Problem 10 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Suppose there is an infinitely long tube containing water lying along the \(X\) -axis from \(-\infty\) to \(\infty\) and at time \(t=0\) a bolus injection of one gram of salt is made at the origin. Let \(u(x, t)\) be the concentration of salt at position \(x\) in the tube at time \(t\). Considering \(t=0\) is a bit of stressful: \(u(x, 0)=0\) for \(x \neq 0 ;\) but the bolus injection of one \(\mathrm{gm}\) at the origin causes the concentration at \(x=0\) and \(t=0\) to be rather large; \(u(0,0)=\infty\). Moving on, for \(t>0\) we may assume that $$ u_{t}(x, t)=k u_{x x}(x, t) $$ where the diffusion coefficient, \(k,\) describes the rate at which salt diffuses in water. a. Show that $$ u(x, t)=\frac{1}{\sqrt{4 \pi k t}} e^{-x^{2} /(4 k t)} $$ is a solution to Equation 13.38 . b. Suppose \(k=1 / 4\). Sketch the graphs of \(u(x, 1), u(x, 4),\) and \(u(x, 8)\). c. Suppose \(k=1 / 4\). Sketch the graphs of \(u(x, 1), u(x, 1 / 2),\) and \(u(x, 1 / 4)\). d. Estimate the areas under the previous curves. For any time, \(t_{0},\) what do you expect to be the area under the curve of \(u\left(x, t_{0}\right), \infty

Step-by-Step Solution

Answer

Yes, the given function satisfies the heat equation and the area under the curve always equals 1 for any fixed time, reflecting the conservation of mass.

Step 1: Verify the Solution

To verify that \(u(x, t)=\frac{1}{\sqrt{4\pi kt}} e^{-\frac{x^2}{4kt}}\) satisfies the heat equation \(u_t = k u_{xx}\), we first need to compute the partial derivatives \(u_t\) and \(u_{xx}\). Start by differentiating \(u\) with respect to \(t\) and \(x\).Differentiating \(u\) with respect to \(t\):\[u_{t}(x, t) = \frac{\partial}{\partial t} \left( \frac{1}{\sqrt{4\pi kt}} e^{-\frac{x^2}{4kt}} \right)\]Using the product and chain rule, you get:\[ u_t(x, t) = \left( -\frac{1}{2t} \right) \frac{1}{\sqrt{4\pi kt}} e^{-\frac{x^2}{4kt}} + \frac{x^2}{4kt^2} \frac{1}{\sqrt{4\pi kt}} e^{-\frac{x^2}{4kt}}\]Differentiating \(u\) with respect to \(x\), twice, we have:\[u_{x}(x, t) = -\frac{x}{2kt} \cdot \frac{1}{\sqrt{4\pi kt}} e^{-\frac{x^2}{4kt}}\]\[u_{xx}(x, t) = \left( \frac{x^2}{4k^2t^2} - \frac{1}{2kt} \right) \frac{1}{\sqrt{4\pi kt}} e^{-\frac{x^2}{4kt}}\]Substitute these into the heat equation to verify that \(u_t - k u_{xx} = 0\) holds. Simplify to check equality.

Step 2: Simplify and Validate the Derivatives

Substitute \(u_t\) and \(k u_{xx}\) back into the equation \(u_t - k u_{xx} = 0\) and simplify:Matching coefficients of \(\frac{1}{\sqrt{4\pi kt}} e^{-x^2/(4kt)}\):\[-\frac{1}{2t} + \frac{x^2}{4kt^2} + k\left( \frac{x^2}{4k^2t^2} - \frac{1}{2kt} \right) = 0\]Resolving gives:\[-\frac{1}{2t} + \frac{x^2}{4kt^2} + \frac{x^2}{4kt^2} - \frac{1}{2t} = 0\]Which simplifies to 0. Thus, \(u(x, t)=\frac{1}{\sqrt{4\pi kt}} e^{-x^{2}/(4kt)}\) satisfies the heat equation.

Step 3: Sketching for Different Times with k = 1/4

Substitute \(k = 1/4\) into the solution \(u(x, t)\). For \(t = 1\):\[u(x, 1) = \frac{1}{\sqrt{\pi}} e^{-x^{2}/1}\]For \(t = 4\):\[u(x, 4) = \frac{1}{2\sqrt{\pi}} e^{-x^{2}/4}\]For \(t = 8\):\[u(x, 8) = \frac{1}{\sqrt{8\pi}} e^{-x^{2}/8}\]Sketch each equation considering these different variances, showing peak height and spread.

Step 4: Altering Time Intervals

Now sketch for different times maintaining \(k = 1/4\), focusing on smaller intervals.For \(t = 1\):\[u(x, 1) = \frac{1}{\sqrt{\pi}} e^{-x^{2}}\]For \(t = 1/2\):\[u(x, 1/2) = \frac{1}{\sqrt{2\pi}} e^{-x^{2}/2}\]For \(t = 1/4\):\[u(x, 1/4) = \frac{1}{2\sqrt{\pi}} e^{-x^{2}/4}\]Again, sketch these functions to show their narrower peaks as \(t\) decreases.

Step 5: Estimating Areas Under Curves

The solution \(u(x, t)\) is a Gaussian function. The integral over all \(x\), \(-\infty < x < \infty\), for any fixed \(t\) is:\[\int_{-\infty}^{\infty} \frac{1}{\sqrt{4\pi kt}} e^{-x^{2}/(4kt)} \, dx = 1\]This holds for any fixed \(t\), reflecting conservation of the mass (or area), ensuring the total area remains 1 gram of salt.

Key Concepts

Heat Equation

The heat equation is a fundamental concept in differential equations, often used to describe the distribution of heat (or variation in temperature) in a given region over time. In our provided problem, it is represented as \[ u_t = k u_{xx} \]This partial differential equation (PDE) illustrates how the temperature changes at every point in space due to both time and space partial derivatives. The term \(u_t\) represents the change in temperature with time, while \(u_{xx}\) indicates how the temperature changes spatially. The constant \(k\) is known as the diffusion coefficient, and it influences the rate at which heat diffuses through a medium, such as water.In the context of the exercise, our focus is on a specific solution to the heat equation that describes how a salt concentration diffuses in an infinitely long tube over time. The goal is to demonstrate that a function \(u(x, t)\) is a solution to this heat equation by computing and verifying its partial derivatives with respect to time and space. This forms a foundational understanding of how heat, or any diffusive process, behaves dynamically over time and space.

Gaussian Function

A Gaussian function, often identified by its bell-shaped curve, is crucial in modeling phenomena across various scientific fields, including statistics and physics. In this problem, the solution \[ u(x, t) = \frac{1}{\sqrt{4\pi kt}} e^{-x^{2}/(4kt)} \]takes the form of a Gaussian function, effectively describing the concentration of salt as it diffuses in the tube. This function is a classic example of a probability distribution function. It features prominently due to its properties of symmetry and all values falling within a certain range under the bell curve.The Gaussian function in this context denotes a change from a point source (where all salt is concentrated at \(x = 0\) and \(t = 0\)) into a distribution of concentration over space and time as diffusion progresses. The parameters \(4\pi kt\) in the denominator adjust the spread of the function, with larger \(kt\) values indicating a wider and shorter distribution, signifying more diffusion over time.

Partial Derivatives

In calculus, understanding partial derivatives is essential for grasping how functions change with respect to multiple variables. A partial derivative of a function of several variables is its derivative with respect to one of those variables, while holding the others constant. In our exercise, verifying that the Gaussian function satisfies the heat equation involves computing partial derivatives with respect to both \(x\) and \(t\).To find \(u_t\), the derivative with respect to time, you apply the chain and product rules. This derivative reflects how the salt concentration changes over time. Similarly, the second derivative with respect to \(x\), \(u_{xx}\), shows how the concentration varies spatially. In the process of verifying the solution, it's crucial to ensure that these derivatives satisfy the original partial differential heat equation. Achieving this through calculation helps confirm that the model accurately represents the physical diffusion process being studied.Partial derivatives are vital tools in differential equations, enabling the description of complex physical systems by dissecting their behavior along each dimension separately.