Problem 5 - Calculus for Biology and Medicine

Question

A chemical diffuses in a container that occupies the interval \(0 \leq x \leq 1\). The concentration of the chemical at time \(t\) and at a point \(x\) is given by the diffusion equation: $$ \frac{\partial c}{\partial t}=D \frac{\partial^{2} c}{\partial x^{2}} $$ (a) Suppose that the chemical is allowed to diffuse through the entire container until the concentration reaches an equilibrium value where \(c\) does not change any more with time, that is, \(\partial c / \partial t=0 .\) Suppose that chemical that touches the walls of the container is removed so that $$ c(0, t)=c(1, t)=0 . $$ The steady state concentration of chemical will be a function \(C(x)\) with $$ 0=D \frac{d^{2} C}{d x^{2}} \text { for } x \in(0,1) $$ and \(C(0)=C(1)=0\). Show that \(C(x)=0\) satisfies this differential equation and the constraints as the points \(x=0\) and \(x=1\). (b) Now suppose that chemical is added to the container by a reaction that occurs at the wall \(x=0 .\) This reaction keeps the concentration of chemical at this wall equal to \(c(0, t)=1 . \mathrm{Un}\) der these conditions the steady state distribution of chemical will obey a differential equation: $$ 0=D \frac{d^{2} C}{d x^{2}} \text { for } x \in(0,1) $$ with \(C(0)=1\) and \(C(1)=0 .\) Show that \(C(x)=1-x\) satisfies both the differential equation and the boundary conditions at \(x=0\) and \(x=1\). (c) Notice that the steady state distributions in (a) and (b) do not depend on \(D .\) Can you explain why?

Step-by-Step Solution

Answer

(a) \(C(x)=0\); (b) \(C(x)=1-x\); (c) Solutions are independent of \(D\) because \(D\) is a multiplicative constant.

Step 1: Understanding the steady-state condition

At equilibrium, the time-dependent part of the concentration equation vanishes, leading to \( \frac{\partial c}{\partial t} = 0 \). This simplifies the diffusion equation to \( 0 = D \frac{d^{2} C}{d x^{2}} \). Hence, the equation becomes \( \frac{d^{2} C}{d x^{2}} = 0 \). The problem now involves solving this second-order differential equation.

Step 2: Solving the differential equation for part (a)

For part (a), we have \( \frac{d^{2} C}{d x^{2}} = 0 \) subject to boundary conditions \( C(0) = 0 \) and \( C(1) = 0 \). Integrating the differential equation twice yields the general solution \( C(x) = Ax + B \). Applying the boundary conditions, we find that \( B = 0 \) and since \( C(1) = 0 = A \cdot 1 + 0 \), it follows \( A = 0 \). Thus, \( C(x) = 0 \).

Step 3: Checking solution for part (a)

The function \( C(x) = 0 \) satisfies the boundary conditions \( C(0) = 0 \) and \( C(1) = 0 \) and the differential equation \( \frac{d^{2} C}{d x^{2}} = 0 \) holds because the second derivative of a constant is zero.

Step 4: Solving the differential equation for part (b)

For part (b), we solve \( \frac{d^{2} C}{d x^{2}} = 0 \) with \( C(0) = 1 \) and \( C(1) = 0 \). Starting from\( C(x) = Ax + B \), the boundary condition \( C(0) = 1 \) gives \( B = 1 \). The boundary condition \( C(1) = 0 \) gives us \( A + 1 = 0 \), so \( A = -1 \). Therefore, \( C(x) = 1 - x \).

Step 5: Checking solution for part (b)

The function \( C(x) = 1 - x \) satisfies \( C(0) = 1 \) and \( C(1) = 0 \) as required by the boundary conditions. Additionally, \( \frac{d^{2} C}{d x^{2}} = 0 \) is satisfied because the second derivative of a linear function is zero.

Step 6: Explaining independence from D

The solutions in parts (a) and (b) do not depend on the diffusion constant \( D \) because it is a multiplyer in the differential equation \( 0 = D \frac{d^{2} C}{d x^{2}} \). Setting the second derivative to zero eliminates \( D \) from the equation, making the solution independent of its value.

Key Concepts

Steady State

In the context of differential equations, particularly the diffusion equation, a steady state is a condition where the system's variables, such as concentration in our case, are no longer changing over time. This occurs when the rate of change (time-derivative) of the concentration becomes zero, i.e., \( \frac{\partial c}{\partial t} = 0 \). This simplification leads to the equation \( 0 = D \frac{d^2 C}{d x^2} \), indicating that the second spatial derivative must equal zero because the diffusion coefficient \( D \) is assumed to be non-zero.

At steady state, the challenge transitions into solving the associated stationary differential equation, where equilibrium conditions or constant values are achieved. For instance, in our exercise, after applying the steady-state condition, we are left with \( \frac{d^2 C}{d x^2} = 0 \), where \( C(x) \) is the steady-state concentration. It's crucial to note that in the steady state, any dynamic changes in concentration have balanced out, leaving a stable, unchanging distribution.

Boundary Conditions

Boundary conditions are essential constraints that a solution to a differential equation must satisfy at the edges of the domain. In physical systems, these conditions represent real-world limits or interactions. They dictate how the solution behaves at specific points, such as the walls of a container in diffusion problems.

In our problem:

• Part (a) specifies \( C(0) = 0 \) and \( C(1) = 0 \), implying that the chemical is absorbed at the container walls, resulting in zero concentration at both ends.
• Part (b) alters this by setting \( C(0) = 1 \) and \( C(1) = 0 \). Here, a reaction at \( x = 0 \) maintains a constant concentration of 1, while the wall at \( x = 1 \) still absorbs all the chemical.

These conditions are pivotal in determining the form of the solution, ensuring it is consistent with the physical behaviour of the system. By setting specific concentrations at the boundary, we're able to uniquely define the steady-state distribution of the chemical within the container.

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are instrumental in modeling various natural phenomena, including the diffusion process we are examining. In this context, the diffusion equation \( \frac{\partial c}{\partial t} = D \frac{\partial^2 c}{\partial x^2} \) governs how the concentration \( c \) changes over time and space.

In steady-state analysis, we focus on the simplified form \( \frac{d^2 C}{d x^2} = 0 \), derived from setting the time derivative to zero. This second-order differential equation tells us that the second derivative of the concentration \( C(x) \) with respect to \( x \) should be zero, indicating no curvature or linearity in the solution.

To solve such an equation, we typically integrate twice, deriving a linear solution: \( C(x) = Ax + B \). The constants \( A \) and \( B \) are determined via boundary conditions, offering a complete solution specific to the system's parameters. Understanding differential equations is crucial because they provide the mathematical framework underlying many scientific problems, enabling us to predict and explain real-world behaviors.