定義より， \[ \mathcal{F}\{f(x - a)\} = \int_{-\infty}^\infty f(x - a) e^{-ikx}\,dx. \] \(x - a = u\)とおく，\(x = u + a\)，\(dx = du\) \[ \mathcal{F}\{f(u)\} = \int_{-\infty}^\infty f(u) e^{-ik(u + a)}\,du = e^{-ika} \int_{-\infty}^\infty f(u) e^{-iku}\,du = e^{-ika} F(k). \] よって， \[ \boxed{\mathcal{F}\{f(x - a)\} = e^{-ika} F(k).} \]
\[ \mathcal{F}\{f(x)\} = \int_{-x_0}^{x_0} e^{-ikx}\,dx = \Bigl[-\frac{1}{ik} e^{-ikx} \Bigr]_{-x_0}^{x_0} = \frac{i}{k}\Bigl(e^{-ikx_0} - e^{ikx_0}\Bigr). \] \[ \boxed{\mathcal{F}\{f(x)\} = \frac{i}{k}\Bigl(e^{-ikx_0} - e^{ikx_0}\Bigr).} \]
\[ \mathcal{F}\{f(x)*g(x)\} = \int_{-\infty}^\infty f(x)*g(x) e^{-ikx}\,dx = \int_{-\infty}^\infty \Bigl(\int_{-\infty}^\infty f(x - y)g(y)dy \Bigr) e^{-ikx}\,dx \] \[ = \int_{-\infty}^\infty g(y)e^{-iky} \Bigl(\int_{-\infty}^\infty f(x - y)e^{-ik(x - y)}dx \Bigr) \,dy \] \[ = \int_{-\infty}^\infty g(y)e^{-iky}dy \Bigl(\int_{-\infty}^\infty f(x')e^{-ikx'}dx' \Bigr) = F(k)G(k). \] \[ \boxed{\mathcal{F}\{f(x)*g(x)\} = F(k)G(k).} \]
\[ U(k,t) = \int_{-\infty}^\infty u(x,t) e^{-ikx} dx. \] \[ \mathcal{F}\Bigl\{\frac{\partial u}{\partial x}\Bigr\}(x,t) = \int_{-\infty}^\infty \frac{\partial u}{\partial x} e^{-ikx} \,dx \] \[ = \Bigl[u(x,t)e^{-ikx}\Bigr]_{-\infty}^\infty + ik \int_{-\infty}^\infty u(x,t) e^{-ikx} \,dx = ikU(k,t) \] \[ \mathcal{F}\Bigl\{\frac{\partial^2 u}{\partial x^2}\Bigr\}(x,t) = \int_{-\infty}^\infty \frac{\partial^2 U}{\partial x^2} e^{-ikx} \,dx \] \[ = \Bigl[\frac{\partial u}{\partial x}e^{-ikx}\Bigr]_{-\infty}^\infty + ik \int_{-\infty}^\infty \frac{\partial u}{\partial x} e^{-ikx} \,dx = ik \cdot ikU(k,t) = -k^2 U(k,t) \] ここで，最初に定義した式を時間微分する． \[ \frac{\partial U}{\partial t}(k,t) = \frac{\partial}{\partial t} \int_{-\infty}^\infty u(x,t) e^{-ikx} \,dx \] \[ = \int_{-\infty}^\infty \frac{\partial u}{\partial t}(x,t) e^{-ikx} \,dx = \mathcal{F}\Bigl\{\frac{\partial u}{\partial t}\Bigr\}(x,t) \] \[ \Leftrightarrow \frac{\partial U}{\partial t}(k,t) = \mathcal{F}\Bigl\{\frac{\partial u}{\partial t}\Bigr\}(x,t). \] 与式より， \[ \frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t} \] であるから， \[ \frac{\partial U}{\partial t}(k,t) = \mathcal{F}\Bigl\{\frac{\partial u}{\partial t}\Bigr\}(x,t) = \mathcal{F}\Bigl\{\frac{\partial^2 u}{\partial x^2}\Bigr\}(x,t) = -k^2 U(k,t). \] \[ \boxed{\frac{\partial U}{\partial t}(k,t) = -k^2 U(k,t).} \]
式変形して， \[ \frac{1}{U(k,t)} \frac{\partial U(k,t)}{\partial t} = -k^2. \] 両辺時間積分して， \[ \int_0^t \frac{1}{U(k,t)} \frac{\partial U(k,t)}{\partial t} \,dt = -\int_0^t k^2 \,dt, \] \[ \Leftrightarrow \Bigl[ \ln \Bigl|U(k,t)\Bigr|\Bigr]_0^t = -k^2t, \] \[ \Leftrightarrow \Biggl[ \ln \Biggl|\frac{U(k,t)}{U_0(k)}\Biggr|\Biggr]_0^t = -k^2t, \] \[ \Leftrightarrow \ln \Biggl|\frac{U(k,t)}{U_0(k)}\Biggr| = \exp(-k^2t), \] \[ \Leftrightarrow U(k,t) = U_0(k) \exp(-k^2t). \] \[ \boxed{U(k,t) = U_0(k) \exp(-k^2t).} \]
定義より，
\[
u(x,t) = \frac{1}{2\pi} \int_{-\infty}^\infty U(k,t) e^{ixk} \,dk
= \frac{1}{2\pi} \int_{-\infty}^\infty U_0(k) \exp(-k^2t) e^{ixk} \,dk,
\]
ここで，畳み込み定理を利用する．
\[
G(x,t) = \frac{1}{2\pi} \int_{-\infty}^\infty \exp(-k^2t) e^{ixk} \,dk
\]
とする．すると，
\[
u(x,t) = \int_{-\infty}^\infty G(x - y) U_0(y) \,dy.
\]
\[
\mathcal{F}\{e^{-ax^2}\} = \sqrt{\frac{\pi}{a}} \exp{\Bigl(-\frac{k^2}{4a}\Bigr)},
\]
より，
\[
\int_{-\infty}^\infty e^{-ax^2} e^{-ikx} \,dx = \sqrt{\frac{\pi}{a}} \exp{\Bigl(-\frac{k^2}{4a}\Bigr)}.
\]
逆変換すると，
\[
\frac{1}{2\pi} \int_{-\infty}^\infty \exp{\Bigl(-\frac{k^2}{4a}\Bigr)} e^{ixk} \,dk
= \frac{1}{2\pi} \sqrt{\frac{\pi}{a}} e^{-ax^2}
= \frac{1}{2\sqrt{\pi a}} e^{-ax^2}.
\]
ここで，
\[
\frac{1}{4a} = t, a = \frac{1}{4t}
\]
とおくと，
\[
\frac{1}{2\pi} \int_{-\infty}^\infty \exp{\Bigl(-\frac{k^2}{4a}\Bigr)} e^{ixk} \,dk
= G(x,t)
= \sqrt{\frac{t}{\pi}} \exp\Bigl(-\frac{x^2}{4t}\Bigr).
\]
以上より，
\[
\boxed{u(x,t) = \sqrt{\frac{t}{\pi}} \int_{-\infty}^\infty \exp\Bigl\{-\frac{(x - y)^2}{4t}\Bigr\} u_0(y) \,dy.}
\]