Страница сетевой поддержки учеников Тимура Юрьевича Альпина

Помимо того, что в ответе многого не хватало, неясно, почему там не взяли второй интеграл — а он тоже Эйлера-Пуассона. Ответ получается вполне компактный.

Решение содержит много копипасты из решения №77, так что смотреть можно с середины.

Условия задачи:
$$U=U\left(x,t\right),\quad-\infty < x < \infty,\quad t\geqslant0.$$ $$\begin{equation} U_{t}=a^{2}U_{xx}\label{eq:ur} \end{equation}$$ $$\begin{equation} \lim_{x\to\pm\infty}U=0\label{eq:usfur} \end{equation}$$ $$\begin{equation} \left.U\right|_{t=0}=U_{0}e^{-\frac{x^{2}}{l^{2}}}\label{eq:na4} \end{equation}$$ Преобразования Фурье туда и оттуда: $$\begin{equation} \bar{f}\left(\lambda\right)=\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}f\left(\xi\right)e^{-i\lambda\xi}d\xi\qquad f\left(x\right)=\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}\bar{f}\left(\lambda\right)e^{i\lambda x}d\lambda,\quad\lim_{x\to\pm\infty}f\left(x\right)=0 \end{equation}$$ $$\begin{equation} \bar{U}\left(\lambda,t\right)=\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}Ue^{-i\lambda x}dx \end{equation}$$ Применим их к ($\ref{eq:ur}$): $$\begin{equation} U_{t}e^{-i\lambda x}=a^{2}U_{xx}e^{-i\lambda x} \end{equation}$$ $$\begin{equation} \frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}U_{t}e^{-i\lambda x}dx=\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}a^{2}U_{xx}e^{-i\lambda x}dx \end{equation}$$ $$\frac{\partial}{\partial t}\left(\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}Ue^{-i\lambda x}dx\right)=\frac{a^{2}}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}\frac{\partial}{\partial x}U_{x}e^{-i\lambda x}dx=\left.\frac{a^{2}}{\sqrt{2\pi}}U_{x}e^{-i\lambda x}\right|_{-\infty}^{\infty}-\frac{a^{2}}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}U_{x}\frac{\partial}{\partial x}e^{-i\lambda x}dx=$$ $$\begin{equation} =\frac{i\lambda a^{2}}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}U_{x}e^{-i\lambda x}dx=\left.\frac{i\lambda a^{2}}{\sqrt{2\pi}}Ue^{-i\lambda x}\right|_{-\infty}^{\infty}-\frac{i\lambda a^{2}}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}U\frac{\partial}{\partial x}e^{-i\lambda x}dx= \end{equation}$$ $$-\lambda^{2}a^{2}\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}Ue^{-i\lambda x}dx$$ $$\begin{equation} \bar{U}_{t}=-\lambda^{2}a^{2}\bar{U}. \end{equation}$$ Найдём Фурье-образ $\bar{U}$: $$\begin{equation} \ln\left|\bar{U}\right|=-\lambda^{2}a^{2}t+\tilde{\varphi}\left(\lambda\right), \end{equation}$$ $$\begin{equation} \bar{U}=\varphi\left(\lambda\right)e^{-\lambda^{2}a^{2}t}. \end{equation}$$ Чтобы найти $\varphi\left(\lambda\right)$, преобразуем ($\ref{eq:na4}$): $$\left.\bar{U}\right|_{t=0}=\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}\left.U\right|_{t=0}e^{-i\lambda x}dx=$$ $$\begin{equation} =\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}U_{0}e^{-\frac{x^{2}}{l^{2}}}e^{-i\lambda x}dx= \end{equation}$$ $$=\frac{U_{0}}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}e^{-i\lambda x-\frac{x^{2}}{l^{2}}}dx.$$ С другой стороны, $$\begin{equation} \left.\bar{U}\right|_{t=0}=\left.\varphi\left(\lambda\right)e^{-\lambda^{2}a^{2}t}\right|_{t=0}=\varphi\left(\lambda\right), \end{equation}$$ $$\begin{equation} \varphi\left(\lambda\right)=\frac{U_{0}}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}e^{-i\lambda\xi-\frac{\xi^{2}}{l^{2}}}d\xi. \end{equation}$$ $$\begin{equation} \bar{U}=\frac{U_{0}}{\sqrt{2\pi}}e^{-\lambda^{2}a^{2}t}\intop_{-\infty}^{\infty}e^{-i\lambda\xi-\frac{\xi^{2}}{l^{2}}}d\xi. \end{equation}$$ Найдём саму функцию $U$: $$U=\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}\bar{U}e^{i\lambda x}d\lambda=$$ $$\begin{equation} =\frac{U_{0}}{2\pi}\intop_{-\infty}^{\infty}\intop_{-\infty}^{\infty}e^{-\lambda^{2}a^{2}t}e^{-i\lambda\xi-\frac{\xi^{2}}{l^{2}}}e^{i\lambda x}d\xi d\lambda= \end{equation}$$ $$=\frac{U_{0}}{2\pi}\intop_{-\infty}^{\infty}\intop_{-\infty}^{\infty}e^{-\lambda^{2}a^{2}t-i\lambda\xi-\frac{\xi^{2}}{l^{2}}+i\lambda x}d\lambda d\xi.$$ Отдельно вычислим $$\intop_{-\infty}^{\infty}e^{-\lambda^{2}a^{2}t-i\lambda\xi-\frac{\xi^{2}}{l^{2}}+i\lambda x}d\lambda=\intop_{-\infty}^{\infty}e^{-\left[\left(\lambda a\sqrt{t}\right)^{2}-i\left(x-\xi\right)\lambda+\left(\frac{i\left(x-\xi\right)}{2a\sqrt{t}}\right)^{2}-\left(\frac{i\left(x-\xi\right)}{2a\sqrt{t}}\right)^{2}\right]-\frac{\xi^{2}}{l^{2}}}d\lambda=e^{-\left(\frac{x-\xi}{2a\sqrt{t}}\right)^{2}-\frac{\xi^{2}}{l^{2}}}\intop_{-\infty}^{\infty}e^{-\left[\lambda a\sqrt{t}-\frac{i\left(x-\xi\right)}{2a\sqrt{t}}\right]^{2}}d\lambda=$$ $\lambda a\sqrt{t}-\frac{i\left(x-\xi\right)}{2a\sqrt{t}}=\gamma$, $d\lambda=\frac{d\gamma}{a\sqrt{t}}$ $$=e^{-\left(\frac{x-\xi}{2a\sqrt{t}}\right)^{2}-\frac{\xi^{2}}{l^{2}}}\intop_{-\infty}^{\infty}e^{-\gamma^{2}}\frac{d\gamma}{a\sqrt{t}}=\frac{\sqrt{\pi}}{a\sqrt{t}}e^{-\left(\frac{x-\xi}{2a\sqrt{t}}\right)^{2}-\frac{\xi^{2}}{l^{2}}}.$$ Тогда $$U=\frac{U_{0}}{2\pi}\intop_{-\infty}^{\infty}\intop_{-\infty}^{\infty}e^{-\lambda^{2}a^{2}t-i\lambda\xi-\frac{\xi^{2}}{l^{2}}+i\lambda x}d\lambda d\xi=$$ $$=\frac{U_{0}}{2\pi}\intop_{-\infty}^{\infty}\frac{\sqrt{\pi}}{a\sqrt{t}}e^{-\left(\frac{x-\xi}{2a\sqrt{t}}\right)^{2}-\frac{\xi^{2}}{l^{2}}}d\xi=\frac{U_{0}}{2\sqrt{\pi}a}\frac{1}{\sqrt{t}}\intop_{-\infty}^{\infty}e^{-\left(\frac{x-\xi}{2a\sqrt{t}}\right)^{2}-\frac{\xi^{2}}{l^{2}}}d\xi.$$ Примерно это было получено в методичке, но мы на этом не остановимся. Преобразуем показатель $$-\left(\frac{x-\xi}{2a\sqrt{t}}\right)^{2}-\frac{\xi^{2}}{l^{2}}=-\frac{x^{2}-2x\xi+\xi^{2}}{4a^{2}t}-\frac{\xi^{2}}{l^{2}}=-\frac{x^{2}}{4a^{2}t}-\left[\left(\frac{1}{4a^{2}t}+\frac{1}{l^{2}}\right)\xi^{2}-2\frac{x}{4a^{2}t}\xi+\left(\frac{\frac{x}{4a^{2}t}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}\right)^{2}-\left(\frac{\frac{x}{4a^{2}t}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}\right)^{2}\right]=$$ $$=-\left[\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}\xi-\frac{\frac{x}{4a^{2}t}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}\right]^{2}+\frac{\frac{x^{2}}{\left(4a^{2}t\right)^{2}}}{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}-\frac{x^{2}}{4a^{2}t}=$$ $$=-\left[\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}\xi-\frac{\frac{x}{4a^{2}t}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}\right]^{2}+\frac{l^{2}}{l^{2}+4a^{2}t}\frac{x^{2}}{4a^{2}t}-\frac{x^{2}}{4a^{2}t}=$$ $$=-\left[\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}\xi-\frac{\frac{x}{4a^{2}t}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}\right]^{2}-\frac{x^{2}}{l^{2}+4a^{2}t}$$ Подставим в U и сведём к интегралу Эйлера-Пуассона $$U=\frac{U_{0}}{2\sqrt{\pi}a}\frac{1}{\sqrt{t}}\intop_{-\infty}^{\infty}e^{-\left(\frac{x-\xi}{2a\sqrt{t}}\right)^{2}-\frac{\xi^{2}}{l^{2}}}d\xi=\frac{U_{0}}{2\sqrt{\pi}a}\frac{1}{\sqrt{t}}e^{-\frac{x^{2}}{l^{2}+4a^{2}t}}\intop_{-\infty}^{\infty}e^{-\left[\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}\xi-\frac{\frac{x}{4a^{2}t}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}\right]^{2}}d\xi=$$ $\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}\xi-\frac{\frac{x}{4a^{2}t}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}=\gamma$, $\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}d\xi=d\gamma$ $$=\frac{U_{0}}{2\sqrt{\pi}a}\frac{1}{\sqrt{t}}e^{-\frac{x^{2}}{l^{2}+4a^{2}t}}\frac{1}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}\intop_{-\infty}^{\infty}e^{-\gamma^{2}}d\gamma=\frac{U_{0}}{2a\sqrt{t}}\frac{e^{-\frac{x^{2}}{l^{2}+4a^{2}t}}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}.$$ Ещё немного упростим: $$\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}=\sqrt{\frac{l^{2}+4a^{2}t}{4a^{2}tl^{2}}}=\frac{\sqrt{l^{2}+4a^{2}t}}{2a\sqrt{t}l},$$ откуда $$U=\frac{U_{0}}{2a\sqrt{t}}\frac{e^{-\frac{x^{2}}{l^{2}+4a^{2}t}}}{\sqrt{\frac{1}{4a^{2}t}+\frac{1}{l^{2}}}}=\frac{U_{0}}{2a\sqrt{t}}\frac{e^{-\frac{x^{2}}{l^{2}+4a^{2}t}}}{\frac{\sqrt{l^{2}+4a^{2}t}}{2a\sqrt{t}l}}=\frac{U_{0}l}{\sqrt{l^{2}+4a^{2}t}}e^{-\frac{x^{2}}{l^{2}+4a^{2}t}}.$$