . We denote p to be the sum of interpolation points along the space and time coordinates respectively. That is ,bgp+= where g is the number of interpolation points along the space axis, while b is the number of interpolation points along the time coordinate [7]. The bases functions rrsqr,sr1,...p-1rare the Taylor’s and Legendre’s polynomials which are known, raare the constants to be determined. The interpolation values are assumed to have been determined from previous steps, while the method seeks to obtain
Applying the above interpolation conditions on eqn. (2.0) we obtain
We let arbitrarily and k=0, then by Cramer’s rule, eqn. (2.1) becomes
and
Where exist. Hence, from eqn. (2.2), we obtain
The vector is now determined in terms of known parameters in , then
Substituting eqn. (2.4) into eqn. (2.5) we obtain
We reverse the roles of x and t in eqn. (2.1) and we arbitrarily set again by Cramer’s rule, eqn. (2.1) become
and
where exist.
Hence, from eqn. (2.7) we obtain
The vector is now determined in terms of known parameters in row of ,L then
Also, eqn. (2.9) determines the values of ar explicitly.
Taking the first derivatives of eqn. (2.0) with respect to t we obtain
Substituting eqn. (2.9) into eqn. (2.10) we obtain
But by eqns. (1.0) and (1.1) it is obvious that eqn. (2.11) is equal to eqn. (2.6), therefore,
Collocating eqn. (2.12) at ixx= and jtt= produces a new numerical scheme that solves equations (1.0) and (1.1) explicitly.
Numerical examples
In this section, we will test the numerical accuracy of the new method by using the new scheme to solve two examples. That is, we compute an approximate solutions of eqns. (1.0) and (1.1) at each time level. To achieve this, we truncate the polynomials after second degree and the average is used as the basis function in the computation [8-10]. The resultant scheme is used to solve the following problems.
Example 1
A hollow tube 20 cm long is initially filled with air containing 2% of ethyl alcohol vapors [2]. At the bottom of the tube is a pool of alcohol which evaporates into the stagnant gas above. (Heat transfers to the alcohol from the surroundings to maintain a constant temperature of 30 °C, at which temperature the vapour pressure is 0.1 atm.) At the upper end of the tube, the alcohol vapors dissipate to the outside air, so the concentration is essentially zero. Considering only the effects of molecular diffusion, determine the concentration of alcohol as a function of time and the distance x measured from the top of the tube.
Molecular diffusion follows the law
Where, D is the diffusion coefficient, with units of cm2/sec. (This is the same as for the ratio k/cp, which is often termed thermal diffusivity.) For ethyl alcohol, D = 0.119 cm2/sec at 30 °C , and the vapor pressure is such that 10 volume percent alcohol in air is present at the surface [11-15]. The initial condition is ()0.20,=xc, and boundary conditions are ()()10,20,0,0==tctc. Subdivide the length of the tube into five intervals, so that xΔ = 4cm. Using the maximum value permitted for tΔ yields
Taking 32,4==αβ implies that For and by taking two interpolation points along space coordinate and one interpolation point along time coordinate implies that 3,1,2=⇒==pbg, and this simply means that then the calculated concentration of alcohol is as shown in Table 1 [15-20].
Example 2
Solve for the temperature in a copper rod 1.2 cm long, with the outer curved surface insulated so that heat flows in on one direction [21-24]. If the initial temperature ()Co within the rod are given by
Find the temperature as a function of x and t if both faces are maintained at 0° C. For copper, 433.0,37.0==cpk. We use ,20.0cmx=Δ we then find tΔ by the relation
Taking 32,4==αβ implies that .For and by taking two interpolation points along space coordinate and one interpolation point along time coordinate implies that ,1,2==bgand 3=p, this simply gives 81,041,0,41=−=kandh, then the calculated temperatures are as shown in Table 2 [25-27].
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