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Fix typos in matphys seminars
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SeTSeR committed Feb 18, 2020
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26 changes: 15 additions & 11 deletions MatPhys/matphys_sem1.org
Original file line number Diff line number Diff line change
Expand Up @@ -1612,17 +1612,17 @@ u(x, t) = \frac1{2\sqrt{\pi a^2 t}}e^{\alpha x + a^2t + a^2t\alpha}\int_{\frac{-
** Задача 6.1
#+BEGIN_EXPORT latex
Проверить, являются ли функции гармоническими:
1. u_1 = \frac1{\sqrt{x^2 + y^2 + z^2}}, x^2 + y^2 + z^2
2. u_2 = e^{xyz}
3. u_3 = \sin x\sin y\sin z
4. u_4 = \sin 3x\sin 4y\sh 5z
1. $u_1 = \frac1{\sqrt{x^2 + y^2 + z^2}}, x^2 + y^2 + z^2$
2. $u_2 = e^{xyz}$
3. $u_3 = \sin x\sin y\sin z$
4. $u_4 = \sin 3x\sin 4y\sh 5z$
#+END_EXPORT
*** Решение
#+BEGIN_EXPORT latex
1.
$$\frac{\partial u}{\partial x} = -\frac{x}{(x^2 + y^2 + z^2)^{\frac32}}$$
$$\operatorname{grad}u = -\left\{\frac{x}{(x^2 + y^2 + z^2)^{\frac32}}, \frac{y}{(x^2 + y^2 + z^2)^{\frac32}}, \frac{z}{(x^2 + y^2 + z^2)^{\frac32}}\right\}$$
$$\Delta u = \operatorname{div}(\operatorname{grad}u) = \frac{2x^2 - y^2 - z^2 + 2y^2 - x^2 - z^2 + 2z^2 - x^2 - y^2}{\frac{x^2 + y^2 + z^2}^{\frac52}} = 0$$
$$\Delta u = \operatorname{div}(\operatorname{grad}u) = \frac{2x^2 - y^2 - z^2 + 2y^2 - x^2 - z^2 + 2z^2 - x^2 - y^2}{(x^2 + y^2 + z^2)^{\frac52}} = 0$$
$$\frac{\partial^2u}{\partial x^2} = -\frac{(x^2 + y^2 + z^2)^{\frac32}} + x\frac32(x^2 + y^2 + z^2)^{\frac12}\cdot2x = \frac{2x^2 - y^2 - z^2}{(x^2 + y^2 + z^2)^{\frac52}}$$
Гармоническая\\
2.
Expand Down Expand Up @@ -1916,10 +1916,12 @@ u(r, \varphi) = \frac12 + \left(\frac{a}r\right)^2\frac12\cos2\varphi
#+END_EXPORT
** Задача 7.4
#+BEGIN_EXPORT latex
\begin{equation}
\begin{cases}
\Delta u = 0, r < a, 0 \leq \varphi \leq 2\pi, \\
\frac{\partial u}{\partial r}|_{r = a} = f(\varphi), \int_0^{2\pi}f(\varphi)d\varphi = 0.
\end{cases}
\end{equation}
Последнее условие является необходимым для наличия решения.
#+END_EXPORT
*** Решение
Expand Down Expand Up @@ -2128,10 +2130,10 @@ G(P, Q) = \frac1{2\pi}\ln\frac1{r_{PQ}} - \frac1{2\pi}\ln\frac1{r_{PQ_1}}
Пусть $Q(x_0, y_0)$. Введём точки $Q_1(x_0, -y_0), Q_2(-x_0, y_0), Q_3(-x_0, -y_0)$. Тогда функция
Грина имеет вид:
\begin{multline}
G = \frac1{2\pi}\left(\ln\frac1{\sqrt{(x - x_0)^2 + (y - y_0)^2}} -
G = \frac1{2\pi}(\ln\frac1{\sqrt{(x - x_0)^2 + (y - y_0)^2}} -
\ln\frac1{\sqrt{(x - x_0)^2 + (y + y_0)^2}} - \\
- \ln\frac1{\sqrt{(x + x_0)^2 + (y - y_0)^2}} +
\ln\frac1{\sqrt{(x + x_0)^2 + (y + y_0)^2}}\right)
\ln\frac1{\sqrt{(x + x_0)^2 + (y + y_0)^2}})
\end{multline}
#+END_EXPORT
** Задача 8.13
Expand Down Expand Up @@ -2324,7 +2326,7 @@ f_1'(x)\cdot(-a) + f_2'(x)\cdot a = \psi(x)
\end{equation}
или, интегрируя второе уравнение от $x_0$ до $x$:
\begin{equation}
-a\int_{x_0}^xf_1'(\xi)d\xi + a\int_{x_0}^xf_2'(\xi)d\xi = \int_{x_0}^x\psi(\xi)\d\xi
-a\int_{x_0}^xf_1'(\xi)d\xi + a\int_{x_0}^xf_2'(\xi)d\xi = \int_{x_0}^x\psi(\xi)d\xi
\Rightarrow
f_2(x) - f_1(x) = \frac1a\int_{x_0}^x\psi(\xi)d\xi
\end{equation}
Expand Down Expand Up @@ -2612,7 +2614,7 @@ u_t(x, 0) = 0, 0 \leq x \leq l.
Собственные значения и собственнные функции ЗШЛ:
\begin{equation}
\begin{cases}
\lambda_n = \left(\frac{\pi n}l)^2, \\
\lambda_n = \left(\frac{\pi n}l\right)^2, \\
X_n = \sin\frac{\pi n}lx, \\
T_n = C_{1n}\sin\frac{\pi n}lat + C_{2n}\cos\frac{\pi n}lat
\end{cases}
Expand Down Expand Up @@ -2766,7 +2768,7 @@ u(x, t) = \sum_{n = 1}^{\infty}\sin\frac{\pi n}lx\frac{2l^2}{(\pi n a)^2}(1 - (-
\begin{cases}
u_{tt} = 4u_{xx}, \\
u(0, t) = t, \\
u_x\left(\frac{pi}2, x\right) = \pi, \\
u_x\left(\frac{pi}2, t\right) = \pi, \\
u(x, 0) = 2\sin5x + \pi x, \\
u_t(x, 0) = 1.
\end{cases}
Expand All @@ -2778,7 +2780,7 @@ u_t(x, 0) = 1.
\begin{equation}
\begin{cases}
u(0, t) = t = b(t),
u_x\left(\frac{\pi}2\right, t) = a(t) = \pi
u_x\left(\frac{\pi}2, t\right) = a(t) = \pi
\end{cases}
\Rightarrow
\begin{cases}
Expand Down Expand Up @@ -2814,11 +2816,13 @@ u(x, t) = \pi x + t + 2\sin5x\cos10t
#+END_EXPORT
** Задача 13.11
#+BEGIN_EXPORT latex
\begin{equation}
\begin{cases}
u_{tt} = u_{xx} + 3u + \sin t\cos2x, 0 < x < \pi, t > 0, \\
u_x(0, t) = u_x(\pi, t) = 0, t > 0, \\
u(x, 0) = u_t(x, 0) = 0, 0 \leq x \leq \pi.
\end{cases}
\end{equation}
#+END_EXPORT
*** Решение
#+BEGIN_EXPORT latex
Expand Down
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