What is the essential difference bet ween first and second kind equations. Is it some aspect of the numerical technique or are these two equations really that different. In the next slides, we will try to answer this question 3.5 Example Problems 3.5.1 1D First Kind Equation Di culty LIDE 16 Denote the integral operator as K x-x|(x)dS→K The integral operator is singular: K has a null space 斗(x)=0,x≠0,0(0)=1 Ko0=/|x-r|o(x)s′=0 IfK=重 then K(aa+00)=业 Note 16 On the top of the above slide, we introduce the abstract notion that ar-r'o(a)dsy an operator on the function o, which we denote with the symbol K. As shown n the top of the slide this notation makes writing the integral equation look Ist like writing The key problem is that the operator K is singular. And if were a matrix equation with a singular K, one would not be surpr ised to discover the system of equations is hard, or impossible, to solve We will not try, in this lecture to be formal about the concept of a singular operator. To do so, we would necessarily be examining details about certain types of function spaces. Instead, we will try to develop Some intuition. In particular, we will draw an analogy to matrices and note that if an operator is lar, it must have a null space o see that K does have a null space, consider the spike function ao(a) depicted n the middle of the slide. This spike function is one at a=0 and zero other wise Note, this function is not an impulse function. Unlike the impulse function, the Spike's value at =0 is finite and the area under its curve is obviously zero■✶t❆❈❋❏á❁⑤❣✬❏Pt★✾❂✾✹❣r❣✐✾✹❃✏❏P❁⑤❈❋▼❭❅✰❁✁￾✚✾❲❇r✾❲❃✤❱✠✾➃❊✴✾✠❏⑥ÝÞ✾❲✾✹❃❍✚❆❇❤❣✐❏▲❈■❃❆❅❚❣P✾✹❱✠●❥❃❆❅s❖◗❁❄❃❆❅❚✾✜♠❥❍✤❈✦❏P❁❄●■❃✤❣❲ã▲➦⑥❣á❁é❏ ❣P●■❀q✾❂❈❥❣✐♣✴✾✹❱✠❏á●■ßê❏Pt❆✾❂❃◗❍★❀q✾✹❇P❁⑤❱❲❈■▼✞❏r✾✹❱❤t★❃❆❁❄♠✏❍★✾❂●■❇▲❈❋❇r✾➃❏rt★✾✹❣P✾❹❏⑥Ý❯●☞✾✹♠✏❍❆❈❋❏P❁❄●■❃❆❣✬❇P✾✜❈❋▼❄▼◆⑦❪❏rt❆❈✦❏ ❅✰❁￾✴✾✹❇P✾✹❃✏❏✹ã➐➦➧❃❘❏Pt★✾❼❃★✾❲å◗❏❵❣✐▼❄❁❄❅★✾✹❣✹Ü✰Ý❯✾❼Ý✿❁❄▼◆▼✚❏r❇P⑦✆❏P●✇❈❋❃✤❣✐ÝÞ✾❲❇Þ❏Pt❆❁❄❣❉♠✏❍★✾✜❣⑥❏r❁◆●❥❃✻ã ✢✪➊✁￾ ✣✒✤✿➍✦✥➛★✧✞✩✫✪è➝★↔✭✬★✧✞✩✮✥➏ ☎✔➡✄✂✻➡ ❝ ❝◗❡ ❅✿✱➯✣✦✼✹✙✰✯s✱❜✘✞✺➜✩á✫✮✭✞✥★✙✹✱✳✲✴✘✶❡☞✱✄✂❿❴■✭✯✧➯✙❤⑨ ✌ ✍✏✎✒✑✔✓✖✕❄☎ ♥✬✾❲❃★●■❏P✾á❏Pt★✾❼❁❄❃✏❏P✾✹❧■❇❤❈❋▼✮●■♣✴✾❲❇❤❈✦❏r●■❇✪❈■❣✆☎ ☎è➼✿✍ ✒ ➴ ✳ ➴ ✴ ➭➜✏➥➭➻ ✴ ➼➐➫➯➭➻ ➲✐➽✏➾➻ ❋✝☎è➼☞➳ ➩ ä✪t★✾▲❁❄❃❥❏r✾❲❧❥❇r❈■▼✮●■♣✴✾❲❇❤❈✦❏P●❥❇❯❁⑤❣❉✼✜✱✳✘✯✢✤✭✯✧✳✥★✣❦➨✞☎ ✸✯✥★✼✇✥➥✘✮✭✞✧❜✧❯✼✹❩❭✥★❴■✛ SMA-HPC ©1999 MIT Convergence Analysis Example Problems 1-D First Kind Equation Difficulty ( ) 1 1 σ σ x x x dS σ − Κ ≡ " − ′ ′ ′ # Κ = Ψ Denote the integral operator as K The integral operator is singular Κ has a null space σ σ 0 0 ( ) x x = ≠ 0, 0, (0) =1 -1 1 ( ) 1 0 0 1 σ σ x x x dS 0 − Κ = − ′ ′ ′ = " ( ) 0 If a a Κ = σ σ Ψ # Κ +σ = Ψ 0 ☎è➼✑á➳ ✒ ➴ ✳ ➴ ✴ ➭➜✏➥➭➻ ✴ ➼✑◗➫➯➭➻ ➲✐➽✏➾➻ ➳✭✵ ➦③ß✟☎Ú➼❒ ➳æ➩ ✖✙✘✚❉✜ ☎✖➫✳➼❒ ✌❿➼✑ ➲➞➳➶➩ Ô➥Õ✮ÖØ×ÚÙ✗❘ ✰❃❘❏Pt★✾▲❏r●■♣❪●❋ß✞❏Pt★✾✉❈■❊✤●✦♦❥✾á❣P▼◆❁⑤❅✰✾❥Ü★Ý❯✾▲❁❄❃✏❏P❇r●◗❅★❍❆❱✠✾á❏Pt❆✾✉❈❋❊❆❣✐❏P❇❤❈■❱✠❏✪❃❆●❋❏P❁❄●■❃❘❏Pt✤❈✦❏ ✒ ➴ ✳ ➴ ✴ ➭➜✏è➭➻ ✴ ➼➐➫➯➭➻ ➲⑥➽◗➾➻ ❁⑤❣➐❈❋❃➃●❥♣✤✾✹❇r❈❋❏P●❥❇✞●❥❃➃❏rt★✾Þß➯❍★❃❆❱Ø❏r❁◆●❥❃✇➼❭Ü❋Ý✿t★❁⑤❱❤t❹ÝÞ✾✿❅✰✾❲❃★●■❏P✾ÞÝ✿❁é❏rt❂❏Pt★✾✿❣P⑦✏❀❹❊✤●❥▼✠☎❿ã Û❣➐❣Pt★●✦Ý✿❃ ●■❃❚❏Pt★✾❹❏P●❥♣Ú●❋ß➞❏Pt★✾q❣✐▼❄❁⑤❅✰✾■Ü✚❏rt★❁❄❣á❃★●■❏r❈❋❏P❁❄●■❃❚❀✇❈❋❖❥✾✹❣✬Ý✿❇P❁◆❏P❁❄❃★❧✆❏Pt★✾❂❁◆❃✏❏r✾❲❧■❇❤❈❋▼✯✾✹♠✏❍❆❈❋❏P❁❄●■❃❚▼❄●◗●■❖ ￾⑥❍❆❣✐❏❉▼◆❁❄❖■✾▲Ý✿❇r❁é❏r❁◆❃❆❧q❈❂❀✇❈✦❏r❇P❁◆å✆✾✜♠❥❍✤❈✦❏P❁❄●■❃✔ã ä✪t★✾▲❖■✾✹⑦❦♣★❇r●■❊★▼❄✾❲❀ ❁❄❣✪❏rt❆❈✦❏✿❏rt★✾▲●■♣✴✾❲❇❤❈✦❏r●■❇✆☎ ❁❄❣❉❣P❁❄❃★❧■❍★▼⑤❈❋❇✜ã Û❃❆❅❪❁◆ß ☎Ú➼s➳ ✂ ÝÞ✾❲❇r✾➞❈✿❀✇❈✦❏r❇P❁◆åá✾✜♠✏❍❆❈✦❏r❁◆●❥❃❼Ý✿❁é❏rt❹❈❉❣P❁◆❃❆❧■❍★▼⑤❈❋❇✡☎❿Ü②●■❃★✾➞Ý❯●❥❍★▼❄❅✉❃★●❋❏✔❊✴✾✪❣✐❍★❇r♣★❇r❁❄❣P✾✹❅✬❏r●✬❅✰❁⑤❣P❱❲●✦♦■✾✹❇ ❏Pt❆✾✉❣✐⑦✰❣✐❏P✾❲❀ ●❋ß❭✾✹♠✏❍❆❈✦❏r❁◆●❥❃❆❣Þ❁❄❣❉t❆❈❋❇❤❅✮Ü✰●■❇✪❁❄❀q♣✤●✏❣P❣P❁◆❊❆▼◆✾❥Ü✰❏P●✇❣P●■▼❄♦■✾■ã ■➥✾❦Ý✿❁◆▼❄▼➞❃❆●❋❏✉❏r❇P⑦❥Ü✯❁❄❃➎❏Pt❆❁❄❣❹▼◆✾✜❱Ø❏r❍★❇P✾❥Ü✔❏r●❚❊✴✾✇ß➯●■❇r❀✇❈❋▼❯❈■❊✤●❥❍✰❏✉❏rt★✾✆❱✠●❥❃❆❱✠✾✹♣✰❏➃●■ß✿❈s❣P❁❄❃★❧■❍★▼⑤❈❋❇ ●■♣✴✾❲❇❤❈✦❏r●■❇✜ãsä✔●➥❅✰●Ú❣P●❆Ü❭Ý❯✾✆Ý❯●❥❍★▼❄❅❿❃★✾✹❱❲✾✹❣r❣P❈■❇P❁❄▼◆⑦è❊✤✾❘✾✠å★❈❋❀q❁❄❃★❁❄❃★❧Ú❅✰✾❲❏r❈❋❁❄▼⑤❣➃❈❋❊✴●■❍★❏❂❱✠✾✹❇✐❏❤❈❋❁❄❃ ❏⑥⑦◗♣✤✾✜❣❂●❋ß❵ß➯❍❆❃❆❱Ø❏r❁◆●❥❃➜❣P♣❆❈❥❱✠✾✹❣✹ã❿➦➧❃❆❣✐❏P✾✜❈■❅✮Ü❯Ý❯✾❪Ý✿❁❄▼◆▼✪❏r❇P⑦➎❏r●❬❅✰✾❲♦❥✾❲▼❄●■♣✖❣P●■❀q✾☞❁◆❃✏❏r❍★❁é❏r❁◆●❥❃✻ã ➦➧❃ ♣❆❈■❇✐❏r❁❄❱❲❍★▼❄❈■❇✹Ü✰ÝÞ✾❼Ý✿❁❄▼◆▼✔❅★❇r❈②Ý ❈❋❃s❈❋❃❆❈■▼◆●❥❧■⑦✇❏r●✇❀✇❈✦❏P❇r❁⑤❱✠✾✹❣✿❈■❃❆❅☞❃★●❋❏r✾▲❏Pt❆❈❋❏❵❁◆ß➐❈■❃❪●■♣✴✾❲❇❤❈✦❏P●❥❇✪❁❄❣ ❣P❁◆❃★❧❥❍★▼⑤❈❋❇✜Ü✏❁◆❏❉❀➃❍✤❣⑥❏❉t❆❈②♦❥✾á❈❂❃◗❍★▼❄▼✔❣P♣❆❈■❱❲✾■ã ä✞●✉❣✐✾✹✾❯❏rt❆❈✦❏☛☎â❅✰●◗✾✹❣❭t❆❈②♦❥✾✪❈▲❃◗❍★▼❄▼❆❣P♣❆❈■❱❲✾■Ü✏❱✠●■❃✤❣✐❁⑤❅✰✾❲❇❭❏rt★✾❉❣P♣★❁◆❖❥✾✪ß➯❍★❃❆❱✠❏P❁❄●■❃❦➼✑✏➫➯➭✚➲ê❅✰✾✹♣★❁❄❱✠❏P✾✜❅ ❁❄❃✉❏Pt❆✾❯❀q❁⑤❅★❅✰▼❄✾Þ●■ß✰❏Pt★✾✿❣P▼◆❁⑤❅✰✾❥ã✞ä✪t★❁⑤❣❭❣✐♣★❁❄❖■✾➞ß➯❍★❃❆❱✠❏P❁❄●■❃❂❁❄❣✯●■❃❆✾❯❈✦❏❭➭❘➳ ✵✬❈■❃❆❅ ❑❲✾✹❇P●✬●❋❏rt★✾❲❇rÝ✿❁❄❣P✾■ã ✔❉●■❏P✾❥Ü❥❏rt★❁⑤❣êß➯❍★❃✤❱Ø❏P❁❄●■❃❘❁❄❣ ✯❴❃❁✱✚❈❋❃❦❁❄❀q♣★❍★▼⑤❣✐✾✬ß➯❍★❃❆❱✠❏P❁❄●■❃✻ã ☎❉❃❆▼◆❁❄❖■✾✬❏Pt★✾á❁❄❀❂♣❆❍★▼❄❣P✾❉ß➯❍❆❃❆❱Ø❏r❁◆●❥❃✻Ü◗❏Pt★✾ ❣P♣★❁◆❖❥✾❆❅ ❣Þ♦②❈■▼◆❍❆✾✉❈✦❏✿➭☞➳ ✵❹❁⑤❣◆✚❆❃★❁◆❏P✾✉❈■❃❆❅❘❏Pt★✾✉❈■❇P✾✜❈➃❍❆❃❆❅✰✾❲❇✿❁◆❏r❣❵❱✠❍★❇r♦■✾á❁❄❣❉●■❊◗♦◗❁◆●❥❍❆❣✐▼❄⑦ ❑❲✾✹❇P●✤ã ë ç