Consider a dilute solution of glucose in water at 25 oC. The solution was in a horizontal tube of length 10 cm, at was prepared such that there was a linear gradation of glucose concentration from cglucose(0, 0) = 0.1M at x = 0 cm to cglucose(0, 10cm) = 0.05M at x = 10 cm. (A) What is the force (and direction) acting at the center of the tube (x = 5 cm)? Derive a formula for cglucose(0, x) for x รขยย [0, 10]cm. (B) We wish to solve the time time dependence of cglucose(t, x) for all times t > 0. The next steps of this question will lead you through how to solve such a problem. Consider Fickรขยยs second law of diffusion in one dimension, รขยยc รขยยt = D รขยย 2 c รขยยx2 (2) Show that a function of the form c(t, x) = exp(รยปt) รย (Aeikx + Beรขยยikx) satisfies Eq. (2) when รยป is related to k and D. Determine รยป for such a solution. (C) For the problem like that considered in (A) for particles confined to a a region of length L where x รขยย (0, L) (in (A), L = 10 cm, for example) The flux at the end of the containers is equal to zero (no particles leave or enter the container). Mathematically, this imposes a boundary condition that J(x รขยย 0รขยย) = J(x รขยย L+) = 0 (x รขยย 0รขยย means you approach x = 0 from the left and x รขยย L+ means you approach x = L from the right). Show that these boundary conditions imply that A = B and k รขยยก kn = nรย L for n = 0, 1, 2, ...รขยย. That is, show that solutions look like 2Aeรยปnt cos nรยx L (where we now relate รยปn to kn). (D) Spatially normalize the solutions. Letรขยยs denote the solutions to Eq. (2) with the above boundary conditions as cn(t, x) = Aรย ne รยปnt cos nรยx d for n = 0, 1, 2, ...รขยย. We wish to verify that at t = 0, these solutions are orthonormal. What the means is we want to show: Z L 0 dxcn(0, x)cm(0, x) = 0 (3) for n 6= m and Z L 0 dxc0(0, x)c0(0, x) = L(Aรย 0) 2 = 1 Z L 0 dxcn6=0(0, x)cn6=0(0, x) = L 2 (Aรย n) 2 = 1 (4) 3 (Hint: it may be easier to verify the above integrals using the relationships that cos(รยธ) = e iรยธ+eรขยยiรยธ 2 and sin(รยธ) = e iรยธรขยยeรขยยiรยธ 2i ). From the above orthonormality conditions, determine what An for n = 0, 1, 2, ...รขยย. (E) Superposition principle. Verify that c(t, x) = Pรขยย n=0 bncn(t, x) satisfies Eq. (2) and also the boundary condition that รขยยc(t,x) รขยยx xรขยย0รขยย = รขยยc(t,x) รขยยx xรขยยL+ = 0. The coefficients, bn are simple numbers telling you how much of cn(t, x) is part of your solution for c(t, x). (F) Imposing the initial condition at t = 0. Suppose that at t = 0, c(0, x) = F +Gx for x รขยย (0, L) and c(0, x) = 0 for x รขยยค 0 and x รขยยฅ L, i.e., there is a linear gradation to the concentration within the sample region. By the superposition principle, we can write c(t, x) as c(t, x) = X n=0 bncn(t, x) (5) At t = 0 we must have: c(0, x) = F + Gx = X n=0 bncn(0, x) (6) Using orthonormality conditions you derived above, show that bn = Z L 0 dxcn(0, x)(F + Gx) (7) Explicitly determine formulas for b0 and bn6=0. Using these expressions, finally write out your solution to c(t, x). (G) Finally, letรขยยs go back to (A) and determine how the glucose concentration changes with time. The diffusion coefficient for glucose in water is D = 5.7 รย 10รขยย10 m2 s . Numerically calculate cglucose(t, x) for x รขยย (0, L) at t = 0.5 s, t = 1s, t = 10s, and t = 1000 s. You will need to approximate your formulas for c(t, x) by taking only a finite number of terms.