عرض تقديمي في powerpoint - weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1...
TRANSCRIPT
![Page 1: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/1.jpg)
:
:
𝑢(𝑥,𝑦) = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑦
:
(𝑥1,𝑦1) , (𝑥2,𝑦2) , (𝑥3,𝑦3)
:
𝑢1(𝑥, 𝑦) = 𝑎0 + 𝑎1𝑥1 + 𝑎2𝑦1
𝑢2(𝑥, 𝑦) = 𝑎0 + 𝑎1𝑥2 + 𝑎2𝑦2
𝑢3(𝑥, 𝑦) = 𝑎0 + 𝑎1𝑥3 + 𝑎2𝑦3
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1 𝑥1 𝑦111
𝑥2𝑥3
𝑦2𝑦3
𝑎0
𝑎1
𝑎2
=
𝑢1
𝑢2
𝑢3
𝑎𝑖 :
𝑎0 =1
2𝐴[𝑢1(𝑥2𝑦3 − 𝑥3𝑦2) + 𝑢2(𝑥3𝑦1 − 𝑥1𝑦3) + 𝑢3(𝑥1𝑦2 −
𝑥2𝑦1)]
:انر ذأخز انشكم انصفف انران
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𝑎1 =1
2𝐴[𝑢1(𝑦2 − 𝑦3) + 𝑢2(𝑦3 − 𝑦1) + 𝑢3(𝑦1 − 𝑦2)]
𝑎2 =1
2𝐴[𝑢1(𝑥3 − 𝑥2) + 𝑢2(𝑥1 − 𝑥3) + 𝑢3(𝑥2 − 𝑥1)]
𝐴 :
𝐴 =1
2 𝑑𝑒𝑡
1 𝑥1 𝑦111
𝑥2𝑥3
𝑦2𝑦3
=
=1
2[(𝑥2𝑦3 − 𝑥3𝑦2) + (𝑥3𝑦1 − 𝑥1𝑦3) + (𝑥1𝑦2 − 𝑥2𝑦1)
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انر أعا ك انرؼثش ػا تؼذ انرجغ تذالنح ذاتغ :انشكم انر ركشاا عاتقا تانشكم
𝑢(𝑥, 𝑦) = 𝑁1𝑢1 + 𝑁2𝑢2 + 𝑁3𝑢3
𝑁1 =1
2𝐴[(𝑥2𝑦3 − 𝑥3𝑦2) + (𝑦2 − 𝑦3)𝑥 + (𝑥3 − 𝑥2)𝑦]
𝑁2 =1
2𝐴[(𝑥3𝑦1 − 𝑥1𝑦3) + (𝑦3 − 𝑦1)𝑥 + (𝑥1 − 𝑥3)𝑦]
𝑁3 =1
2𝐴[(𝑥1𝑦2 − 𝑥2𝑦1) + (𝑦1 − 𝑦2)𝑥 + (𝑥2 − 𝑥1)𝑦]
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𝑁1:تانشكمانر غرطغ كراترا =𝛼1:𝛽1𝑥 :𝛾1𝑦
2𝐴
𝑁2 =𝛼2 + 𝛽2𝑥 + 𝛾2𝑦
2𝐴
𝑁3 =𝛼3 + 𝛽3𝑥 + 𝛾3𝑦
2𝐴
: دس أ 𝛼1 = 𝑥2𝑦3 − 𝑥3𝑦2 , 𝛽1 = (𝑦2 − 𝑦3) , 𝛾1 = (𝑥3 − 𝑥2) 𝛼2 = 𝑥3𝑦1 − 𝑥1𝑦3 , 𝛽2 = (𝑦3 − 𝑦1) , 𝛾2 = (𝑥1 − 𝑥3) 𝛼3 = 𝑥1𝑦2 − 𝑥2𝑦1 , 𝛽3 = (𝑦1 − 𝑦2) , 𝛾3 = (𝑥2 − 𝑥1)
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يالدظج
ايح
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ذكها ػ اإلدذاشاخ انطثؼح كع يا دسعا إدذاشاخ انغادح
ξ1 = A1/A , 𝜉2 = 𝐴2/𝐴 , 𝜉3 = 𝐴3/𝐴 𝐴,𝐴1,𝐴2,𝐴3
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(: 1)ػذ انؼقذج 𝜉1 = 1 𝜉2 = 𝜉3 = 0
(: 2)ػذ انؼقذج 𝜉2 = 1 𝜉1 = 𝜉3 = 0
(: 3)ػذ انؼقذج 𝜉3 = 1 𝜉1 = 𝜉2 = 0
انر ذذقق
والسؤال ملاذا إحداثيات املساحة ؟
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غرطغ ي خالل ز اإلدذاشاخ إجاد ذقاتم ت جهح اإلدذاشاخ انؼح انجهح انجذذج يا ذرض ت ز انجهح أ يرغشاذا
0 ≤ 𝜉𝑖 ≤ . يشذثطح تكم ػصش 1
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انر ك أ انذصل ػها ي خالل انؼالقاخ
𝜉1𝜉2𝜉3
=(1/2 A)
𝑥2𝑦3 − 𝑥3𝑦2 𝑦23 𝑥32𝑥3𝑦1 − 𝑥1𝑦3 𝑦31 𝑥13𝑥1𝑦2 − 𝑥2𝑦1 𝑦12 𝑥21
1𝑥𝑦
:أ دس
2𝐴 =1 1 1𝑥1 𝑥2 𝑥3𝑦1 𝑦2 𝑦3
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تانران غرطغ انرؼثش ػ دال انشكم تذالنح إدذاشاخ انغادح
:ال غ أ دال انشكم ذذقق𝑁1 + 𝑁1 + 𝑁3 = 1
𝜉1: كا أ + 𝜉2 + 𝜉3 = 1
𝜉1 =𝑁1: األغة أ ك ػذ انؼقذ 𝑁2 = 𝜉2
𝑁3 = 𝜉3
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:ف دال أخزا ػصش انصهس تغد قاغ فإ
1 2
3
4
5 6
𝜉1 = 1 , 𝜉2 = 𝜉3 = 0
𝜉3 = 1 , 𝜉1 = 𝜉2 = 0
𝜉2 = 1 , 𝜉1 = 𝜉3 = 0
0.5 0.5
𝜉2 = 0 , 𝜉1 = 𝜉3 = 0.5
𝜉1 = 0 , 𝜉2 = 𝜉3 = 0.5
𝜉3 = 0 , 𝜉1 = 𝜉2 = 0.5
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𝐴1
3
1 2
𝐴2
𝐴3
0.5
𝜉1 = 0, 𝜉2 = 𝜉3 = 0.5
عظخ األيش ػذ انؼقذ اإلظافح
تفظ األعهب تاق انؼقذ
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:دال انشكم كا قها إا ػذ انقاغ ذغا إدذاشاخ انغادح أ 𝑁𝑖 = 𝜉𝑖 𝑖 = 1,2,3
:عفشض أ , نك ػذ جد انؼقذ اإلظافح ذراض إن ػم أكصش 𝑁𝑖 = 𝑐𝑖𝜓(𝜉𝑖 , 𝜉𝑗)
انر جة أ ذؼذو ف تاق ػقذ 𝑁1ػذ انؼقذج األن شذ دغاب ذؼذو ػذ 𝜉1تالدظح أ ,انؼصش ذغا انادذ ف انؼقذج األن
4,6ػذ انؼقذ 0.5 ذغا 𝜉1فذراض إن انؼايم 2,3,5انؼقذ 𝜉1)فذراض إن انؼايم − :تانران (0.5
𝑁1 = 𝑐1𝜉1(𝜉1 − 0.5)
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:غرفذ ي ك انذانح ذغا انادذ ػذ انؼقذج األن 𝑐1نرؼ انصاتد 1 = 𝑐1 1 1 − 0.5 ⇒ 𝑐1 = 2
أ أ
𝑁1 = 2𝜉1(𝜉1 − 0.5)
تانصم جذ 𝑁2 = 2𝜉2(𝜉2 − 0.5) 𝑁3 = 2𝜉3(𝜉3 − 0.5)
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انذانح انر 𝑁4كا أ ػذ انؼقذج انشاتؼح ذراض إن جة أ
, ذؼذو ػ انؼقذ ذغا انادذ ػذ انؼقذج انشاتؼح تالدظح ا
:فك 𝜉2أ 𝜉1ؼذو أيا 1,2,3,5,6ػذ انؼقذ
𝑁4 = 𝑐4𝜉1𝜉2
:أعا تاالعرفادج ي كا ذغا انادذ ػذ انؼقذج انشاتؼح
𝑁4 = 𝑐4𝜉1𝜉2 ⟹ 1 = 𝑐4 0.5 0.5 ⟹ 𝑐4 = 4
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: أ أ 𝑁4 = 4𝜉1𝜉2
:تانصم جذ أ
𝑁5 = 4𝜉3𝜉2 𝑁6 = 4𝜉1𝜉3
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ف دال جد قػ إظافح أخش ن رغش االيش كصشا
1 4 2
5 6
3
𝜉1 = 0 , 𝜉2 = 𝜉3 = 0.5 𝜉2 = 0 , 𝜉1 = 𝜉3 = 0.5
𝜉3 = 0 , 𝜉1 = 𝜉2 = 0.5
7 8
9
𝜉3 = 0.5, 𝜉2 = 𝜉1 = 0.25
𝜉2 = 0.5, 𝜉3 = 𝜉1 = 0.25 𝜉1 = 0.5, 𝜉2 = 𝜉3 = 0.25
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أعا جة غشح ذغاؤل جذذ كف رؼايم ,𝜉1يغ شالز إدذاشاخ 𝜉2, 𝜉3 ذ رؼايم يغ
.يغأنح تؼذ
أ زا اإلشكال تغػ تالدظح أ دال انشكم :تاألصم ذرثغ نرغش فقػ
𝑁1 = 2𝜉1(𝜉1 − 0.5)
𝑁2 = 2𝜉2 𝜉2 − 0.5
𝑁3 = 2𝜉3(𝜉3 − 0.5)
𝑁4 = 4𝜉1𝜉2
𝑁5 = 4𝜉3𝜉2
𝑁6 = 4𝜉1𝜉3
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أن هرا األمر قودنا إىل أن نتعامم بشكم تاو مع إحداثني دون أن كون هناك أي
إشكال يف انتعامم حتى يف انعنصر املستطه إذ سنتعامم مع إحداثني خمتهفني
𝜉, 𝜂 .
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: األيش ن خرهف ف دانح ػصش انغرطه تأستغ قاغ 𝑢 𝑥, 𝑦 = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑦 + 𝑎3𝑥𝑦
:رذقق ػذ انؼقذ
𝑢1 𝑥, 𝑦 = 𝑎0 + 𝑎1𝑥1 + 𝑎2𝑦1 + 𝑎3𝑥1𝑦1 𝑢2 𝑥, 𝑦 = 𝑎0 + 𝑎1𝑥2 + 𝑎2𝑦2 + 𝑎3𝑥2𝑦2 𝑢3 𝑥, 𝑦 = 𝑎0 + 𝑎1𝑥3 + 𝑎2𝑦3 + 𝑎3𝑥3𝑦3 𝑢4 𝑥, 𝑦 = 𝑎0 + 𝑎1𝑥4 + 𝑎2𝑦4 + 𝑎3𝑥4𝑦4
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:انر ذكرة تانشكم
𝑢1𝑢2𝑢3𝑢4
=
1 𝑥1 𝑦1 𝑥1𝑦11 𝑥2 𝑦2 𝑥2𝑦211
𝑥3𝑥4
𝑦3𝑦4
𝑥3𝑦3𝑥4𝑦4
𝑎0𝑎1𝑎2𝑎3
𝑛1(𝑥1, 𝑦1)
𝑛4(𝑥1, 𝑦4)
𝑛3(𝑥2, 𝑦4)
𝑛2(𝑥2, 𝑦1)
:نك
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:تذم انجهح انغاتقح يشاػاج انشكم ذرؼ دال انشكم ػه انذ انران
𝑁1 =(𝑥 − 𝑥2)(𝑦 − 𝑦4)
(𝑥1 − 𝑥2)(𝑦1 − 𝑦4)
𝑁2 =(𝑥 − 𝑥1)(𝑦 − 𝑦4)
(𝑥2 − 𝑥1)(𝑦1 − 𝑦4)
𝑁3 =(𝑥 − 𝑥1)(𝑦 − 𝑦1)
(𝑥2 − 𝑥1)(𝑦4 − 𝑦1)
𝑁4 =(𝑥 − 𝑥2)(𝑦 − 𝑦1)
(𝑥1 − 𝑥2)(𝑦4 − 𝑦1)
23
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غرطغ ي خالل إدذاشاخ انغادح قم أ شكم ي
اإلدذاشاخ انؼح إن إدذاشاخ
انغادح
24
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25
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, 𝜉 تانران ذك دال انشكم تذالنح 𝜂 :
𝑁1 𝜉, 𝜂 =1
41 − 𝜉 1 − 𝜂
𝑁2 𝜉, 𝜂 =1
41 + 𝜉 1 − 𝜂
𝑁3 𝜉, 𝜂 =1
41 + 𝜉 1 + 𝜂
𝑁4 𝜉, 𝜂 =1
41 − 𝜉 1 + 𝜂
26
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:ػثاسج انذم ي انشكم
𝑢 𝑥, 𝑦 = 𝑁𝑗
𝑚
𝑗<1
𝑢𝑗
.ػذد انؼقذ 𝑚دس أ تفظ األعهب اقش يغأنح انؼصش انغرطه ترغغ ػقذ
27
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28
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:دال انشكم ػذئز ذك
𝑁1 𝜉, 𝜂 =1
4𝜉𝜂 1 − 𝜉 1 − 𝜂
𝑁2 𝜉, 𝜂 =1
4𝜉𝜂 1 + 𝜉 1 − 𝜂
𝑁3 𝜉, 𝜂 =1
4𝜉𝜂 1 + 𝜉 1 + 𝜂
𝑁4 𝜉, 𝜂 =1
4𝜉𝜂 1 − 𝜉 1 + 𝜂
29
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𝑁5 𝜉, 𝜂 = −1
2𝜂 1 − 𝜉2 1 − 𝜂
𝑁6 𝜉, 𝜂 =1
2𝜉 1 + 𝜉 1 − 𝜂2
𝑁7 𝜉, 𝜂 =1
2𝜂 1 − 𝜉2 1 + 𝜂
𝑁8 𝜉, 𝜂 = −1
2𝜉 1 − 𝜉 1 − 𝜂2
𝑁9 𝜉, 𝜂 = 1 − 𝜉
2 1 − 𝜂2
30
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31
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32
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ثق تا كفح إجاد انشكم انصفف نهصل إن نزنك , انذم انائ
عغرؼشض دم يؼادنح ذفاظهح جضئح ي انشذثح انصاح :تشكها انؼاو
𝜕
𝜕𝑥𝛼𝑥𝜕𝑢
𝜕𝑥+𝜕
𝜕𝑦𝛼𝑦𝜕𝑢
𝜕𝑦+ 𝛽𝑢 = 𝑓
ذصم ػه انصغح , كا ركشا عاتقا انصغح انقح انعؼفح
:تعشب انطشف تراتغ ياعة انكايهح ػه يطقح انذم
𝑤𝜕
𝜕𝑥𝛼𝑥𝜕𝑢
𝜕𝑥+𝜕
𝜕𝑦𝛼𝑦𝜕𝑢
𝜕𝑦+ 𝛽𝑢 𝑑𝑥𝑑𝑦 = 𝑤𝑓 𝑑𝑥𝑑y
(∗) 33
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: دس أ 𝜕
𝜕𝑥𝑤𝛼𝑥
𝜕𝑢
𝜕𝑥=𝜕𝑤
𝜕𝑥𝛼𝑥𝜕𝑢
𝜕𝑥+ 𝑤
𝜕
𝜕𝑥(𝛼𝑥𝜕𝑢
𝜕𝑥)
:تانران
𝑤𝜕
𝜕𝑥𝛼𝑥𝜕𝑢
𝜕𝑥=𝜕
𝜕𝑥𝑤𝛼𝑥𝜕𝑢
𝜕𝑥−𝜕𝑤
𝜕𝑥𝛼𝑥𝜕𝑢
𝜕𝑥
=𝜕
𝜕𝑥𝑤𝛼𝑥𝜕𝑢
𝜕𝑥− 𝛼𝑥𝜕𝑤
𝜕𝑥
𝜕𝑢
𝜕𝑥
:تانصم
𝑤𝜕
𝜕𝑦𝛼𝑦𝜕𝑢
𝜕𝑦=𝜕
𝜕𝑦𝑤𝛼𝑦𝜕𝑢
𝜕𝑦− 𝛼𝑦𝜕𝑤
𝜕𝑦
𝜕𝑢
𝜕𝑦
34
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:جذ (∗)تانرثذم ف
𝜕
𝜕𝑥𝑤𝛼𝑥𝜕𝑢
𝜕𝑥+𝜕
𝜕𝑦𝑤𝛼𝑦𝜕𝑢
𝜕𝑦𝑑𝑥𝑑𝑦 −
(𝛼𝑥𝜕𝑤
𝜕𝑥
𝜕𝑢
𝜕𝑥+ 𝛼𝑦𝜕𝑤
𝜕𝑦
𝜕𝑢
𝜕𝑦)𝑑𝑥𝑑𝑦 + 𝛽𝑢𝑤 𝑑𝑥𝑑𝑦 =
= 𝑤𝑓 𝑑𝑥𝑑y
(∗∗) 35
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غش ( صغح ) ذراض إن يثشح (∗∗)إلذاو انركايم انغاتق
(:غش غاص)
Ω ∈ 𝑅2 يطقح يذطاΓ , 𝑛(𝑛𝑥 , 𝑛𝑦) شؼاع ادذ
Γاناظى ػه :فإ
𝛻.𝐴 𝑑𝐴Ω
= 𝐴 . 𝑛𝑑𝑙Γ
:تشكم أتغػ
(𝜕𝐴𝑥𝜕𝑥 +𝜕𝐴𝑦
𝜕𝑦)𝑑𝑥𝑑𝑦 = 𝐴𝑥𝑛𝑥 + 𝐴𝑦𝑛𝑦 𝑑𝑙 3
6
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𝐴𝑥تظغ = 𝑤𝛼𝑥𝜕𝑢
𝜕𝑥
𝐴𝑦 = 𝑤𝛼𝑦𝜕𝑢
𝜕𝑦
: جذ
(𝜕
𝜕𝑥(𝑤𝛼𝑥𝜕𝑢
𝜕𝑥) +𝜕
𝜕𝑦𝑤𝛼𝑦𝜕𝑢
𝜕𝑦𝑑𝑥𝑑𝑦 =
(𝑤𝛼𝑥𝜕𝑢
𝜕𝑥)𝑛𝑥 + (𝑤𝛼𝑦
𝜕𝑢
𝜕𝑦)𝑛𝑦 𝑑𝑙
37
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:جذ أ ∗∗ترؼط انؼالقح انغاتقح ف
− (𝛼𝑥𝜕𝑤
𝜕𝑥
𝜕𝑢
𝜕𝑥+ 𝛼𝑦𝜕𝑤
𝜕𝑦
𝜕𝑢
𝜕𝑦)𝑑𝑥𝑑𝑦 + 𝛽𝑢𝑤 𝑑𝑥𝑑𝑦 =
= 𝑤𝑓 𝑑𝑥𝑑𝑦 − (𝑤𝛼𝑥𝜕𝑢
𝜕𝑥)𝑛𝑥 + (𝑤𝛼𝑦
𝜕𝑢
𝜕𝑦)𝑛𝑦 𝑑𝑙
=𝑤𝑖 : ك دال انشكم ذصهخ أ ذك دال ص فغعغ 𝑁𝑖
:نذا انذم انرقشث
𝑢 𝑥, 𝑦 = 𝑁𝑗(𝑥, 𝑦)𝑢𝑗
𝑚
𝑗<1
38
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:تانرؼط ف انشكم األخش جذ أ
− (𝛼𝑥𝜕𝑁𝑖𝜕𝑥
𝜕 𝑁𝑗 𝑥, 𝑦 𝑢𝑗𝑚𝑗<1
𝜕𝑥+𝛼𝑦𝜕𝑁𝑖𝜕𝑦
𝜕( 𝑁𝑗(𝑥, 𝑦)𝑢𝑗)𝑚𝑗<1
𝜕𝑦)𝑑𝑥𝑑𝑦
+ 𝛽𝑁𝑖 𝑁𝑗(𝑥, 𝑦)𝑢𝑗
𝑚
𝑗<1
𝑑𝑥𝑑𝑦 = 𝑁𝑖𝑓 𝑑𝑥𝑑𝑦
− (𝑁𝑖𝛼𝑥𝜕
𝜕𝑥( 𝑁𝑗(𝑥, 𝑦)𝑢𝑗
𝑚
𝑗<1
)𝑛𝑥 + (𝑁𝑖𝛼𝑦𝜕
𝜕𝑦( 𝑁𝑗(𝑥, 𝑦)𝑢𝑗
𝑚
𝑗<1
)𝑛𝑦 𝑑𝑙
𝑖 = 1,2, . . . , 𝑚
.ػذد انؼقذ 𝑚دس
39
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: تاإلصالح جذ
− [𝛼𝑥𝜕𝑁𝑖𝜕𝑥 ( 𝑢𝑗
𝜕𝑁𝑗
𝜕𝑥) + 𝛼𝑦
𝜕𝑁𝑖𝜕𝑦( 𝑢𝑗
𝜕𝑁𝑗
𝜕𝑦)] 𝑑𝑥𝑑𝑦
𝑚
𝑗<1
𝑚
𝑗<1
+ 𝛽𝑁𝑖 𝑁𝑗𝑢𝑗
𝑚
𝑗<1
𝑑𝑥𝑑𝑦 = 𝑁𝑖𝑓 𝑑𝑥𝑑𝑦 −
− 𝑁𝑖 𝛼𝑥𝜕𝑢
𝜕𝑥𝑛𝑥 + 𝛼𝑦
𝜕𝑢
𝜕𝑦𝑛𝑦 𝑑𝑙 ; 𝑖 = 1,2, . . . 𝑚
: انز غرطغ كراتر تانشكم انصفف انران
40
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𝑀11 𝑀12 … 𝑀1𝑛𝑀21 𝑀22 … 𝑀2𝑛⋮𝑀𝑛1
⋮𝑀𝑛2
⋱…⋮𝑀𝑛𝑛
𝑢1𝑢2⋮𝑢𝑛
+
𝑇11 𝑇12 … 𝑇1𝑛𝑇21 𝑇22 … 𝑇2𝑛⋮𝑇𝑛1
⋮𝑇𝑛2
⋱…⋮𝑇𝑛𝑛
𝑢1𝑢2⋮𝑢𝑛
=
=
𝑓1𝑓2⋮𝑓𝑛
+
𝑝1𝑝2⋮𝑝𝑛
:دس أ
𝑀𝑖𝑗 = − 𝛼𝑥𝜕𝑁𝑖𝜕𝑥
𝜕𝑁𝑗
𝜕𝑥+ 𝛼𝑦
𝜕𝑁𝑖𝜕𝑦
𝜕𝑁𝑗
𝜕𝑦𝑑𝑥𝑑𝑦
41
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𝑇𝑖𝑗 = 𝛽𝑁𝑖𝑁𝑗𝑑𝑥𝑑𝑦 𝑓𝑖 = 𝑁𝑖 𝑓 𝑑𝑥𝑑𝑦
𝑝𝑖 = − 𝑁𝑖 𝛼𝑥𝜕𝑢
𝜕𝑥𝑛𝑥 + 𝛼𝑦
𝜕𝑢
𝜕𝑦𝑛𝑦 𝑑𝑙
: ك أ ذصم ػه انشكم انطهب إرا اػرثشا أ K𝑖𝑗 = 𝑀𝑖𝑗 + 𝑇𝑖𝑗
𝑏𝑖 = 𝑓𝑖 + 𝑝𝑖
42
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:فك
𝐾11 𝐾12 … 𝐾1𝑛𝐾21 𝐾22 … 𝐾2𝑛⋮𝐾𝑛1
⋮𝐾𝑛2
⋱…⋮𝐾𝑛𝑛
𝑢1𝑢2⋮𝑢𝑛
=
𝑏1𝑏2⋮𝑏𝑛
43
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− [𝛼𝑥𝜕𝑁𝑖𝜕𝑥 ( 𝑢𝑗
𝜕𝑁𝑗
𝜕𝑥) + 𝛼𝑦
𝜕𝑁𝑖𝜕𝑦( 𝑢𝑗
𝜕𝑁𝑗
𝜕𝑦)] 𝑑𝑥𝑑𝑦
𝑚
𝑗<1
𝑚
𝑗<1
+ 𝛽𝑁𝑖 𝑁𝑗𝑢𝑗
𝑚
𝑗<1
𝑑𝑥𝑑𝑦 = 𝑁𝑖𝑓 𝑑𝑥𝑑𝑦 −
− 𝑁𝑖 𝛼𝑥𝜕𝑢
𝜕𝑥𝑛𝑥 + 𝛼𝑦
𝜕𝑢
𝜕𝑦𝑛𝑦 𝑑𝑙 ; 𝑖 = 1,2, . . . 𝑚
44
![Page 45: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/45.jpg)
:تانؼدج إن انغأنح ف انذاظشج انغاتقح
𝛻2 𝑢 =𝜕2𝑢
𝜕𝑥2+𝜕2𝑢
𝜕𝑦2= 1 ; 0 ≤ 𝑥 ≤ 1 , 0 ≤ 𝑦 ≤ 1
)ك دس إ انششغ اإلظافح ششغ يا 𝜕𝑢
𝜕𝑥𝑖= 0)
عادح انرؼشف يظذح تانشكم دس ذظش يقغح إن ػقذ أعاعح يشقح تانرشقى ذغغفا ػاصش شا
كا ظش ذشقا انذه تانغثح نكم ػصش , انؼى
45
![Page 46: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/46.jpg)
46
e2
e4
9
6
3 2 1
5
3
2 1 1 2
3 1
2
3
1 2
3
4
7 8
e1 e3
e5 e6 e7
e8
1
1
1 1
2 2 2
2
3
3
3 3
![Page 47: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/47.jpg)
: ي أجم انؼصش األل
𝑛3(0,0)
𝑛2(0, 0.5) 𝑛1(0.5,0.5)
𝛼1 = 𝑥2𝑦3 − 𝑥3𝑦2 = 0 𝛼2 = 𝑥3𝑦1 − 𝑥1𝑦3 = 0 𝛼3 = 𝑥1𝑦2 − 𝑥2𝑦1 = 0.25 𝛽1 = (𝑦2 − 𝑦3) = 1/2 𝛽2 = 𝑦3 − 𝑦1 = −0.5 𝛽3 = 𝑦1 − 𝑦2 = 0 𝛾1 = (𝑥3 − 𝑥2) = 0 𝛾2 = (𝑥1 − 𝑥3) = 1/2 𝛾3 = 𝑥2 − 𝑥1 = −1/2
47
![Page 48: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/48.jpg)
𝐴 =1
2𝑑𝑒𝑡1 0.5 0.51 0 0.51 0 0
= 1/8
𝑁𝑖: دال انشكم =1
2𝐴(𝛼𝑖 + 𝛽𝑖𝑥 + 𝛾𝑖𝑦)
𝑁1 = 2𝑥 , 𝑁2 = 2𝑦 − 2𝑥 , 𝑁3 = 1 − 2𝑦
𝐾11 = − [ 𝜕𝑁1𝜕𝑥
𝜕𝑁1𝜕𝑥+𝜕𝑁1𝜕𝑦
𝜕𝑁1𝜕𝑦 ] 𝑑𝑥𝑑𝑦
+ 𝑁1𝑁1𝑑𝑥𝑑𝑦
:يصففح انصالتح نهؼصش
48
![Page 49: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/49.jpg)
49
= − [ 2 2 + 0 + 0 𝑑𝑦]𝑑𝑥 +0.5
𝑥
0.5
0
[ 2𝑥 2𝑥 𝑑𝑦]𝑑𝑥 = −23
48
0.5
𝑥
0.5
0
, 𝐾13 = 𝐾31 =1
96 , 𝐾21 =
49
96 , 𝐾22 = −
47
48
, 𝐾23 = 𝐾32 =49
96 ,𝐾33 = −
11
24𝐾12 =
49
96
𝐾1 =
−23
48
49
96
1
9649
96−47
48
49
961
96
49
96−11
24
![Page 50: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/50.jpg)
𝑏𝑖: يصففح انذنح = 𝑓𝑖 + 𝑝𝑖
𝑏1 = 𝑁1𝑓 𝑑𝑥𝑑𝑦 −
− 𝑁1 𝛼𝑥𝜕𝑢
𝜕𝑥𝑛𝑥 + 𝛼𝑦
𝜕𝑢
𝜕𝑦𝑛𝑦 𝑑𝑙
50
𝑝𝑖: تذغة انششغ انذذح فإ = :تانران 0
𝑏1 = [ 𝑁1𝑓 𝑑𝑦]𝑑𝑥 =1
24
0.5
𝑥
0.5
0
𝑏2 = 𝑏3 =1
24
![Page 51: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/51.jpg)
:انؼصش انصا
𝑛1(0,0) 𝑛2(0.5,0)
𝑛3(0.5,0.5)
𝛼1 = 𝑥2𝑦3 − 𝑥3𝑦2 = 1/4 𝛼2 = 𝑥3𝑦1 − 𝑥1𝑦3 = 0 𝛼3 = 𝑥1𝑦2 − 𝑥2𝑦1 = 0 𝛽1 = (𝑦2 − 𝑦3) = −1/2 𝛽2 = (𝑦3 − 𝑦1) = 1/2 𝛽3 = (𝑦1 − 𝑦2) = 0 𝛾1 = (𝑥3 − 𝑥2) = 0 𝛾2 = (𝑥1 − 𝑥3) = −1/2 𝛾3 = (𝑥2 − 𝑥1) = 1/2
51
![Page 52: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/52.jpg)
:ؼض ا ف دال انشكم
𝑁1 =1/4 − 1/2𝑥
2/8= 1 − 2𝑥
𝑁2 =1/2𝑥 − 1/2𝑦
2/8= 2𝑥 − 2𝑦
𝑁3 = 1/2𝑦
2/8= 2𝑦 :يصففح انصالتح نهؼصش
𝐾11 = − [ 𝜕𝑁1𝜕𝑥
𝜕𝑁1𝜕𝑥+𝜕𝑁1𝜕𝑦
𝜕𝑁1𝜕𝑦 ] 𝑑𝑥𝑑𝑦
+ 𝑁1𝑁1𝑑𝑥𝑑𝑦
52
𝐾11 = − [ [𝜕𝑁1𝜕𝑥
𝜕𝑁1𝜕𝑥+𝜕𝑁1𝜕𝑦
𝜕𝑁1𝜕𝑦 ]𝑑𝑦]𝑑𝑥
𝑦<𝑥
𝑦<0
0.5
0
+ [ 𝑁1𝑁1𝑑𝑦]𝑑𝑥𝑦<𝑥
𝑦<0
0.5
0
![Page 53: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/53.jpg)
53
𝐾2 =
−11
24
49
96
1
9649
96−47
48
49
961
96
49
96−23
48
: يصففح انذنح
𝑏2 =
1
241
241
24
![Page 54: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/54.jpg)
: ي أجم انؼصش انصانس
𝑛3(0.5,0)
𝑛2(0.5,0.5) 𝑛1(1,0.5)
𝛼1 = 𝑥2𝑦3 − 𝑥3𝑦2 = −0.25 𝛼2 = 𝑥3𝑦1 − 𝑥1𝑦3 = 0.25 𝛼3 = 𝑥1𝑦2 − 𝑥2𝑦1 = 0.25
𝛽1 = (𝑦2 − 𝑦3) = 1/2 𝛽2 = (𝑦3 − 𝑦1) = −1/2 𝛽3 = (𝑦1 − 𝑦2) = 0
𝛾1 = (𝑥3 − 𝑥2) = 0 𝛾2 = (𝑥1 − 𝑥3) = 1/2 𝛾3 = (𝑥2 − 𝑥1) = −1/2
54
![Page 55: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/55.jpg)
𝑁1 =−14 + 1/2𝑥
2/8= −1 + 2𝑥
𝑁2 =1/4 − 1/2𝑥 + 1/2𝑦
2/8= 1 − 2𝑥 + 2𝑦
𝑁3 =1/4 − 1/2𝑦
2/8= 1 − 2𝑦
:دال انشكم
55
𝐾3 =
−23
48
49
96
1
9649
96−47
48
49
961
96
49
96−23
48
𝑏3 =
1
241
241
24
𝐾11 = − [ [𝜕𝑁1𝜕𝑥
𝜕𝑁1𝜕𝑥+𝜕𝑁1𝜕𝑦
𝜕𝑁1𝜕𝑦 ]𝑑𝑦]𝑑𝑥
𝑦<1/2
𝑦<𝑥;1/2
1
0.5
+ [ 𝑁1𝑁1𝑑𝑦]𝑑𝑥𝑦<1/2
𝑦<𝑥;1/2
1
0.5
![Page 56: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/56.jpg)
:ي أجم انؼصش انشاتغ
𝑛1(0.5,0) 𝑛2(1,0)
𝑛3(1,0.5)
𝛼1 = 𝑥2𝑦3 − 𝑥3𝑦2 = 1/2 𝛼2 = 𝑥3𝑦1 − 𝑥1𝑦3 = −1/4 𝛼3 = 𝑥1𝑦2 − 𝑥2𝑦1 = 0
𝛽1 = (𝑦2 − 𝑦3) = −1/2 𝛽2 = (𝑦3 − 𝑦1) = 1/2 𝛽3 = (𝑦1 − 𝑦2) = 0
𝛾1 = (𝑥3 − 𝑥2) = 0 𝛾2 = (𝑥1 − 𝑥3) = −1/2 𝛾3 = (𝑥2 − 𝑥1) = 1/2
56
![Page 57: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/57.jpg)
:ذك تزنك دال انشكم 𝑁1 = 2 − 2𝑥
𝑁2 = −1 + 2𝑥 − 2𝑦 𝑁3 = 2𝑦
57
𝐾4 =
−23
48
49
96
1
9649
96−47
48
49
961
96
49
96−23
48
𝑏4 =
1
241
241
24
![Page 58: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/58.jpg)
: ي أجم انؼصش انخايظ 𝑛1(0,1)
𝑛2(0,0.5)
𝑛3(0.5,1)
𝛼1 = 𝑥2𝑦3 − 𝑥3𝑦2 = −0.25 𝛼2 = 𝑥3𝑦1 − 𝑥1𝑦3 = 0.5 𝛼3 = 𝑥1𝑦2 − 𝑥2𝑦1 = 0
𝛽1 = (𝑦2 − 𝑦3) = −0.5 𝛽2 = (𝑦3 − 𝑦1) = 0 𝛽3 = (𝑦1 − 𝑦2) = 0.5
𝛾1 = (𝑥3 − 𝑥2) = 0.5 𝛾2 = (𝑥1 − 𝑥3) = −0.5 𝛾3 = (𝑥2 − 𝑥1) = 0
58
![Page 59: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/59.jpg)
:ذك تزنك دال انشكم 𝑁1 = −1 − 2𝑥 + 2𝑦 𝑁2 = 2 − 2𝑦 𝑁3 = 2𝑥
59
𝐾5 =
−47
48
49
96
49
9649
96−23
48
1
9649
96
1
96−23
48
𝑏5 =
1
241
241
24
![Page 60: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/60.jpg)
: انؼصش انغادط
𝑛1(0,0.5) 𝑛2(0.5,0.5)
𝑛3(0.5,1)
𝛼1 = 𝑥2𝑦3 − 𝑥3𝑦2 = 0.25 𝛼2 = 𝑥3𝑦1 − 𝑥1𝑦3 = 0.25 𝛼3 = 𝑥1𝑦2 − 𝑥2𝑦1 = −0.25
𝛽1 = (𝑦2 − 𝑦3) = −0.5 𝛽2 = (𝑦3 − 𝑦1) = 0.5 𝛽3 = (𝑦1 − 𝑦2) = 0
𝛾1 = (𝑥3 − 𝑥2) = 0 𝛾2 = (𝑥1 − 𝑥3) = −0.5 𝛾3 = (𝑥2 − 𝑥1) = 0.5
: انؼصش انغادط
60
![Page 61: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/61.jpg)
:ذك تزنك دال انشكم 𝑁1 = 1 − 2𝑥 𝑁2 = 1 + 2𝑥 − 2𝑦 𝑁3 = −1 + 2𝑦
61
𝐾6 =
−23
48
49
96
1
9649
96−47
48
49
961
96
49
96−23
48
𝑏6 =
1
241
241
24
![Page 62: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/62.jpg)
𝑛1 :انؼصش انغاتغ 0.5,1
𝑛2 0.5,0.5
𝑛3 1,1
𝛼1 = 𝑥2𝑦3 − 𝑥3𝑦2 = 0 𝛼2 = 𝑥3𝑦1 − 𝑥1𝑦3 = 0. 5 𝛼3 = 𝑥1𝑦2 − 𝑥2𝑦1 = −0.25
𝛽1 = (𝑦2 − 𝑦3) = −0.5 𝛽2 = (𝑦3 − 𝑦1) = 0 𝛽3 = (𝑦1 − 𝑦2) = 0.5
𝛾1 = (𝑥3 − 𝑥2) = 0.5 𝛾2 = (𝑥1 − 𝑥3) = −0.5 𝛾3 = (𝑥2 − 𝑥1) = 0
62
![Page 63: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/63.jpg)
:ذك تزنك دال انشكم 𝑁1 = −2𝑥 + 2𝑦 𝑁2 = 2 − 2𝑦 𝑁3 = −1 + 2𝑥
63
𝐾7 =
−47
48
49
96
49
9649
96−23
48
1
9649
96
1
96−23
48
𝑏7 =
1
241
241
24
![Page 64: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/64.jpg)
أخشا ي أجم انؼصش انصاي
𝑛1(0.5,0.5) 𝑛2(1,0.5)
𝑛3(1,1)
𝛼1 = 𝑥2𝑦3 − 𝑥3𝑦2 = 1/2 𝛼2 = 𝑥3𝑦1 − 𝑥1𝑦3 = 0 𝛼3 = 𝑥1𝑦2 − 𝑥2𝑦1 = −1/4
𝛽1 = (𝑦2 − 𝑦3) = −1/2 𝛽2 = (𝑦3 − 𝑦1) = 1/2 𝛽3 = (𝑦1 − 𝑦2) = 0
𝛾1 = (𝑥3 − 𝑥2) = 0 𝛾2 = (𝑥1 − 𝑥3) = −1/2 𝛾3 = (𝑥2 − 𝑥1) = 1/2
64
![Page 65: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/65.jpg)
:ذك دال انشكم 𝑁1 = 2 − 2𝑥 𝑁2 = 2𝑥 − 2𝑦 𝑁3 = −1 + 2𝑦
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𝐾8 =
−23
48
49
96
1
9649
96−47
48
49
961
96
49
96−23
48
𝑏8 =
1
241
241
24
![Page 66: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/66.jpg)
:ترجغ انصففاخ تانذم جذ 𝑢1 = 0. 27 𝑢2 ≈ 0.315789 𝑢3 ≈ −0.384556 𝑢4 ≈ −0.26794 𝑢5 ≈ 0.34256 𝑢6 ≈ −0.42673 𝑢7 ≈ 0.325648 𝑢8 = 0.45289 𝑢9 ≈ −0.24538
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![Page 67: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/67.jpg)
:أخشا تانرؼط ف ػثاسج انذم
𝑢 𝑥, 𝑦 = 𝑁𝑗
𝑚
𝑗<1
𝑢𝑗
𝑢 𝑥, 𝑦 = [2𝑥 0.34256 + 2𝑦 − 2𝑥 −0.26794 + 1 − 2𝑦 0.272727 + 1 − 2𝑥 0.272727 + 2𝑥 − 2𝑦 0.315789 + 2𝑦 0.34256
+ 2𝑥 − 1 −0.42673 + 1 − 2𝑥 + 2𝑦 0.34256 + 1 − 2𝑦 0.315789 + 2 − 2𝑥 0.315789 + 2𝑥 − 2𝑦 + 1 −0.384556 + 2𝑦 −0.42673 + 2𝑦 − 2𝑥 − 1 0.325648 + 2 − 2𝑦 −0.26792 + 2𝑥 0.45289 + 1 − 2𝑥 −0.26794 + 1 + 2𝑥 − 2𝑦 0.34256 + 2𝑦 − 1 0.45289 + 2𝑦 − 2𝑥 0.45289 + 2 − 2𝑦 0.34256 + 2𝑥 − 1 −0.24538 + 2 − 2𝑥 0.34256 + 2𝑥 − 2𝑦 −0.42673 + (2𝑦 − 1)(−0.24538)]
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![Page 68: عرض تقديمي في PowerPoint - Weeblynumericalanalysis.weebly.com/uploads/1/3/8/6/... · e 1 t 1 u 1 1 1 t 2 t 3 u 2 u 3 i e = 0 = 1 = 2 i = e q 1 q 2 q 3 i ÝÛ·ß = e ßíì](https://reader034.vdocuments.pub/reader034/viewer/2022050215/5f60e6361021853b503eabdb/html5/thumbnails/68.jpg)
:تانران فإ
𝑢 𝑥, 𝑦 = 0.93259 𝑦 − 3.09177 𝑥 + 0.26352
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