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2. Two-Point Boundary Value Problems
        40
        our attention to numerical methods for solving this simple problem, and we
        will carefully study how well the numerical solutions mimic the properties
        of the exact solutions. Finally, we will study eigenvalue problems associated
        with the boundary value problem (2.1). The results of this analysis will be
        a fundamental tool in later chapters.
          Although the equations investigated in this chapter are very simple and
        allow analytical solution formulas, we find it appropriate to start our study
        of numerical methods by considering these problems. Clearly, numerical
        values of the solutions of these problems could have been generated without
        the brute force of finite difference schemes. However, as we will encounter
        more complicated equations later on, it will be useful to have a feeling for
        how finite difference methods handle the very simplest equations.
        2.1 Poisson’s Equation in One Dimension
        In this section we will show that the problem (2.1) has a unique solution.
        Moreover, we will find a representation formula for this solution.
          We start by recalling a fundamental theorem of calculus: There is a
        constant c 1 such that
                                            x
                              u(x)= c 1 +   u (y) dy,                (2.2)

                                          0
        and similarly, there is a constant c 2 such that
                                            y


                              u (y)= c 2 +  u (z) dz.                (2.3)
                                          0
        This is true for any twice continuously differentiable function u. Suppose
        now that u satisfies the differential equation (2.1). Then (2.3) implies that
                                            y

                              u (y)= c 2 −   f(z) dz.                (2.4)
                                          0
        Then, inserting this equation into (2.2), we obtain

                                          x       y
                       u(x)= c 1 + c 2 x −      f(z) dz dy.          (2.5)
                                        0    0
        In order to rewrite this expression in a more convenient form, we define

                                          y
                                F(y)=     f(z) dz,
                                        0
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