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𝜕𝑢 𝜕𝑡
∴
𝜕2 𝑢 𝜕𝑡 2
=
=
𝑢𝑖,𝑗+1 − 𝑢𝑖,𝑗 𝑘
𝑢𝑖+1,𝑗 − 2𝑢𝑖,𝑗 +𝑢𝑖−1,𝑗 ℎ2
From equation 1 𝑢𝑖+1,𝑗 − 2𝑢𝑖,𝑗 +𝑢𝑖−1,𝑗 ℎ2
∴ 𝑢𝑖,𝑗+1 − 𝑢𝑖,𝑗 = Put r =
=α
𝑢𝑖,𝑗+1 − 𝑢𝑖,𝑗 𝑘
𝑘 − 2𝑢𝑖,𝑗 + 𝑢𝑖−1,𝑗 ] [𝑢 𝛼ℎ2 𝑖+1,𝑗
𝑘 𝛼ℎ 2
∴ 𝑢𝑖,𝑗+1 − 𝑢𝑖,𝑗 = 𝑟[𝑢𝑖+1,𝑗 − 2𝑢𝑖,𝑗 + 𝑢𝑖−1,𝑗 ] ∴ 𝑢𝑖,𝑗+1 − 𝑢𝑖,𝑗 = 𝑟𝑢𝑖+1,𝑗 + (1 − 2𝑟)𝑢𝑖,𝑗 + 𝑟𝑢𝑖−1,𝑗 … … ..(*) (𝑎 < 𝑟 ≤ 0.5) 𝑃𝑢𝑡 𝑟 =
1 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (∗) 𝑤𝑒 𝑔𝑒𝑡 2 ∴ 𝑢𝑖,𝑗+1 = 𝑟[𝑢𝑖+1,𝑗 + 𝑢𝑖−1,𝑗 ]
∴
𝜕2 𝑢 𝜕𝑡 2
=
=
𝑢𝑖,𝑗+1 − 𝑢𝑖,𝑗 𝑘
𝑢𝑖+1,𝑗 − 2𝑢𝑖,𝑗 +𝑢𝑖−1,𝑗 ℎ2
From equation 1 𝑢𝑖+1,𝑗 − 2𝑢𝑖,𝑗 +𝑢𝑖−1,𝑗 ℎ2
∴ 𝑢𝑖,𝑗+1 − 𝑢𝑖,𝑗 = Put r =
=α
𝑢𝑖,𝑗+1 − 𝑢𝑖,𝑗 𝑘
𝑘 − 2𝑢𝑖,𝑗 + 𝑢𝑖−1,𝑗 ] [𝑢 𝛼ℎ2 𝑖+1,𝑗
𝑘 𝛼ℎ 2
∴ 𝑢𝑖,𝑗+1 − 𝑢𝑖,𝑗 = 𝑟[𝑢𝑖+1,𝑗 − 2𝑢𝑖,𝑗 + 𝑢𝑖−1,𝑗 ] ∴ 𝑢𝑖,𝑗+1 − 𝑢𝑖,𝑗 = 𝑟𝑢𝑖+1,𝑗 + (1 − 2𝑟)𝑢𝑖,𝑗 + 𝑟𝑢𝑖−1,𝑗 … … ..(*) (𝑎 < 𝑟 ≤ 0.5) 𝑃𝑢𝑡 𝑟 =
1 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (∗) 𝑤𝑒 𝑔𝑒𝑡 2 ∴ 𝑢𝑖,𝑗+1 = 𝑟[𝑢𝑖+1,𝑗 + 𝑢𝑖−1,𝑗 ]
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