Fix boundary point in Laplace in a strip tests (#927)

Author: jishnub

examples/NonlinearBVP1.jl

@@ -3,18 +3,18 @@ using LinearAlgebra
 
 # Define the vector that collates the differential equation and
 # the boundary conditions
-N(u, x = Fun()) = [u(-1.)-1., u(1.)+0.5, 0.001u'' + 6(1-x^2)u' + u^2 - 1];
+N1(u, x = Fun()) = [u(-1.)-1., u(1.)+0.5, 0.001u'' + 6(1-x^2)u' + u^2 - 1];
 
 # Solve the equation using Newton iteration
 function nbvpsolver()
     x = Fun()
     u0 = 0 * x # starting value
 
-    newton(N, u0)
+    newton(N1, u0)
 end
 
 u = nbvpsolver();
 
 #src # We verify that the solution satisfies the differential equation and the boundary conditions
 using Test #src
-@test norm(N(u)) ≤ 1000eps() #src
+@test norm(N1(u)) ≤ 1000eps() #src

examples/NonlinearBVP2.jl

@@ -4,7 +4,7 @@
 using ApproxFun
 using LinearAlgebra
 
-N(u1, u2) = [u1'(0) - 0.5*u1(0)*u2(0);
+N2(u1, u2) = [u1'(0) - 0.5*u1(0)*u2(0);
                 u2'(0) + 1;
                 u1(1) - 1;
                 u2(1) - 1;
@@ -15,6 +15,6 @@ function nbvpsolver2()
     x = Fun(0..1)
     u10 = one(x)
     u20 = one(x)
-    newton(N, [u10,u20])
+    newton(N2, [u10,u20])
 end
 u1,u2 = nbvpsolver2();

test/runtests.jl

@@ -93,7 +93,7 @@ end
     b = rangespace(A)
 
     @test Fun(component(v[1],1), component(b[1],1))(0.1,-1.0) ≈ v(0.1,-1.0)[1]
-    @test Fun(component(v[1],2), component(b[1],2))(0.1,-1.0) ≈ v(0.1,-1.0)[1]
+    @test Fun(component(v[1],2), component(b[1],2))(0.1,1.0) ≈ v(0.1,1.0)[1]
     @test ApproxFun.default_Fun(v[1] , b[1])(0.1,1.0) ≈ v(0.1,1.0)[1]
 end