Discussion
For the Fredholm integral equation
\(u(t) = \lambda\displaystyle\int_{a}^{b} k(t, s)\, u(s)\, ds\)
with separable kernel, Fredholm determinant \(D(\lambda) \ne 0\). Then the integral equation has :
Answer: A
The equation has only the zero solution.
For a homogeneous Fredholm equation with separable kernel \(k(t, s) = \sum a_{i}(t)b_{i}(s)\), any solution has the form \(u = \lambda\sum c_{i}a_{i}(t)\), and the constants satisfy a homogeneous linear system \((I - \lambda A)c = 0\). Its determinant is the Fredholm determinant \(D(\lambda)\).
If \(D(\lambda) \ne 0\), the system has only \(c = 0\), so \(u \equiv 0\) is the unique solution. Non-zero solutions, infinitely many of them, exist only when \(D(\lambda) = 0\), that is, when \(\lambda\) is an eigenvalue. A homogeneous equation always has the zero solution, so 'no solution' is impossible.
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