Discussion
Consider the partial differential equation \(\frac{\partial^{2}f}{\partial x^{2}}+\frac{\partial^{2}f}{\partial y^{2}}=0\) where \(x, y\) are real.
If \(f(x, y) = a(x)b(y)\), where \(a(x)\) and \(b(y)\) are real functions, which one of the following statements can be true?
Answer: C
The possible case is a(x) periodic and b(y) exponential.
Substituting \(f = a(x)b(y)\) into Laplace's equation and dividing by ab separates the variables:
\[ \frac{a''(x)}{a(x)} = -\frac{b''(y)}{b(y)} = -k^{2} \]
A negative constant gives \(a'' = -k^{2}a\), so a is a sine or cosine (periodic), and \(b'' = k^{2}b\), so b is a combination of \(e^{ky}\) and \(e^{-ky}\) (exponential). This is the familiar separated solution such as \(\sin kx\, e^{ky}\).
The two functions always have opposite-signed constants, so they cannot both be exponential or both periodic. A linear b(y) needs the constant to be zero, and then a(x) is linear too, not periodic.
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