РЕШЕНИЕ ДИФФУЗИОННОГО УРАВНЕНИЯ - Студенческий научный форум

# VI Международная студенческая научная конференция Студенческий научный форум - 2014

## РЕШЕНИЕ ДИФФУЗИОННОГО УРАВНЕНИЯ

Омарова Б.Ж. 1, Есеркепов Е.Ж. 2
1Актюбинский региональный государственный университет им.К.Жубанова
2Актюбинский региональный государственный университет имени К.Жубанова
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1. Homogeneous boundary conditions.

We consider one dimensional diffusion in a pipe of length L, and solve the diffusion equation for the concentration

(1)

Both initial and boundary conditions are required for a unique solution. That is, we assume the initial concentration distribution in the pipe is given by

(2)

Furthermore, we assume that boundary conditions are given at the ends of the pipes. When the concentration value is specified at the boundaries, the boundary conditions are called Dirichlet boundary conditions. As the simplest example, we assume here homogeneous Dirichlet boundary conditions, that is zero concentration of dye at the ends of the pipe, which could occur if the ends of the pipe open up into large reservoirs of clear solution,

(3)

We will later also discuss inhomogeneous Dirichlet boundary conditions and homogeneous Neumann boundary conditions, for which the derivative of the concentration is specified to be zero at the boundaries. Note that if is identically zero, then the trivial solution satisfies the differential equation and the initial and boundary conditions and is therefore the unique solution of the problem. In what follows, we will assume that is not identically zero so that we need to find a solution different than the trivial solution.

The solution method we use is called separation of variables. We assume that can be written as a product of two other functions, one dependent only on position x and the other dependent only on time t. That is, we make the ersatz

(4)

Whether this ersatz will succeed depends on whether the solution indeed has this form. Substituting (4) into (1), we obtain

which we rewrite by separating the x and t dependence to opposite sides of the equation:

The left hand side of this equation is independent of t and the right hand side is independent of x. Both sides of this equation are therefore independent of both x and t and equal to a constant. Introducing –λ as the separation constant, we have

and we obtain the two ordinary differential equations

(5)

Because of the boundary conditions, we must first consider the equation for . To solve, we need to determine the boundary conditions at and . Now, from (3) and (4),

Since T(t) is not identically zero for all t (which would result in the trivial solution for u, we must have . Similarly, the boundary condition at requires . We therefore consider the two-point boundary value problem

(6)

The equation given by (6) is called an o de eigenvalue problem. The allowed values of λ and the corresponding functions are called the eigenvalues and eigenfunctions of the differential equation. Since the form of the general solution of the ode depends on the sign of λ, we consider in turn the cases and For we write and determine the general solution of

to be

Applying the boundary condition at , we find . The boundary condition at x=L then yields

The solution B=0 results in the trivial solution for u and can be ruled out. Therefore, we must have

which is an equation for the eigenvalue μ. The solutions are

where n is an integer. We have thus determined the eigenvalues to be

(7)

with corresponding eigenfunctions

(8)

For we write and determine the general solution of

to be

where we have previously intro diced the hyperbolic sine and cosine functions. Applying the boundary condition at , we find . The boundary condition at x=L then yields

which for has only the solution . Therefore, there is no nontrivial solution for u with Finally, for we have

with general solution

The boundary condition at and yields again there is no nontrivial solution for u with

We now turn to the equation for The equation corresponding to the eigenvalue using (7), is given by

which has solution proportional to

(9)

Therefore, multiplying the solutions given by (8) and (9), we conclude that the functions

(10)

satisfy the pde given by (1) and the boundary conditions given by (3) for every positive integer n.

The principle of linear superposition for homogeneous linear differential equations then states that the general solution to (1) and (3) is given by

(11)

The final solution step is to satisfy the initial conditions given by (2). At , we have

(12)

We immediately recognize (12) as a Fourier sine for an odd function with period 2L. Equation (12) is a periodic extension of our original defined on and is an odd function because of the boundary condition From our solution for the coefficient of a Fourier sine, we determine

(13)

Thus the solution to the diffusion equation with homogeneous Dirichlet boundary conditions defined by (1), (2) and (3) is given by (11) with the coefficient computed from (13).

2. Pipe with closed ends

There is no diffusion of dye through the ends of a sealed pipe. Accordingly, the mass flux of dye through the pipe ends, given by (1), is zero so that the boundary conditions on the dye concentration becomes

(14)

which are known as homogeneous Neumann boundary conditions. Again, we apply the method of separation of variables and as before, we obtain the two ordinary differential equations given by (5). Considering first the equation for , the appropriate boundary conditions are now on the first derivative of , and we must solve

(15)

Again, we consider in turn the cases and For we write and determine the general solution of (15) to be

so that taking the derivative

Applying the boundary condition at we find The boundary condition at then yields

The solution results in the trivial solution for u and can be ruled out. Therefore, we must have

with solutions

where n is an integer. We have thus determined the eigenvalues to be

(16)

with corresponding eigenfunctions

(17)

For we write and determine the general solution of (15) to be

so that taking the derivative

Applying the boundary condition at we find The boundary condition at then yields

which for has only the solution Therefore, there is no nontrivial solution for u with Finally, for the general solution of (15) is

so that taking the derivative

The boundary condition at yields is then trivially satisfied. Therefore, we have an additional eigenvalue and eigenfunction given by

which can be seen as extending the formula obtained for eigenvalues and eigenvectors for positive λ given by (16) and (17) to

We now turn to the equation for The equation corresponding to the eigenvalue using (16), is given by

which has solution proportional to

(18)

valid for n=0, 1, 2,…. Therefore, multiplying the solutions given by (17) and (18), we conclude that the functions

(19)

satisfy the pde given by (1) and the boundary conditions given by (14) for every positive integer n.

The principle of linear superposition then yields the general solution as

(20)

where we have redefined the constants so that and The final solution step is to satisfy the initial conditions given by (2). At we have

(21)

which we recognize as a Fourier cosine for an even function with period 2L. We have obtained a cosine series for the periodic extension of because of the boundary condition , which is satisfied by an even function. From our solution for the coefficients of a Fourier cosine series, we determine

(22)

Thus the solution to the diffusion equation with homogeneous Neumann boundary conditions defined by (1), (2) and (14) is given by (20) with the coefficient computed from (22).

References

1. Abell, M. L. and J. P. Braselton. Differential Equations with Mathematica, 3rd ed.

Elsevier Academic Press, 2004.

2.Jeffrey R. Chasnov. Introduction to Differential Equations_The Hong Kong University of

Science and Technology, 2012.

3.Тихонов А.Н., Самарский А.А., Уравнения математической физики. Москва - 1972г

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