Method of External Potential in Solution of Cauchy Mixed Problem for the Heat Equation

Abstract

Numerous research works are devoted to study Cauchy mixed problem for model heat equations because of its theoretical and practical importance. Among them we can notice monographers Vladimirov (1988), Ladyzhenskaya (1973), and Tikhonov and Samarskyi (1980) which demonstrate main research methods, such as Fourier method, integral equations method, and the method of a priori estimates. But at the same time to represent the solution of Cauchy mixed problem in integral form by given and known functions has not been achieved up to now. This paper completes this omission for the one-dimensional heat equation.

1. Introduction

Partial differential equations of parabolic type are widely represented in the study of heat conductivity and diffusion process. Numerous research works are devoted to study Cauchy mixed problem for model heat equations because of its theoretical and practical importance. Among them we can notice monographers [1–3] which demonstrate main research methods, such as Fourier method, integral equations method, and method of a priori estimates. But at the same time to represent the solution of Cauchy mixed problem in integral form by given and known functions has not been achieved up to now. This paper completes this omission for the one-dimensional heat equation.

Exterior potential method as a special continuation of a solution for all half-space is widely used under the solution of Cauchy mixed problem. Our idea is based on a representation possibility of general solution only in the form of volume potential excluding surface integrals. Thus the system of integral equations obtained by this method allows us to construct the solution in quadrature.

2. Material and Methods

Consider the following problem in a plane domain Ω={0<𝑥<1,0<𝑡<𝑇}.

Cauchy Mixed Problem. To find a regular solution of the following equation in Ω()≡𝜕◊𝑢𝑥,𝑡−𝜕𝜕𝑡𝜕𝑥²²𝑢(𝑥,𝑡)=𝑓(𝑥,𝑡), (1) with the initial condition 𝑢(𝑥,0)=0 (2) and boundary conditions 𝑢(0,𝑡)=𝑢(1,𝑡)=0. (3)

Our goal is to construct a classical solution of the problem (1)–(3) in a quadrature. We will seek a solution in the form of sum of three volume potentials: 𝑢(𝑥,𝑡)=𝑢_𝑓(𝑥,𝑡)+𝑢_𝑔0(𝑥,𝑡)+𝑢_𝑔1(𝑥,𝑡), (4) where 𝑢_𝑓(𝑥,𝑡)=𝜀₁()=∗𝑓𝑥,𝑡_𝑡0𝑑𝜏¹₀exp−(𝑥−𝜉)²/4(𝑡−𝜏)2𝜋()𝑓()𝑢𝑡−𝜏𝜉,𝜏𝑑𝜉,_𝑔0(𝑥,𝑡)=𝜀₁∗𝑔₀()=𝑥,𝑡_𝑡0𝑑𝜏⁰_−∞exp−(𝑥−𝜉)²/4(𝑡−𝜏)2𝜋(𝑡−𝜏)×𝑔₀()𝑢𝜉,𝜏𝑑𝜉,_𝑔1(𝑥,𝑡)=𝜀₁∗𝑔₁()=𝑥,𝑡_𝑡0𝑑𝜏₁^+∞exp−(𝑥−𝜉)²/4(𝑡−𝜏)2𝜋()𝑔𝑡−𝜏¹(𝜉,𝜏)𝑑𝜉. (5) Here 𝜀₁(𝑥,𝑡)=𝜃(𝑡)(2√𝜋𝑡)⁻¹exp(−|𝑥|²/4𝑡) is a fundamental solution of the heat equation (1) and 𝜃(𝑡)=1,𝑡>0,0,𝑡≤0 (6) is Heaviside theta-function.

It should be noticed that the heat potential 𝑢_𝑓(𝑥,𝑡) satisfies the following boundary condition: −𝑢(𝑥,𝑡)2+^𝑡₀(𝑥−𝜉)4√𝜋()−()𝑡−𝜏×exp𝑥−𝜉⁻¹4()𝑢()−1𝑡−𝜏𝜉,𝜏2𝜋()−()𝑡−𝜏×exp𝑥−𝜉⁻¹4()()𝑡−𝜏𝜕𝑢𝜉,𝜏𝜕𝜉^𝜉=1_𝜉=0∀()𝑑𝜏=0,𝑥,𝑡∈𝜕Ω, (7) where 𝜕Ω is a boundary of the domain Ω. Note that in works [4, 5] differential operators with nonlocal boundary conditions are investigated as above.

It is easy to verify that the first term in representation (4), 𝑢_𝑓(𝑥,𝑡), is a solution of nonhomogeneous equation (1) and the second and third ones, 𝑢_𝑔0(𝑥,𝑡),𝑢_𝑔1(𝑥,𝑡), are solutions of homogeneous equation. Consequently, representation (4) gives a solution of non-homogeneous equation (1) satisfying initial condition (2) for arbitrary 𝑔₀(𝑥,𝑡), 𝑔₁(𝑥,𝑡).

Our aim is to choose unknown functions 𝑔₀(𝑥,𝑡) and 𝑔₁(𝑥,𝑡) such that the solution will satisfy boundary condition (3).

We will seek functions 𝑔₀(𝑥,𝑡), 𝑔₁(𝑥,𝑡) in the forms 𝑔₀(𝑥,𝑡)=𝑥𝜃(−𝑥)𝑔₀(𝑡),𝑔₁(𝑥,𝑡)=(𝑥−1)𝜃(𝑥−1)𝑔₁(𝑡). (8) Let 𝜑(𝑥) be periodical function with period 2𝜋 defined on the interval (0,2𝜋) by the formula 𝜑(𝑥)=(𝑥/2)−(𝑥²/4𝜋). This function can be represented by the formula 𝜑(𝑥)=(𝜋/6)−(1/2𝜋)∑^𝑘=+∞_𝑘=−∞(1/𝑘²)exp(𝑖𝑘𝑥) out of the interval [0, 2𝜋] .

We introduce the following notations: 𝐺₀(𝑡)=^+∞_−∞𝜑₀(𝑥)+𝑥²𝑑4𝜋𝑑𝑥²²−|𝑥|exp²4(𝜋)²𝑡𝐺𝑑𝑥,₁(𝑡)=^+∞_−∞𝜑₀()+()𝑥−𝜋𝑥−𝜋²𝑑4𝜋𝑑𝑥²²−|𝑥|×exp²4(𝜋)²𝑡𝐹𝑑𝑥,₀(𝑡)=𝑢_𝑓(𝑥,𝑡)_𝑥=0=^𝑡₀𝑑𝜏¹₀exp−𝜉²/4(𝑡−𝜏)2𝜋()()𝐹𝑡−𝜏×𝑓𝜉,𝜏𝑑𝜉,₁(𝑡)=𝑢_𝑓(𝑥,𝑡)_𝑥=1=^𝑡₀𝑑𝜏¹₀exp−(1−𝜉)²/4(𝑡−𝜏)2𝜋()()𝑡−𝜏×𝑓𝜉,𝜏𝑑𝜉. (9)(10)

3. Results and Discussion

The main result of this paper is as follows.

Theorem 1. Let 𝑓(𝑥,𝑡)∈𝐶^𝛼,𝛼/2(Ω),0<𝛼<1, and then the solution of the problem (1)–(3) is represented by formula (4), where 𝑔₀(𝑥,𝑡) and 𝑔₁(𝑥,𝑡) are given by formulae (8): 𝑔₀(𝑡)=𝑑²𝑑𝑡^2𝑡₀𝐺₀(𝑡−𝜏)𝐹₀(𝜏)−𝐺₁(𝑡−𝜏)𝐹₁(𝜏𝑔)𝑑𝜏,₁(𝑡)𝑑=−²𝑑𝑡^2𝑡₀𝐺₀(𝑡−𝜏)𝐹₁(𝜏)−𝐺₁(𝑡−𝜏)𝐹₀(𝜏)𝑑𝜏. (11)

To prove the theorem, the main role plays the following.

Lemma 2. Let unknown functions be given by the formulae (8). Then 𝑔₀(𝑡) and 𝑔₁(𝑡) satisfy the system of equations ^𝑡₀𝑔₀(𝜏)2𝜋()−𝑡−𝜏𝑑𝜏_𝑡0exp(−1/(4𝜋(𝑡−𝜏)))2𝜋()𝑔𝑡−𝜏¹(𝜏)𝑑𝑑𝜏=𝐹𝑑𝑡⁰(𝑡),_𝑡0𝑔₁(𝜏)2𝜋()−𝑡−𝜏𝑑𝜏_𝑡0exp(−1/(4𝜋(𝑡−𝜏)))2𝜋()𝑔𝑡−𝜏⁰(𝜏)𝑑𝑑𝜏=−𝐹𝑑𝑡¹(𝑡), (12) where 𝐹₀(𝑡), 𝐹₁(𝑡) are defined by formula (10).

Proof. By substituting function (4) in boundary condition (3) and taking account of (10), we get a system of interval equations with respect to unknown functions 𝑔₀(𝑡) and 𝑔₁(𝑡): ^𝑡₀𝑔₀(𝜏)𝑑𝜏⁰_−∞exp−𝜉²/4𝜋(𝑡−𝜏)2𝜋()+𝑡−𝜏𝜉𝑑𝜉_𝑡0𝑔₁(𝜏)𝑑𝜏₁^+∞exp−𝜉²/4𝜋(𝑡−𝜏)2𝜋()()𝑡−𝜏𝜉−1𝑑𝜉=−𝐹₀(𝑡),_𝑡0𝑔₀(𝜏)𝑑𝜏⁰_−∞exp−(𝜉−1)²/4𝜋(𝑡−𝜏)2𝜋()+𝑡−𝜏𝜉𝑑𝜉_𝑡0𝑔₁(𝜏)𝑑𝜏₁^+∞exp−(𝜉−1)²/4𝜋(𝑡−𝜏)2𝜋()×()𝑡−𝜏𝜉−1𝑑𝜉=−𝐹₁(𝑡). (13) Integrating by part it is easy to verify correctness of the following equations: ⁰_−∞exp−𝜉²/4𝜋(𝑡−𝜏)2𝜋()2𝑡−𝜏𝜉𝑑𝜉=−𝜋(𝑡−𝜏)^0−∞𝜕−𝜉𝜕𝜉exp²4𝜋(𝑡−𝜏)𝑑𝜉=−(𝑡−𝜏)√𝜋_∞1,exp−𝜉²/4𝜋(𝑡−𝜏)2𝜋()()=𝑡−𝜏𝜉−1𝑑𝜉(𝑡−𝜏)√𝜋−1√𝜋^∞_{1/2()𝑡−𝜏}exp−𝜉²𝑑𝜉. (14)
After putting 𝜉∶=1−𝜉 in integrals (13), our system can be transformed to ^𝑡₀(𝑡−𝜏)√𝜋𝑔₀(𝜏)−𝑑𝜏_𝑡0(𝑡−𝜏)√𝜋−1exp4()−1𝑡−𝜏√𝜋^∞_{1/2()𝑡−𝜏}exp−𝜉²𝑑𝜉×𝑔₁(𝜏)𝑑𝜏=𝐹₀(𝑡),_𝑡0(𝑡−𝜏)√𝜋−1exp4()−1𝑡−𝜏√𝜋^∞_{1/2()𝑡−𝜏}exp−𝜉²𝑑𝜉×𝑔₀(𝜏)𝑑𝜏−^𝑡₀(𝑡−𝜏)√𝜋𝑔₁(𝜏)𝑑𝜏=𝐹₁(𝑡). (15) By differentiating these relations we obtain the system (12). The lemma is proved.
Now we will construct the solution of the system (12) by using the Laplace transformation properties: 𝐹exp−𝛼²/4𝑡√𝑡=√𝜋exp𝛼√𝑝√𝑝,𝐹=𝑓𝑝𝑔𝑝,𝐹𝑑𝑓∗𝑔𝑓𝑓𝑝.𝑑𝑡=𝑝 (16) The system (12) can be transformed in the next form: 𝑔₀𝑝√𝑝−𝑔₁𝑝√𝑝exp−√𝑝=𝑝𝐹₀𝑝,−𝑔₀𝑝√𝑝exp−√𝑝+𝑔₁𝑝√𝑝=−𝑝𝐹₁𝑝, (17) where 𝑔₀(𝑝)=𝐹(𝑔₀)(𝑝), 𝑔₁(𝑝)=𝐹(𝑔₁)(𝑝) are images of the Laplace transform. After solving this system with respect to 𝑔₀(𝑝) and 𝑔₁(𝑝) we find 𝑔₀𝑝=𝑝²√𝑝𝐹₀𝑝−𝐹₁𝑝exp−√𝑝1−exp−2√𝑝,𝑔₁𝑝𝑝=−²√𝑝𝐹₁𝑝−𝐹₀𝑝exp−√𝑝1−exp−2√𝑝. (18)
Since Laplace inverse transformation of function (1/√𝑝)(1/(1−exp(−2√𝑝))) has no table form, so by using the expansions 1√𝑝11−exp−2√𝑝=^∞_𝑘=01√𝑝exp−2√,1𝑝𝑘√𝑝exp−√𝑝1−exp−2√𝑝=^∞_𝑘=01√𝑝exp−(2𝑘+1)√𝑝 (19) and table values of the Laplace inverse transformation, 𝐹⁻¹exp−2𝑘√𝑝√𝑝=1√𝜋exp−(2𝑘)²/4𝑡√𝑡,𝐹⁻¹exp−(2𝑘+1)√𝑝√𝑝=1√𝜋exp−(2𝑘+1)²/4𝑡√𝑡. (20)
Then from (18) we obtain 𝑔₀(𝑡)=𝑑²𝑑𝑡^2𝑡₀𝐺₀(𝑡−𝜏)𝐹₀(𝜏)−𝐺₁(𝑡−𝜏)𝐹₁(𝜏𝑔)𝑑𝜏,₁(𝑡)𝑑=−²𝑑𝑡^2𝑡₀𝐺₀(𝑡−𝜏)𝐹₁(𝜏)−𝐺₁(𝑡−𝜏)𝐹₀(𝜏)𝑑𝜏, (21) where 𝐺₀(𝑡)=𝐹^−1∞_𝑘=0exp−2𝑘√𝑝√𝑝=^∞_𝑘=0𝐹⁻¹exp−2𝑘√𝑝√𝑝=1√𝜋^∞_𝑘=0exp−(2𝑘)²/4𝑡√𝑡,𝐺₁(𝑡)=𝐹^−1∞_𝑘=0exp−(2𝑘+1)√𝑝√𝑝=1√𝜋^∞_𝑘=0exp−(2𝑘+1)²/4𝑡√𝑡. (22)(23)
To complete the proof of the theorem we represent functions 𝐺₀(𝑡) and 𝐺₁(𝑡) by integrals of known functions.
As long as 𝛿(𝑥−2𝑘𝜋), (𝑘=0,±1,±2,…) is a tempered distribution [6, page 112], and then we have 𝛿()−𝑥𝑥−2𝑘𝜋,exp²4𝜋²𝑡−()=exp2𝑘².4𝑡 (24)
Therefore, the following relations hold: ^∞_𝑘=−∞𝛿()−𝑥𝑥−2𝑘𝜋,exp²4𝜋²𝑡=^∞_𝑘=−∞−()exp2𝑘²,4𝑡_{∞𝑘=−∞}𝛿(()𝜋)−𝑥𝑥−2𝑘+1,exp²4𝜋²𝑡=^∞_𝑘=−∞−()exp2𝑘+1².4𝑡 (25)(26)
We define function 𝜑₀(𝑥)=(𝑥/2)−(𝑥²/4𝜋) on the interval [0,2𝜋] and then continue it periodically on all axes (−∞,∞). It is shown in [6, page 113] ^∞_𝑘=−∞𝛿()=𝑑𝑥−2𝑘𝜋²𝑑𝑥²𝜑₀(𝑥)+𝑥²,4𝜋 (27) where (𝑑²/𝑑𝑥²) is regarded in the general function sense.
From here and relation (25) we conclude that ^∞_𝑘=0−()exp2𝑘²=14𝑡2𝑑²𝑑𝑥²𝜑₀(𝑥)+𝑥²−𝑥4𝜋,exp²4𝜋²𝑡=12^∞_−∞𝜑₀(𝑥)+𝑥²𝑑4𝜋𝑑𝑥²²−𝑥exp²4𝜋²𝑡𝑑𝑥. (28) Analogically, the relation ^∞_𝑘=0−()exp2𝑘+1²=14𝑡2^∞_−∞𝜑₀()+()𝑥−𝜋𝑥−𝜋²𝑑4𝜋𝑑𝑥²²−𝑥×exp²4𝜋²𝑡𝑑𝑥 (29) follows from (26).
We obtain formula (9) after substituting the last relations in (22) and (23).
From (23) we have 𝐺₁(𝑡)∈𝐶^∞[0,1], and from (22) (𝐺₀(𝑡)−(1/√𝜋𝑡))∈𝐶^∞[0,1].
Since𝑓(𝑥,𝑡)∈𝐶^{1+𝛼,1+(𝛼/2)}(Ω), 0<𝛼<1, then 𝐹₀(𝑡), 𝐹₁(𝑡)∈𝐶^1+(𝛼/2)(0,𝑇). Therefore it is not difficult to establish 𝑔₀(𝑡),𝑔₁(𝑡)∈𝐶^1+(𝛼/2)[0,𝑇].
Consequently, representation of the solution 𝑢(𝑥,𝑡)∈𝐶^{2+𝛼,1+(𝛼/2)}(Ω) in the form (4) is established. The theorem is fully proved.