{"title":"边值问题导论","authors":"D. Gleich","doi":"10.1142/9789813274037_0001","DOIUrl":null,"url":null,"abstract":"1 # We use a grid representation of the function u(x) on [0, 1] 2 # so we divide [0, 1] up into N + 1 increments of 1/N each. 3 4 N = 10 5 xgrid = collect(0:1/N:1) 6 7 # This means that u is a vector with n = N+1 elements, 8 # but two of them are fixed at alpha and beta. 9 10 n = length(xgrid) 11 12 example_u = zeros(n); 13 example_u[1] = 1 # suppose alpha = 1 14 example_u[end] = 1/2 # suppose beta = 1/2 15 16 example_u[2:end-1] = 1.06-1*(xgrid[2:end-1]-0.25).^2; 17 18 plot(xgrid, example_u)","PeriodicalId":140610,"journal":{"name":"Ordinary Differential Equations and Boundary Value Problems","volume":"14 1","pages":"0"},"PeriodicalIF":0.0000,"publicationDate":"2018-09-18","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":"{\"title\":\"Introduction to Boundary Value Problems\",\"authors\":\"D. Gleich\",\"doi\":\"10.1142/9789813274037_0001\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"1 # We use a grid representation of the function u(x) on [0, 1] 2 # so we divide [0, 1] up into N + 1 increments of 1/N each. 3 4 N = 10 5 xgrid = collect(0:1/N:1) 6 7 # This means that u is a vector with n = N+1 elements, 8 # but two of them are fixed at alpha and beta. 9 10 n = length(xgrid) 11 12 example_u = zeros(n); 13 example_u[1] = 1 # suppose alpha = 1 14 example_u[end] = 1/2 # suppose beta = 1/2 15 16 example_u[2:end-1] = 1.06-1*(xgrid[2:end-1]-0.25).^2; 17 18 plot(xgrid, example_u)\",\"PeriodicalId\":140610,\"journal\":{\"name\":\"Ordinary Differential Equations and Boundary Value Problems\",\"volume\":\"14 1\",\"pages\":\"0\"},\"PeriodicalIF\":0.0000,\"publicationDate\":\"2018-09-18\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"0\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"Ordinary Differential Equations and Boundary Value Problems\",\"FirstCategoryId\":\"1085\",\"ListUrlMain\":\"https://doi.org/10.1142/9789813274037_0001\",\"RegionNum\":0,\"RegionCategory\":null,\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"\",\"JCRName\":\"\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"Ordinary Differential Equations and Boundary Value Problems","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.1142/9789813274037_0001","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
1 # We use a grid representation of the function u(x) on [0, 1] 2 # so we divide [0, 1] up into N + 1 increments of 1/N each. 3 4 N = 10 5 xgrid = collect(0:1/N:1) 6 7 # This means that u is a vector with n = N+1 elements, 8 # but two of them are fixed at alpha and beta. 9 10 n = length(xgrid) 11 12 example_u = zeros(n); 13 example_u[1] = 1 # suppose alpha = 1 14 example_u[end] = 1/2 # suppose beta = 1/2 15 16 example_u[2:end-1] = 1.06-1*(xgrid[2:end-1]-0.25).^2; 17 18 plot(xgrid, example_u)