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5. Выберите правильное решение задачи определения прогоночных коэффициентов на 1-м шаге по координате х
(т.е. \(\displaystyle \alpha_1,\, \beta_1\)), если задано левое граничное условие 3-го рода:
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\(\displaystyle \frac{\partial u}{\partial x}(t,x=0)=3tu(t,x=0)-1{,}5e^t.\) |
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\(\displaystyle u_1^{n+1}=\alpha_1 u_2^{n+1}+\beta_1,\\ \frac{u_2^{n+1}-u_1^{n+1}}{h}=3\Delta t n u_1^{n+1}+1{,}5e^{\Delta t n},\\ u_1^{n+1}=\frac{u_2^{n+1}-1{,}5he^{\Delta t n}}{1+3\Delta t n h}\ \Rightarrow\ \alpha_1=\frac{1}{1+3\Delta t n h},\quad \beta_1=-\frac{1{,}5he^{\Delta t n}}{1+3\Delta t n h}.\) |
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\(\displaystyle u_1^{n+1}=\alpha_1 u_2^{n+1}+\beta_1,\\ \frac{u_2^{n+1}-u_1^{n+1}}{h}=3\Delta t n u_1^{n+1}-1{,}5e^{\Delta t n},\\ u_1^{n+1}=\frac{u_2^{n+1}+1{,}5he^{\Delta t n}}{1+3\Delta t n h}\ \Rightarrow\ \alpha_1=\frac{1}{1+3\Delta t n h},\quad \beta_1=\frac{1{,}5he^{\Delta t n}}{1+3\Delta t n h}.\) |
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\(\displaystyle u_1^{n+1}=\alpha_1 u_2^{n+1}+\beta_1,\\ \frac{u_2^{n+1}-u_1^{n+1}}{h}=3\Delta t n u_1^{n+1}-1{,}5e^{\Delta t n},\\ u_1^{n+1}=\frac{u_2^{n+1}+1{,}5he^{\Delta t n}}{1+3\Delta t n h}\ \Rightarrow\ \alpha_1=\frac{1}{1+3\Delta t n h},\quad \beta_1=\frac{1{,}5he^{\Delta t n}}{1+3\Delta t n h}.\) |
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\(\displaystyle u_1^{n+1}=\alpha_1 u_2^{n+1}+\beta_1,\\ \frac{u_2^{n+1}-u_1^{n+1}}{h}=3\Delta t n u_1^{n+1}-1{,}5e^{\Delta t n},\\ u_2^{n+1}=(1+3\Delta t n h)u_1^{n+1}-1{,}5he^{\Delta t n}\ \Rightarrow\ \alpha_1=1+3\Delta t n h,\quad \beta_1=-1{,}5he^{\Delta t n}.\) |
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