d.5.2 1d dirt Ki)d i quatio) Difficulty from the 0 atrix SLID 17 CoLocation generates a iscrete form of K Ka=业5KnOn=亚1 ∫-x1 -4- The matrix Kn is the not the operator Kn Note 1E constant charge ' ensity representation with coLocation at subintervaMcenters results in a system of equations which relates the subintervaMTi's to the cold cation point potential From this perspective, the matrix on the above slil hought of as a iscrete representation of the operator K. We enote he matrix with K, to in'icate the matrix was generate using a'iscretization ith n basis functions Below, we wihave to be more precise about the iscrete represent ation of the perator K, but for the moment, the matrix is sufficient 3. K ff umerical Results with Increasing n n=40 n=20 Eigenvalues accumulating at zero Note 18 If the operator K is singular, one might expect to see that reflecte in the eigen valles of a matrix generate by iscretizing K. In particular, one wouM expect the matrix to have eigenvalues that are near zero. In the above eigen- values of matrices generate by'iscretizing K for the 1-D problem are plotte Discretizing using 10 subintervals generates a matrix with 10 eigenvalues plot ein ble. The ble eigenvalue closest to zero is N.01. As the'iscretization☎✔➡✄✂✻➡⑤➠ ❝◗❡ ❅✿✱➯✣✦✼✹✙✰✯s✱❜✘✞✺➜✩á✫✮✭✞✥★✙✹✱✳✲✴✘✶❡☞✱✄✂❿❴■✭✯✧➯✙❤⑨ ❫Ò✣✦✲✴❨ ✙✹✸✯✛❚✵✷✥★✙❲✣✦✱➯❳ ✌ ✍✏✎✒✑✔✓✖✕ ✄ ❙❯●■▼❄▼◆●✰❱❲❈❋❏P❁❄●■❃☞❧■✾✹❃★✾❲❇❤❈✦❏r✾✹❣✪❈q❅✰❁❄❣r❱✠❇r✾✠❏r✾áß➯●■❇r❀✄●❋ß✟☎ ☎Ú➼s➳æ➩ ￾ ☎❦➚✰➼✤➚❦➳æ➩á➚ Convergence Analysis Example Problems ( ) ( ) 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 Kn ä✪t❆✾á❀✇❈❋❏P❇r❁éå ☎➚ ❁⑤❣Þ❏Pt❆✾❼❃★●❋❏✿❏rt★✾❼●■♣✴✾❲❇❤❈✦❏r●■❇ ☎➚✂✁ Ô➥Õ✮ÖØ×ÚÙ✮✓ Û❣✆❃★●❋❏r✾✹❅➜❈❋❊✴●✦♦■✾❥ÜÞ❅★❁❄❣r❱✠❇r✾✠❏r❁✧❑✹❁◆❃★❧➎❏rt★✾❚❁❄❃✏❏P✾❲❧❥❇r❈■▼❉✾✹♠✏❍❆❈❋❏P❁❄●■❃➜❊◗⑦✖❱❲●■❀➃❊❆❁◆❃★❁❄❃★❧❿❈❬♣❆❁◆✾✜❱✠✾❲Ý✿❁⑤❣P✾ ❱✠●❥❃❆❣✐❏r❈❋❃✏❏➃❱❤t❆❈■❇P❧❥✾❦❅✰✾❲❃✤❣✐❁◆❏⑥⑦è❇r✾❲♣❆❇P✾✜❣✐✾✹❃❥❏❤❈✦❏r❁◆●❥❃➥Ý✿❁◆❏Pt✖❱✠●■▼❄▼❄●◗❱✹❈✦❏r❁◆●❥❃➎❈❋❏❹❣P❍★❊★❁❄❃❥❏r✾❲❇r♦✦❈❋▼❯❱❲✾❲❃✏❏P✾✹❇r❣ ❇r✾✹❣P❍★▼é❏❤❣❉❁❄❃Ú❈❦❣P⑦✰❣⑥❏r✾❲❀➙●❋ßê✾✹♠✏❍❆❈❋❏P❁❄●■❃❆❣❵Ý✿t★❁⑤❱❤ts❇P✾✹▼❄❈❋❏P✾✹❣❉❏Pt★✾❂❣✐❍❆❊★❁◆❃✏❏r✾❲❇r♦②❈■▼✞➼➹ ❅ ❣❉❏r●✇❏Pt❆✾❹❱✠●❥▼◆▼❄●❋à ❱❲❈❋❏P❁❄●■❃➎♣✴●■❁❄❃❥❏✉♣✴●❋❏r✾❲❃✏❏P❁⑤❈❋▼⑤❣✹ã❘❢★❇r●■❀ ❏Pt★❁⑤❣✉♣✴✾❲❇❤❣P♣✤✾✜❱Ø❏P❁❄♦■✾❥Ü✻❏rt★✾❦❀✇❈✦❏r❇P❁◆åè●❥❃è❏rt★✾✆❈■❊✤●✦♦❥✾q❣P▼❄❁❄❅✰✾ ❱❲❈■❃è❊✴✾✇❏rt★●■❍★❧❥t✏❏✉●❋ß✿❈■❣❼❈s❅✰❁⑤❣P❱❲❇P✾❲❏P✾✇❇r✾❲♣★❇r✾✹❣P✾❲❃✏❏❤❈✦❏P❁❄●■❃è●❋ßÞ❏Pt★✾✇●❥♣✤✾✹❇r❈❋❏P●❥❇ ☎❿ã☎■➥✾❦❅✰✾❲❃❆●❋❏P✾ ❏Pt❆✾✬❀✇❈✦❏r❇P❁◆å✇Ý✿❁é❏rt ☎➚ ❏P●➃❁❄❃❆❅✰❁⑤❱❲❈❋❏P✾✬❏Pt★✾á❀✇❈✦❏r❇P❁◆å✇ÝÞ❈❥❣ê❧■✾✹❃★✾❲❇❤❈✦❏r✾✹❅✇❍✤❣✐❁❄❃★❧❹❈❹❅✰❁⑤❣P❱❲❇P✾❲❏P❁✕❑✹❈✦❏r❁◆●❥❃ Ý✿❁◆❏Pt❇✜Ú❊❆❈❥❣✐❁⑤❣Þß➯❍★❃❆❱✠❏P❁❄●■❃❆❣✹ã ✄Þ✾❲▼❄●✦Ý❼Ü◗Ý❯✾áÝ✿❁◆▼❄▼✻t❆❈②♦❥✾❉❏r●❂❊✴✾❼❀q●■❇r✾✬♣❆❇P✾✜❱✠❁⑤❣✐✾▲❈❋❊✴●■❍✰❏✪❏rt★✾❼❅✰❁❄❣r❱✠❇r✾✠❏r✾▲❇P✾✹♣★❇P✾✜❣✐✾✹❃✏❏r❈✦❏r❁◆●❥❃❦●❋ß✞❏Pt★✾ ●■♣✴✾❲❇❤❈✦❏r●■❇✆☎❿Ü✰❊❆❍✰❏✿ß➯●■❇✪❏rt★✾❼❀q●■❀q✾❲❃✏❏✹Ü◗❏rt★✾❼❀✇❈✦❏r❇P❁◆å✆❁⑤❣❉❣✐❍★✈✇❱❲❁◆✾✹❃✏❏✹ã ✢✪➊✆☎ ￾➓✥✩✻➝❆➑P→✚➍✦✧✂✁✩✔➏◗➓✧➧➣★➏☎✄✖➑⑥➣★➟✝✆✏➔❵→✚➝✩✔➍❭➏◗➑P➔✟✞ ➔ ✌ ✍✏✎✒✑✔✓✖✕ ✱ Numerical Results with increasing n n = 10 n = 20 n = 40 Eigenvalues accumulating at zero. Ô➥Õ✮ÖØ×ÚÙ✒￾ ➦③ß✴❏Pt★✾✿●❥♣✤✾✹❇r❈❋❏P●■❇✟☎➙❁⑤❣➐❣✐❁❄❃★❧■❍❆▼❄❈■❇✹Ü❋●❥❃★✾✪❀q❁◆❧❥t❥❏➐✾❲å◗♣✴✾✹❱✠❏➐❏P●✉❣P✾❲✾Þ❏Pt✤❈✦❏ê❇r✾✞✝❆✾✜❱Ø❏P✾✜❅➃❁❄❃❂❏Pt★✾✿✾✹❁◆❧❥✾❲❃✰à ♦✦❈❋▼❄❍★✾✹❣❯●❋ß✔❈➃❀✇❈✦❏P❇r❁◆åq❧❥✾❲❃★✾✹❇r❈❋❏P✾✜❅q❊✏⑦✇❅✰❁⑤❣r❱✠❇r✾✠❏P❁✕❑❲❁❄❃★❧ ☎ã❭➦➧❃✆♣❆❈■❇✐❏r❁❄❱❲❍★▼❄❈■❇✹Ü❥●■❃★✾✬Ý❯●❥❍★▼⑤❅✇✾✠å✰♣✴✾✹❱Ø❏ ❏Pt❆✾❵❀✇❈✦❏r❇P❁◆åq❏P●➃t✤❈②♦■✾❉✾✹❁◆❧❥✾❲❃◗♦✦❈❋▼❄❍★✾✹❣ê❏Pt❆❈❋❏Þ❈❋❇r✾❵❃❆✾✹❈❋❇✱❑❲✾✹❇P●✤ã✞➦➧❃✇❏rt★✾▲❈❋❊✴●✦♦■✾❵❣P▼◆❁⑤❅✰✾❥Ü■❏rt★✾✬✾✹❁◆❧❥✾❲❃✰à ♦✦❈❋▼❄❍★✾✹❣❯●❋ß✞❀q❈❋❏P❇r❁❄❱❲✾✹❣❯❧■✾✹❃★✾❲❇❤❈✦❏r✾✹❅✇❊◗⑦❦❅✰❁⑤❣P❱❲❇P✾❲❏P❁✕❑❲❁❄❃★❧ ☎ ß➯●■❇Þ❏Pt❆✾ ë à③♥➶♣★❇r●■❊★▼❄✾❲❀✄❈■❇P✾✬♣★▼◆●■❏✐❏r✾✹❅✮ã ♥✬❁❄❣r❱✠❇r✾✠❏r❁✧❑✹❁◆❃★❧✆❍❆❣✐❁❄❃★❧❍✳❉✵✇❣✐❍❆❊★❁◆❃✏❏r✾❲❇r♦②❈■▼❄❣❵❧■✾✹❃★✾❲❇❤❈✦❏r✾✹❣❉❈✇❀✇❈❋❏P❇r❁éå❪Ý✿❁é❏rt ë✏✎ ✾✹❁◆❧❥✾❲❃◗♦✦❈❋▼❄❍★✾✹❣❉♣★▼◆●■❏✐à ❏P✾✜❅✆❁❄❃❪❊★▼❄❍★✾■ãêä✪t★✾▲❊★▼❄❍★✾▲✾❲❁❄❧■✾✹❃◗♦②❈■▼◆❍❆✾▲❱❲▼◆●✏❣✐✾✜❣⑥❏Þ❏P●✗❑✹✾❲❇r●❹❁⑤❣✍✤✭✵ ✓ ✵ ✳■ã Û❣✪❏rt★✾❼❅✰❁⑤❣P❱❲❇P✾❲❏P❁✕❑✹❈✦❏r❁◆●❥❃ ë ❴