����������� ������ 55 (2003), 97–106

UDK 519.633

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SPECTRAL PROBLEMS FOR PARABOLIC EQUATIONS WITH CONDITIONS OF CONJUGATION AND DYNAMICAL

BOUNDARY CONDITIONS

Sonja Gegovska-Zajkova

Abstract. The initial boundary value problems for the heat equation with discontinuous heat flow and concentrated heat capacity in the interior points or at the boundary are considered. The corresponding spectral problems, where the eigenvalues appear in the boundary or interface conditions, are derived and studied.

1. Abstract setting of the problem

Let H be a real separable Hilbert space endowed with the inner product (·, ·), and norm ∥ · ∥, and A is a linear selfadjoint unbounded positive definite linear operator with domain D(A) dense in H. The product (u, v)_{A} = (Au, v), (u, v ∈ D(A)) satisfies the inner product axioms. Reinforcing D(A) in the norm ∥u∥_{A} = (u, u)_{A}^{1/2}, we obtain a Hilbert space H_{A} ⊂ H. Operator A extends to mapping A: H_{A} → H_{A^{−1}}, where H_{A^{−1}} is the adjoint space for H_{A}, and H_{A} ⊂ H ⊂ H_{A^{−1}} form a Gel’fand triple. We also define the Sobolev spaces W_{2}^{s}(a, b; H), W_{2}^{0}(a, b; H) = L_{2}(a, b; H) of the functions u = u(t) mapping the interval (a, b) ⊂ R into H. The problems solved below can ve written as an abstract Cauchy problem

\frac{du}{dt} + Au = f(t),  0 < t < T; u(0) = u_{0},  (1.1)

where u_{0} is a given element of H, f(t) ∈ L_{2}(0, T; H_{A^{−1}}) and u(t) is the unknown function from (0, T) into H_{A}. Let us define λ by

\frac{1}{λ} = sup_{U∈H_{A}} \frac{∥U∥^{2}}{∥U∥_{A}^{2}} .  (1.2)

The solution of the variational problem is the solution of the following spectral problem [2]:

AU = λU.  (1.3)

AMS Subject Classification: 35 K 20, 35 P 99

Keywords and phrases: Parabolic equations, conjugation conditions, dynamical boundary conditions, spectral problems.

The spectrum of (1.3) is discrete, all eigenvalues λ = λ_{n}, n = 1, 2,… are positive: 0 < λ_{1} ⩽ λ_{2} ⩽ ⋯, λ_{n} → ∞, while the normalized eigenfunctions U = U_{n}, n = 1, 2,… satisfy the condition of orthogonality (U_{n}, U_{m}) = δ_{nm} and represent a basis of the spaces H and H_{A}. Thus, the solution of the problem (1.1) can be obtained in the form

u(t) = ∑_{n=1}^{∞} e^{−λ_{n}t}[c_{n} + ∫_{0}^{t} e^{λ_{n}τ} f_{n}(τ) dτ]U_{n},  (1.4)

where c_{n} = (u_{0}, U_{n}), f_{n}(t) = (f(t), U_{n}).

Setting f(t) = 0 in (1.1), an a priori estimate can be done. We take inner product in H of (1.1) with u and apply the inequality λ_{1}(x, x) ⩽ (Ax, x) to obtain

∥u(t)∥ ⩽ e^{−λ_{1}t}∥u_{0}∥.  (1.5)

Now, let B be a linear selfadjoint positive definite operator with domain D(B) dense in H, A is a linear selfadjoint unbounded positive definite linear operator with domain D(A) dense in H_{B} and A ⩾ B. We consider the abstract Cauchy problem

B \frac{du}{dt} + Au = f(t),  t ∈ (0, T); u(0) = u_{0},  (1.6)

where u_{0} is a given element of H_{B}, f(t) ∈ L_{2}(0, T; H_{A^{−1}}) and u(t) is the unknown function from (0, T) into H_{A}. It is easy to see that problem (1.6) can be written in the form

\frac{dũ}{dt} = Ãũ = f̃, t ∈ (0, T); ũ(0) = ũ_{0},

if one sets ũ = B^{1/2}u, B^{−1/2}AB^{−1/2} = Ã and B^{−1/2}f(t) = f̃. Then, the following spectral problem can be obtained:

ÃŨ = λ̃Ũ.

Operator Ã is positive definite and unbounded in H. Therefore, its spectrum is discrete, all eigenvalues λ̃_{n}, n = 1, 2,… are positive, λ̃_{1} ⩽ λ̃_{2} ⩽ ⋯, λ̃_{n} → ∞, while the eigenfunctions U_{n}, n = 1, 2,… satisfy the condition of orthogonality (Ũ_{j}, Ũ_{k}) = δ_{jk} and represent a basis of the space H. If we set Ũ = B^{1/2}U, Ũ_{n} = B^{1/2}U_{n}, the spectral problem takes the form

AU = λ̃BU.  (1.7)

In such a way, the spectrum of (1.7) is discrete, all eigenvalues are positive, λ̃_{1} ⩽ λ̃_{2} ⩽ ⋯, λ̃_{n} → ∞, while the eigenfunctions U_{n}, n = 1, 2,… satisfy the condition of orthogonality

(U_{j}, U_{k})_{B} = (BU_{j}, U_{k}) = (Ũ_{j}, Ũ_{k}) = δ_{jk}

and represent a basis of the space H_{B}.

The solution of problem (1.6) can be written in the form

u(t) = ∑_{n=1}^{∞} e^{−λ̃_{n}t}[c_{n} + ∫_{0}^{t} e^{λ̃_{n}τ} f_{n}(τ) dτ]U_{n},

where c_{n} = (u_{0}, U_{n})_{B}, f_{n}(t) = (f(t), U_{n}).

In order to obtain an energy estimate for the solution of the problem, when f(t) = 0, we take an inner product of (1.6) with 2u,

\frac{d}{dt}[(BU, u)] + 2∥u∥_{A}^{2} = 0.

Takimg into account the inequality ∥u∥_{A}^{2} ⩾ λ̃_{1}∥u∥_{B}^{2}, we get

\frac{d(∥u∥_{B}^{2})}{∥u∥_{B}^{2}} ⩽ −2λ̃_{1} dt.

Integrating the result from 0 to t , the following estimate can be done:

∥u(t)∥_{B} ⩽ ∥u_{0}∥_{B}e^{−λ̃_{1}t}.

2. Heat equation with concentrated capacity

2.1. The first boundary value problem

Let us consider the first initial boundary value problem for the heat equation with concentrated capacity at the interior point x = ξ:

[1 + Kδ(x − ξ)]\frac{∂u}{∂t} = \frac{∂^{2}u}{∂x^{2}} + f(x, t),  (x, t) ∈ (0, 1) × (0, T),  (2.1)
u(0, t) = u(1, t) = 0,  0 < t < T,  (2.2)
u(x, 0) = u_{0}(x),  x ∈ (0, 1),  (2.3)

where K > 0 and δ(x) is the Dirac distribution. Similar problems are already mentioned in [3], [4]. The derivations in (2.1) are taken in the sense of the distribution theory. It follows from (2.1), that the solution of this problem satisfies at (x, t) ∈ (0, ξ) × (0, T) and (x, t) ∈ (ξ, 1) × (0, T) the equation

\frac{∂u}{∂t} = \frac{∂^{2}u}{∂x^{2}} + f(x, t), x ∈ (0, ξ) ∪ (ξ, 1), t ∈ (0, T),

u(0, t) = u(1, t) = 0, u(x, 0) = u_{0}(x),

and at x = ξ the conditions of conjugation

[u]_{x=ξ} ≡ u(ξ + 0, t) − u(ξ − 0, t) = 0, K \frac{∂u}{∂t}(ξ, t) = [\frac{∂u}{∂x}]_{x=ξ} .

The initial boundary value problem (2.1)–(2.3) can be reduced to the form (1.6) letting H = L_{2}(0, 1),

Au = −\frac{∂^{2}u}{∂x^{2}}, Bu = [1 + Kδ(x − ξ)]u

and

(u, v)_{B} = ∫_{0}^{1} u(x)v(x) dx + Ku(ξ)v(ξ),

while H_{A} = W^{∘} _{2}^{1}(0, 1). Thus, the following spectral problem can be obtained:

−\frac{d^{2}U}{dx^{2}} = λ[1 + Kδ(x − ξ)]U(x), x ∈ (0, 1),

U(0) = U(1) = 0,

or

−\frac{d^{2}U}{dx^{2}} = λU(x), x ∈ (0, ξ) ∪ (ξ, 1),

U(0) = U(1) = 0,  (2.4)

[U]_{x=ξ} ≡ U(ξ + 0) − U(ξ − 0) = 0, − [\frac{dU}{dx} ]_{x=ξ} = λKU(ξ).

The solution of this problem can be written in the following explicit form:

U(x) = \{ \begin{matrix}A sin αx, & x ∈ (0, ξ), \\ B sin α(1 − x), & x ∈ (ξ, 1), \end{matrix}

It is obvious that U(x) satisfies the boundary conditions. The values of the constants A and B can be obtained using the first condition of conjugation, and we get A = C sin α(1 − ξ), B = C sin αξ, where C is a multiplicative constant, so we can set C = 1. From (2.4) and the second condition of conjugation, we find that

α = \frac{1}{K} [cot α(1 − ξ) + cot αξ].  (2.5)

If ξ = 1/2, the equation (2.5) takes the

form α = \frac{2}{K} cot \frac{α}{2}. There exists a countable set of its solutions: α_{n}, n = 1, 2,…. Using the condition λ = α^{2}, we obtain the eigenvalues λ_{n} = α_{n}^{2}, n = 1, 2,…. The graphical solution of this equation is shown in Fig. 1, and the numerical values of α_{n} and λ_{n} are shown in Table 1. In Fig. 2 the first three eigenfunc-

tions U_{1}, U_{2} and U_{3} are presented. \begin{matrix}i & α_{i} & λ_{i} \\ 1 & 1.07687 & 1.15965 \\ 2 & 3.64360 & 13.27582 \\ 3 & 6.57833 & 43.27442 \\ 4 & 9.62956 & 92.72843 \\ 5 & 12.72230 & 161.85692 \end{matrix}

Table 1

Figure 1 Figure 2

If ξ ≠ 1/2 the right-hand side of (2.5) is a sum of two periodical functions which have different periods. Transcendent equation (2.5) has a countable set of solutions α = α_{n}, n = 1, 2,…. So we can obtain the eigenvalues

λ = λ_{n} = α_{n}^{2}, 0 < λ_{1} < λ_{2} < ⋯, λ_{n} → ∞,

and the corresponding system of eigenfunctions U = U_{n}(x), n = 1, 2,….

2.2. The second boundary value problem

Let us consider the heat equation with concentrated capacity (2.1) with inital value (2.3) and Neumann’s boundary conditions

\frac{∂u}{∂x}(0, t) = \frac{∂u}{∂x}(1, t) = 0,  0 < t < T.  (2.6)

The problem (2.1), (2.3), (2.6) can be written in the form

\frac{∂u}{∂t} = \frac{∂^{2}u}{∂x^{2}} + f(x, t), x ∈ (0, ξ) ∪ (ξ, 1), t ∈ (0, T),

\frac{∂u}{∂x}(0, t) = \frac{∂u}{∂x}(1, t) = 0, u(x, 0) = u_{0}(x),

[u]_{x=ξ} = 0, [\frac{∂u}{∂x}]_{x=ξ} = K \frac{∂u}{∂t}(ξ, t).

Using the same procedure as it was done in the section 2.1, the following spectral problem can be obtained

−\frac{d^{2}U}{dx^{2}} = λU(x), x ∈ (0, ξ) ∪ (ξ, 1),

U′(0) = U′(1) = 0,

[U]_{x=ξ} = 0, − [\frac{dU}{dx} ]_{x=ξ} = λKU(ξ).

The solution of this spectral problem can be written in the form

U(x) = \{ \begin{matrix}A cos αx, & x ∈ (0, ξ), \\ B cos α(1 − x), & x ∈ (ξ, 1). \end{matrix}

It is obvious that it automatically satisfies the boundary conditions. We obtain the values of the constants A and B using the first condition of conjugation: A = C cos α(1 − ξ), B = C cos αξ. Here, C is a multiplicative constant, and we can set C = 1. By the second condition of conjugation, taking into account that λ = α^{2}, we obtain

α = −\frac{1}{K} [tan αξ + tan α(1 − ξ)].  (2.7)

If ξ = 1/2, the equation (2.7) takes the

form α = −\frac{2}{K} tan \frac{α}{2} and has a countable set of solutions α_{n}, n = 1, 2,…. Now, using the condition λ = α^{2}, we can obtain the eigenvalues λ_{n} = α_{n}^{2}, n = 1, 2,…. The graphical solution of equation (2.7) is shown in Fig. 3, and the numerical values of α_{n} and λ_{n} are shown in Table 2. In Fig. 4 the first three

eigenfunctions U_{1}, U_{2} and U_{3} are presented. \begin{matrix}i & α_{i} & λ_{i} \\ 1 & 4.05752 & 16.46347 \\ 2 & 9.82636 & 96.55735 \\ 3 & 15.95730 & 254.63542 \\ 4 & 22.17110 & 491.55768 \\ 5 & 28.41490 & 807.40654 \end{matrix}

Table 2

Figure 3 Figure 4

If ξ ≠ 1/2 the right-hand side of (2.7) is a sum of two periodical functions which have different periods. Again, there exists a countable set of solutions α = α_{n}, n = 1, 2,…. From here we obtain the eigenvalues λ_{n} = α_{n}^{2}, 0 < λ_{1} < λ_{2} < ⋯, λ_{n} → ∞, and the corresponding eigenfunctions U = U_{n}(x), n = 1, 2,….

2.3. The third boundary value problem

We consider the heat equation with concentrated capacity (2.1) with inital value (2.3) and the third boundary conditions

−\frac{∂u}{∂x}(0, t) + αu(0, t) = 0,  \frac{∂u}{∂x}(1, t) + bu(1, t) = 0,  0 < t < T.  (2.8)

The problem (2.1), (2.3), (2.8) can be written in the form

\frac{∂u}{∂t} = \frac{∂^{2}u}{∂x^{2}} + f(x, t), x ∈ (0, ξ) ∪ (ξ, 1), t ∈ (0, T),

−\frac{∂u}{∂x}(0, t) + au(0, t) = 0, \frac{∂u}{∂x}(1, t) + bu(1, t) = 0,

u(x, 0) = u_{0}(x),

[u]_{x=ξ} = 0, [\frac{∂u}{∂x}]_{x=ξ} = K \frac{∂u}{∂t}(ξ, t).

Thus, the following spectral problem can be obtained

−\frac{d^{2}U}{dx^{2}} = λU(x), x ∈ (0, ξ) ∪ (ξ, 1),

−U′(0) + aU(0) = 0,  U′(1) + bU(1) = 0,  (2.9)

[U]_{x=ξ} = 0, − [\frac{dU}{dx} ]_{x=ξ} = λKU(ξ).

The solution of this spectral problem can be written in the form

U(x) = \{ \begin{matrix}A sin(αx + β), & x ∈ (0, ξ), \\ B sin[α(1 − x) + γ], & x ∈ (ξ, 1). \end{matrix}

It is easy to see that λ = α^{2}. By the boundary conditions we obtain

tan β = \frac{α}{a}, tan γ = \frac{α}{b}.

Using the first condition of conjugation we get the values of the constants: A = D sin[α(1 − ξ) + γ] and B = D sin(αξ + β). Here, D is a multiplicative constant, and we can set D = 1. Takimg into account the second condition of conjugation we obtain

α = \frac{1}{K} [\frac{1 − \frac{α}{a} tan αξ}{\frac{α}{a} + tan αξ} + \frac{1 − \frac{α}{b} tan α(1 − ξ)}{\frac{α}{b} + tan α(1 − ξ)}] .  (2.10)

If we consider the special case ξ = 1/2 and a = b, the equation (2.10) takes

the form α = \frac{2}{K} F (α), F (α) = \frac{a − α tan \frac{α}{2}}{α + a tan \frac{α}{2}} .

It is easy to see that there exists a countable set of solutions α_{n}, n = 1, 2,…. Now, using the condition λ = α^{2}, we can obtain the eigenvalues λ_{n} = α_{n}^{2}, n = 1, 2,…. The graphical solution of this equation for α = 1 is shown in Fig. 5, and the numerical values of α_{n} and λ_{n} are shown in Table 3. In Fig. 6 the first three

eigenfunctions U_{1}, U_{2} and U_{3} are presented. \begin{matrix}i & α_{i} & λ_{i} \\ 1 & 0.87638 & 0.76804 \\ 2 & 4.43297 & 19.65122 \\ 3 & 10.01790 & 100.35832 \\ 4 & 16.07970 & 258.55675 \\ 5 & 22.26010 & 495.51205 \end{matrix}

Table 3

Figure 5 Figure 6

If ξ = 1/2 and a ≠ b, the equation (2.10) takes the form

α = \frac{1}{K} [\frac{a − α tan \frac{α}{2}}{α + a tan \frac{α}{2}} + \frac{b − α tan \frac{α}{2}}{α + b tan \frac{α}{2}} ] .

Its graphical and numerical sopultions, setting a = 1, b = 4, are presented in Fig. 7 and Tab. 4 respectively. In Fig. 8 the first three

eigenfunctions U_{1}, U_{2} and U_{3} are presented. \begin{matrix}i & α_{i} & λ_{i} \\ 1 & 1.13814 & 1.29536 \\ 2 & 3.97213 & 15.77782 \\ 3 & 5.15003 & 26.52281 \\ 4 & 9.78116 & 95.67109 \\ 5 & 10.54330 & 111.16117 \end{matrix}

Table 4

Figure 7 Figure 8

If ξ ≠ 1/2 on the right-hand side of (2.10) we have a sum of two functions that have the same form as F (α). Once again, there exists a countable set of solutions α = α_{n}, n = 1, 2,…. So, we can obtain the eigenvalues

λ_{n} = α_{n}^{2}, 0 < λ_{1} < λ_{2} < ⋯, λ_{n} → ∞,

and the corresponding eigenfuctions U = U_{n}(x), n = 1, 2,….

3. Heat equation with dynamical boundary condition

Let us consider the heat equation

\frac{∂u}{∂t} = \frac{∂^{2}u}{∂x^{2}} + f(x, t),  (x, t) ∈ (0, 1) × (0, T),  (3.1)

with initial value (2.3) and dynamical boundary conditions at the point x = 0 (see [1], [4]):

K \frac{∂u}{∂t}(0, t) = \frac{∂u}{∂x}(0, t),  0 < t < T,  (3.2)

where K > 0. At x = 1 we consider one of the standard boundary conditions:

u(1, t) = 0,  0 < t < T,  (3.3)
\frac{∂u}{∂x}(1, t) = 0,  0 < t < T,  (3.4)

Spectral problems for parabolic equations 105

or

\frac{∂u}{∂x}(1, t) + bu(1, t) = 0, 0 < t < T. (3.5)

Using odd extension of the solution and the input data these problems can be reduced to the above mentioned. Indeed, setting

f̃(x, t) = f(−x, t), ũ_{0}(x) = u_{0}(−x), ũ(x) = u(−x), x ∈ (−1, 0) the problem (3.1), (2.3), (3.2), (3.3) reduces to

\frac{∂ũ}{∂t} = \frac{∂^{2}ũ}{∂x^{2}} + f̃(x, t), x ∈ (−1, 0) ∪ (0, 1), t ∈ (0, T),

ũ(−1, t) = ũ(1, t) = 0, ũ(x, 0) = ũ_{0}(x),

[ũ]_{x=0} = 0, 2K \frac{∂ũ}{∂t}(0, t) = [\frac{∂ũ}{∂x}]_{x=0} ,

or equivalently

[1 + 2Kδ(x)]\frac{∂ũ}{∂t} = \frac{∂^{2}ũ}{∂x^{2}} + f̃(x, t), (x, t) ∈ (−1, 1) × (0, T);

ũ(−1, t) = ũ(1, t) = 0, 0 < t < T,

ũ(x, 0) = ũ_{0}(x), x ∈ (−1, 1).

Analogously, (3.1), (2.3), (3.2), (3.4) reduces to

[1 + 2Kδ(x)]\frac{∂ũ}{∂t} = \frac{∂^{2}ũ}{∂x^{2}} + f̃(x, t), (x, t) ∈ (−1, 1) × (0, T);

\frac{∂ũ}{∂x}(−1, t) = \frac{∂ũ}{∂x}(1, t) = 0, 0 < t < T,

ũ(x, 0) = ũ_{0}(x), x ∈ (−1, 1),

while (3.1), (2.3), (3.2), (3.5) reduces to

[1 + 2Kδ(x)]\frac{∂ũ}{∂t} = \frac{∂^{2}ũ}{∂x^{2}} + f̃(x, t), (x, t) ∈ (−1, 1) × (0, T);

−\frac{∂ũ}{∂x}(−1, t) + bũ(−1, t) = \frac{∂ũ}{∂x}(1, t) + bũ(1, t) = 0,

ũ(x, 0) = ũ_{0}(x), x ∈ (−1, 1).

The corresponding spectral problems

−\frac{d^{2}U}{dx^{2}} = λU, x ∈ (0, 1),  −\frac{dU}{dx} (0) = λKU(0),  U(1) = 0, (3.6)
−\frac{d^{2}U}{dx^{2}} = λU, x ∈ (0, 1),  −\frac{dU}{dx} (0) = λKU(0),  \frac{dU}{dx} (1) = 0, (3.7)

and

−\frac{d^{2}U}{dx^{2}} = λU, x ∈ (0, 1),  −\frac{dU}{dx} (0) = λKU(0),  \frac{dU}{dx} (1) + bU(1) = 0 (3.8)

reduce respectively to

−\frac{d^{2}Ũ}{dx^{2}} = λ[1 + 2Kδ(x)]Ũ, x ∈ (−1, 1),  Ũ(−1) = Ũ(1) = 0,  (3.9)

−\frac{d^{2}Ũ}{dx^{2}} = λ[1 + 2Kδ(x)]Ũ, x ∈ (−1, 1),  \frac{dŨ}{dx} (−1) = \frac{dŨ}{dx} (1) = 0, (3.10)

and

\begin{matrix}−\frac{d^{2}Ũ}{dx^{2}} = λ[1 + 2Kδ(x)]Ũ,  x ∈ (−1, 1), \\ −\frac{dŨ}{dx} (−1) + bŨ(−1) = \frac{dŨ}{dx} (1) + bŨ(1) = 0. \end{matrix}  (3.11)

Note that the singular point in all three cases is the midpoint of the interval (−1, 1). From the previous results we immediately obtain eigenvalues and eigenfunc-

tions of (3.6) in the form

λ_{n} = α_{n},  U_{n}(x) = sin α_{n}(1 − x),  n = 1, 2,…,

where α_{n} are solutions of the equation α = \frac{1}{K} cot α. Analogously, eigenvalues and eigenfunctions of (3.7) are

λ_{n} = α_{n},  U_{n}(x) = cos α_{n}(1 − x),  n = 1, 2,…,

where α_{n} are solutions of the equation α = −\frac{1}{K} tan α, while eigenvalues and eigenfunctions of (3.8) are

λ_{n} = α_{n},  U_{n}(x) = sin [α_{n}(1 − x) + arctan \frac{α_{n}}{b} ] ,  n = 1, 2,…,

where α_{n} are solutions of the equation α = −\frac{1}{K} \frac{b − α tan α}{α + b tan α}.

REFERENCES

[1] Braianov, I., Convergence of a Crank-Nicolson difference scheme for heat equations with interface in the heat flow and concentrated heat capacity, Lect. Notes Comput. Sci. 1196 (1997), 58–65.

[2] Jovanović, B.S., Energy stability for a class of interface parabolic problems, XV Conference on Applied Mathematics Prim 2002.

[3] Tihonov, A.N., Samarskii A.A., Equations of Mathematical Physics, GITTL, Moscow 1953 (in Russian).

[4] Vulkov, L., Applications of Steklov-type eigenvalue problems to convergence of difference schemes for parabolic and hyperbolic equations with dynamical boundary conditions, Lect. Notes Comput. Sci. 1196 (1997), 557–564.

(received 15.12.2002)

Faculty of Electrical Engineering, P.O.Box 574, Skopje, Macedonia E-mail: szajkova@etf.ukim.edu.mk