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Two-point BVP

The initial-value problems of Chapter 6 are characterized by an ordinary differential equation plus a value of the solution’s state at one value of the independent variable (typically, time).

In a boundary-value problem, the state is not entirely given at any point. Instead, partial information is given at multiple values of the independent variable. We will focus on the most common type.

10.1.1Definition

An IVP for the same ODE as in (10.1.1) would specify values for both u(a)u(a) and u(a)u'(a). Although this may look like a minor difference, the characters of IVPs and BVPs are substantially different. In a typical IVP, in which the independent variable is often time, the initial value determines everything about the future course of the solution. In a TPBVP, however, complete information is not given at any one location in the domain, and there may be more than one way (or no way) to satisfy the boundary conditions.

Certain special cases of the boundary conditions have their own nomenclature.

While time can be the independent variable in a TPBVP, as in Example 10.1.1, it is often space, which has no intrinsic direction of information flow.

10.1.2Numerical solution

We can solve the TPBVP (10.1.1) by recasting the problem as a first-order system, as introduced in Section 6.3.2.

Characterizing the conditioning of a TPBVP theoretically is difficult. There are some numerical tools going by the name sensitivity analysis, but the details are too lengthy for us to explore here.

10.1.3Exercises

  1. ✍ In each case, explain whether the TPBVP is linear or nonlinear.

    (a) x2u+xu+(x21)u=0,u(0)=1,  u(4)=0x^2 u'' +xu' + (x^2 - 1) u = 0, \quad u(0) =1,\; u(4) =0 \qquad (Bessel equation)

    (b) uuu=1,u(0)=u(1)=1u'' - u u' = 1, \quad u(0) = u'(1) = 1

    (c) u+u=1,u(0)=u(1)u(1)=1u'' + u = 1, \quad u(0) = u(1) u'(1) = 1

    (d) ϵu+2(1x2)u+u2=1,u(1)=u(1)=0\epsilon u'' +2(1-x^2) u +u^2 = 1, \quad u(-1) = u(1) = 0 \qquad (Carrier equation Carrier (1970))

    (e) uu=3(u)2,u(1)=1,  u(2)=12u u'' = 3(u')^2, \quad u(-1) = 1, \; u(2) = \frac{1}{2}

  1. ✍ For each BVP, verify that the given solution is valid, i.e., check that the differential equation and the boundary conditions are satisfied.

    (a) u2xu+8u=0,u(0)=1,  u(1)=53u'' - 2xu' + 8 u = 0, \quad u(0) = 1, \; u(1) = -\frac{5}{3}\qquad (Hermite equation)

    Solution: u(x)=43x44x2+1u(x) = \frac{4}{3}x^4-4x^2+1

    (b) xu+(1x)u+3u=0,u(0)=1,  u(1)=12xu'' + (1-x) u' + 3 u = 0, \quad u(0) =1, \; u'(1) =-\frac{1}{2} \qquad (Laguerre equation)

    Solution: u(x)=16(618x+9x2x3)u(x) = \frac{1}{6}(6-18x+9x^2-x^3)

    (c) (1x2)uxu+25u=0,u(0)=5,  u(12)=12(1-x^2)u'' - xu' + 25u = 0, \quad u'(0) = 5, \; u\left(\frac{1}{2}\right) = \frac{1}{2} \qquad (Chebyshev equation)

    Solution: u(x)=T5(x)=16x520x3+5xu(x) = T_5(x) = 16x^5-20x^3+5x

    (d) (1x2)u2xu+12u=0,u(0)+2u(0)=3,  u(1)=1(1-x^2) u'' -2xu' + 12 u = 0, \quad u(0)+2u'(0) = -3, \; u(1) = 1 \qquad (Legendre equation)

    Solution: u(x)=P3(x)=12(5x33x)u(x) = P_3(x) = \frac{1}{2}(5x^3-3x)

    (e) uu=3(u)2,u(1)=1,  u(2)=12u u'' = 3(u')^2, \quad u(-1) = 1, \; u(2) = \frac{1}{2}

    Solution: u(x)=(x+2)1/2u(x) = ( x+2 )^{-1/2}

  2. ⌨ For each TPBVP in Exercise 2, use solve/bvp4c/solve_bvp to find the solution, following the method in Example 10.1.3. Plot the solution and separately plot the error as a function of xx. (Important: In some cases, you will need to truncate the domain slightly to avoid division by zero.)

  1. ⌨ Consider the pendulum from Example 10.1.1 with g=L=1g=L=1. Suppose we want to release the pendulum from rest such that θ(5)=π/2\theta(5)=\pi/2. Find one solution that passes through θ=0\theta=0 and another solution that does not. Plot θ(t)\theta(t) for both cases together. (Hint: Vary the initial estimate for the solution.)

  1. ⌨ The stationary Allen–Cahn equation is a model of phase changes, such as the change from liquid to solid. In one spatial dimension it can be written as

    ϵu=u3u,0x1,u(0)=1,u(1)=1.\epsilon u'' = u^3-u, \qquad 0 \le x \le 1, \qquad u(0)=-1, \quad u(1)=1.

    As ϵ0\epsilon\rightarrow 0, the solution tends toward a step function transition between -1 and 1. By symmetry, u(x)=u(1x)u'(x)=-u'(1-x).

    (a) Use Function 10.2.1 with initial solution estimates (u=1,u=0)(u=-1,u'=0) to solve the equation for ϵ=0.2\epsilon=0.2. Plot the solution.

    (b) Repeat part (a) for ϵ=0.02\epsilon=0.02.

    (c) Repeat part (a) for ϵ=0.002\epsilon=0.002. (This is a difficult problem for the default method.) Try a few different initializations for uu, and plot all the results. Do any seem to be valid solutions of the BVP?

References
  1. Pelesko, J. A., & Driscoll, T. A. (2006). The Effect of the Small-Aspect-Ratio Approximation on Canonical Electrostatic MEMS Models. Journal of Engineering Mathematics, 53(3–4), 239–252. 10.1007/s10665-005-9013-2
  2. Carrier, G. F. (1970). Singular Perturbation Theory and Geophysics. SIAM Review, 12(2), 175–193. 10.1137/1012041