Exercise 10.5.7

The following nonlinear BVP was proposed by Carrier:

\[ \epsilon u'' + 2(1-x^2)u + u^2 = 1, \quad u(-1) = u(1) = 0. \]

In order to balance the different components of the residual, it’s best to implement the boundary condition numerically as \(u/\epsilon = 0\).

Use bvp to solve the problem with \(\epsilon = 0.003, n=200\), and an initial estimate of all zeros. Plot the result; you should get a solution with 9 local maxima.
Starting with result of (a) continue the parameter sequence of \(\epsilon\) using most recent solution as initialization for the next value. Plot end result for \(\epsilon = 0.3\).

domain = [-1,1];
epsilon = 0.003;
phi = (x,u,du) -> (1 - u^2 - 2*(1-x^2)*u)/epsilon
g1(u,du) = 0;
g2(u,du) = 0;
init = zeros(201);
x,u1 = FNC.bvp(phi,domain,g1,g2,init);

plot(x,u1,xaxis=(L"x"),yaxis=(L"u(x)"),
     title="Solution of Carrier problem")

## (b)
u_ee = init; #trick to access u_ee globally

for ee in -3:0.2:-1
  epsilon = 3*10^ee
  x,u_ee = FNC.bvp(phi,domain,g1,g2,u1)
end

plot(x,u_ee,xaxis=(L"x"),yaxis=(L"u(x)"),
   title="Solution of Carrier problem with eps = 0.3")

## (c) now do b in reverse
init = zeros(201);

for ee in -1.0:-0.2:-3
  epsilon = 3*10^ee
  x,u_x = FNC.bvp(phi,domain,g1,g2,init)
  init = u_x
end

plot(x,u_x,xaxis=(L"x"),yaxis=(L"u(x)"),
   title="Solution of Carrier problem with eps = 0.003")