当前位置:高等教育资讯网  >  中国高校课件下载中心  >  大学文库  >  浏览文档

麻省理工学院:《偏微分方程式数字方法》(英文版)Lecture 22 notes

资源类别:文库,文档格式:PDF,文档页数:25,文件大小:803.45KB,团购合买
Integral Equation Methods Reminder about Galerkin and collocation Example of convergence issues in 1D First and second kind integral equations Develop some intuition about the difficulties Convergence for second kind equations Consistency and stability issues Nystrom Method
点击下载完整版文档(PDF)

Numerical methods for pdes 3 Notes by s De and. white

￾✂✁☎✄✝✆✟✞✡✠☞☛✍✌✏✎✒✑✓✆✕✔✗✖✙✘✛✚✢✜✤✣✥✘✦✞★✧✪✩✬✫✭✜ ✮✰✯✲✱✴✳✶✵✸✷✺✹✼✻✾✽✟✿☞❀✲✹❁✱✶❂❄❃❁✯✙❅❆✳☞✱❈❇✲❃❊❉❁❋❍●❏■❑✳❍▲☞✱▼❀◆✷❖✳◗P ❘❙❂❚❋✰▲☞✷✺✳☞✱▼❂❱❯❲✹✼✱▼❂❄❃✼✯❨❳❩❃❁✯❭❬❪✳☞✷▼✵❫✳✰✯❴▲❖✳☎❵❴❇❛✳❍❃✼✷❝❜ ❞✛❡✾❢❪❣✥❤✛✐❦❥✒❧❑♠✗♥❑♦❪♣✾qr♠ts✉❣✈♣❭q❏✇t①✗②✡③★④❏⑤▼❢❲❣ ⑥⑧⑦♦❲⑤✶⑨❶⑩❴❷✗❸❩❹✾❷✾❷❴⑩

1 Outline SLIDE 1 Integral Equation Methods Reminder about Galerkin and collocation Example of convergence issues in 1D First and second kind integral equations Develop some intuition about the difficulties Convergence for second kind equations Consistency and stability issues Nystrom Method 2 Integral Equation Basics 2.1 Basis Function Approach 2.1.1 Basic Idea SLIDE 2 ntegral equation: y(a)=G(a, ro(a')ds i=10i Basis function Example Basis Represent circle with straight lines Assume o is constant along each line Note 1 As mentioned earlier, we are investigating methods for solving integral equa tions based on representing the solution as a weighted sum of basis functions Then, the original problem is replaced with the problem of determining the basis function weights In the next few slides we review the basis function approach th g replace the original boundary curve with a collection of straig ht sections, like the example circle in the slide. For the circle example, the result of using this basis is to replace the with a polygon. Then, the charge density is assumed constant on each edge of the polygon. The result is a piecewise constant representation of the charge density on a polygon, not a representation of the charge density on the circle

￾ ✁✄✂✆☎✞✝✠✟✠✡☞☛ ✌ ✍✏✎✒✑✔✓✖✕ ✗✠✘✚✙✜✛✏✢✤✣✦✥★✧✪✩✬✫✮✭✯✥✰✙✜✱✳✲✴✘✶✵✷✛✏✙✹✸✯✲✻✺✞✼ ✽✿✾✹❀❂❁❄❃❆❅✰✾✹❇❉❈❋❊✴●■❍✰❏✬❑▲❈■▼◆✾✹❇P❖◗❁❄❃❘❈■❃❆❅❚❙❯●■▼❄▼◆●✰❱❲❈❋❏P❁❄●■❃ ✩✬❳✻✥★❨❬❩❭✧✳✛❪✲✤❫❵❴■✲✴✘✚❛✤✛✏✣✦✢✤✛◗✘✯❴■✛❚✱❜✼✹✼✜✭✞✛◗✼❂✱✳✘❞❝◗❡ ❢✯❁◆❇❤❣✐❏✿❈■❃❆❅❪❣✐✾✜❱✠●❥❃❆❅❦❖◗❁❄❃❆❅☞❁◆❃✏❏P✾✹❧■❇❤❈❋▼✚✾✹♠✏❍❆❈❋❏P❁❄●■❃❆❣ ♥✬✾❲♦■✾✹▼◆●❥♣☞❣P●■❀q✾▲❁❄❃❥❏r❍★❁◆❏P❁❄●■❃s❈❋❊✴●■❍✰❏✪❏rt★✾✉❅✰❁◆✈✇❱❲❍★▼◆❏P❁❄✾✹❣ ①✲✤✘✚❛✴✛✏✣②✢✴✛◗✘✞❴❥✛❚❫③✲❆✣❘✼✹✛◗❴■✲✴✘✯✺✖④✻✱❜✘✞✺✶✛◗✫✮✭✯✥✰✙✹✱❜✲✤✘✯✼ ❙❯●■❃✤❣✐❁⑤❣⑥❏r✾❲❃❆❱❲⑦❦❈❋❃❆❅❪❣✐❏r❈■❊★❁◆▼❄❁◆❏⑥⑦❦❁❄❣r❣P❍★✾✹❣ ⑧❪⑨✻✼✹✙❲✣✦✲✤❨⑩✵✷✛✏✙✹✸✯✲✻✺✞✼ ❶❉❁❄❧■t☞●■❇❤❅✰✾✹❇❉❱✠●■❃◗♦❥✾❲❇r❧■✾❲❃✤❱✠✾ ❷ ❸✚✡❹☎✞☛✿❺❼❻❭❽✉✝❿❾➁➀q✂☞❽▲☎✞✟✠➂➃✡➅➄➆❽✉➇✔✟❲➈❯➇ ➉✪➊✐➋ ➌➎➍➐➏◗➑✐➏❿➒❉➓✬➔❵→✴➣★➑⑥↔❯➔➙↕➜➛✬➛❵➝★↔➞➍❭→✤➟ ➠✻➡ ❝ ➡ ❝ ➢❂✥★✼✜✱✳❴❪✗✠✺✯✛✏✥ ✌ ✍✏✎✒✑✔✓➥➤ ➦➧❃✏❏P✾✹❧■❇❤❈❋▼✚✾✹♠✏❍❆❈❋❏P❁❄●■❃✻➨➞➩❂➫➯➭✚➲➞➳➁➵q➸✇➫✳➭✔➺P➭✤➻✒➲⑥➼➐➫✳➭✤➻➯➲⑥➽◗➾➐➻ ✽✿✾✹♣★❇P✾✜❣✐✾✹❃✏❏✿➼✤➚✮➫➯➭✚➲➞➳➶➪➚➹❄➘✞➴ ➼❆➚➹ ➷❭➹ ➫➯➭✴➲ ➬ ➮✠➱ ✃ ❐❭❒✠❮ ➹ ❮✿❰✹Ï➚■ÐÒÑ ➹⑤Ó➚ ❮ SMA-HPC ©1999 MIT Basis Function Approach Basic Idea Example Basis Represent circle with straight lines Assume σ is constant along each line Integral Equation Basics Ô➥Õ✮ÖØ×ÚÙ Û❣❹❀q✾❲❃✏❏P❁❄●■❃❆✾✹❅➎✾✹❈■❇P▼❄❁❄✾❲❇✜Ü✞ÝÞ✾❘❈■❇P✾❦❁❄❃◗♦■✾✜❣⑥❏r❁◆❧✏❈✦❏P❁❄❃★❧s❀q✾✠❏rt★●✰❅★❣➃ß➯●■❇❹❣P●■▼❄♦◗❁◆❃❆❧❚❁◆❃✏❏r✾❲❧■❇❤❈❋▼➞✾✹♠✏❍❆❈✦à ❏P❁❄●■❃✤❣✬❊✤❈■❣P✾✹❅❚●❥❃Ú❇r✾❲♣★❇r✾✹❣P✾❲❃✏❏r❁◆❃★❧❦❏Pt★✾✇❣P●■▼❄❍✰❏P❁❄●■❃➥❈■❣á❈✆ÝÞ✾❲❁❄❧■t✏❏P✾✜❅Ú❣P❍★❀â●❋ß❯❊❆❈■❣P❁❄❣✬ß➯❍★❃✤❱Ø❏P❁❄●■❃✤❣❲ã ä✪t★✾✹❃✻Ü②❏Pt★✾Þ●■❇r❁❄❧■❁❄❃❆❈❋▼❥♣★❇P●❥❊★▼❄✾❲❀❞❁❄❣✞❇r✾❲♣★▼⑤❈■❱❲✾✹❅❼Ý✿❁◆❏Pt➃❏Pt★✾Þ♣★❇r●■❊★▼❄✾❲❀❞●❋ß✴❅✰✾✠❏r✾❲❇r❀❂❁❄❃★❁❄❃★❧❉❏rt★✾Þ❊❆❈■❣P❁❄❣ ß➯❍★❃❆❱✠❏P❁❄●■❃❪ÝÞ✾❲❁❄❧■t✏❏r❣✹ã✞➦➧❃☞❏rt★✾❼❃★✾✠å◗❏✪ß➯✾❲Ý❞❣P▼◆❁⑤❅✰✾✹❣✪ÝÞ✾▲❇P✾✹♦◗❁◆✾✹Ý➁❏Pt❆✾á❊✤❈■❣P❁❄❣Þß➯❍★❃✤❱Ø❏P❁❄●■❃s❈■♣★♣★❇r●❥❈■❱❤t✔ã ä✪t★✾❚✾❲å★❈❋❀q♣★▼❄✾❚❊❆❈■❣P❁⑤❣q❏Pt❆❈❋❏❘ÝÞ✾❚t❆❈②♦❥✾s❊✤✾✹✾❲❃æ❱❲●■❃❆❣P❁❄❅★✾❲❇r❁◆❃★❧❬❁⑤❣✹Ü✪ß➯●■❇❘❈✖ç②à➧♥â♣★❇r●■❊★▼❄✾❲❀❪Ü✪❏P● ❇r✾❲♣★▼⑤❈■❱❲✾➃❏rt★✾❦●■❇r❁❄❧■❁❄❃❆❈❋▼❭❊✴●■❍★❃❆❅❆❈❋❇r⑦Ú❱✠❍★❇r♦■✾❂Ý✿❁◆❏Pt❿❈❪❱✠●❥▼◆▼❄✾✹❱✠❏P❁❄●■❃➥●❋ß✿❣⑥❏r❇r❈■❁◆❧❥t❥❏❼❣P✾✹❱Ø❏r❁◆●❥❃❆❣✹Ü✻▼◆❁❄❖■✾ ❏Pt❆✾è✾❲å✰❈■❀q♣★▼◆✾➥❱✠❁❄❇r❱❲▼◆✾➥❁❄❃➁❏Pt❆✾➥❣P▼◆❁⑤❅✰✾❥ã✄❢★●■❇❘❏Pt❆✾➥❱❲❁◆❇❤❱✠▼❄✾Ú✾❲å✰❈■❀q♣★▼◆✾❥Ü❉❏rt★✾è❇r✾✹❣P❍★▼◆❏☞●■ß✉❍❆❣P❁◆❃★❧ ❏Pt❆❁❄❣❼❊✤❈■❣P❁❄❣▲❁⑤❣á❏P●s❇r✾❲♣★▼⑤❈■❱❲✾➃❏rt★✾❦❱✠❁❄❇❤❱✠▼❄✾❂Ý✿❁◆❏Pt❬❈❪♣✴●■▼❄⑦◗❧■●■❃✔ãqä✪t❆✾❲❃✻Ü✔❏Pt★✾❦❱❤t❆❈■❇P❧❥✾❂❅★✾❲❃❆❣P❁é❏⑥⑦Ú❁❄❣ ❈■❣r❣P❍★❀q✾✹❅❼❱✠●❥❃❆❣✐❏r❈❋❃✏❏✻●❥❃❼✾✹❈■❱❤t❼✾✜❅✰❧■✾➞●❋ß✰❏Pt❆✾ê♣✤●❥▼◆⑦◗❧❥●■❃✻ã✯ä✪t★✾ê❇r✾✹❣P❍★▼é❏✞❁❄❣✯❈✿♣★❁❄✾✹❱❲✾❲Ý✿❁⑤❣✐✾➞❱❲●■❃❆❣✐❏r❈■❃✏❏ ❇r✾❲♣★❇r✾✹❣P✾❲❃✏❏r❈❋❏P❁❄●■❃s●■ß➞❏Pt❆✾❂❱❤t❆❈■❇P❧❥✾❹❅✰✾✹❃❆❣P❁é❏⑥⑦s●❥❃è❈❘♣✤●❥▼◆⑦◗❧■●❥❃✻Ü✮❃★●■❏▲❈❘❇P✾✹♣★❇r✾✹❣P✾❲❃✏❏r❈❋❏P❁❄●■❃s●❋ß❯❏Pt★✾ ❱❤t❆❈■❇P❧❥✾á❅✰✾✹❃❆❣P❁é❏⑥⑦❦●❥❃✆❏rt★✾✉❱✠❁❄❇❤❱✠▼❄✾■ã ë

2.1.2 Piecewise Constant Straight Sections Example .A suRface set of n Points on the onnecting points with n lines (x)=l if x is on line I 2 In the above slide, we give the algorithm for constructing these piecewise con stant straight section basis functions. First, one takes the boundary curve and places points along the curve. These points are labeled a1, a2,.n. If the curve is a closed curve(meaning that there are no end points), then one can define a set of n straight line segments l1, l2,.n where the end points of line segment are ai and i+1 for i< n. Line segment n is a special case and connects I, ith ai and closes the approximation to the curve. In the circle example on the slide, note that the line segments approximate the arcs of the circle Once the n line segments are defined, the n const ant basis functions can be easily determined. If a in on line segment L; then i(a)=1, i(a)=0 otherwise n inscribed polygon with n points, as is shown on the slide, the area inside the polygon will be smaller than the area inside the circle. How does this area error decrease with n? You may assume the points are uniformly placed on the circle' s boundary.■ b Exercise 2 Suppose one wanted to allow the charge density to vary linearly over each line segment, instead of being piecewise const ant. Such a represe tation could be continu the polygonal curve. One approach would be to assign a charge density value to each point ai, and then to determine the value of the charge density on line segment li, one would use a weighted combinatio of the densities at the line segments endpoints. What would the associated basis be?(The basis functions you describe should be nonzero over MORE than one segment).■ Note 3 If we substitute the basis function representation of o into the integral equation is done at the bottom of the slide, the result is to replace the original inte ration of the product of the Greens function and the density with a weighted

➠✻➡ ❝ ➡⑤➠ ￾✱❜✛✏❴❥✛✂✁q✱❜✼✹✛ ①✲✴✘✞✼✹✙✹✥❆✘✚✙☎✄✻✙❲✣✦✥★✱❜✢✤✸✚✙✆✄✔✛◗❴❋✙✹✱❜✲✤✘✯✼q✩❵❳✻✥❆❨❬❩✯✧❜✛ ✌ ✍✏✎✒✑✔✓✞✝ Basis Function Approach 1) Pick a set of n Points on the surface 1 x 2) Define a new surface by connecting points with n lines. 3) Define i ( ) 1 if is on line i ϕ x = x l otherwise, ϕi ( ) x = 0 2 x n x 1 l 2 l n l ! " = ! = = " Ψ = i linel surface approx G x x dS n i ni n i x dS ni i x G x x ( , ') ' 1 1 ( ) ( , ') σ ϕ ( ') ' σ ✟s✲✠✁ ✺✯✲✡✁➃✛❚✺✯✛❥✙✜✛✏✣②❨✱✳✘✯✛✆✙✜✸✞✛❚➼➚ ➹☞☛ ✼✍✌ Ô➥Õ✮ÖØ×✏✎ ➦➧❃s❏rt★✾❹❈■❊✤●✦♦❥✾✉❣✐▼❄❁❄❅★✾■Ü✚Ý❯✾➃❧■❁❄♦■✾❼❏rt★✾❂❈■▼◆❧❥●■❇r❁é❏rt★❀✄ß➯●❥❇á❱❲●■❃❆❣✐❏P❇r❍❆❱✠❏P❁❄❃★❧✇❏Pt❆✾✹❣P✾✉♣★❁❄✾✹❱✠✾✹Ý✿❁❄❣P✾❹❱❲●■❃✰à ❣✐❏r❈❋❃✏❏❵❣✐❏P❇❤❈❋❁❄❧■t✏❏❉❣✐✾✜❱Ø❏r❁◆●❥❃s❊❆❈■❣P❁❄❣Þß➯❍❆❃❆❱Ø❏r❁◆●❥❃❆❣❲ã✪❢✯❁◆❇❤❣✐❏✹Ü❆●❥❃★✾▲❏r❈■❖■✾✜❣✪❏Pt★✾➃❊✤●❥❍★❃❆❅★❈■❇P⑦☞❱✠❍★❇r♦■✾❼❈❋❃✤❅ ♣★▼⑤❈■❱❲✾✹❣❭♣✴●■❁❄❃❥❏❤❣➞❈■▼◆●❥❃★❧á❏Pt★✾❵❱❲❍★❇r♦■✾■ã✯ä✪t★✾✹❣P✾✿♣✴●■❁❄❃❥❏❤❣ê❈❋❇r✾✿▼⑤❈❋❊✴✾❲▼❄✾✹❅❂➭ ➴ ➺✐➭✒✑❋➺✔✓✕✓✕✓ ➭➚ ã❭➦③ß✤❏rt★✾❵❱✠❍❆❇P♦❥✾ ❁⑤❣✿❈❂❱❲▼◆●✏❣✐✾✜❅❘❱❲❍★❇r♦■✾✗✖➯❀q✾✹❈■❃★❁◆❃❆❧q❏Pt❆❈❋❏Þ❏rt★✾❲❇r✾✉❈❋❇r✾á❃★●q✾❲❃❆❅☞♣✴●■❁❄❃❥❏❤❣✙✘✠Ü✰❏Pt❆✾❲❃❪●■❃★✾❼❱✹❈❋❃❪❅✰✾✛✚❆❃❆✾▲❈ ❣P✾✠❏▲●❋ß✢✜ ❣✐❏P❇❤❈❋❁❄❧■t✏❏á▼◆❁❄❃★✾q❣✐✾✹❧■❀q✾❲❃✏❏r❣☞✣ ➴ ➺✤✣✥✑✦➺✦✓✧✓✕✓ ✣➚ Ý✿t★✾❲❇r✾✉❏rt★✾q✾❲❃❆❅Ú♣✤●❥❁◆❃✏❏❤❣✬●■ß➞▼❄❁◆❃★✾✇❣P✾❲❧❥❀❂✾✹❃✏❏ ✣ ➹ ❈■❇P✾➃➭➹ ❈■❃❆❅s➭➹✧★✞➴ ß➯●■❇✪✩✬✫✭✜➐ã✯✮✔❁◆❃★✾❂❣✐✾✹❧■❀q✾❲❃✏❏✯✜❬❁⑤❣❵❈❘❣✐♣✴✾✹❱❲❁❄❈■▼✞❱✹❈■❣P✾➃❈❋❃❆❅s❱❲●■❃★❃❆✾✹❱Ø❏❤❣❉➭➚ Ý✿❁◆❏Pt✇➭ ➴ ❈❋❃✤❅✇❱✠▼❄●❥❣P✾✹❣❭❏rt★✾❵❈❋♣❆♣★❇P●②å✰❁❄❀✇❈✦❏P❁❄●■❃❂❏r●❼❏Pt★✾✬❱✠❍★❇r♦■✾❥ã✞➦➧❃✇❏rt★✾❵❱❲❁◆❇❤❱✠▼❄✾❉✾✠å★❈■❀❂♣❆▼◆✾❉●❥❃q❏Pt★✾ ❣P▼◆❁⑤❅✰✾■Ü✰❃❆●❋❏P✾▲❏rt❆❈✦❏✿❏rt★✾❼▼❄❁◆❃★✾❼❣P✾❲❧❥❀❂✾✹❃✏❏r❣✿❈❋♣❆♣★❇P●②å✰❁❄❀✇❈✦❏P✾✬❏Pt❆✾✉❈❋❇❤❱❲❣Þ●❋ß✯❏Pt★✾✉❱❲❁◆❇❤❱✠▼❄✾■ã ✰❃❆❱❲✾➐❏Pt★✾✱✜✇▼◆❁❄❃★✾➞❣P✾❲❧❥❀q✾❲❃✏❏r❣✞❈❋❇r✾➞❅✰✾✛✚✤❃★✾✹❅✮Ü✜❏rt★✾✢✜❦❱❲●■❃❆❣✐❏r❈■❃❥❏✞❊❆❈■❣P❁❄❣✻ß➯❍★❃❆❱✠❏P❁❄●■❃❆❣✯❱❲❈■❃❼❊✤✾❯✾✹❈■❣P❁❄▼◆⑦ ❅✰✾❲❏P✾❲❇r❀q❁◆❃❆✾✹❅✮ã➐➦③ß❭➭s❁◆❃❪●❥❃❘▼❄❁◆❃❆✾✉❣✐✾✹❧■❀q✾❲❃✏❏✲✣ ➹ ❏Pt★✾✹❃ ➷✯➹ ➫➯➭✚➲➞➳✴✳❥Ü ➷❭➹ ➫✳➭✴➲➞➳✭✵q●❋❏rt★✾❲❇rÝ✿❁❄❣P✾■ã ✶✸✷✪✹×✍✺✼✻✾✽❀✿■×➜Ùè❙❯●❥❃❆❣P❁❄❅✰✾✹❇➃❈❪❇❤❈■❅✰❁❄❍❆❣❼●❥❃★✾✆❱✠❁❄❇❤❱✠▼❄✾■ã✆➦③ß❉●❥❃★✾✇❇r✾❲♣★❇r✾✹❣P✾❲❃✏❏r❣á❏rt★✾✆❱❲❁◆❇❤❱✠▼❄✾✇❍❆❣P❁◆❃★❧ ❈❋❃❬❁❄❃❆❣r❱✠❇r❁◆❊✴✾✹❅❿♣✤●❥▼◆⑦◗❧■●❥❃❬Ý✿❁é❏rt❁✜➜♣✤●❥❁◆❃✏❏❤❣❲Üê❈❥❣✉❁⑤❣❹❣Pt★●✦Ý✿❃❬●■❃➎❏rt★✾❘❣P▼❄❁❄❅✰✾❥Ü❭❏Pt★✾☞❈❋❇r✾✹❈❪❁❄❃❆❣P❁❄❅✰✾ ❏Pt❆✾✉♣✤●❥▼◆⑦◗❧❥●■❃☞Ý✿❁◆▼❄▼✔❊✴✾➃❣P❀q❈■▼◆▼❄✾❲❇❉❏Pt❆❈■❃❘❏rt★✾➃❈■❇P✾✜❈❹❁❄❃❆❣P❁❄❅★✾❼❏Pt★✾❹❱✠❁❄❇r❱❲▼◆✾❥ã❯❶❉●✦Ý❞❅✰●◗✾✹❣❉❏Pt★❁⑤❣❵❈■❇P✾✜❈ ✾❲❇r❇r●■❇❯❅★✾✹❱✠❇r✾✹❈❥❣✐✾✬Ý✿❁é❏rt✆✜❃❂✏❄ê●❥❍✆❀✇❈②⑦✇❈■❣r❣✐❍❆❀❂✾✬❏Pt❆✾✬♣✴●■❁❄❃✏❏r❣✪❈❋❇r✾❵❍★❃❆❁éß➯●❥❇P❀q▼❄⑦✇♣★▼❄❈❥❱✠✾✜❅❦●■❃✆❏Pt★✾ ❱✠❁❄❇❤❱✠▼❄✾❆❅ ❣❯❊✴●■❍❆❃❆❅★❈❋❇r⑦■ã ✶❇✷✯✹×❈✺✼✻❉✽❊✿❋×❋✎❍●◗❍★♣❆♣✤●✏❣✐✾❼●❥❃★✾❼ÝÞ❈■❃✏❏P✾✹❅❘❏r●❦❈❋▼❄▼◆●✦Ýæ❏Pt★✾➃❱❤t✤❈❋❇r❧■✾❼❅✰✾❲❃✤❣✐❁◆❏⑥⑦❦❏r●✇♦②❈■❇P⑦❘▼◆❁❄❃★✾✜❈❋❇r▼◆⑦ ●✦♦■✾✹❇▲✾✹❈■❱❤t➥▼❄❁◆❃★✾❦❣P✾❲❧❥❀q✾❲❃✏❏✹Ü✻❁❄❃❆❣✐❏P✾✜❈■❅➥●❋ß✪❊✤✾✹❁◆❃❆❧☞♣❆❁◆✾✜❱✠✾❲Ý✿❁⑤❣P✾❦❱✠●■❃✤❣⑥❏❤❈❋❃✏❏✹ã☎●◗❍❆❱❤t❬❈❪❇r✾❲♣★❇r✾✹❣P✾❲❃★à ❏r❈❋❏P❁❄●■❃s❱✠●❥❍★▼⑤❅❪❊✤✾➃❱❲●■❃✏❏P❁❄❃◗❍★●■❍✤❣✿●■❃❪❏rt★✾✉♣✴●■▼❄⑦✏❧❥●■❃❆❈■▼✻❱✠❍★❇r♦■✾❥ã ✰❃❆✾➃❈❋♣★♣❆❇P●✏❈■❱❤t☞Ý❯●❥❍★▼❄❅☞❊✴✾❼❏P● ❈■❣r❣P❁◆❧❥❃✆❈q❱❤t❆❈❋❇r❧■✾❵❅★✾❲❃❆❣P❁é❏⑥⑦✆♦②❈■▼◆❍❆✾✬❏r●❹✾✜❈■❱❤t❘♣✤●❥❁◆❃✏❏✿➭➹ Ü★❈❋❃❆❅✆❏Pt❆✾❲❃❘❏P●q❅✰✾✠❏r✾❲❇r❀❂❁❄❃★✾á❏Pt❆✾á♦✦❈■▼◆❍★✾ ●❋ß✔❏Pt★✾❼❱❤t❆❈■❇P❧❥✾✬❅✰✾✹❃❆❣✐❁◆❏⑥⑦✇●■❃❘▼◆❁❄❃★✾❼❣P✾❲❧■❀q✾✹❃❥❏✬✣ ➹ Ü✰●❥❃★✾áÝ❯●❥❍★▼❄❅✆❍❆❣P✾á❈❹Ý❯✾✹❁◆❧❥t✏❏P✾✹❅❘❱✠●❥❀➃❊★❁❄❃❆❈✦❏r❁◆●❥❃ ●❋ß❆❏Pt★✾❉❅✰✾✹❃❆❣P❁é❏r❁◆✾✜❣✯❈❋❏❭❏Pt★✾✪▼❄❁◆❃★✾✿❣P✾❲❧❥❀❂✾✹❃✏❏r❣✯✾❲❃❆❅★♣✤●❥❁◆❃✏❏r❣✹ã❏■✶t❆❈✦❏➐ÝÞ●■❍★▼⑤❅❼❏rt★✾✿❈■❣r❣✐●✰❱✠❁⑤❈✦❏r✾✹❅➃❊❆❈■❣P❁❄❣ ❊✴✾❉❂✸✖❜ä✪t★✾▲❊❆❈■❣P❁❄❣❯ß➯❍★❃✤❱Ø❏P❁❄●■❃✤❣✿⑦■●❥❍❘❅✰✾✜❣P❱❲❇P❁❄❊✤✾▲❣Pt★●■❍★▼⑤❅❘❊✤✾▲❃★●❥❃▲❑❲✾✹❇P●❂●✦♦■✾✹❇◆▼✰✽✲❖ ❏rt❆❈❋❃☞●■❃★✾ ❣P✾❲❧■❀q✾✹❃❥❏P✘Øã Ô➥Õ✮ÖØ×❋◗ ➦③ß✤ÝÞ✾✿❣✐❍❆❊❆❣⑥❏r❁é❏r❍✰❏P✾✪❏rt★✾✿❊❆❈❥❣✐❁⑤❣✞ß➯❍★❃❆❱Ø❏r❁◆●❥❃q❇P✾✹♣★❇P✾✜❣✐✾✹❃✏❏r❈✦❏r❁◆●❥❃✉●❋ß✚➼✆❁❄❃✏❏P●á❏Pt❆✾✿❁◆❃✏❏P✾✹❧■❇❤❈❋▼◗✾✹♠✏❍❆❈❋❏P❁❄●■❃✻Ü ❈■❣❵❁❄❣▲❅✰●❥❃★✾❹❈❋❏❵❏rt★✾❹❊✴●❋❏P❏P●❥❀â●❋ß➐❏Pt❆✾❂❣P▼◆❁⑤❅✰✾❥Ü✤❏rt★✾❂❇P✾✜❣✐❍★▼◆❏✬❁⑤❣❵❏P●✆❇r✾❲♣❆▼❄❈❥❱✠✾✉❏rt★✾❂●■❇r❁◆❧❥❁◆❃❆❈■▼✻❁❄❃❥❏r✾✠à ❧■❇❤❈✦❏r❁◆●❥❃✆●■ß✔❏Pt❆✾❼♣★❇P●✰❅✰❍✤❱Ø❏✿●❋ß✯❏Pt★✾➃❑á❇r✾❲✾✹❃❘❅ ❣➞ß➯❍❆❃❆❱Ø❏r❁◆●❥❃❪❈■❃❆❅❘❏Pt★✾✉❅★✾❲❃❆❣P❁é❏⑥⑦✆Ý✿❁é❏rt❪❈❂ÝÞ✾❲❁❄❧■t✏❏P✾✜❅ ç

euu, fioteshale, reh ethaiSht lioee, f juet the Gheeo'efuocti, o. The oext etea ie heo t, Merel, a ao aaah, ach f, h Meth ioio s the weiShte io thie caee the aie 2.1.3 RestduaFo efiotttoo aod g totmtzattoo a LIMi R(a)=v()- approx G(, a )2onip: (z)s We wtIpptch the onis to mahe R()smap Geoehal aaah, ach: Pick a eet, f teet fuo cti, oe 91, .. n, ao Mf, Ice R()t,be hth, S, oal t, the eet; pi(ar(a)ds=0 for all i Note 4 If the basie fuocti, o heaheeeot i, o haaaeoet, exactlp eaheeeot the Meo etp the the heeiMal R(a)Mefioe M, o the t, a, f the ab, re eline will be zeh, f, h all a weishte the po, t the caee, ao Mio eteaM we will thp t, aick the basie fuocti, o onie t, e u eh, w u io iu ize R(a). Ooe aaah, ach t, u ioi izo S (a)iet, u ake it, hth, S, oal t, a c, llecti,o, f teet fuo cti, oe Aeo, teM, o the b,tt, u, f the eline, eof, hcio s, hth,S,oalitp io thie caee u eao e eo euros that the io teShal, f the ah, Muict, f R(a) ao Mo(), reh the eurface ie zeh, 2.1.4 Restduafg tofmtzattoo Ustol Test Fuocttoos a IM ∫;(x)F(x)dS=0|→ ()(x15-/px(xG(x)∑n9()ds=0 We will generate different methods by choosing theφ1,…,,φn Collocation i(a)=s(r-Tt(point matching Weighted Residual Method i(r)=1 if pi(a)f0(averages) Ae o, teM, o the t, a, f the ab, re dime, eubetitutios f, h the heimdal io the S, dalit aheri, ue eline pielke sir euuati, oe each with tw, ioteSale The firet ioteShal ie, reh the eurface, f the ah, Mact, f the Sire a, teotial with a teet fuo cti, o. The eec, o MioteShaliea Mu ble io teShal, reh the eurface. The ioteShao M, f the Muble io teShal ie a ah, Mict

❣P❍★❀ ●■ß✤❁❄❃❥❏r✾❲❧❥❇r❈■▼❄❣✯●✦♦■✾✹❇➐❣⑥❏r❇r❈■❁◆❧❥t❥❏➐▼❄❁◆❃❆✾✹❣➐●■ß✁￾⑥❍❆❣✐❏ê❏Pt❆✾❵❑á❇P✾✹✾❲❃❘❅ ❣✯ß➯❍★❃❆❱✠❏P❁❄●■❃✻ã➐ä✪t★✾✪❃★✾❲å✏❏➞❣✐❏P✾✹♣q❁❄❣ ❏Pt❆✾❲❃☞❏P●✇❅✰✾✹♦■✾❲▼❄●■♣❪❈❋❃❪❈❋♣❆♣★❇P●✏❈■❱❤t✆ß➯●■❇❉❅✰✾❲❏P✾❲❇r❀q❁◆❃❆❁◆❃★❧❂❏rt★✾❼Ý❯✾✹❁◆❧❥t✏❏r❣✹Ü✰❁◆❃☞❏Pt❆❁❄❣❉❱✹❈■❣P✾❵❏rt★✾✄✂➹ ❅ ❣❲ã ➠✻➡ ❝ ➡✆☎ ✝✛✏✼✜✱✳✺✯✭✯✥★✧❯❡☞✛✟✞✯✘✯✱➯✙✜✱✳✲✴✘✖✥❆✘✞✺➁✵✱✳✘✯✱✳❨❿✱✡✠■✥✰✙✜✱✳✲✴✘ ✌ ✍✏✎✒✑✔✓☞☛ ✌q➫➯➭✚➲✎✍æ➩❂➫✳➭✴➲✑✏✓✒✕✔✗✖✘✖✘✙✛✚✢✜ ✣✥✤✙✧✦★✔✁✩✫✪ ➸✇➫➯➭✞➺✐➭➻ ➲ ➚ ✬➹❄➘✞➴ ➼✤➚➹Ò➷✯➹ ➫➯➭➻ ➲⑥➽◗➾➻ ✭✛❇✁✇✱✳✧❜✧ê❩❭✱✳❴✜④ ✙✹✸✯✛❚➼➚ ➹ ☛ ✼❹✙✜✲❬❨❿✥✰④✴✛✮✌❂➫✳➭✴➲▲✼✜❨❿✥★✧❜✧ ➡ ❑á✾❲❃❆✾❲❇❤❈❋▼✮❈❋♣★♣❆❇P●✏❈■❱❤t✻➨✰✯ê❁⑤❱❤❖✆❈q❣✐✾❲❏❉●❋ß✔❏r✾✹❣✐❏✿ß➯❍★❃❆❱✠❏P❁❄●■❃❆❣✲✱➴ ➺✔✓✔✓✦✓✠➺✛✱➚ Ü✤❈❋❃❆❅✆ß➯●■❇❤❱✠✾✳✌q➫➯➭✴➲❯❏r●✇❊✤✾ ●■❇P❏Pt❆●■❧■●❥❃❆❈❋▼✤❏P●❂❏rt★✾✉❣✐✾❲❏✵✴ ✒ ✱ ➹ ➫➯➭✴➲✥✌q➫➯➭✴➲✐➽✏➾✖➳ ✵ ✶✸✷✢✹✻✺✠✣❊✣❃✩ Ô➥Õ✮ÖØ×✽✼ ➦③ß❆❏rt★✾✿❊❆❈❥❣✐❁⑤❣✞ß➯❍★❃❆❱Ø❏r❁◆●❥❃❹❇r✾❲♣❆❇P✾✜❣✐✾✹❃❥❏❭❁❄●■❃❂t❆❈❋♣★♣✴✾❲❃✤❣✯❏P●▲✾✠å★❈■❱✠❏P▼❄⑦❼❇P✾✹♣★❇r✾✹❣P✾❲❃✏❏✞❏rt★✾❉❅✰✾✹❃❆❣✐❁◆❏⑥⑦■Ü❋❏Pt★✾✹❃ ❏Pt❆✾❦❇P✾✜❣✐❁⑤❅✰❍❆❈■▼✾✌q➫➯➭✴➲➃❅✰✾✛✚✤❃★✾✹❅➎●❥❃è❏rt★✾✇❏r●■♣❬●❋ßÞ❏Pt❆✾✆❈❋❊✴●✦♦■✾✇❣P▼◆❁⑤❅✰✾❦Ý✿❁❄▼◆▼❯❊✤✾✆❑❲✾❲❇r●☞ß➯●■❇➃❈■▼◆▼➞➭✔ã ä✪t★❁⑤❣✿❁⑤❣✿❍❆❣P❍❆❈❋▼❄▼◆⑦❘❃★●❋❏✿❏rt★✾➃❱✹❈■❣P✾■Ü★❈❋❃✤❅❪❁◆❃❆❣✐❏P✾✜❈■❅✮Ü❆Ý❯✾❼Ý✿❁❄▼◆▼✻❏P❇r⑦❦❏P●✇♣❆❁❄❱❤❖✆❏Pt★✾✉❊✤❈■❣P❁❄❣✪ß➯❍❆❃❆❱Ø❏r❁◆●❥❃ ÝÞ✾❲❁❄❧■t✏❏r❣✹Ü✯❏Pt★✾❪➼➚ ➹ ❅ ❣❲Ü➐❏r●è❣P●■❀q✾❲t❆●✦Ý ❀q❁❄❃★❁❄❀❂❁✕❑❲✾✿✌q➫➯➭✚➲Øã ✰❃★✾☞❈❋♣★♣★❇r●❥❈❥❱❤t➥❏P●➥❀❂❁❄❃★❁❄❀q❁✧❑✹❁◆❃★❧ ✌q➫➯➭✚➲➞❁⑤❣Þ❏P●q❀✇❈❋❖❥✾✬❁◆❏❉●■❇P❏Pt❆●■❧■●❥❃❆❈❋▼❆❏P●✇❈q❱✠●❥▼◆▼❄✾✹❱✠❏P❁❄●■❃✆●■ß✔❏r✾✹❣✐❏Þß➯❍❆❃❆❱Ø❏r❁◆●❥❃❆❣❲ã Û❣✪❃★●❋❏r✾✹❅❘●■❃❘❏Pt★✾ ❊✴●❋❏✐❏r●■❀ ●❋ß✪❏rt★✾❘❣P▼◆❁⑤❅✰✾❥Ü✯✾✹❃✰ß➯●■❇❤❱✠❁❄❃★❧❚●❥❇✐❏rt★●■❧❥●■❃❆❈■▼◆❁◆❏⑥⑦❚❁❄❃❬❏Pt❆❁❄❣❹❱❲❈■❣P✾❦❀q✾✹❈■❃❆❣✉✾✹❃❆❣P❍★❇P❁❄❃★❧s❏rt❆❈✦❏ ❏Pt❆✾❼❁◆❃✏❏P✾✹❧■❇❤❈❋▼✚●❋ß✞❏Pt★✾❼♣❆❇P●✰❅✰❍❆❱✠❏❉●❋ß✑✌❂➫✳➭✴➲✿❈■❃❆❅✿✱✞➫✳➭✴➲✪●✦♦❥✾❲❇Þ❏Pt❆✾✉❣✐❍★❇Pß✳❈■❱❲✾▲❁❄❣ ❑✹✾❲❇r●❆ã ➠✻➡ ❝ ➡❁❀ ✝✛✏✼✜✱✳✺✯✭✯✥★✧❯✵✱✳✘✯✱✳❨❿✱✡✠■✥✰✙✜✱✳✲✴✘✓❂❘✼✜✱✳✘✯✢❄❃Þ✛✏✼✹✙❆❅✔✭✯✘✞❴■✙✹✱✳✲✴✘✯✼ ✌ ✍✏✎✒✑✔✓❈❇ ➵❉✱➹ ➫➯➭✚➲❊✌q➫➯➭✚➲⑥➽◗➾ ➳✭✵ ❋ ✒●✱➹ ➫➯➭✚➲P➩❂➫✳➭✴➲⑥➽◗➾❍✏✓✒●✒■✔✗✖✘✖✘✙✛✚✢✜ ✣✥✤✙✧✦★✔✁✩✫✪ ✱ ➹ ➫✳➭✴➲✐➸✇➫➯➭✔➺P➭➻ ➲ ➚ ✬ ❏ ➘✞➴ ➼✤➚❏ ➷ ❏ ➫➯➭➻ ➲✐➽✏➾➻ ➽◗➾❬➳ ✵ ❑✿▲◆▼✾❖◗P❘P❚❙❯▲❲❱❳▲✛❨✥❩✫❬❊▲✲❭✁❖❘❪✘▲✧❨❊▲❲❱❫❬✾❴❵▲✛❬❊❛✟❜✗❭✁❝❡❞✗❢❤❣✥❛✟❜❫❜✵❝✐❖◗❱❳❙✳❬❊❛✟▲❦❥✸❧❲♠✛♥❲♥♦♥❲♠❊❥✘♣ qsr✉t✆t✆r✇✈❯①✟②♦③❁r✘④⑥⑤ ❥✘⑦❊⑧⑩⑨✉❶❸❷❍❹✁⑧⑩⑨✳❺✽⑨❼❻❾❽❊❶➀❿⑩➁✉❜❯❖❘❱❫❬✎❴✄❩✫❬❊❣✥❛✟❖❘❱✟❙✢➂ ➃❤①✟t✆➄✗➅✵➆✇③⑩④❍➇➈➄✗②➊➉✇r❚➋➌⑤ ❥⑦ ⑧⑩⑨✘❶➍❷➌➎⑦ ⑧⑩⑨✘❶➀❿⑩❞✟❩✵❝✐❖❘❝➀➏➐❬❊▲❲❝➑❬✧➂ ➒➄✁③✆➓✘➉✉②❲➄✁➋❈➔❵➄✁→♦③❾➋↔➣✇①❼t❡➇➈➄✗②➊➉❚r✇➋➐⑤ ❥⑦ ⑧⑩⑨✉❶❸❷➈↕❦➙✡➛➜➎⑦ ⑧⑩⑨✘❶◆➝❷❄➞❉❿➟❩♦➠✢▲✛❨✥❩✵❙✵▲❲❝✥➂ Ô➥Õ✮ÖØ×➢➡ Û❣✆❃★●❋❏r✾✹❅➜●❥❃➜❏Pt★✾Ú❏P●■♣✶●■ß▲❏Pt★✾➥❈❋❊✴●✦♦■✾❚❣P▼◆❁⑤❅✰✾■Ü❉❣P❍★❊❆❣✐❏P❁◆❏P❍★❏P❁❄❃★❧❿ß➯●❥❇✇❏Pt★✾è❇P✾✜❣✐❁⑤❅✰❍❆❈■▼❉❁❄❃✶❏Pt★✾ ●■❇P❏Pt❆●■❧■●❥❃❆❈❋▼❄❁◆❏⑥⑦Ú✾✹♠✏❍❆❈❋❏P❁❄●■❃❿●❥❃❬❏Pt❆✾❘♣★❇r✾❲♦◗❁❄●■❍❆❣❹❣✐▼❄❁⑤❅✰✾❘⑦◗❁❄✾❲▼⑤❅★❣✹Ü➐❧■❁❄♦■✾❲❃ ✜➜❏P✾✹❣✐❏❂ß➯❍★❃❆❱Ø❏r❁◆●❥❃❆❣✹Ü ✜ ✾✹♠✏❍❆❈❋❏P❁❄●■❃❆❣✬✾✹❈❥❱❤t❚Ý✿❁◆❏Ptè❏⑥Ý❯●☞❁◆❃✏❏r✾❲❧■❇❤❈❋▼⑤❣✹ã▲ä✪t❆✾ ✚❆❇❤❣✐❏á❁❄❃✏❏P✾❲❧❥❇r❈■▼✞❁⑤❣á●✦♦■✾✹❇❵❏rt★✾❂❣P❍★❇Pß✳❈■❱❲✾❹●■ßê❏Pt★✾ ♣★❇r●✰❅✰❍❆❱Ø❏❭●❋ß❆❏Pt★✾✪❧■❁❄♦■✾✹❃✉♣✴●❋❏P✾✹❃✏❏P❁⑤❈❋▼◗Ý✿❁é❏rt❂❈❵❏P✾✜❣⑥❏✯ß➯❍★❃❆❱✠❏P❁❄●■❃✻ã➐ä✪t★✾✪❣P✾✹❱✠●❥❃❆❅✉❁❄❃✏❏P✾✹❧■❇❤❈❋▼✏❁❄❣❭❈á❅✰●■❍✰à ❊★▼❄✾❼❁◆❃✏❏r✾❲❧■❇❤❈❋▼✮●✦♦❥✾❲❇Þ❏Pt❆✾➃❣✐❍❆❇✐ß✳❈❥❱✠✾■ã❯ä✪t★✾❼❁◆❃✏❏r✾❲❧■❇❤❈❋❃✤❅✆●■ß✯❏rt★✾✉❅✰●❥❍★❊★▼❄✾❼❁◆❃✏❏P✾✹❧■❇❤❈❋▼✮❁⑤❣❉❈q♣★❇r●✰❅✰❍❆❱Ø❏ ➤

of a test function, the Greens function, and the charge density represent ation It is also possible to use fewer or more than n test functions. In that case, the resulting systems of equations is not square and must be solved using some kind of least-squares technique. Such methods are used occasionally, but it is very difficult to analyze their convergence Choosing different test functions generates methods with different names. If the est functions are impulses, the resulting method is called a collocation scheme If the test functions are the same as the basis functions, the method is referred to as a galerkin method. One can also choose test functions with the same support as the basis functions(a function's support is the set of a values fo hich the function is nonzero), but which only take on the valt n that case, the test functions serve to average the residual over the support of he basis function 2.1.5 Collocation SLIDE 6 Collocation: i (a)=8(Er-Tti)(point matching appro G(ti, r)p; (a')dS=y(ot:) A A The collocation method described in the above slide uses shifted impulse func tions as test functions, pi()=8(a-Ti). As the summation equation in the middle of the above slide indicates, testing with impulse functions is equivalent to insisting that R()=0. Or equivalently, that the potential produced by the approximated charge density should match the given potential at n test points That the potentials match at the test points gives rise to the method, s name he points where the potential is exactly matched is"co-located"with a set of The n x n matrix equation at the bottom of the above slide has as its right hand side the potentials at the test points. The unknowns are the basis function eights. The j matrix element for the i row is the potential produced at test point ti by a charge density equal to basis function pj

●❋ß➐❈❂❏P✾✜❣⑥❏❵ß➯❍★❃❆❱Ø❏r❁◆●❥❃✻Ü★❏rt★✾➃❑á❇r✾❲✾❲❃❃❅ ❣✪ß➯❍❆❃❆❱Ø❏r❁◆●❥❃✻Ü✤❈■❃❆❅❘❏Pt★✾➃❱❤t✤❈❋❇r❧■✾❼❅✰✾❲❃✤❣✐❁◆❏⑥⑦✆❇r✾❲♣★❇r✾✹❣P✾❲❃✏❏❤❈✦❏P❁❄●■❃✔ã ➦③❏❵❁⑤❣❉❈❋▼⑤❣P●✇♣✤●✏❣P❣P❁◆❊❆▼◆✾▲❏r●❦❍❆❣P✾▲ß➯✾❲ÝÞ✾❲❇❉●❥❇❉❀q●■❇r✾▲❏Pt❆❈■❃❋✜Ú❏r✾✹❣✐❏❉ß➯❍★❃❆❱✠❏P❁❄●■❃❆❣✹ã✪➦➧❃❪❏Pt❆❈❋❏✬❱❲❈❥❣✐✾❥Ü✰❏Pt★✾ ❇r✾✹❣P❍★▼é❏r❁◆❃❆❧▲❣P⑦✰❣⑥❏r✾❲❀✇❣➐●■ß✴✾✜♠❥❍✤❈✦❏P❁❄●■❃✤❣❭❁❄❣➐❃❆●❋❏➞❣r♠❥❍✤❈❋❇r✾✪❈■❃❆❅❂❀➃❍❆❣✐❏ê❊✴✾❵❣✐●❥▼◆♦❥✾✹❅❹❍❆❣P❁◆❃★❧❼❣P●■❀q✾✿❖◗❁◆❃✤❅ ●❋ßÞ▼◆✾✜❈■❣✐❏✐à➧❣P♠✏❍❆❈■❇P✾✜❣❉❏r✾✹❱❤t★❃❆❁❄♠✏❍★✾❥ã✗●◗❍✤❱❤t➥❀q✾❲❏Pt★●✰❅★❣❼❈■❇P✾❂❍❆❣P✾✹❅è●◗❱✹❱❲❈❥❣✐❁❄●■❃❆❈■▼◆▼❄⑦■Ü✮❊❆❍✰❏✉❁é❏❼❁⑤❣▲♦■✾✹❇P⑦ ❅✰❁◆✈❦❱✠❍★▼◆❏✿❏P●❦❈■❃❆❈❋▼❄⑦✠❑❲✾✬❏Pt★✾✹❁◆❇❉❱❲●■❃◗♦■✾✹❇P❧❥✾❲❃❆❱❲✾■ã ❙❯t★●◗●❥❣P❁❄❃★❧á❅★❁✁￾✚✾❲❇r✾❲❃✏❏➐❏P✾✜❣⑥❏➐ß➯❍❆❃❆❱Ø❏r❁◆●❥❃❆❣➐❧■✾✹❃★✾❲❇❤❈✦❏r✾✹❣✯❀q✾✠❏rt★●✰❅★❣❭Ý✿❁é❏rt✇❅✰❁✁￾✚✾❲❇r✾❲❃✏❏ê❃❆❈■❀q✾✹❣✹ã✞➦③ß✴❏Pt★✾ ❏P✾✜❣⑥❏❯ß➯❍★❃❆❱✠❏P❁❄●■❃❆❣✪❈❋❇r✾❉❁❄❀q♣★❍★▼⑤❣✐✾✜❣❲Ü❥❏Pt★✾á❇r✾✹❣P❍★▼é❏r❁◆❃❆❧✉❀q✾✠❏Pt❆●◗❅✆❁❄❣Þ❱❲❈■▼◆▼❄✾✹❅✆❈➃❱✠●❥▼◆▼❄●✰❱❲❈❋❏P❁❄●■❃❦❣r❱❤t★✾❲❀q✾■ã ➦③ß✔❏rt★✾á❏P✾✜❣⑥❏✿ß➯❍★❃✤❱Ø❏P❁❄●■❃✤❣✿❈❋❇r✾❵❏rt★✾❼❣P❈■❀❂✾❼❈❥❣➞❏rt★✾▲❊❆❈■❣P❁⑤❣➞ß➯❍❆❃❆❱Ø❏r❁◆●❥❃❆❣❲Ü★❏Pt★✾▲❀q✾✠❏rt★●✰❅✆❁⑤❣✪❇P✾❲ß➯✾❲❇r❇P✾✜❅ ❏P●➎❈■❣❂❈➎❑▲❈❋▼❄✾❲❇r❖✏❁❄❃❿❀q✾✠❏rt★●✰❅✮ã ✰❃★✾❪❱❲❈■❃✖❈■▼❄❣P●è❱❤t★●◗●✏❣✐✾✆❏P✾✹❣✐❏❂ß➯❍★❃❆❱Ø❏r❁◆●❥❃❆❣qÝ✿❁◆❏Pt❿❏rt★✾s❣P❈■❀q✾ ❣P❍★♣★♣✴●■❇P❏➃❈■❣❼❏rt★✾❦❊❆❈❥❣✐❁⑤❣▲ß➯❍★❃✤❱Ø❏P❁❄●■❃✤❣✆✖✳❈sß➯❍★❃❆❱Ø❏r❁◆●❥❃❘❅ ❣➃❣P❍★♣★♣✴●■❇P❏➃❁❄❣✉❏Pt★✾❘❣✐✾❲❏➃●❋ß✿➭❿♦✦❈❋▼❄❍★✾✜❣❼ß➯●■❇ Ý✿t★❁⑤❱❤tÚ❏rt★✾❂ß➯❍★❃❆❱✠❏P❁❄●■❃➥❁❄❣▲❃★●❥❃▲❑❲✾✹❇P●✠✘ØÜ✮❊★❍★❏✉Ý✿t★❁❄❱❤tè●■❃★▼❄⑦s❏❤❈❋❖■✾❂●■❃è❏Pt❆✾❂♦✦❈■▼◆❍★✾q●❥❃★✾❹●❥❇☞❑❲✾❲❇r●❆ã ➦➧❃✇❏rt❆❈✦❏✪❱❲❈■❣P✾■Ü❥❏Pt❆✾❉❏P✾✜❣⑥❏Þß➯❍★❃❆❱✠❏P❁❄●■❃❆❣Þ❣P✾❲❇r♦■✾✪❏P●❂❈②♦■✾✹❇r❈■❧■✾Þ❏Pt★✾á❇r✾✹❣P❁❄❅✰❍✤❈❋▼❆●✦♦❥✾❲❇➐❏rt★✾✬❣P❍★♣★♣✴●■❇P❏❯●■ß ❏Pt❆✾❼❊❆❈■❣P❁❄❣❯ß➯❍★❃✤❱Ø❏P❁❄●■❃✔ã ➠✻➡ ❝ ➡✄✂ ①✲✴✧✳✧✳✲✔❴■✥✰✙✜✱✳✲✴✘ ✌ ✍✏✎✒✑✔✓✆☎ ✝❜❯P❘P❘❜❫❣♦❩✫❬❊❖❘❜✵❱✟✞❡❥✘⑦✐⑧⑩⑨✘❶➍❷❍❹❳⑧⑩⑨✳❺ ⑨✉❻❾❽❊❶➀❿⑩➁✉❜✵❖❘❱✗❬✎❴✄❩➊❬❊❣✧❛❳❖◗❱❳❙❫➂ ➵✡✠☞☛✍✌✏✎✑✌✓✒✆❽✕✔✄✖✗☛✍✌✘✔✁✙☞✚✜✛✢✖✣☛✤✌✓✒✆❽✕✔✄✛✦✥ ✧ ➪♣ ★ ✛ ❧✪✩♣ ★ ✫❽✭✬ ✮ ➱ ✃Ø➬ ➮ ✒✰✯✲✱✪✱✴✳✕✵✓✶ ✷✹✸✳✹✺✭✯✼✻✾✽ ✿ ⑧⑩⑨ ❻ ❽✛♠✐⑨✪❀❾❶✐➎★ ⑧⑩⑨✑❀✆❶❂❁❄❃❅❀❼❷❇❆ ⑧⑩⑨ ❻ ❽❊❶ ❈❉ ❉ ❉ ❊ ❋ ❧✹● ❧■❍☞❍☞❍❏❍☞❍☞❍ ❋❧✹● ♣ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❋♣❄● ❧▲❍☞❍☞❍❏❍☞❍☞❍ ❋♣✜● ♣ ▼✁◆◆ ◆ ❖ ❈❉ ❉ ❉ ❊ ✩♣✁❧ ❑ ❑ ❑ ❑ ❑ ❑ ✩♣✵♣ ▼✁◆◆ ◆ ❖ ❷ ❈❉ ❉ ❉ ❊ ❆ ⑧⑩⑨✉❻✭P➊❶ ❑ ❑ ❑ ❑ ❑ ❑ ❆ ⑧⑩⑨✉❻✄◗❼❶ ▼✁◆◆ ◆ ❖ Ô➥Õ✮ÖØ×❙❘ ä✪t★✾✉❱❲●■▼❄▼◆●✰❱❲❈❋❏P❁❄●■❃☞❀q✾✠❏Pt❆●◗❅s❅★✾✹❣r❱✠❇r❁◆❊✴✾✹❅☞❁❄❃❪❏Pt★✾➃❈■❊✤●✦♦❥✾❼❣✐▼❄❁❄❅★✾❼❍❆❣✐✾✜❣❉❣Pt★❁éß✒❏r✾✹❅❪❁❄❀q♣★❍★▼⑤❣✐✾❼ß➯❍❆❃❆❱Øà ❏P❁❄●■❃✤❣❼❈■❣á❏P✾✜❣⑥❏▲ß➯❍★❃✤❱Ø❏P❁❄●■❃✤❣❲Ü✑✱➹ ➫➯➭✚➲✉➳❯❚✰➫➯➭✮✏❿➭➹ ➲✠ã Û❣á❏rt★✾❦❣✐❍❆❀❂❀✇❈❋❏P❁❄●■❃➥✾✹♠✏❍❆❈✦❏r❁◆●❥❃è❁❄❃➥❏Pt★✾ ❀q❁❄❅❆❅✰▼◆✾▲●■ß✻❏Pt★✾▲❈❋❊✴●✦♦■✾á❣✐▼❄❁⑤❅✰✾▲❁◆❃❆❅★❁❄❱✹❈✦❏P✾✜❣❲Ü✏❏r✾✹❣✐❏P❁❄❃★❧❹Ý✿❁◆❏Pt☞❁❄❀❂♣❆❍★▼❄❣P✾áß➯❍★❃❆❱✠❏P❁❄●■❃❆❣Þ❁⑤❣❯✾✜♠✏❍★❁◆♦✦❈■▼◆✾✹❃❥❏ ❏P●➃❁◆❃❆❣P❁⑤❣⑥❏r❁◆❃★❧➃❏Pt❆❈❋❏➀✌q➫➯➭➹ ➲ê➳ ✵★ã ✰❇➞✾✜♠✏❍★❁◆♦✦❈■▼◆✾✹❃❥❏r▼◆⑦❥Ü■❏rt❆❈✦❏❯❏Pt★✾✬♣✤●■❏P✾✹❃❥❏r❁❄❈■▼✤♣❆❇P●✰❅✰❍❆❱❲✾✹❅q❊◗⑦❂❏Pt★✾ ❈❋♣❆♣★❇P●②å✰❁❄❀✇❈✦❏P✾✜❅✇❱❤t❆❈❋❇r❧■✾✬❅✰✾❲❃❆❣P❁◆❏⑥⑦❦❣✐t★●❥❍★▼⑤❅✇❀✇❈✦❏r❱❤t✆❏Pt★✾á❧■❁❄♦■✾✹❃❦♣✴●❋❏P✾✹❃✏❏P❁⑤❈❋▼✮❈✦❏◆✜s❏P✾✜❣⑥❏✪♣✤●❥❁◆❃✏❏❤❣❲ã ä✪t❆❈❋❏✬❏rt★✾q♣✤●■❏P✾❲❃✏❏r❁❄❈■▼❄❣á❀✇❈✦❏❤❱❤tè❈✦❏á❏Pt❆✾➃❏r✾✹❣✐❏▲♣✴●■❁❄❃✏❏r❣á❧■❁❄♦■✾✜❣❵❇r❁⑤❣✐✾❹❏P●❘❏Pt★✾q❀q✾✠❏rt★●✰❅ ❅ ❣✬❃✤❈❋❀q✾■Ü ❏Pt❆✾❹♣✴●■❁❄❃✏❏r❣✬Ý✿t★✾❲❇r✾✉❏rt★✾❹♣✴●❋❏r✾❲❃✏❏P❁⑤❈❋▼❭❁⑤❣á✾✠å★❈■❱✠❏P▼❄⑦☞❀✇❈❋❏r❱❤t★✾✜❅s❁❄❣❲❱P❱✠●❋à③▼❄●◗❱✹❈✦❏r✾✹❅✪❳✉Ý✿❁◆❏Ptè❈❘❣✐✾❲❏á●■ß ❏P✾✜❣⑥❏❉♣✴●■❁❄❃✏❏r❣✹ã ä✪t★✾ ✜❩❨ ✜➜❀✇❈✦❏P❇r❁◆åè✾✹♠✏❍❆❈❋❏P❁❄●■❃➎❈✦❏✉❏rt★✾❦❊✴●❋❏P❏P●■❀ ●❋ß✪❏Pt❆✾❦❈❋❊✴●✦♦■✾✇❣✐▼❄❁⑤❅✰✾❦t❆❈❥❣✉❈■❣❼❁◆❏r❣✉❇P❁❄❧■t✏❏✐à t❆❈■❃❆❅❹❣P❁⑤❅✰✾Þ❏Pt★✾✿♣✴●❋❏r✾❲❃✏❏P❁⑤❈❋▼⑤❣➐❈✦❏❭❏Pt❆✾✪❏P✾✹❣✐❏➐♣✤●❥❁◆❃✏❏❤❣❲ã❭ä✪t★✾✿❍★❃★❖◗❃★●✦Ý✿❃✤❣✯❈■❇P✾❯❏Pt❆✾✿❊❆❈■❣P❁❄❣✯ß➯❍❆❃❆❱Ø❏r❁◆●❥❃ ÝÞ✾❲❁❄❧■t✏❏r❣✹ã✆ä✪t★✾❭❬ Ñ❫❪ ❀✇❈❋❏P❇r❁éå➥✾❲▼❄✾❲❀q✾❲❃✏❏✉ß➯●❥❇▲❏Pt★✾☎✩ Ñ❫❪ ❇P●✦Ý➆❁❄❣▲❏rt★✾❦♣✴●❋❏r✾❲❃✏❏P❁⑤❈❋▼ê♣❆❇P●✰❅✰❍❆❱❲✾✹❅➎❈✦❏ ❏P✾✜❣⑥❏❉♣✴●■❁❄❃✏❏✿➭➹ ❊◗⑦✆❈✇❱❤t❆❈■❇P❧❥✾✬❅✰✾✹❃❆❣P❁é❏⑥⑦❘✾✹♠✏❍❆❈❋▼✚❏r●q❊❆❈■❣P❁❄❣Þß➯❍❆❃❆❱Ø❏r❁◆●❥❃ ➷ ❏ ã ❴

2.1.6 Galefhto a IM Galerkin: i r)=pi(a)(test=basis) If G(c, r )=G(, r)then A ,j= Aj, i= A is symmetric Note he eline ab, re, we Sire the euuati, oef, h the gale Akio u eth, M io which the teet fuocti, oe ahe equal t, the badie fuo cti, oe Io aahticulah,, oe Seo ehatee equati,oef, h the baeie fuocti, o weiShte bp io eietioS that R(z)ie, hth,S, oal t, each, f the badie fuo cti, oe i of, cios, hth,S,oalitp c, heea, oMet, eettio S p(a)R(ds=0 Meubetitutio S the Ne oiti,o, f R(a)iot, the, hth,S, oalitp C, o Miti the eluati, o io the ceoteh, f the ab, re elite. N, te that the Galerkin u eth, MpielMe epeteu fy equati, oe, oe f, h each hth, S, oalitp c, o Miti, 0, aoM uoko, woe, oe f, h each basie fuo cti, o weiSht Ale, the epeteu Meeo, t hare the a, teotial exalicitlp ae the iSht hao Meine. Io eteaM the hiSht-hao MeiMe eothpie the areasE, f the ah, Muict, f the a, teotial basie fuocti, o 3 Convcrgcncc Analysis 3.1 I ampli proble ms 3.1.1 lo Ftfst ktod leiattoo W(x)=/1|x-r|(x)dSx∈[-1,1 The potential is given The density must be computed

➠✻➡ ❝ ➡✁￾ ✂✥★✧❜✛✏✣✜④✻✱❜✘ ✌ ✍✏✎✒✑✔✓☎✄ ✆❩✵P❘▲✛❨✞✝✗❖❘❱✟✞ ❥✘⑦✐⑧⑩⑨✘❶➍❷✓➎✸⑦❊⑧⑩⑨✉❶➀❿❾❬❊▲❲❝➑❬✧➏➀❞❼❩✫❝✐❖◗❝✥➂ SMA-HPC ©1999 MIT Laplace’s Equation in 2-D Basis Function Approach Galerkin ( ) ( ) ( ) ( ) ( ) 1 , , n i j i j approx j approx approx surface surface surface i i j x x dS G x x x x dS dS b A ϕ α ϕ ϕ = ′ ′ Ψ = ! ′ ′ ′ " " " #$$$%$$$& #$$$$$$%$$$$$$$& ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 , 0 n i i i j j app j rox surface ϕ ϕ x R x dS x x dS ϕ x G x x α ϕ x dS dS = = Ψ − ′ ′ ! ′ = " " " " nj σ nj σ ❈❉ ❉ ❉ ❊ ❋❧✹● ❧ ❍✾❍☞❍ ❍☞❍✓❍ ❋ ❧✹● ♣ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❑ ❋♣✜● ❧▲❍✾❍☞❍ ❍☞❍✓❍ ❋♣❄● ♣ ▼✁◆◆ ◆ ❖ ❈❉ ❉ ❉ ❊ ✩♣✁❧ ❑ ❑ ❑ ❑ ❑ ❑ ✩♣❯♣ ▼✁◆◆ ◆ ❖ ❷ ❈❉ ❉ ❉ ❊ ✟ ❧ ❑ ❑ ❑ ❑ ❑ ❑ ✟♣ ▼✁◆◆ ◆ ❖ ✠☛✡ ✿ ⑧⑩⑨ ♠➑⑨ ❀ ❶➍❷ ✿ ⑧⑩⑨ ❀ ♠➑⑨✘❶ ❬❊❛✟▲❲❱ ❋⑦✭● ★ ❷ ❋ ★ ● ⑦ ✧✌☞ ③❾→❵→✎✍✑✏✒✏➄✗②❲➅✫③❾✈ Ô➥Õ✮ÖØ×☎✓ ➦➧❃❦❏rt★✾▲❣✐▼❄❁❄❅★✾á❈■❊✤●✦♦❥✾■Ü✏Ý❯✾á❧■❁❄♦■✾❵❏Pt★✾á✾✹♠✏❍❆❈❋❏P❁❄●■❃❆❣➞ß➯●■❇❯❏Pt★✾✉❑▲❈■▼◆✾✹❇P❖◗❁❄❃❦❀q✾✠❏Pt❆●◗❅✻Ü✰❁◆❃❘Ý✿t★❁❄❱❤t✆❏Pt★✾ ❏P✾✜❣⑥❏❼ß➯❍❆❃❆❱Ø❏r❁◆●❥❃❆❣❼❈❋❇r✾q✾✹♠✏❍❆❈❋▼❭❏r●❪❏Pt★✾✇❊✤❈■❣P❁❄❣✬ß➯❍★❃❆❱✠❏P❁❄●■❃❆❣✹ã❦➦➧❃➥♣❆❈❋❇P❏P❁⑤❱✠❍❆▼❄❈■❇✹Ü✻●❥❃★✾q❧■✾✹❃★✾❲❇❤❈✦❏r✾✹❣ ✜ ✾✹♠✏❍❆❈❋❏P❁❄●■❃❆❣❯ß➯●❥❇✪❏Pt★✾❼❊❆❈❥❣✐❁⑤❣Þß➯❍★❃❆❱✠❏P❁❄●■❃❪ÝÞ✾❲❁❄❧■t✏❏r❣Þ❊◗⑦❦❁❄❃❆❣✐❁⑤❣✐❏P❁❄❃★❧q❏Pt❆❈❋❏✲✌❂➫✳➭✴➲✪❁⑤❣✪●■❇P❏Pt❆●■❧■●❥❃❆❈❋▼✤❏P● ✾✹❈❥❱❤t✆●■ß✞❏rt★✾❼❊❆❈■❣P❁⑤❣Þß➯❍★❃❆❱Ø❏r❁◆●❥❃❆❣✹ã ❖ê❃✰ß➯●■❇❤❱✠❁❄❃★❧q●■❇P❏Pt❆●■❧■●❥❃❆❈❋▼❄❁◆❏⑥⑦q❱❲●■❇r❇P✾✜❣✐♣✴●■❃✤❅★❣❯❏P●✇❣✐✾❲❏✐❏r❁◆❃★❧ ✒ ➷ ➫➯➭✴➲✥✌q➫➯➭✴➲✐➽✏➾✖➳ ✵ ❈❋❃✤❅Ú❣P❍★❊❆❣✐❏P❁◆❏P❍✰❏r❁◆❃★❧☞❏rt★✾❂❅★✾✛✚❆❃★❁◆❏P❁❄●■❃➥●❋ß➀✌❂➫✳➭✴➲á❁❄❃✏❏P●❘❏Pt★✾❂●■❇P❏Pt★●❥❧■●❥❃❆❈❋▼❄❁é❏⑥⑦❪❱✠●❥❃❆❅✰❁◆❏P❁❄●■❃è⑦✏❁❄✾❲▼⑤❅★❣ ❏Pt❆✾❼✾✹♠✏❍❆❈✦❏r❁◆●❥❃❘❁❄❃❘❏Pt★✾✉❱❲✾❲❃✏❏P✾✹❇✿●❋ß✯❏Pt★✾✉❈■❊✤●✦♦❥✾á❣P▼◆❁⑤❅✰✾❥ã ✔❉●■❏P✾❘❏Pt❆❈❋❏q❏Pt★✾s❑▲❈❋▼❄✾❲❇r❖◗❁◆❃✖❀❂✾❲❏Pt★●✰❅✖⑦◗❁◆✾✹▼❄❅★❣q❈➥❣✐⑦✰❣✐❏P✾✹❀ ●❋ß✯✜➁✾✜♠✏❍❆❈✦❏r❁◆●❥❃❆❣❲Ü➞●■❃★✾❘ß➯●■❇q✾✹❈❥❱❤t ●■❇P❏Pt❆●■❧■●❥❃❆❈❋▼❄❁◆❏⑥⑦❚❱✠●■❃✤❅✰❁é❏r❁◆●❥❃✻Ü✞❈■❃❆❅✡✜ ❍★❃★❖◗❃★●✦Ý✿❃❆❣✹Ü✔●❥❃★✾qß➯●■❇✉✾✜❈■❱❤t➎❊❆❈❥❣✐❁⑤❣▲ß➯❍★❃❆❱✠❏P❁❄●■❃❬ÝÞ✾❲❁❄❧■t✏❏✹ã Û▼⑤❣P●❆Ü✜❏Pt★✾Þ❣✐⑦✰❣✐❏P✾✹❀❞❅★●✏✾✜❣✔❃❆●❋❏✞t✤❈②♦■✾➐❏rt★✾➞♣✴●❋❏r✾❲❃✏❏P❁⑤❈❋▼✏✾✠å✰♣★▼❄❁❄❱❲❁é❏r▼◆⑦❼❈❥❣✮❏Pt❆✾❯❇r❁◆❧❥t✏❏✔t❆❈❋❃✤❅✉❣✐❁⑤❅✰✾❥ã✔➦➧❃✰à ❣✐❏P✾✹❈❥❅✮Ü✦❏rt★✾✬✩ Ñ❫❪ ❇P❁❄❧■t✏❏✐à③t❆❈■❃❆❅❹❣P❁⑤❅✰✾✪✾❲❃✏❏P❇r⑦✉❁⑤❣❭❏Pt★✾❉❈②♦❥✾❲❇❤❈❋❧❥✾➞●❋ß✤❏Pt★✾✿♣❆❇P●✰❅✰❍❆❱✠❏ê●❋ß✤❏Pt★✾✿♣✴●❋❏r✾❲❃✏❏P❁⑤❈❋▼ ❈❋❃✤❅❦❏rt★✾ ✩ Ñ❫❪ ❊❆❈■❣P❁❄❣Þß➯❍❆❃❆❱Ø❏r❁◆●❥❃✻ã ✕ ✖➂➃✡✘✗✉☛✪❻❭❺✉☛✿✡☞➈❯☛✚✙✡☞❽✉✝✜✛✇➇✔✟❲➇ ✢✪➊✐➋ ✣✒✤✿➍✦✥➛★✧✞✩✫✪è➝★↔✭✬★✧✞✩✮✥➏ ☎✔➡ ❝ ➡ ❝ ❝◗❡ ❅✿✱➯✣✦✼✹✙✰✯s✱❜✘✞✺➜✩á✫✮✭✞✥★✙✹✱✳✲✴✘ ✌ ✍✏✎✒✑✔✓✲✱ ➩❂➫✳➭✴➲➞➳➶➵ ➴ ✳ ➴✵✴ ➭➜✏➥➭✤➻ ✴ ➼➐➫✳➭✤➻✒➲⑥➽◗➾➐➻ ➭✷✶✹✸❁✏ ✳■➺✦✳✻✺ Convergence Analysis Example Problems 1-D First Kind Equation σ ( ) x is unknown ( ) 3 Ψ = x x x − The potential is given The density must be computed Solution = 3x Ψ x x σ ✼

n the next several slides we will investigate the convergence properties of these discretization methods. How these methods converge depends on what kind of Itegral equation is being solving. Examining this issue will introduce one of o begin, consider the example one-dimensional first-kind integral equation on the top of the above slide. For this equation, we assume that the potential, 4(a), is known and that the charge density o(a) is unknown. Here, a is in the terval [-1, 1, and the integration is over that same interval. Note that fo this example, the Green's function is giv n the left plot below the equation, an example given potential, a 3-a is plotted as a function of a. On the right is a plot of a charge density as a function of x ich might be a solution to the this problem the ques tion of what is the solution is not so easy to ans wer 3.1.2 Collocation Discretization of 1D Equation 业(x)=广1|x-xp(x)dsx∈[-1,1 Centroid Collocated Piecewise Constant Scheme |+++ To compute the numerical solution to this one-dimensional problem, conside solving the integral equation at the top of the slide using a piecewise-constant allocation scheme. In such a scheme, we first select n+l points on the interval, in this case [-1, 1]. We denote those points as ao, Ii, Inb, as shown in the figure in the middle of the slide. For this example, ao=-1 and n=1 Then, we can define a set of basis functions on the subintervals, pi(e),2(a),,Pn(a)l [x;-1,xi]y;(x) The charge density a can then be represented approximately

Ô➥Õ✮ÖØ×✁￾ ➦➧❃✇❏rt★✾✬❃❆✾✠å◗❏Þ❣P✾❲♦❥✾❲❇❤❈❋▼❆❣P▼◆❁⑤❅✰✾✜❣➞ÝÞ✾❵Ý✿❁❄▼◆▼✚❁❄❃◗♦■✾✹❣✐❏P❁❄❧❥❈❋❏P✾✪❏rt★✾á❱❲●■❃◗♦■✾✹❇P❧❥✾❲❃❆❱❲✾✿♣★❇r●■♣✴✾❲❇P❏P❁❄✾✹❣ê●■ß✮❏Pt★✾✜❣✐✾ ❅✰❁⑤❣P❱❲❇P✾❲❏P❁✕❑✹❈❋❏P❁❄●■❃☞❀❂✾❲❏Pt★●✰❅★❣✹ã✪❶❉●✦Ý➶❏Pt★✾✜❣✐✾✉❀q✾❲❏Pt★●✰❅★❣✬❱✠●■❃◗♦❥✾❲❇r❧■✾▲❅✰✾❲♣✴✾❲❃❆❅❆❣❉●■❃❪Ý✿t❆❈❋❏❉❖◗❁◆❃❆❅❪●■ß ❁❄❃❥❏r✾❲❧❥❇r❈■▼➐✾✹♠✏❍❆❈✦❏r❁◆●❥❃➎❁❄❣➃❊✤✾✹❁◆❃★❧è❣✐●❥▼◆♦◗❁❄❃★❧❆ã ❖➐å★❈❋❀q❁◆❃❆❁◆❃★❧❪❏rt★❁⑤❣✉❁⑤❣P❣P❍★✾❦Ý✿❁❄▼◆▼❯❁◆❃✏❏r❇P●✰❅✰❍❆❱❲✾✇●■❃★✾✇●■ß ❏Pt❆✾✉❣✐❍★❊★❏P▼❄✾❼♣✤●❥❁◆❃✏❏r❣✿❈■❊✤●❥❍✰❏✿❁◆❃✏❏r✾❲❧■❇❤❈❋▼✮✾✜♠✏❍❆❈✦❏r❁◆●❥❃❆❣❲ã ä✞●✇❊✤✾✹❧■❁❄❃✻Ü✤❱❲●■❃❆❣P❁❄❅★✾❲❇✪❏Pt❆✾✉✾✠å★❈❋❀q♣★▼❄✾❼●■❃❆✾✠à➧❅✰❁◆❀q✾✹❃❆❣✐❁❄●■❃✤❈❋▼ ✚❆❇❤❣⑥❏PàÒ❖◗❁❄❃❆❅❘❁◆❃✏❏P✾✹❧■❇❤❈❋▼✻✾✜♠✏❍❆❈✦❏r❁◆●❥❃❘●❥❃ ❏Pt❆✾✆❏r●■♣ ●■ß❉❏Pt❆✾❪❈■❊✤●✦♦❥✾✆❣P▼◆❁⑤❅✰✾■ã❿❢❆●■❇❂❏Pt❆❁❄❣❂✾✹♠✏❍❆❈❋❏P❁❄●■❃✻Ü➞Ý❯✾☞❈■❣r❣✐❍★❀q✾✆❏Pt❆❈❋❏❂❏rt★✾☞♣✤●■❏P✾❲❃✏❏r❁❄❈■▼❜Ü ✂á➫➯➭✚➲ØÜ✤❁❄❣❵❖✏❃❆●✦Ý✿❃☞❈■❃❆❅❘❏Pt❆❈❋❏❉❏Pt★✾❹❱❤t❆❈❋❇r❧■✾▲❅✰✾❲❃✤❣✐❁◆❏⑥⑦❘➼➐➫✳➭✴➲✿❁⑤❣✿❍★❃❆❖✏❃❆●✦Ý✿❃✻ã❯❶❵✾❲❇r✾■Ü★➭❚❁⑤❣✿❁❄❃❪❏Pt★✾ ❁❄❃❥❏r✾❲❇r♦✦❈❋▼ ✸❁✏ ✳■➺✦✳✻✺③Ü✯❈■❃❆❅➥❏Pt★✾❘❁◆❃✏❏r✾❲❧■❇❤❈✦❏r❁◆●❥❃➎❁❄❣➃●✦♦■✾❲❇▲❏rt❆❈✦❏q❣P❈■❀q✾✇❁◆❃✏❏P✾✹❇P♦✦❈■▼❜ã ✔❵●❋❏P✾❦❏Pt❆❈❋❏➃ß➯●■❇ ❏Pt❆❁❄❣✿✾❲å★❈❋❀q♣★▼❄✾■Ü◗❏Pt❆✾➃❑á❇P✾✹✾❲❃❘❅ ❣Þß➯❍★❃❆❱✠❏P❁❄●■❃❪❁❄❣✪❧❥❁◆♦❥✾❲❃☞❊◗⑦✆➸✇➫➯➭✞➺✐➭✤➻➯➲ê➳ ✴ ➭➜✏è➭✴➻ ✴ ã ➦➧❃❂❏Pt★✾❵▼◆✾❲ß✒❏➞♣★▼❄●❋❏ê❊✴✾❲▼❄●✦Ý ❏rt★✾✿✾✹♠✏❍❆❈❋❏P❁❄●■❃✻Ü✏❈❋❃q✾✠å★❈❋❀q♣★▼❄✾✿❧■❁❄♦■✾✹❃❹♣✴●❋❏r✾❲❃✏❏P❁⑤❈❋▼ÒÜ■➭☎✄❸✏✇➭✇❁⑤❣➐♣★▼❄●❋❏P❏P✾✜❅ ❈■❣✿❈❂ß➯❍❆❃❆❱Ø❏r❁◆●❥❃s●❋ß❭➭✔ã ✰❃☞❏rt★✾❼❇P❁❄❧■t✏❏❉❁⑤❣❉❈❂♣★▼❄●❋❏❵●■ß❭❈q❱❤t❆❈■❇P❧❥✾▲❅★✾❲❃❆❣P❁é❏⑥⑦☞❈■❣✿❈❂ß➯❍★❃✤❱Ø❏P❁❄●■❃❪●■ß❭➭ Ý✿t★❁⑤❱❤t❚❀q❁◆❧❥t✏❏✬❊✴✾❂❈❘❣✐●❥▼◆❍✰❏r❁◆●❥❃❪❏P●❘❏Pt★✾❹❁◆❃✏❏r✾❲❧■❇❤❈❋▼✞✾✹♠✏❍❆❈❋❏P❁❄●■❃✻ã Û❣✬Ý❯✾❹Ý✿❁◆▼❄▼❭❣✐✾✹✾❹❣Pt★●■❇P❏P▼❄⑦■Ü✴ß➯●■❇ ❏Pt❆❁❄❣✿♣❆❇P●❥❊★▼◆✾✹❀ ❏Pt❆✾✉♠❥❍❆✾✹❣✐❏P❁❄●■❃☞●❋ß✞Ý✿t✤❈✦❏❉❁⑤❣Þ❏Pt★✾✉❣P●■▼❄❍✰❏P❁❄●■❃☞❁⑤❣✿❃★●❋❏❉❣P●q✾✹❈■❣P⑦✇❏r●q❈■❃❆❣PÝ❯✾✹❇✹ã ☎✔➡ ❝ ➡⑤➠ ①✲✴✧✳✧✳✲✔❴■✥✰✙✜✱✳✲✴✘✖❡☞✱✳✼✜❴❋✣✦✛✏✙✹✱✡✠■✥✰✙✜✱✳✲✴✘✶✲✤❫❼❝◗❡ ✩✬✫✮✭✯✥✰✙✜✱✳✲✴✘ ✌ ✍✏✎✒✑✔✓✝✆ ➩❂➫✳➭✴➲➞➳➶➵ ➴ ✳ ➴ ✴ ➭➜✏➥➭➻ ✴ ➼➐➫✳➭➻ ➲⑥➽◗➾➻ ➭✷✶✹✸❁✏ ✳■➺✦✳✻✺ ①✛✏✘✚✙✹✣✦✲✤✱❜✺ ①✲✴✧✳✧❜✲✻❴■✥✰✙✜✛◗✺ ￾✱❜✛◗❴■✛✂✁✇✱✳✼✜✛ ①✲✴✘✞✼✹✙✹✥❆✘✚✙☎✄✔❴②✸✞✛◗❨❿✛ SMA-HPC ©1999 MIT Convergence Analysis Example Problems Collocation Discretization of 1-D Equation ( ) ( ) 1 1 x x x σ x dS − Ψ = − ′ ′ ′ " x∈ −[ 1,1] x0 = −1 xn =1 1 x n 1 x x2 − 1c x 2c x nc x ( ) 1 1 j i i j x n c j c j x x σ x x dS − = Ψ = ! − ′ ′ " n1 σ nn σ ➩❂➫➯➭✴Ð ❽ ➲ê➳➶➪➚ ❏ ➘✞➴ ➼✤➚❏ ➵✟✞ ✮ ✞ ✮✡✠ P ✴ ➭✴Ð ❽ ✏➎➭✤➻ ✴ ➽✏➾➐➻ Ô➥Õ✮ÖØ×☞☛ ä✞●❚❱✠●■❀q♣★❍★❏P✾q❏Pt❆✾❦❃◗❍★❀q✾❲❇r❁❄❱✹❈❋▼ê❣P●■▼❄❍✰❏P❁❄●■❃➥❏P●❪❏rt★❁⑤❣✉●■❃❆✾✠à➧❅✰❁◆❀q✾✹❃❆❣✐❁❄●■❃✤❈❋▼➐♣★❇r●■❊★▼❄✾❲❀❪Ü✯❱✠●■❃✤❣✐❁⑤❅✰✾❲❇ ❣P●■▼❄♦✏❁❄❃★❧✇❏rt★✾➃❁❄❃✏❏P✾✹❧■❇❤❈❋▼✻✾✜♠❥❍✤❈✦❏P❁❄●■❃è❈✦❏❉❏rt★✾➃❏r●■♣❚●■ß➐❏Pt★✾❂❣✐▼❄❁⑤❅✰✾➃❍❆❣P❁❄❃★❧✆❈❦♣❆❁◆✾✜❱✠✾❲Ý✿❁⑤❣P✾✠à➧❱✠●■❃✤❣⑥❏❤❈❋❃✏❏ ❱✠●❥▼◆▼❄●✰❱❲❈❋❏P❁❄●■❃❹❣r❱❤t★✾✹❀❂✾❥ã✞➦➧❃✇❣P❍❆❱❤tq❈á❣r❱❤t★✾✹❀❂✾❥Ü❋ÝÞ✾✱✚✤❇r❣✐❏➞❣✐✾✹▼◆✾✜❱Ø❏ ✜✍✌ ✳❯♣✴●■❁❄❃✏❏r❣❭●■❃❂❏rt★✾✿❁◆❃✏❏r✾❲❇r♦②❈■▼❜Ü ❁❄❃❪❏Pt★❁⑤❣❵❱✹❈■❣P✾ ✸◗✏ ✳❥➺✔✳✻✺③ã✢■➥✾➃❅✰✾✹❃★●❋❏r✾❼❏Pt❆●❥❣P✾❼♣✤●❥❁◆❃✏❏r❣✬❈■❣✏✎✜➭✒✑❥➺✐➭ ➴ ➺✔✓✕✓✕✓◆➺P➭✤➚✔✓✏Ü❆❈■❣✬❣✐t★●✦Ý✿❃❪❁❄❃❪❏Pt★✾ ✚❆❧❥❍★❇P✾➞❁◆❃➃❏Pt★✾Þ❀q❁❄❅★❅★▼◆✾❯●❋ß★❏rt★✾❯❣P▼◆❁⑤❅✰✾❥ã✯❢❆●■❇✻❏rt★❁⑤❣✞✾❲å★❈❋❀q♣★▼❄✾■Ü②➭✒✑á➳ ✏ ✳➞❈■❃❆❅❼➭✤➚❦➳ ✳➞ä✪t★✾✹❃✻Ü✦Ý❯✾ ❱❲❈■❃❦❅✰✾✛✚✤❃★✾✬❈❹❣✐✾❲❏❯●❋ß✻❊✤❈■❣P❁❄❣êß➯❍★❃❆❱Ø❏r❁◆●❥❃❆❣❯●■❃✇❏rt★✾✬❣P❍★❊★❁❄❃✏❏P✾❲❇r♦✦❈❋▼⑤❣❲Ü✒✎ ➷ê➴ ➫✳➭✴➲✠➺ ➷ ✑ ➫➯➭✴➲✠➺✔✓✕✓✧✓❄➺ ➷ ➚✻➫✳➭✴➲✕✓❥Ü Ý✿t★✾✹❇P✾ ➷➹ ➫➯➭✴➲❯➳ ✳ ➭✷✶☎✸➭ ➹ ✳ ➴ ➺✐➭➹ ✺ ➷➹ ➫✳➭✴➲➞➳✭✵ ✷✗✖✙✘✒✚❯✹✗✛✲✩✢✜✣✚✂✓ ✖ ë ✘ ä✪t★✾❼❱❤t❆❈■❇P❧❥✾á❅✰✾✹❃❆❣P❁é❏⑥⑦✆➼➎❱✹❈❋❃❘❏Pt★✾✹❃☞❊✴✾❼❇r✾❲♣★❇r✾✹❣P✾❲❃✏❏P✾✜❅✆❈■♣★♣★❇r●②å◗❁❄❀✇❈✦❏r✾❲▼❄⑦❦❈■❣ ➼➐➫➯➭✚➲✥✤✶➼✤➚✻➫➯➭✚➲✎✍ ➚ ✬➹◆➘✯➴ ➼✤➚➹③➷✯➹ ➫➯➭✚➲Ø➺ ✦

where oni is the weight associated with the ith basis function. It may seem odd that we used the same letter to represent the density and the basis function weights, but there is a reason. The above basis set is such that only one basis function is nonzero for a given a, and basis functions only take on the value zero or one. Therefore, Oni will be equal to the approximate charge density whe Plugging the basis function representation of the charge density into the integral equation at the top of the slide yields 4(a)=/|-x1∑n(x)d, which can be simplified by exploiting the specific basis functions to fa)=4(x)-∑on where we have introduced the residual, R(a) If collocation is used to solve this equation, then R(ac: )=0 for all i, where c is the i collocation point. The collocation points shown in slide are the subinterval center points, ac: =0.5*ai-1+ai). There are other choices for Using the fact that R(ci=0 leads t R(xc;)=重(xc;) l ci -r'dS=0 which can be reorg anized into the equation at the bottom of the slide 3.1.3 Collocation Discretization of 1D Equation- The Matrix SLIDE 10 2-x1 平(x) 5.-x14…j-x1 Note 10 In the above slide, we generate a system of equations that can be used to solve for the ani s, the piece wise constant charge densities for each of the subinter vals

Ý✿t★✾✹❇P✾❵➼➚ ➹ ❁❄❣➞❏Pt❆✾✬ÝÞ✾❲❁❄❧■t✏❏❯❈❥❣P❣P●✰❱✠❁⑤❈✦❏P✾✜❅qÝ✿❁é❏rt❦❏rt★✾✪✩ Ñ❫❪ ❊❆❈■❣P❁⑤❣➐ß➯❍★❃❆❱✠❏P❁❄●■❃✻ã➐➦③❏Þ❀✇❈②⑦q❣P✾❲✾✹❀ ●◗❅❆❅ ❏Pt✤❈✦❏qÝ❯✾❘❍❆❣✐✾✜❅❬❏rt★✾☞❣r❈❋❀q✾❘▼◆✾❲❏✐❏P✾✹❇❂❏P●Ú❇r✾❲♣❆❇P✾✜❣✐✾✹❃❥❏➃❏Pt★✾s❅✰✾✹❃❆❣✐❁◆❏⑥⑦❬❈❋❃❆❅❿❏Pt★✾☞❊❆❈❥❣✐❁⑤❣✉ß➯❍❆❃❆❱Ø❏r❁◆●❥❃ ÝÞ✾❲❁❄❧■t✏❏r❣✹Ü❆❊★❍★❏❵❏Pt❆✾❲❇r✾➃❁⑤❣❵❈❦❇r✾✹❈❥❣✐●❥❃✻ã✿ä✪t★✾❹❈■❊✤●✦♦❥✾✉❊❆❈❥❣✐❁⑤❣❵❣P✾✠❏❵❁⑤❣á❣✐❍❆❱❤t❪❏rt❆❈✦❏á●■❃★▼❄⑦❪●■❃★✾➃❊❆❈■❣P❁❄❣ ß➯❍★❃❆❱✠❏P❁❄●■❃q❁⑤❣➐❃★●■❃❑❲✾❲❇r●✬ß➯●■❇➞❈á❧■❁❄♦■✾✹❃❹➭✞Ü❥❈■❃❆❅❂❊❆❈■❣P❁❄❣✯ß➯❍❆❃❆❱Ø❏r❁◆●❥❃❆❣➐●■❃❆▼◆⑦➃❏r❈❋❖❥✾✪●■❃❹❏Pt★✾❉♦✦❈■▼◆❍★✾ ❑✹✾❲❇r● ●■❇❼●❥❃★✾■ã❪ä✪t★✾✹❇P✾❲ß➯●■❇r✾■Ü✞➼✤➚➹ Ý✿❁❄▼❄▼ê❊✤✾✆✾✹♠✏❍❆❈❋▼➞❏P●❪❏rt★✾✆❈❋♣❆♣★❇P●②å✰❁❄❀✇❈✦❏P✾✇❱❤t❆❈■❇P❧❥✾✇❅✰✾❲❃❆❣P❁◆❏⑥⑦ÚÝ✿t★✾✹❃ ➭✷✶☎✸➭ ➹ ✳ ➴ ➺P➭➹ ✺Òã ✯ê▼❄❍★❧■❧❥❁◆❃❆❧❵❏rt★✾❯❊✤❈■❣P❁❄❣✞ß➯❍★❃❆❱✠❏P❁❄●■❃❂❇P✾✹♣★❇r✾✹❣P✾❲❃✏❏r❈❋❏P❁❄●■❃✉●■ß★❏Pt★✾✿❱❤t❆❈■❇P❧❥✾➞❅✰✾✹❃❆❣✐❁◆❏⑥⑦❼❁❄❃❥❏r●á❏Pt★✾Þ❁◆❃✏❏r✾❲❧■❇❤❈❋▼ ✾✹♠✏❍❆❈❋❏P❁❄●■❃❪❈✦❏✪❏Pt❆✾▲❏P●■♣❪●■ß✔❏Pt❆✾✉❣✐▼❄❁❄❅★✾❼⑦✏❁❄✾❲▼⑤❅★❣ ✂á➫➯➭✚➲➞➳ ✒ ➴ ✳ ➴ ✴ ➭➜✏➥➭➻ ✴ ➚ ✬➹◆➘✞➴ ➼➚ ➹ ➷➹ ➫➯➭➻ ➲✐➽✏➾➻ ➺ ✖❜ç❆✘ Ý✿t★❁⑤❱❤t☞❱✹❈❋❃☞❊✤✾✉❣P❁❄❀❂♣❆▼◆❁✧✚❆✾✹❅☞❊◗⑦✆✾❲å✰♣★▼◆●❥❁é❏r❁◆❃❆❧❹❏rt★✾✉❣P♣✤✾✜❱✠❁✧✚✤❱á❊❆❈■❣P❁❄❣Þß➯❍❆❃❆❱Ø❏r❁◆●❥❃❆❣✪❏P● ✌q➫➯➭✴➲❯➳ ✂á➫✳➭✴➲✑✏ ➚ ✬ ❏ ➘✯➴ ➼➚ ❏ ✒✞ ✮ ✞ ✮✡✠ P ✴ ➭➜✏➥➭➻ ✴ ➽◗➾➻ ✖ ➤ ✘ Ý✿t★✾✹❇P✾▲ÝÞ✾át✤❈②♦■✾á❁◆❃✏❏P❇r●✰❅✰❍❆❱❲✾✹❅❘❏Pt★✾▲❇r✾✹❣P❁❄❅★❍❆❈❋▼ÒÜ✘✌q➫➯➭✴➲✠ã ➦③ß❵❱✠●❥▼◆▼❄●✰❱❲❈❋❏P❁❄●■❃❬❁⑤❣❹❍✤❣✐✾✜❅➎❏P●➥❣✐●❥▼◆♦❥✾✇❏Pt★❁⑤❣❹✾✹♠✏❍❆❈✦❏r❁◆●❥❃✻Ü❭❏Pt❆✾❲❃➐✌q➫✳➭Ð ❽ ➲q➳ ✵sß➯●❥❇❹❈❋▼❄▼Þ❁❜ÜêÝ✿t❆✾❲❇r✾ ➭Ð ❽ ❁⑤❣▲❏Pt★✾✆✩ Ñ❫❪ ❱✠●■▼❄▼❄●◗❱✹❈✦❏r❁◆●❥❃è♣✴●■❁❄❃✏❏✹ã✆ä✪t★✾❦❱❲●■▼❄▼◆●✰❱✹❈✦❏P❁❄●■❃➥♣✴●■❁❄❃❥❏❤❣✉❣Pt★●✦Ý✿❃è❁❄❃❿❣P▼◆❁⑤❅✰✾❦❈❋❇r✾❂❏Pt★✾ ❣P❍★❊★❁❄❃❥❏r✾❲❇r♦✦❈❋▼ê❱✠✾✹❃✏❏P✾❲❇❼♣✴●■❁❄❃✏❏r❣✹Ü✻➭✴Ð ❽ ➳ ✵▲✓ ✁✄✂ ➫✳➭➹ ✳ ➴ ✌ ➭ ➹ ➲✠ã✆ä✪t★✾✹❇P✾✇❈❋❇r✾❂●❋❏Pt❆✾❲❇✉❱❤t★●❥❁❄❱❲✾✹❣áß➯●■❇ ❱✠●❥▼◆▼❄●✰❱❲❈❋❏P❁❄●■❃✆♣✴●■❁❄❃✏❏r❣✹Ü❆❣P❍❆❱❤t☞❈❋❏✿➭✴Ð ❽ ➳✶➭➹ ã ☎❵❣P❁❄❃★❧❂❏Pt★✾▲ß✳❈❥❱Ø❏✿❏Pt✤❈✦❏s✌q➫➯➭✚Ð ❽ ➲ê➳✭✵❂▼❄✾✹❈■❅❆❣Þ❏P● ✌❂➫✳➭Ð ❽ ➲ê➳ ✂á➫➯➭ Ð ❽ ➲✑✏ ➚ ✬ ❏ ➘✞➴ ➼➚ ❏ ✒✞ ✮ ✞ ✮✡✠ P ✴ ➭Ð ❽ ✏è➭➻ ✴ ➽◗➾➻ ➳ ✵ ✖❴✘ Ý✿t★❁⑤❱❤t☞❱✹❈❋❃☞❊✤✾❼❇r✾❲●❥❇P❧✏❈❋❃★❁✕❑❲✾✜❅✇❁❄❃❥❏r●q❏Pt★✾❼✾✜♠✏❍❆❈✦❏r❁◆●❥❃☞❈❋❏Þ❏rt★✾❼❊✴●❋❏✐❏r●■❀ ●❋ß✔❏rt★✾✉❣P▼◆❁⑤❅✰✾■ã ☎✔➡ ❝ ➡✆☎ ①✲✴✧✳✧✳✲✔❴■✥✰✙✜✱✳✲✴✘✖❡☞✱✳✼✜❴❋✣✦✛✏✙✹✱✡✠■✥✰✙✜✱✳✲✴✘✶✲✤❫❼❝◗❡ ✩✬✫✮✭✯✥✰✙✜✱✳✲✴✘✝✆❲❃✉✸✞✛Ú✵✷✥✰✙❲✣✦✱✳❳ ✌ ✍✏✎✒✑✔✓✖✕✟✞ Convergence Analysis Example Problems One row for each collocation point One column for each density value ( ) ( ) 1 1 1 0 1 1 1 0 1 1 n n n n n n n x x c c x x c x x n c c c x x x x dS x x dS x x x x dS x x dS σ σ − − $ % − − ′ ′ ′ ′ & ' $Ψ % $ % & ' & ' = & ' & ' & ' & ' ( ) &Ψ ' − − ′ ′ ′ ′ ( ) & ' ( ) " " " " ' ( ) ( ( ( ' n1 σ nn σ Ô➥Õ✮ÖØ×ÚÙ✡✠ ➦➧❃❘❏Pt★✾✉❈■❊✤●✦♦❥✾✬❣P▼◆❁⑤❅✰✾■Ü✰ÝÞ✾▲❧■✾✹❃★✾❲❇❤❈✦❏r✾á❈q❣P⑦◗❣✐❏P✾✹❀ ●❋ß✯✾✹♠✏❍❆❈✦❏r❁◆●❥❃❆❣Þ❏Pt✤❈✦❏❉❱❲❈■❃❘❊✴✾▲❍❆❣✐✾✜❅❦❏r●✇❣✐●❥▼◆♦❥✾ ß➯●■❇✯❏rt★✾✿➼❆➚➹ ❅ ❣✹Ü✦❏rt★✾✪♣★❁◆✾✜❱✠✾✹Ý✿❁❄❣P✾Þ❱❲●■❃❆❣✐❏r❈■❃✏❏❭❱❤t❆❈■❇P❧❥✾❯❅✰✾✹❃❆❣P❁é❏r❁◆✾✜❣✯ß➯●■❇❭✾✹❈❥❱❤t➃●❋ß✤❏Pt★✾✿❣P❍★❊★❁❄❃✏❏P✾❲❇r♦✦❈❋▼⑤❣❲ã ☛

The right-han' sie of this system of equations is a vector of known potentials at interval centers(the collocation points). The ith row of the matrix correspon's 重(xe) an'the entries in the jth column correspon's to how much the charge on the jtn interval contributes to the ith potential Note that the matrix is square an' ense b Exercise 3 Is the above matrix symmetric? If we use'zc: =Ti, woul the atrix still be 3.2 Numerical Results with Increasing n Answers Are Getting Worse!l! One usually believes that a'iscretization scheme shoul'pro'uce progressively more accurate answers as the iscret ization is refine,. in this case, as we i- vie the interval into progressively finer subintervals, one might expect that he piecewise constant represent ation of the charge 'ensity given by on(a)N onipPi(a)woul become more accurate as n inc Unfortunately, the plot in the above sli'e in'icates that the metho' is not converging. In the plot, which is har'to'ecipher without looking at a co version, shows the onis pro uce'using n=10, n=z0 ann=40 subintervals For each'iscretization, a point is plot teat oni, ai for i=l,, 7, so there are en points plotte for the coarsest'iscretization an forty points plot te for the tization, but all sets of points span the interval a E[1, 1 What is clear from comparing the blue points(n R1o) to the re' points(nR20) the green points(nRcO), is that the charge proaching infinity as the iscretization is refine. The results are certainly not converging

ä✪t★✾Þ❇P❁❄❧■t✏❏PàÒt❆❈■❃❆❅❼❣P❁❄❅✰✾Þ●❋ß✰❏rt★❁⑤❣➐❣✐⑦✰❣✐❏P✾❲❀❞●■ß❆✾✹♠✏❍❆❈❋❏P❁❄●■❃❆❣✞❁❄❣✯❈✬♦■✾✹❱✠❏P●❥❇✔●❋ß✤❖✏❃❆●✦Ý✿❃✉♣✤●■❏P✾✹❃❥❏r❁❄❈■▼❄❣✯❈✦❏ ❁❄❃❥❏r✾❲❇r♦✦❈❋▼✮❱✠✾✹❃❥❏r✾❲❇❤❣✪✖➯❏Pt★✾❼❱❲●■▼❄▼◆●✰❱❲❈❋❏P❁❄●■❃❦♣✴●■❁❄❃✏❏r❣ ✘Øãêä✪t★✾☞✩ Ñ❫❪ ❇P●✦Ý✶●■ß✻❏Pt❆✾á❀✇❈❋❏P❇r❁éå✆❱❲●■❇r❇P✾✜❣✐♣✴●■❃❆❅❆❣ ❏P●q❍★❃★ß➯●■▼⑤❅✰❁◆❃❆❧q❏Pt★✾✉❣P❍★❀✄❁❄❃☞❏Pt★✾✉❱❲●■▼❄▼◆●✰❱❲❈❋❏P❁❄●■❃☞✾✹♠✏❍❆❈❋❏P❁❄●■❃ ✂á➫✳➭✴Ð ❽ ➲➞➳ ➚ ✬ ❏ ➘✯➴ ➼❆➚❏ ✒✞ ✮ ✞ ✮✡✠ P ✴ ➭✴Ð ❽ ✏➥➭➻ ✴ ➽◗➾➻ ➺ ❈❋❃✤❅❪❏Pt★✾❂✾❲❃✏❏r❇P❁❄✾✹❣✬❁◆❃Ú❏Pt★✾ ❬ Ñ❫❪ ❱✠●■▼❄❍★❀q❃è❱❲●■❇r❇P✾✜❣✐♣✴●■❃✤❅★❣✪❏P●☞t★●✦Ý ❀❹❍❆❱❤ts❏Pt❆✾❂❱❤t❆❈■❇P❧❥✾✉●■❃Ú❏Pt★✾ ❬ Ñ❫❪ ❁◆❃✏❏r✾❲❇r♦②❈■▼✻❱✠●❥❃❥❏r❇P❁❄❊★❍✰❏r✾✹❣Þ❏r●q❏Pt★✾ ✩ Ñ❫❪ ♣✴●❋❏P✾✹❃✏❏P❁⑤❈❋▼Òã ✔❉●■❏P✾á❏Pt✤❈✦❏✿❏Pt❆✾❼❀q❈❋❏P❇r❁éå❘❁❄❣❉❣r♠✏❍❆❈❋❇r✾▲❈❋❃❆❅☞❅✰✾❲❃✤❣✐✾❥ã ✶❍✷✪✹×❈✺❆✻✾✽❀✿■× ◗❪➦⑥❣✬❏Pt★✾❂❈❋❊✴●✦♦■✾✉❀✇❈❋❏P❇r❁éå❚❣P⑦✏❀q❀q✾✠❏r❇P❁⑤❱✦❂ ➦③ß➞ÝÞ✾➃❍❆❣P✾✹❅❚➭Ð ❽ ➳❞➭➹ Ü✚Ý❯●❥❍★▼⑤❅❪❏Pt★✾ ❀✇❈✦❏r❇P❁◆å❘❣✐❏P❁❄▼◆▼✻❊✴✾✉❣P⑦✏❀q❀q✾✠❏r❇P❁⑤❱✦❂ ✢✪➊❜➉ ￾➓✥✩✻➝❆➑P→✚➍✦✧✂✁✩✔➏◗➓✧➧➣★➏☎✄✖➑⑥➣★➟✝✆✏➔❵→✚➝✩✔➍❭➏◗➑P➔✟✞ ➔ ✌ ✍✏✎✒✑✔✓✖✕★✕ n = 10 n = 20 n = 40 Answers Are Getting Worse!!! σ x Ô➥Õ✮ÖØ×ÚÙ✴Ù ✰❃★✾❼❍❆❣P❍❆❈■▼◆▼❄⑦✆❊✴✾❲▼❄❁◆✾✹♦■✾✜❣✪❏Pt❆❈❋❏✬❈✇❅✰❁⑤❣P❱❲❇P✾❲❏P❁✕❑✹❈❋❏P❁❄●■❃❪❣P❱❤t★✾✹❀q✾✉❣✐t★●❥❍★▼⑤❅❘♣❆❇P●✰❅✰❍❆❱❲✾❼♣★❇r●■❧■❇r✾✹❣r❣P❁◆♦❥✾❲▼❄⑦ ❀q●■❇r✾✇❈■❱❲❱❲❍★❇❤❈✦❏P✾✇❈■❃❆❣✐ÝÞ✾❲❇❤❣▲❈■❣▲❏rt★✾❦❅✰❁⑤❣P❱❲❇P✾❲❏P❁✕❑✹❈❋❏P❁❄●■❃➎❁❄❣✉❇P✾✔✚❆❃★✾✹❅✻ã❦➦➧❃➎❏Pt★❁⑤❣✉❱✹❈■❣P✾■Ü✞❈❥❣▲Ý❯✾❦❅★❁éà ♦◗❁❄❅★✾❪❏Pt★✾s❁❄❃✏❏P✾❲❇r♦✦❈❋▼❉❁❄❃✏❏P●❬♣★❇r●■❧❥❇P✾✜❣P❣P❁❄♦■✾❲▼❄⑦✞✚❆❃★✾✹❇✆❣✐❍❆❊★❁◆❃✏❏r✾❲❇r♦②❈■▼❄❣✹ÜÞ●■❃★✾❪❀q❁◆❧❥t✏❏❦✾✠å✰♣✤✾✜❱Ø❏❦❏rt❆❈✦❏ ❏Pt❆✾❦♣★❁❄✾✹❱✠✾✹Ý✿❁❄❣P✾✆❱❲●■❃❆❣✐❏r❈■❃❥❏✉❇r✾❲♣❆❇P✾✜❣✐✾✹❃❥❏❤❈✦❏r❁◆●❥❃è●■ßÞ❏rt★✾✆❱❤t❆❈■❇P❧❥✾✇❅✰✾❲❃❆❣P❁◆❏⑥⑦è❧❥❁◆♦❥✾❲❃➎❊◗⑦➥➼➚ ➫➯➭✚➲ ✤ ➪➚➹❄➘✞➴ ➼➚ ➹ ➷➹ ➫✳➭✴➲✪ÝÞ●■❍★▼⑤❅❘❊✤✾✜❱✠●■❀q✾▲❀q●■❇r✾❼❈■❱❲❱❲❍★❇❤❈✦❏P✾▲❈■❣ ✜➥❁❄❃❆❱✠❇r✾✹❈❥❣✐✾✜❣❲ã ☎❉❃★ß➯●■❇P❏P❍★❃❆❈❋❏P✾✹▼◆⑦❥Ü✿❏Pt★✾➥♣★▼❄●❋❏❪❁◆❃æ❏Pt★✾➥❈❋❊✴●✦♦■✾➥❣✐▼❄❁❄❅★✾Ú❁❄❃❆❅✰❁⑤❱❲❈✦❏r✾✹❣✆❏Pt❆❈❋❏❘❏rt★✾➥❀❂✾❲❏Pt★●✰❅æ❁❄❣❘❃★●❋❏ ❱✠●❥❃◗♦■✾❲❇r❧■❁❄❃★❧✤ã❂➦➧❃➥❏rt★✾✇♣★▼❄●❋❏✹Ü✯Ý✿t★❁⑤❱❤t➎❁❄❣❼t✤❈❋❇❤❅❚❏r●❚❅✰✾✹❱❲❁◆♣★t❆✾❲❇✉Ý✿❁◆❏Pt★●❥❍✰❏➃▼❄●◗●■❖◗❁◆❃❆❧❪❈❋❏➃❈s❱✠●❥▼◆●❥❇ ♦■✾✹❇r❣P❁❄●■❃✻Ü✦❣Pt★●✦Ý❉❣✞❏Pt★✾❉➼✤➚➹ ❅ ❣❭♣★❇r●✰❅✰❍❆❱✠✾✜❅➃❍✤❣✐❁❄❃★❧☞✜❪➳ ✳✦✵❆Ü✍✜s➳✡✠❈✵á❈❋❃❆❅ ✜❚➳☞☛✂✵✬❣P❍★❊★❁❄❃✏❏P✾❲❇r♦✦❈❋▼⑤❣❲ã ❢★●❥❇✿✾✹❈❥❱❤t❪❅★❁❄❣r❱✠❇r✾✠❏r❁✧❑✜❈✦❏P❁❄●■❃✔Ü★❈q♣✤●❥❁◆❃✏❏❵❁⑤❣✿♣★▼❄●❋❏P❏P✾✹❅s❈❋❏❵➼❆➚➹ Ü★➭➹ ß➯●■❇✯✩➐➳ ✳■➺✦✓✧✓❄➺ ✜➐Ü❆❣P●q❏Pt❆✾❲❇r✾✉❈❋❇r✾ ❏P✾✹❃q♣✤●❥❁◆❃✏❏r❣➞♣★▼❄●❋❏✐❏r✾✹❅❹ß➯●■❇ê❏Pt★✾❵❱❲●❥❈■❇r❣P✾✹❣✐❏❭❅★❁❄❣r❱✠❇r✾✠❏r❁✧❑✜❈✦❏P❁❄●■❃q❈❋❃✤❅❂ß➯●■❇P❏⑥⑦✉♣✤●❥❁◆❃✏❏❤❣➐♣★▼❄●❋❏✐❏r✾✹❅❂ß➯●❥❇➐❏Pt★✾ ✚❆❃★✾✜❣⑥❏❵❅★❁❄❣r❱✠❇r✾✠❏r❁✧❑✜❈✦❏P❁❄●■❃✔Ü✏❊❆❍✰❏❵❈❋▼❄▼✔❣P✾✠❏r❣✪●■ß✞♣✴●■❁❄❃✏❏r❣❉❣P♣❆❈❋❃❘❏Pt❆✾❼❁◆❃✏❏P✾✹❇P♦✦❈■▼✮➭✷✶✹✸❁✏ ✳■➺✦✳✻✺③ã ■✶t❆❈❋❏✪❁⑤❣✿❱✠▼❄✾✹❈■❇Þß➯❇P●❥❀ ❱✠●❥❀❂♣✤❈❋❇r❁◆❃★❧➃❏Pt★✾❼❊❆▼◆❍★✾▲♣✴●■❁❄❃✏❏r❣ ✖➯❃✍✌ ë✏✎ ✘➞❏r●❹❏rt★✾▲❇P✾✜❅✆♣✴●■❁❄❃✏❏r❣ ✖✳❃✑✌▲ç ✎ ✘ ❈❋❃✤❅ ❏r●❬❏Pt❆✾❚❧■❇r✾❲✾✹❃ ♣✴●■❁❄❃✏❏r❣❍✖✳❃✑✌❴✎ ✘ØÜ✪❁⑤❣✇❏Pt✤❈✦❏❦❏rt★✾Ú❱❤t❆❈■❇P❧❥✾❪❅★✾❲❃❆❣P❁é❏⑥⑦➜❣P✾❲✾❲❀✇❣q❏r●❿❊✴✾Ú❈❋♣✰à ♣★❇r●❥❈❥❱❤t★❁◆❃❆❧❂❁❄❃✚❆❃❆❁é❏⑥⑦❪❈■❣✪❏rt★✾➃❅✰❁⑤❣r❱✠❇r✾✠❏P❁✕❑✹❈❋❏P❁❄●■❃❪❁⑤❣✿❇P✾✔✚❆❃★✾✜❅✮ãÞä✪t❆✾✉❇P✾✜❣✐❍❆▼é❏❤❣❉❈❋❇r✾✉❱✠✾✹❇✐❏❤❈❋❁❄❃★▼◆⑦❘❃★●❋❏ ❱✠●❥❃◗♦■✾❲❇r❧■❁❄❃★❧✤ã ✒

happening the numerical technique at fault, or is the integral equation a problem? 3.3 Example ProbLems 3.3.1 1D Secon P in Equation (x)=0(x)+/1|x-1(x)dsx∈[-1,1 The potential is gⅳen The density must be computed y(x=r-x o(x) is unknown Note 1I We are going to post pone examining what went wrong in the first-kind example nd instead look at a oecond Kind equation. For this equation, we assume that the potential, (a), is known and that the charge density a(a) is unknown Here, a is in the interval [1, 1], and the integration is over that same interval Once again, the Greend function is given by G(, r)=r-a. What makes this equation oecond-Kind instead of first is circled in the equation on the top of the slide, The unknown c s both inside and outside of he integral. In the first-kind equation, the density appeared only inside the ntegral. This seemly small difference has enormous numerical ramifications n the left plot below the equation, an example given potential, as-a is plotted function of a. On the right is a plot of a charge density as a function of hich satisfies this second kind integral equation. As we will see below, this equation is easily solved numerically 3.3.2 Collocation Discretization of 1D Equation SLIDE 13 业(x)=(x)+1x-x1(x)dsx∈[-1,1 CentroilCollocatel Piecewise Constant Sc eme (ci)=oni+2t=ant 2i _ lac: -a'ldsr

■✶t◗⑦s❁❄❣✬❏rt★❁❄❣▲t✤❈❋♣★♣✴✾❲❃★❁❄❃★❧✠❂➜➦⑥❣á❏Pt★✾❂❃◗❍★❀q✾❲❇r❁❄❱✹❈❋▼✞❏P✾✜❱❤t★❃★❁⑤♠❥❍❆✾q❈❋❏✬ß✳❈■❍★▼◆❏✹Ü✮●■❇▲❁⑤❣✬❏Pt★✾❂❁◆❃✏❏r✾❲❧■❇❤❈❋▼ ✾✹♠✏❍❆❈❋❏P❁❄●■❃❪❈❂♣★❇P●❥❊★▼❄✾❲❀☎❂ ✢✪➊☛✢ ✣✒✤✿➍✦✥➛★✧✞✩✫✪è➝★↔✭✬★✧✞✩✮✥➏ ☎✔➡✆☎✔➡ ❝ ❝◗❡ ✄✻✛◗❴❥✲✤✘✯✺ ✯s✱✳✘✯✺➜✩✬✫✮✭✯✥✰✙✹✱❜✲✤✘ ✌ ✍✏✎✒✑✔✓✖✕■➤ ➩❂➫➯➭✚➲➞➳➁➼➐➫✳➭✴➲ ✌ ➵ ➴ ✳ ➴ ✴ ➭➜✏➥➭✤➻ ✴ ➼➐➫➯➭✤➻➯➲✐➽✏➾ê➻ ➭✷✶☎✸◗✏ ✳❥➺✔✳ ✺ Convergence Analysis Example Problems x∈ −[ 1,1] σ ( ) x is unknown ( ) 3 Ψ = x x x − The potential is given The density must be computed Solution = 3x ( ) ( ) ( ) 1 1 x σ σ x x x x dS − Ψ = + − ′ ′ ′ " Ô➥Õ✮ÖØ×ÚÙ ✎ ■➥✾✪❈■❇P✾Þ❧■●❥❁◆❃❆❧❵❏r●✬♣✴●❥❣✐❏P♣✴●■❃★✾✪✾❲å✰❈■❀q❁◆❃★❁❄❃★❧▲Ý✿t❆❈❋❏✯ÝÞ✾❲❃✏❏➐Ý✿❇P●❥❃★❧á❁◆❃❹❏Pt★✾◆✚❆❇❤❣✐❏✐à③❖✏❁❄❃❆❅❹✾✠å★❈❋❀q♣★▼❄✾■Ü ❈❋❃✤❅✇❁◆❃✤❣⑥❏r✾✹❈■❅❦▼◆●◗●■❖q❈✦❏✿❈ ●◗✾✜❱✠●❥❃❆❅✁￾▲❁❄❃❆❅❦✾✜♠❥❍✤❈✦❏P❁❄●■❃✔ã✯❢❆●■❇❯❏Pt★❁⑤❣➞✾✜♠✏❍❆❈✦❏r❁◆●❥❃✻Ü❥ÝÞ✾✬❈❥❣P❣P❍★❀q✾❉❏rt❆❈✦❏ ❏Pt❆✾s♣✤●■❏P✾❲❃✏❏r❁❄❈■▼❜Ü ✂á➫➯➭✴➲✠Ü✪❁❄❣✇❖◗❃★●✦Ý✿❃➜❈❋❃✤❅❿❏rt❆❈✦❏✇❏rt★✾s❱❤t❆❈❋❇r❧■✾❪❅✰✾❲❃✤❣✐❁◆❏⑥⑦ ➼➐➫➯➭✚➲q❁⑤❣✇❍★❃★❖◗❃★●✦Ý✿❃✻ã ❶❉✾✹❇P✾❥Ü✰➭❪❁❄❣✪❁❄❃☞❏Pt★✾▲❁❄❃❥❏r✾❲❇r♦✦❈❋▼✭✸❁✏ ✳■➺✔✳ ✺ÒÜ✰❈■❃❆❅❦❏rt★✾▲❁◆❃✏❏r✾❲❧■❇❤❈✦❏r❁◆●❥❃✆❁⑤❣✪●✦♦■✾✹❇❯❏rt❆❈✦❏❉❣r❈❋❀q✾▲❁◆❃✏❏r✾❲❇r♦②❈■▼❜ã ✰❃❆❱❲✾❹❈❋❧✏❈❋❁❄❃✻Ü✤❏Pt★✾q❑á❇P✾✹✾❲❃❘❅ ❣❉ß➯❍❆❃❆❱Ø❏r❁◆●❥❃Ú❁⑤❣❵❧❥❁◆♦❥✾❲❃❚❊◗⑦s➸✇➫➯➭✞➺✐➭✤➻✒➲❉➳ ✴ ➭✽✏➎➭✤➻ ✴ ã ■✶t✤❈✦❏✬❀✇❈■❖■✾✹❣ ❏Pt❆❁❄❣❵✾✹♠✏❍❆❈✦❏r❁◆●❥❃❍●◗✾✹❱❲●■❃❆❅◗à✂￾▲❁❄❃❆❅❪❁◆❃✤❣⑥❏r✾✹❈■❅❪●■ß ✚❆❇r❣✐❏❉❁⑤❣❵❱✠❁❄❇❤❱✠▼❄✾✹❅❪❁❄❃❪❏Pt★✾❹✾✹♠✏❍❆❈✦❏r❁◆●❥❃❪●■❃❪❏Pt❆✾❼❏P●❥♣ ●❋ßÞ❏Pt❆✾❦❣✐▼❄❁❄❅★✾■ã❘ä✪t❆✾✇❍★❃★❖◗❃★●✦Ý✿❃❬❱❤t❆❈❋❇r❧■✾q❅✰✾✹❃❆❣✐❁◆❏⑥⑦è❈■♣★♣✴✾✹❈❋❇❤❣▲❊✴●❋❏Pt❬❁❄❃❆❣P❁❄❅★✾❦❈❋❃❆❅➥●■❍★❏r❣P❁❄❅✰✾✇●■ß ❏Pt❆✾✆❁❄❃❥❏r✾❲❧❥❇r❈■▼❜ã❚➦➧❃ ❏Pt★✾✆✚❆❇❤❣⑥❏PàÒ❖◗❁❄❃❆❅❬✾✹♠✏❍❆❈❋❏P❁❄●■❃✻Ü✯❏rt★✾☞❅★✾❲❃❆❣P❁é❏⑥⑦❬❈■♣★♣✤✾✜❈❋❇r✾✹❅➎●❥❃★▼◆⑦➎❁❄❃❆❣P❁❄❅✰✾✆❏Pt★✾ ❁❄❃❥❏r✾❲❧❥❇r❈■▼❜ã❭ä✪t★❁⑤❣❉❣✐✾✹✾❲❀q▼❄⑦✆❣✐❀✇❈■▼◆▼✻❅✰❁￾✚✾❲❇r✾❲❃❆❱❲✾❼t❆❈■❣✪✾✹❃★●■❇r❀q●■❍❆❣Þ❃◗❍★❀q✾❲❇r❁❄❱✹❈❋▼✮❇❤❈❋❀q❁✧✚✤❱❲❈❋❏P❁❄●■❃❆❣✹ã ➦➧❃❂❏Pt★✾❵▼◆✾❲ß✒❏➞♣★▼❄●❋❏ê❊✴✾❲▼❄●✦Ý ❏rt★✾✿✾✹♠✏❍❆❈❋❏P❁❄●■❃✻Ü✏❈❋❃q✾✠å★❈❋❀q♣★▼❄✾✿❧■❁❄♦■✾✹❃❹♣✴●❋❏r✾❲❃✏❏P❁⑤❈❋▼ÒÜ■➭☎✄❸✏✇➭✇❁⑤❣➐♣★▼❄●❋❏P❏P✾✜❅ ❈■❣▲❈❘ß➯❍★❃❆❱✠❏P❁❄●■❃➥●❋ßÞ➭✔ã ✰❃Ú❏rt★✾q❇P❁❄❧■t✏❏▲❁❄❣❼❈☞♣★▼❄●❋❏❼●❋ß✪❈❘❱❤t✤❈❋❇r❧■✾❹❅★✾❲❃❆❣P❁é❏⑥⑦è❈■❣▲❈❘ß➯❍★❃❆❱Ø❏r❁◆●❥❃è●■ß ➭ÚÝ✿t❆❁❄❱❤t❚❣r❈✦❏r❁❄❣✚❆✾✹❣❉❏Pt★❁⑤❣❵❣P✾✹❱❲●■❃❆❅❪❖◗❁❄❃❆❅❪❁❄❃❥❏r✾❲❧❥❇r❈■▼✔✾✹♠✏❍❆❈❋❏P❁❄●■❃✻ã Û❣❵ÝÞ✾✉Ý✿❁❄▼◆▼✯❣✐✾✹✾✉❊✤✾✹▼◆●✦Ý❼Ü✤❏Pt❆❁❄❣ ✾✹♠✏❍❆❈❋❏P❁❄●■❃☞❁❄❣✪✾✜❈■❣P❁◆▼❄⑦✆❣P●■▼❄♦■✾✜❅❦❃◗❍★❀q✾❲❇r❁❄❱✹❈❋▼❄▼◆⑦❥ã ☎✔➡✆☎✔➡⑤➠ ①✲✴✧✳✧✳✲✔❴■✥✰✙✜✱✳✲✴✘✖❡☞✱✳✼✜❴❋✣✦✛✏✙✹✱✡✠■✥✰✙✜✱✳✲✴✘✶✲✤❫❼❝◗❡ ✩✬✫✮✭✯✥✰✙✜✱✳✲✴✘ ✌ ✍✏✎✒✑✔✓✖✕❆✝ ➩❂➫➯➭✚➲➞➳➁➼➐➫✳➭✴➲ ✌ ➵ ➴ ✳ ➴✭✴ ➭➜✏➥➭✤➻ ✴ ➼➐➫➯➭✤➻➯➲✐➽✏➾ê➻ ➭✷✶☎✸◗✏ ✳❥➺✔✳ ✺ ①✛✏✘✚✙✹✣✦✲✤✱❜✺ ①✲✴✧✳✧❜✲✻❴■✥✰✙✜✛◗✺ ￾✱❜✛◗❴■✛✂✁✇✱✳✼✜✛ ①✲✴✘✞✼✹✙✹✥❆✘✚✙☎✄✔❴②✸✞✛◗❨❿✛ SMA-HPC ©1999 MIT Convergence Analysis Example Problems Collocation Discretization of 1-D Equation ( ) ( ) 1 1 x x x σ x dS − Ψ = − ′ ′ ′ " x∈ −[ 1,1] x0 = −1 xn =1 1 x n 1 x x2 − 1c x 2c x nc x ( ) 1 1 j i i j x n c j c j x x σ x x dS − = Ψ = ! − ′ ′ " n1 σ nn σ ➩❂➫➯➭✚Ð ❽ ➲ê➳➁➼✤➚➹ ✌ ➪➚ ❏ ➘✯➴ ➼❆➚❏ ➵✞ ✮ ✞ ✮✡✠ P ✴ ➭✴Ð ❽ ✏➥➭✤➻ ✴ ➽◗➾➐➻ ✄

点击下载完整版文档(PDF)VIP每日下载上限内不扣除下载券和下载次数;
按次数下载不扣除下载券;
24小时内重复下载只扣除一次;
顺序:VIP每日次数-->可用次数-->下载券;
共25页,试读已结束,阅读完整版请下载
相关文档

关于我们|帮助中心|下载说明|相关软件|意见反馈|联系我们

Copyright © 2008-现在 cucdc.com 高等教育资讯网 版权所有