Department of Mathematics, College of Science and Arts at Tabrjal, AlJouf University, Kingdom of Saudi Arabia
Received Date: March 01, 2016; Accepted Date: March 22, 2017; Published Date: April 02, 2017
Citation: Hassan AA (2017) Green’s Function for the Heat Equation. Fluid Mech Open Acc 4: 152. doi: 10.4172/2476-2296.1000152
Copyright: © 2017 Hassan AA. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
Visit for more related articles at Fluid Mechanics: Open Access
The solution of problem of non-homogeneous partial differential equations was discussed using the joined Fourier- Laplace transform methods in finding the Green’s function of heat equation in different situations.
The Green's function is a powerful tool of mathematics method is used in solving some linear non-homogenous PDEs, ODEs. So Green’s functions are derived by the specially development method of separation of variables, which uses the properties of Dirac’s function. This method was considerable more efficient than the others well known classical methods.
The series solution of differential equation yields an infinite series which often converges slowly. Thus it is difficult to obtain an insight into over-all behavior of the solution. The Green’s function approach would allow us to have an integral representation of the solution instead of an infinite series.
To obtain the filed u, caused by distributed source we calculate the effect of each elementary portion of source and add (integral) them all. If G(r, r0) is the field at the observers point r caused by a unit source at the source point r0, then the field at r caused by distribution f(r0) is the integral of f(r0) G(r, r0) over the whole range of r0 occupied by the course. The function G is called Green’s function.
Consider a linear second order differential equation:
Where λ is a parameter to be determined by the boundary conditions. A(x) is positive continuous function, then by dividing every term by A(x), eqn (1) can be written as:
Let us define integrating factor p(x) by:
Multiplying eqn (2) by p(x), we have:
Thus eqn (3) can be written as:
Regular Sturm-Liouville problem: In case p(a)≠0 and p(b)≠0, p(x),q(x),r(x) are continuous, the Sturm-Liouville eqn (4) can be expressed as:
If the above equation is associated with the following boundary condition:
Where 1+α2≠0 and β1+β2≠0
The eqn (4) and the boundary condition (7) are called regular Sturm-Liouville problem (RSLP).
Singular Sturm-Liouville problem: Consider the equation:
Where L is defined by eqn (6), p(x) is smooth and r(x)is positive, then the Sturm-Liouville problem is called singular if one of the following situations is occurred.
(i) If p(a)=0 or p(b)=0 or both
(ii)The interval (a,b) is infinite.
Eigenvalue and Eigenfunction: The Eigenvalue from eqn (4) defining by a Sturm-Liouville operator can be expressed as:
The non-trival solutions that satisfy the equation and boundary conditions are called eigenfunctions. Therefore the eigenfunction of the Sturm-Liouville problem from complete sets of orthogonal bases for the function space is which the weight function is r(x).
The delta function is defined as:
But such that the integral of δ (x−ξ) is normalized to unit;
In fact the first operator where Dirac used the delta function is the integration:
Where f(x) is a continuous function, we have to find the value of the integration eqn (12).
Since δ (x−ξ) is zero for x≠ξ, the limit of integration may be change to ζ− and ζ+ε, where ε is a small positive number, f(x) is continuous at x−ξ, it’s values within the interval (ζ−ε,ζ−) will not different much from f(ζ), approximately that:
With the approximation improving as ε approaches zero.
From eqn (11), we have:
From all values ofε, then by letting ε→0, we can exactly have:
Despite the delta function considered as fundamental role in electrical engineering and quantum mechanics, but no conventional could be found that satisfies eqns (10) and (11), then the delta function sought to be view as the limit of the sequence of strongly peaked function δn(x) such that:
Some important properties of Dirac delta function
Property (1): Symmetry
Proof: Let ζ=−x, then dx=−dζ
We can write:
Therefore, from eqns (19) and (20), we conclude that δ (−x)=δ (x).
Property (2): Scaling
Proof: Let ζ=ax, then
If a>0, then,
Similarly for a<0, , then,
We can write:
Proof: The argument of this function goes to zero when x=a and x=−a, wherefore:
Only at the zero of the argument of the delta function that is:
Near the two zeros x2−a2 can be approximated as:
In the limit as ε0 the integral eqn (23) becomes:
Heat equation over infinite or semi-infinite domains
Consider one dimensional heat equation:
Subject to boundary conditions |u(x,t)|<∞ as |x|<∞ and initial condition. Let G(x, t, ζ, τ) be the Green’s function for the onedimensional heat equation, then:
Subject to the boundary condition |G(x, t, ζ, τ)|<∞ as |x|<∞, and the initial condition G(x, t, ζ, τ)=0. Let us find G(x, t, ζ, τ).
We begin by taking the Laplace transform of eqn (24) with respect to t, we have:
Where g(x, s, ζ, τ) the Laplace transform of G(x, t, ζ, τ)
Now by taking the Fourier transform of eqn (26) with respect to x, so that:
Where is Fourier transform of g(x, s, ζ, τ), now let
To find g(x, s, ζ, τ), we use the inversion integral:
Transforming eqn (27) into a closed contour, we evaluate it by the residue theorem and find that:
Now substituting for we have:
Taking Laplace transform of eqn (29) we obtain:
Example (5.1.1) Consider one dimensional heat equation:
Subject to boundary condition and initial condition U(x,0)=0
Solution: Let us find the Green’s function for the following problem:
Subject to the boundary conditions and initial condition
From the boundary condition G(0, t, ζ, τ)=0, we deduce that G(x, t, ζ, τ) Green’s function by introducing an image source of −δ(x−ζ) and resolving eqn (25) with the source (x−ζ)δ (t−τ)−δ (x+ζ)δ (t−τ). Eqn (30) gives the solution for each delta function and the green’s function for eqn (32) can be written:
In a similar manner, if the boundary condition at x=0 changes to Gx(0, t, ζ, τ)=0, then eqns (31) and (32) become:
Heat Equation within a finite Cartesian Domain
In this section we find Green’s function for the heat equation within finite Cartesian domains. These solutions can be written as series involving orthonormal eigenfunction form regular Sturm- Liouville problem [6-10].
Here we find the Green’s function for the one-dimensional heat equation over the interval 0<x<L.
Where a2 is the diffusivity constant,
To find the Green’s function for this problem, consider the following problem,
With the boundary conditions,
And the initial condition,
We begin by taking the Laplace transform of eqn (38) and find that:
Applying the technique of the eigenfunction expansions, we have that:
Where φn(x) is the nth orthonormal eigenfunction to the regular Sturm-Liouville problem:
Subject to the boundary conditions:
And α2φ(L)+β2φ′(L)=0 (48)
Taking the inverse of eqn (45), we have that:
Heat equation within cylinder
In this section, we turn our attention to cylindrical domains. The techniques used here have much in common with those used in the previous section. However, in place of sines cosines form in Sturm- Liouville problem, we will encounter Bessel functions.
Find the solution of the following problem by construction the Green’s function:
Subject to the boundary conditions , and the initial condition U(r,0)=0.
Solution: In this problem, we find Green’s function for head equation in cylindrical coordinates:
Subject to the boundary conditions G(b,t,ρ,τ)=0, and the initial condition G(b,0,ρ,τ)=0
Now we begin by taking the Laplace transform to eqn (51) with respect to t:
Next we re-express as the Fourier-Bessel expansion: (53)
Where kn is the nth root of J0(kn)=0, and
So that (55)
By applying Fourier transform:
Taking the inverse of eqn (56) and applying second shifting theorem,
If we modify the boundary condition at r=b,
Where h0, our analysis now leads to:
Where kn are the positive roots of kJ1(k)+hbJ0(k)=0.
Find the solution for the heat equation in cylindrical by construction the Green’s function:
Subject to the boundary conditions U(α,t)=U(β,t)=0, and the initial condition U(r, 0)=0
Let G(b,t,ρ,τ) Green’s function, then:
Subject to the boundary condition that G(α,t,ρ,τ)= G(β,t,ρ,τ)=0, and the initial condition G(r,0,ρ,τ)=0.
We begin by taking Laplace transform of eqn (61), we obtain:
Now, we express as a Fourier series by considering the regular Sturm-Liouville problem:
The eigenfunction that satisfy eqn (63) are:
Provided kn is the nth zero of therefore, the expansion for the delta function in term of φn(r) is:
Using the orthogonality condition that:
with the wronskian relationship:
We find that:
Inverting the Laplace transform, we obtain:
In similar manner, we can solve the general problem eqn (61) with Robin boundary conditions:
And b1Gr(β,t, ρ,τ)−b2G(β,t,ρ,τ)=0 (75)
Where a1, a2, b1, b2≥0. The solution for this problem is:
Heat equation within a sphere
Let us find Green’s function for the radically symmetric heat equation within a sphere of radius b. mathematically, we must solve:
With the boundary condition and the initial condition G(r,0,ρ,τ)=0.
We begin by introducing the dependent variable u(r,t,ρ,τ)=rG(r,t,ρ,τ), so that condition eqn (69) becomes:
With the boundary conditions u(0,t,ρ,τ)=u(b,t,ρ,τ)=0, and the initial condition u(r,0,ρ,τ)=0. Taking the Laplace transform of eqn (80), we obtain:
Let us expand in the Fourier sine series:
Therefore we write eqn (81) as:
Because this particular solution also satisfies the boundary condition, we do not require any homogeneous solution so that the sum of the particular and homogeneous solution satisfies the boundary conditions.
Taking the inverse of eqn (85), using the second shifting theorem and substituting for U(r,t,ρ,τ), we finally obtain:
For the case of a sphere α<r<β, the Green’s function can be bound by introducing the new independent variable x=r−α into the partial differential equation:
With the boundary conditions G(α,t,ρ,τ)=G(β,t,ρ,τ)=0, and the initial condition G(r,0,ρ,τ)=0. Therefore the Green’s function in this particular case is: