Discussion
Let \(u(x, t)\) be the solution of the initial value problem for the heat equation on the real line:
\(\dfrac{\partial u}{\partial t}=k\dfrac{\partial^2u}{\partial x^2}\),
\(-\infty \lt x \lt \infty\), \(t \gt 0\), \(k \in \mathbb{R}\)
with the initial condition
\[ u(x, 0) = e^{-a|x|}, \quad a \gt 0. \]
If \(\hat{u}(w, t) = \int_{-\infty}^{\infty} u(x, t) e^{iwx}\, dx\) is the Fourier transform of \(u(x, t)\) with respect to x, then \(\hat{u}(w, t)\) is equal to
Answer: A
The answer is \(\hat{u}(w, t) = \frac{2a}{a^2+w^2} e^{-kw^2 t}\).
Transforming the equation in x turns the second derivative into multiplication by \(-w^2\), leaving an ordinary differential equation in t:
\[ \begin{aligned} \frac{\partial \hat{u}}{\partial t} &= -k w^2 \hat{u} \\ \hat{u}(w, t) &= \hat{u}(w, 0)\, e^{-k w^2 t} \end{aligned} \]
The transform of the initial condition is
\[ \begin{aligned} \hat{u}(w, 0) &= \int_{-\infty}^{\infty} e^{-a|x|} e^{iwx} dx \\ &= \frac{2a}{a^2 + w^2}. \end{aligned} \]
The option with a in the numerator has lost the factor 2 from the two halves of the line; the one with \(e^{-ka^2 t}\) decays at the wrong rate; and the Gaussian is the solution kernel in x, not this transform.
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