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Numerical Analysis
Method of Successive Over Relaxation
SOR Method
Formula: π‘₯𝑖
π‘˜+1
= 1 βˆ’ 𝑀 π‘₯𝑖
π‘˜
+
𝑀
π‘Ž 𝑖𝑖
{𝑏 𝑛 βˆ’ 𝑗=1
π‘–βˆ’1
π‘Žπ‘–π‘— π‘₯𝑖
π‘˜+1
βˆ’ 𝑗=𝑖+1
𝑛
π‘Žπ‘–π‘— π‘₯𝑗
π‘˜
}
To Prove the Formula of Succesive over relaxation (SOR).
Proof: We prove this formula from Gauss
seidal Formula which implies:
π‘₯𝑖
π‘˜+1
=
1
π‘Žπ‘–π‘–
{𝑏𝑖 βˆ’
𝑗=1
π‘–βˆ’1
π‘Žπ‘–π‘— π‘₯𝑗
π‘˜+1
βˆ’
𝑗=𝑖+1
𝑛
π‘Žπ‘–π‘— π‘₯𝑗
π‘˜
}
π‘₯𝑖
π‘˜+1
= π‘₯𝑖
π‘˜
+
1
π‘Žπ‘–π‘–
{𝑏 𝑛 βˆ’
𝑗=1
π‘–βˆ’1
π‘Žπ‘–π‘— π‘₯𝑖
π‘˜+1
βˆ’
𝑗=𝑖+1
𝑛
π‘Žπ‘–π‘— π‘₯𝑗
π‘˜
βˆ’ π‘Žπ‘–π‘– π‘₯𝑖
π‘˜
}
Now Add and Subtract π‘₯𝑖
π‘˜
π‘₯𝑖
π‘˜+1 = 𝑀π‘₯𝑖
π‘˜ +
𝑀
π‘Žπ‘–π‘–
{𝑏 𝑛 βˆ’
𝑗=1
π‘–βˆ’1
π‘Žπ‘–π‘— π‘₯𝑖
π‘˜+1 βˆ’
𝑗=𝑖+1
𝑛
π‘Žπ‘–π‘— π‘₯𝑗
π‘˜ βˆ’ π‘Žπ‘–π‘– π‘₯𝑖
π‘˜}
Now Multiplying the Equation by β€œW”
π‘₯𝑖
π‘˜+1
= 𝑀π‘₯𝑖
π‘˜
+
𝑀
π‘Ž 𝑖𝑖
{βˆ’π‘Žπ‘–π‘– π‘₯𝑖
π‘˜}{𝑏 𝑛 βˆ’ 𝑗=1
π‘–βˆ’1
π‘Žπ‘–π‘— π‘₯𝑖
π‘˜+1 βˆ’ 𝑗=𝑖+1
𝑛
π‘Žπ‘–π‘— π‘₯𝑗
π‘˜}
Now By further Steps We can write it as
π‘₯𝑖
π‘˜+1
= π‘₯𝑖
π‘˜
βˆ’ 𝑀π‘₯𝑖
π‘˜
+
𝑀
π‘Ž 𝑖𝑖
{𝑏𝑖 βˆ’ 𝑗=1
π‘–βˆ’1
π‘Žπ‘–π‘— π‘₯𝑗
π‘˜+1
βˆ’ 𝑗=𝑖+1
𝑛
π‘Žπ‘–π‘— π‘₯𝑗
π‘˜
}
Now taking π‘₯𝑖
π‘˜
as common
π‘₯𝑖
π‘˜+1
= 1 βˆ’ 𝑀 π‘₯𝑖
π‘˜
+
𝑀
π‘Žπ‘–π‘–
{𝑏𝑖 βˆ’
𝑗=1
π‘–βˆ’1
π‘Žπ‘–π‘— π‘₯𝑗
π‘˜+1
βˆ’
𝑗=𝑖+1
𝑛
π‘Žπ‘–π‘— π‘₯𝑗
π‘˜
}
Proved.
π»π‘’π‘Ÿπ‘’ "𝑀" 𝑖𝑠 π‘‘β„Žπ‘’ π‘…π‘’π‘™π‘Žπ‘₯π‘Žπ‘‘π‘–π‘œπ‘› π‘ƒπ‘’π‘Ÿπ‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ.
𝑖𝑓 𝑀 = 1 π‘‘β„Žπ‘’π‘› π‘‘β„Žπ‘–π‘  π‘šπ‘’π‘‘β„Žπ‘œπ‘‘ 𝑖𝑠 π‘Ÿπ‘’π‘‘π‘’π‘π‘’π‘‘ π‘‘π‘œ πΊπ‘Žπ‘’π‘ π‘  π‘†π‘’π‘–π‘‘π‘Žπ‘™ π‘€π‘’π‘‘β„Žπ‘œπ‘‘.
𝑖𝑓 𝑀 > 1 π‘‘β„Žπ‘’π‘› 𝑖𝑑 𝑖𝑠 π‘π‘Žπ‘™π‘™π‘’π‘‘ π‘œπ‘£π‘’π‘Ÿ π‘…π‘’π‘™π‘Žπ‘₯π‘Žπ‘‘π‘–π‘œπ‘›.
𝑖𝑓 𝑀 < 1 π‘‘β„Žπ‘’π‘› 𝑖𝑑 𝑖𝑠 π‘π‘Žπ‘™π‘™π‘’π‘‘ π‘ˆπ‘›π‘‘π‘’π‘Ÿ π‘…π‘’π‘™π‘Žπ‘₯π‘Žπ‘‘π‘–π‘œπ‘›.
π‘‡β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ "𝑀" 𝑙𝑖𝑒𝑠 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 0 < 𝑀 < 2.
𝑖𝑓 "𝑀"𝑖𝑠 π‘”π‘Ÿπ‘’π‘Žπ‘‘π‘’π‘Ÿ π‘‘β„Žπ‘Žπ‘› 2 π‘‘β„Žπ‘’π‘› π‘‘β„Žπ‘’ π‘šπ‘’π‘‘β„Žπ‘œπ‘‘ π‘‘π‘–π‘£π‘’π‘Ÿπ‘”π‘’π‘ .
𝑀𝑒 π‘π‘Žπ‘› 𝑓𝑖𝑛𝑑 π‘Š 𝑏𝑦 π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž: 𝑀 =
2
1 + 1 βˆ’ 𝑝(𝑑 𝑗)2
𝐴𝑠 𝑑𝑗 = π·βˆ’1
(𝐿 + π‘ˆ)
Best Of Luck
By: Khushdil Ahmad
BS-Mathematics
Govt: P.G Jahanzeb
College Swat
03428978608

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Methods of successive over relaxation

  • 1. Numerical Analysis Method of Successive Over Relaxation SOR Method
  • 2. Formula: π‘₯𝑖 π‘˜+1 = 1 βˆ’ 𝑀 π‘₯𝑖 π‘˜ + 𝑀 π‘Ž 𝑖𝑖 {𝑏 𝑛 βˆ’ 𝑗=1 π‘–βˆ’1 π‘Žπ‘–π‘— π‘₯𝑖 π‘˜+1 βˆ’ 𝑗=𝑖+1 𝑛 π‘Žπ‘–π‘— π‘₯𝑗 π‘˜ } To Prove the Formula of Succesive over relaxation (SOR).
  • 3. Proof: We prove this formula from Gauss seidal Formula which implies: π‘₯𝑖 π‘˜+1 = 1 π‘Žπ‘–π‘– {𝑏𝑖 βˆ’ 𝑗=1 π‘–βˆ’1 π‘Žπ‘–π‘— π‘₯𝑗 π‘˜+1 βˆ’ 𝑗=𝑖+1 𝑛 π‘Žπ‘–π‘— π‘₯𝑗 π‘˜ }
  • 4. π‘₯𝑖 π‘˜+1 = π‘₯𝑖 π‘˜ + 1 π‘Žπ‘–π‘– {𝑏 𝑛 βˆ’ 𝑗=1 π‘–βˆ’1 π‘Žπ‘–π‘— π‘₯𝑖 π‘˜+1 βˆ’ 𝑗=𝑖+1 𝑛 π‘Žπ‘–π‘— π‘₯𝑗 π‘˜ βˆ’ π‘Žπ‘–π‘– π‘₯𝑖 π‘˜ } Now Add and Subtract π‘₯𝑖 π‘˜
  • 5. π‘₯𝑖 π‘˜+1 = 𝑀π‘₯𝑖 π‘˜ + 𝑀 π‘Žπ‘–π‘– {𝑏 𝑛 βˆ’ 𝑗=1 π‘–βˆ’1 π‘Žπ‘–π‘— π‘₯𝑖 π‘˜+1 βˆ’ 𝑗=𝑖+1 𝑛 π‘Žπ‘–π‘— π‘₯𝑗 π‘˜ βˆ’ π‘Žπ‘–π‘– π‘₯𝑖 π‘˜} Now Multiplying the Equation by β€œW”
  • 6. π‘₯𝑖 π‘˜+1 = 𝑀π‘₯𝑖 π‘˜ + 𝑀 π‘Ž 𝑖𝑖 {βˆ’π‘Žπ‘–π‘– π‘₯𝑖 π‘˜}{𝑏 𝑛 βˆ’ 𝑗=1 π‘–βˆ’1 π‘Žπ‘–π‘— π‘₯𝑖 π‘˜+1 βˆ’ 𝑗=𝑖+1 𝑛 π‘Žπ‘–π‘— π‘₯𝑗 π‘˜} Now By further Steps We can write it as
  • 7. π‘₯𝑖 π‘˜+1 = π‘₯𝑖 π‘˜ βˆ’ 𝑀π‘₯𝑖 π‘˜ + 𝑀 π‘Ž 𝑖𝑖 {𝑏𝑖 βˆ’ 𝑗=1 π‘–βˆ’1 π‘Žπ‘–π‘— π‘₯𝑗 π‘˜+1 βˆ’ 𝑗=𝑖+1 𝑛 π‘Žπ‘–π‘— π‘₯𝑗 π‘˜ } Now taking π‘₯𝑖 π‘˜ as common π‘₯𝑖 π‘˜+1 = 1 βˆ’ 𝑀 π‘₯𝑖 π‘˜ + 𝑀 π‘Žπ‘–π‘– {𝑏𝑖 βˆ’ 𝑗=1 π‘–βˆ’1 π‘Žπ‘–π‘— π‘₯𝑗 π‘˜+1 βˆ’ 𝑗=𝑖+1 𝑛 π‘Žπ‘–π‘— π‘₯𝑗 π‘˜ } Proved.
  • 8. π»π‘’π‘Ÿπ‘’ "𝑀" 𝑖𝑠 π‘‘β„Žπ‘’ π‘…π‘’π‘™π‘Žπ‘₯π‘Žπ‘‘π‘–π‘œπ‘› π‘ƒπ‘’π‘Ÿπ‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ. 𝑖𝑓 𝑀 = 1 π‘‘β„Žπ‘’π‘› π‘‘β„Žπ‘–π‘  π‘šπ‘’π‘‘β„Žπ‘œπ‘‘ 𝑖𝑠 π‘Ÿπ‘’π‘‘π‘’π‘π‘’π‘‘ π‘‘π‘œ πΊπ‘Žπ‘’π‘ π‘  π‘†π‘’π‘–π‘‘π‘Žπ‘™ π‘€π‘’π‘‘β„Žπ‘œπ‘‘. 𝑖𝑓 𝑀 > 1 π‘‘β„Žπ‘’π‘› 𝑖𝑑 𝑖𝑠 π‘π‘Žπ‘™π‘™π‘’π‘‘ π‘œπ‘£π‘’π‘Ÿ π‘…π‘’π‘™π‘Žπ‘₯π‘Žπ‘‘π‘–π‘œπ‘›. 𝑖𝑓 𝑀 < 1 π‘‘β„Žπ‘’π‘› 𝑖𝑑 𝑖𝑠 π‘π‘Žπ‘™π‘™π‘’π‘‘ π‘ˆπ‘›π‘‘π‘’π‘Ÿ π‘…π‘’π‘™π‘Žπ‘₯π‘Žπ‘‘π‘–π‘œπ‘›. π‘‡β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ "𝑀" 𝑙𝑖𝑒𝑠 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 0 < 𝑀 < 2. 𝑖𝑓 "𝑀"𝑖𝑠 π‘”π‘Ÿπ‘’π‘Žπ‘‘π‘’π‘Ÿ π‘‘β„Žπ‘Žπ‘› 2 π‘‘β„Žπ‘’π‘› π‘‘β„Žπ‘’ π‘šπ‘’π‘‘β„Žπ‘œπ‘‘ π‘‘π‘–π‘£π‘’π‘Ÿπ‘”π‘’π‘ . 𝑀𝑒 π‘π‘Žπ‘› 𝑓𝑖𝑛𝑑 π‘Š 𝑏𝑦 π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž: 𝑀 = 2 1 + 1 βˆ’ 𝑝(𝑑 𝑗)2 𝐴𝑠 𝑑𝑗 = π·βˆ’1 (𝐿 + π‘ˆ)
  • 9. Best Of Luck By: Khushdil Ahmad BS-Mathematics Govt: P.G Jahanzeb College Swat 03428978608