A Uniqueness Result for a Coupled System of Elliptic PDEs

Let $D\subset {ℝ}^3$ be a bounded domain with a smooth boundary S,

$Lu+av\mathrm{<mo>=</mo>0,}-bu+Mv\mathrm{<mo>=</mo>0}\text{in}D\mathrm{,}$

$u\mathrm{<mo>=</mo>}v\mathrm{<mo>=</mo>0}\text{on }S\mathrm{.}$

Assume that L and M are positive elliptic Dirichlet operators of second order, a > 0 and b > 0 are constants. We prove that under these assumptions, the unique solution is u = v = 0 in D.

1. Introduction

$Lu+av\mathrm{<mo>=</mo>0}\text{in}D\mathrm{,}\text{ (1)}$

$-bu+Mv\mathrm{<mo>=</mo>0}\text{in}D\mathrm{,}\text{ (2)}$

$u\mathrm{<mo>=</mo>}v\mathrm{<mo>=</mo>0}\text{on}S\mathrm{,}\text{ (3)}$

Assume that L and M are second order Dirichlet elliptic operators with real-valued coefficients, the bilinear form (Lu, u) is positive-definite for u ≠ 0.

$\mathrm{<mo\; stretchy="false">(</mo>}Lu\mathrm{,}u\mathrm{<mo\; stretchy="false">)</mo><mo>></mo>0}\text{for}u\cancel{\equiv }\mathrm{0,}\text{ (4)}$

$\mathrm{<mo\; stretchy="false">(</mo>}Mv\mathrm{,}v\mathrm{<mo\; stretchy="false">)</mo><mo>></mo>0}\text{for}v\cancel{\equiv }\mathrm{0,}\text{ (5)}$

a and b are constants,

$a\mathrm{<mo>></mo>0,}b\mathrm{<mo>></mo>0.}\text{ (6)}$

Theorem 1. If assumptions (1)–(6) hold, then u = v = 0 in D.

This problem concerns the uniqueness of solutions to coupled elliptic PDE systems with variable coefficients, a fundamental topic in mathematical analysis [1]. The full problem allows a(x), b(x) to vary spatially, increasing its generality and complexity. In the general formulation, the coupling coefficients a(x) and b(x) are positive continuous functions. a = a(x) > 0 and b = b(x) > 0.

2. Proofs

Lets define the inner product (u, v) over the domain D as $\mathrm{<mo\; stretchy="false">(</mo>}u\mathrm{,}v\mathrm{<mo\; stretchy="false">)</mo><mo>:</mo><mo>=</mo>}{\displaystyle {\int }_D}uvdx$ . From our assumptions one derives, multiplying (1) by u and integrating by parts, the relation:

$\mathrm{<mo\; stretchy="false">(</mo>}Lu\mathrm{,}u\mathrm{<mo\; stretchy="false">)</mo>}+a\mathrm{<mo\; stretchy="false">(</mo>}v\mathrm{,}u\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>0,}\text{ (7)}$

and multiplying (2) by v and integrating by parts, the relation:

$-b\mathrm{<mo\; stretchy="false">(</mo>}u\mathrm{,}v\mathrm{<mo\; stretchy="false">)</mo>}+\mathrm{<mo\; stretchy="false">(</mo>}Mv\mathrm{,}v\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>0.}\text{ (8)}$

Combining the previous inequalities and assumptions yields from the assumptions (4)–(6) and from the inequalities (7) and (8) one gets, which can lead to the inequality:

$\mathrm{<mo\; stretchy="false">(</mo>}v\mathrm{,}u\mathrm{<mo\; stretchy="false">)</mo>}\le \mathrm{0,}\mathrm{<mo\; stretchy="false">(</mo>}u\mathrm{,}v\mathrm{<mo\; stretchy="false">)</mo>}\ge 0.\text{ (9)}$

As u and v are real-valued functions, it follows that, one has

$\mathrm{<mo\; stretchy="false">(</mo>}v\mathrm{,}u\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo><mo\; stretchy="false">(</mo>}u\mathrm{,}v\mathrm{<mo\; stretchy="false">)</mo>.}\text{ (10)}$

Therefore

$\mathrm{<mo\; stretchy="false">(</mo>}v\mathrm{,}u\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo><mo\; stretchy="false">(</mo>}u\mathrm{,}v\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>0}\text{ (11)}$

Consequently,

$\mathrm{<mo\; stretchy="false">(</mo>}Lu\mathrm{,}u\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo><mo\; stretchy="false">(</mo>}Mv\mathrm{,}v\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>0.}\text{ (12)}$

From (12) and our assumptions (4)–(5) it follows that u = v = 0 in D. Theorem 1 is proved. $\mathrm{<mo>|</mo>}a-b\mathrm{<mo>|</mo>}$

Remark. Note that in Theorem 1 there are no restrictions on the size of the constants a and b or .

It is possible to show that the solution to problem (1)–(5) with a = a(x) and b = b(x), where a(x) and b(x) are continuous functions in the closure of D and

$\underset{x\in D}{{\displaystyle \sup }}\mathrm{<mo>|</mo>}a\mathrm{<mo\; stretchy="false">(</mo>}x\mathrm{<mo\; stretchy="false">)</mo>}-b\mathrm{<mo\; stretchy="false">(</mo>}x\mathrm{<mo\; stretchy="false">)</mo><mo>|</mo><mo><</mo>}\nu \mathrm{,}\text{ (13)}$

where V is a sufficiently small constant, is equal to zero in D.

Let us prove this conclusion. From (1)–(2) it follows that

$\mathrm{<mo\; stretchy="false">(</mo>}Lu\mathrm{,}u\mathrm{<mo\; stretchy="false">)</mo>}+{\displaystyle {\int }_S}\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{<mo\; stretchy="false">(</mo>}x\mathrm{<mo\; stretchy="false">)</mo>}-b\mathrm{<mo\; stretchy="false">(</mo>}x\mathrm{<mo\; stretchy="false">)</mo><mo\; stretchy="false">)</mo>}u\mathrm{<mo\; stretchy="false">(</mo>}x\mathrm{<mo\; stretchy="false">)</mo>}v\mathrm{<mo\; stretchy="false">(</mo>}x\mathrm{<mo\; stretchy="false">)</mo>}dx+\mathrm{<mo\; stretchy="false">(</mo>}Mv\mathrm{,}v\mathrm{<mo\; stretchy="false">)</mo><mo>=</mo>0.}\text{ (14)}$

From (4)–(5) it follows that

$\mathrm{<mo\; stretchy="false">(</mo>}Lu\mathrm{,}u\mathrm{<mo\; stretchy="false">)</mo><mo>></mo>}{c}_L\parallel u{\parallel }^2\mathrm{,}\mathrm{<mo\; stretchy="false">(</mo>}Mv\mathrm{,}v\mathrm{<mo\; stretchy="false">)</mo>}{c}_M\parallel v{\parallel }^2\mathrm{,}\text{ (15)}$

where $\parallel u{\parallel }^2\mathrm{<mo>=</mo><mo\; stretchy="false">(</mo>}u\mathrm{,}u\mathrm{<mo\; stretchy="false">)</mo>.}$

From our assumption (13) it follows that

$\mathrm{<mo>|</mo>}{\displaystyle {\int }_S}\mathrm{<mo\; stretchy="false">(</mo>}a\mathrm{<mo\; stretchy="false">(</mo>}x\mathrm{<mo\; stretchy="false">)</mo>}-b\mathrm{<mo\; stretchy="false">(</mo>}x\mathrm{<mo\; stretchy="false">)</mo><mo\; stretchy="false">)</mo>}u\mathrm{<mo\; stretchy="false">(</mo>}x\mathrm{<mo\; stretchy="false">)</mo>}v\mathrm{<mo\; stretchy="false">(</mo>}x\mathrm{<mo\; stretchy="false">)</mo>}dx\mathrm{<mo>|</mo>}\le \nu \parallel u\parallel \parallel v\parallel \mathrm{.}\text{ (16)}$

From (14)–(16) one derives:

$\mathrm{0<mo>></mo>}{c}_L\parallel u{\parallel }^2+c_M\parallel v{\parallel }^2-\nu \parallel u\parallel \parallel v\parallel \mathrm{.}\text{ (17)}$

Choosing $\nu \mathrm{<mo><</mo>}{c}_L+c_M$ , ensures that inequality (17) leads to $\parallel u\parallel \mathrm{<mo>=</mo>}\parallel v\parallel \mathrm{<mo>=</mo>0}$ , so $u\mathrm{<mo>=</mo>}v\mathrm{<mo>=</mo>0}$ in $D$ .

Open problem: We pose the following open question: Does the conclusion of Theorem 1 remain valid if the assumption that a,b are constants is replaced by the assumption $a\mathrm{<mo>=</mo>}a\mathrm{<mo\; stretchy="false">(</mo>}x\mathrm{<mo\; stretchy="false">)</mo>}\ge 0$ that  and $b\mathrm{<mo>=</mo>}b\mathrm{<mo\; stretchy="false">(</mo>}x\mathrm{<mo\; stretchy="false">)</mo>}\ge 0$  are continuous functions?

3. Conclusion

This result confirms that no non-trivial solution exists under constant coupling terms, contributing to PDE theory. Representing a coupled elliptic system with Dirichlet boundary conditions (1)–(3). The basic assumptions are: a > 0 and b > 0 are constants, L and M are positive elliptic operators such that (Lu, u) = 0 implies u =0, and (Mv, v) = 0 implies v = 0.

This result contributes to the broader understanding of stability and uniqueness in PDE modeling across physics and engineering.

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