solve problem #1 depending on the given information Consider the following 1D second order elliptic equation with Dirichlet boundary conditions du(x) (c(x)du ) = f(x) (a $15 b), u(a) = ga, u(b) = gb dr: where u(x) is the unknown function, ga and gb are the Dirichlet boundary values, c(x) is a given coefficient function and f(x) is a given source function. See the theorem 10.1 in the textbook for the existence and uniqueness of the solution. 1.1 Weak Formulation Let's first use the target problem as an example to introduce the weak formulation. First, multiply a test function v(x) and then take the integral fron a to b on both sides of the equation (c(x) du 2) = f(x), we obtain C(x) Here u(z) is called a trial function. Second, using integration by parts, we obtain v dr Hence e(b)u, (b)r(b) _ c(a)u' (a)v(a)-| cu'u, dr = | fu dr. Third, since the Dirichlet boundary conditions u(a) = ga,a(b) = gb directly provide the exact solution at the two ends, we don't need to do any test at a and b. Then we may choose the test function v(x) such that v(a) (b) 0. Therefore, the weak formulation is to find u E Ha, b such that for any v E a, b. Here Ha, b] is the functional space whose functions are first weakly differentiable in [a, b]. And Hola. b] is a functional space whose functions are first weakly differentiable in [a, b] and also vanish at the two ends. Remark: Since the definition of weakly differentiable functions is not introduced in our course, we may use C1a, b] to replace H [a, b] here for simplification. More details about functional spaces will be discussed in Math 6602 (Mathematical Foundation for Finite Element Methods). Here we also recall the following two definitions from the lecture slides of Chapter 1: Let Cla, b denote the function space of continuous functions on a b ; let Ck a de- note the function space of kth differentiable functions on [a.bl that is, if f(x) E Ck[a, b then f(i)(z) (i = 0, 1, . . . , k) are continuous on [a,b The basic idea of the Galerkin method is to use a finite dimensional space to approximate the infinite dimensional space in the weak formulation in order to form a finite system for the computers. For example, H'a, b needs to be approximated by a finite dimensional space V for the above weak formulation. When the space V is chosen to be a finite element space, the Galerkin method is called finite element method Therefore, the finite element methods highly reply on the weak formulation and the corresponding functional spaces, which are varying for different equations with different boundary conditions. In the following let's study and practice on the weak formulations for several fundamental cases Usually we can assume the 0 values of the test function v(x) only at the Dirichlet boundary (a(a) = ga, u(b) = gb), but not at the Neumann boundary (u'(a) Pa, u'(b) = рь) or the Robin boundary (u' (a)+Jau(a) = pa, u, (b)+9ba(b) P>). Instead, the Neumann boundary conditions and the Robin boundary conditions should be imposed in the weak formulation directly and naturally Problem #1: Consider du(z) (a) Derive the weak formulation with the boundary condition u, (a) = pa,a(b) = gb. Consider the following 1D second order elliptic equation with Dirichlet boundary conditions du(x) (c(x)du ) = f(x) (a $15 b), u(a) = ga, u(b) = gb dr: where u(x) is the unknown function, ga and gb are the Dirichlet boundary values, c(x) is a given coefficient function and f(x) is a given source function. See the theorem 10.1 in the textbook for the existence and uniqueness of the solution. 1.1 Weak Formulation Let's first use the target problem as an example to introduce the weak formulation. First, multiply a test function v(x) and then take the integral fron a to b on both sides of the equation (c(x) du 2) = f(x), we obtain C(x) Here u(z) is called a trial function. Second, using integration by parts, we obtain v dr Hence e(b)u, (b)r(b) _ c(a)u' (a)v(a)-| cu'u, dr = | fu dr. Third, since the Dirichlet boundary conditions u(a) = ga,a(b) = gb directly provide the exact solution at the two ends, we don't need to do any test at a and b. Then we may choose the test function v(x) such that v(a) (b) 0. Therefore, the weak formulation is to find u E Ha, b such that for any v E a, b. Here Ha, b] is the functional space whose functions are first weakly differentiable in [a, b]. And Hola. b] is a functional space whose functions are first weakly differentiable in [a, b] and also vanish at the two ends. Remark: Since the definition of weakly differentiable functions is not introduced in our course, we may use C1a, b] to replace H [a, b] here for simplification. More details about functional spaces will be discussed in Math 6602 (Mathematical Foundation for Finite Element Methods). Here we also recall the following two definitions from the lecture slides of Chapter 1: Let Cla, b denote the function space of continuous functions on a b ; let Ck a de- note the function space of kth differentiable functions on [a.bl that is, if f(x) E Ck[a, b then f(i)(z) (i = 0, 1, . . . , k) are continuous on [a,b The basic idea of the Galerkin method is to use a finite dimensional space to approximate the infinite dimensional space in the weak formulation in order to form a finite system for the computers. For example, H'a, b needs to be approximated by a finite dimensional space V for the above weak formulation. When the space V is chosen to be a finite element space, the Galerkin method is called finite element method Therefore, the finite element methods highly reply on the weak formulation and the corresponding functional spaces, which are varying for different equations with different boundary conditions. In the following let's study and practice on the weak formulations for several fundamental cases Usually we can assume the 0 values of the test function v(x) only at the Dirichlet boundary (a(a) = ga, u(b) = gb), but not at the Neumann boundary (u'(a) Pa, u'(b) = рь) or the Robin boundary (u' (a)+Jau(a) = pa, u, (b)+9ba(b) P>). Instead, the Neumann boundary conditions and the Robin boundary conditions should be imposed in the weak formulation directly and naturally Problem #1: Consider du(z) (a) Derive the weak formulation with the boundary condition u, (a) = pa,a(b) = gb.