🎨
Цвет акцента
Синий
Фиолетовый
Пурпурный
Тренажёры · Глава 11

Решение дифференциальных уравнений эллиптического типа

Задания для самоконтроля

     5. Численное решение дифференциального уравнения

  \(\displaystyle 5\left( \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} \right) = 7u - 6xy\)
проводится методом установления с использованием схемы предиктор-корректор. Из представленных ниже разностных схем выберите ту, которая записана без ошибок.

\(\displaystyle \begin{cases} \dfrac{u_{j,k}^{n+1/4} - u_{j,k}^n}{\Delta t} = 5\lambda_{xx} u_{j,k}^{n+1/4} \qquad \dfrac{u_{j,k}^{n+1/2} - u_{j,k}^{n+1/4}}{\Delta t} = 5\lambda_{yy} u_{j,k}^{n+1/2} \\[2ex] \dfrac{u_{j,k}^{n+1} - u_{j,k}^{n+1/2}}{\Delta t} = 5\lambda_{xx} u_{j,k}^{n+1/2} + 5\lambda_{yy} u_{j,k}^{n+1/2} - 7u_{j,k}^{n+1/2} + 6(j-1)(k-1)h^2 \end{cases}\)

\(\displaystyle \begin{cases} \dfrac{u_{j,k}^{n+1/4} - u_{j,k}^n}{\Delta t/2} = 5\lambda_{xx} u_{j,k}^{n+1/4} \qquad \dfrac{u_{j,k}^{n+1/2} - u_{j,k}^{n+1/4}}{\Delta t/2} = 5\lambda_{yy} u_{j,k}^{n+1/2} \\[2ex] \dfrac{u_{j,k}^{n+1} - u_{j,k}^{n+1/2}}{\Delta t} = 5\lambda_{xx} u_{j,k}^{n+1/2} + 5\lambda_{yy} u_{j,k}^{n+1/2} - 7u_{j,k}^{n+1/2} + 6(j-1)(k-1)h^2 \end{cases}\)

\(\displaystyle \begin{cases} \dfrac{u_{j,k}^{n+1/4} - u_{j,k}^n}{\Delta t/2} + 5\lambda_{xx} u_{j,k}^{n+1/4} = 0 \qquad \dfrac{u_{j,k}^{n+1/2} - u_{j,k}^{n+1/4}}{\Delta t/2} + 5\lambda_{yy} u_{j,k}^{n+1/2} = 0 \\[2ex] \dfrac{u_{j,k}^{n+1} - u_{j,k}^n}{\Delta t} + 5\lambda_{xx} u_{j,k}^{n+1/2} + 5\lambda_{yy} u_{j,k}^{n+1/2} = 7u_{j,k}^{n+1/2} - 6(j-1)(k-1)h^2 \end{cases}\)

\(\displaystyle \begin{cases} \dfrac{u_{j,k}^{n+1/4} - u_{j,k}^n}{\Delta t/2} = 5\lambda_{xx} u_{j,k}^{n+1/4} \qquad \dfrac{u_{j,k}^{n+1/2} - u_{j,k}^{n+1/4}}{\Delta t/2} = 5\lambda_{yy} u_{j,k}^{n+1/2} \\[2ex] \dfrac{u_{j,k}^{n+1} - u_{j,k}^n}{\Delta t} = 5\lambda_{xx} u_{j,k}^{n+1/2} + 5\lambda_{yy} u_{j,k}^{n+1/2} - 7u_{j,k}^{n+1/2} + 6(j-1)(k-1)h^2 \end{cases}\)

\(\displaystyle \begin{cases} \dfrac{u_{j,k}^{n+1/4} - u_{j,k}^n}{\Delta t/2} = 5\lambda_{xx} u_{j,k}^{n+1/4} \qquad \dfrac{u_{j,k}^{n+1/2} - u_{j,k}^{n+1/4}}{\Delta t/2} = 5\lambda_{yy} u_{j,k}^{n+1/2} \\[2ex] \dfrac{u_{j,k}^{n+1} - u_{j,k}^{n+1/2}}{\Delta t/2} = 5\lambda_{xx} u_{j,k}^{n+1/2} + 5\lambda_{yy} u_{j,k}^{n+1/2} - 7u_{j,k}^{n+1/2} + 6(j-1)(k-1)h^2 \end{cases}\)

\(\displaystyle \begin{cases} \dfrac{u_{j,k}^{n+1/4} - u_{j,k}^n}{\Delta t/2} = 5\lambda_{xx} u_{j,k}^{n+1/4} \qquad \dfrac{u_{j,k}^{n+1/2} - u_{j,k}^{n+1/4}}{\Delta t/2} = 5\lambda_{yy} u_{j,k}^{n+1/2} \\[2ex] \dfrac{u_{j,k}^{n+1} - u_{j,k}^n}{\Delta t} = 5\lambda_{xx} u_{j,k}^{n+1/2} + 5\lambda_{yy} u_{j,k}^{n+1/2} + 7u_{j,k}^{n+1/2} - 6(j-1)(k-1)h^2 \end{cases}\)