Problem 3 - Calculus for Biology and Medicine

Question

A solution of $$ \frac{\partial c(x, t)}{\partial t}=D \frac{\partial^{2} c(x, t)}{\partial x^{2}} $$ is the function $$ c(x, t)=\frac{1}{\sqrt{4 \pi D t}} \exp \left[-\frac{x^{2}}{4 D t}\right] $$ for \(x \in \mathbf{R}\) and \(t>0\). (a) Show that, as a function of \(x\) for fixed values of \(t>0, c(x, t)\) is (i) positive for all \(x \in \mathbf{R}\), (ii) is increasing for \(x<0\) and decreasing for \(x>0\), (iii) has a local maximum at \(x=0\), and (iv) has inflection points at \(x=\pm \sqrt{2 D t}\). (b) Graph \(c(x, t)\) as a function of \(x\) when \(D=1\) for \(t=0.01\), \(t=0.1\), and \(t=1\)

Step-by-Step Solution

Answer

(a) The function is positive, increases for \(x<0\), decreases for \(x>0\), peaks at \(x=0\), and inflects at \(x=\pm\sqrt{2Dt}\). (b) See symmetrical bell curves centered at \(x=0\) with inflection points at \(x=\pm\sqrt{2Dt}\) for given \(t\).

Step 1: Analyze the Given Function

The function given is:\[ c(x, t) = \frac{1}{\sqrt{4 \pi D t}} \exp \left[-\frac{x^2}{4 D t}\right] \]This is a Gaussian function centered at 0, scaled by time \(t\). We will analyze it for a fixed \(t > 0\) to show the requested properties.

Step 2: Show Positivity of the Function

Since \( \exp \left[-\frac{x^2}{4 D t}\right] > 0 \) for all \(x\), \(c(x, t)\) is positive for any real \(x\) and \(t > 0\). The other terms \(1/\sqrt{4\pi Dt}\) are also positive, so \(c(x, t) > 0\) holds for all \(x \in \mathbb{R}\).

Step 3: Find the Derivative with Respect to \(x\)

To find where the function is increasing or decreasing, compute the derivative:\[ \frac{\partial c(x, t)}{\partial x} = c(x, t) \left( -\frac{x}{2Dt} \right) \]This comes from differentiating the exponential part.

Step 4: Analyze the Derivative for Increasing/Decreasing Behavior

Since \( \frac{\partial c(x, t)}{\partial x} = -\frac{x}{2Dt} c(x, t) \) with \(c(x, t) > 0\):- For \(x < 0\), \(\frac{\partial c}{\partial x} > 0\) (increasing).- For \(x > 0\), \(\frac{\partial c}{\partial x} < 0\) (decreasing).- For \(x = 0\), \(\frac{\partial c}{\partial x} = 0\), indicating a maximum at \(x = 0\).

Step 5: Analyze the Second Derivative for Inflection Points

To find inflection points, compute the second derivative:\[ \frac{\partial^2 c(x, t)}{\partial x^2} = c(x, t) \left( \frac{x^2 - 2Dt}{4D^2t^2} \right) \]Set this to zero to find inflection points:\[ x^2 - 2Dt = 0 \Rightarrow x = \pm\sqrt{2Dt} \]

Step 6: Graph for Given Parameters

Given parameters \(D=1\) and \(t = 0.01, 0.1, 1\):- For \(t = 0.01\), function is narrow and peaks sharply.- For \(t = 0.1\), function is wider.- For \(t = 1\), the curve is much wider with the same qualitative features.Graphs would be symmetrical bell curves centered at \(x=0\), peaking at \(x=0\), and having inflection points shifting with \(t\) at \(x = \pm\sqrt{2Dt}\).

Key Concepts

Gaussian Function

The Gaussian function is often encountered in probability and statistics as part of the normal distribution, but in this exercise, it applies to a physical context. The key thing to notice about it is its characteristic bell shape. It's symmetric about the vertical axis and centered at zero. The function's mathematical form is given by \[ c(x, t) = \frac{1}{\sqrt{4 \pi D t}} \exp \left[-\frac{x^2}{4 D t}\right] \].Here, the term \( \exp \left[-\frac{x^2}{4 D t}\right] \) shows that the exponential decreases rapidly as \( x \) moves away from zero, giving the bell shape. The parameter \( t \) in the formula acts as a scaling factor. As \( t \) changes, the spread of the Gaussian bell changes, making the function either wider or narrower depending on the specific value of \( t \). The constant factor \( \frac{1}{\sqrt{4 \pi D t}} \) ensures that the area under the curve remains constant as \( t \) varies. This preservation of the total area is important because it relates to the probability aspect of Gaussian functions, although here we're using it in a physical diffusion context.

Inflection Points

Inflection points occur in a function where the concavity changes. For the Gaussian function in this task, the concavity transition reveals significant information about where the rate of change is pivoting. To find these points, you compute the second derivative of the function with respect to \( x \). Here's the second derivative:\[ \frac{\partial^2 c(x, t)}{\partial x^2} = c(x, t) \left( \frac{x^2 - 2Dt}{4D^2t^2} \right) \].By setting the second derivative equal to zero, \[ x^2 - 2Dt = 0 \Rightarrow x^2 = 2Dt \],we find the inflection points at \( x = \pm\sqrt{2Dt} \).These points fundamentally outline where this bell shape transitions in curvature. From mathematical and graphical points of view, they help to illustrate how the Gaussian shape moves and changes with different values of \( t \), essentially guiding where it starts opening up wider.

Derivative Analysis

The derivative of a function provides insight into its behaviour, revealing where it is increasing or decreasing. For the given Gaussian function \( c(x, t) \), the first derivative with respect to \( x \) is:\[ \frac{\partial c(x, t)}{\partial x} = c(x, t) \left( -\frac{x}{2Dt} \right) \].This derivative tells us about the slopes on either side of zero. Each term in the product is positive except for \(-\frac{x}{2Dt} \), so the sign of the derivative overall is dictated by \( x \):

• For \( x < 0 \), the derivative is positive, which denotes that the Gaussian function is increasing.
• For \( x > 0 \), it is negative, showing that the function is decreasing.
• At \( x = 0 \), the derivative equals zero, indicating a local maximum.

Understanding where the function grows and shrinks helps to frame the characteristic bell shape of the Gaussian and supports applications in various fields such as physics and engineering.

Time-dependent Solutions

The time dependence in the Gaussian function \( c(x, t) = \frac{1}{\sqrt{4 \pi D t}} \exp \left[-\frac{x^2}{4 D t}\right] \) influences how quickly the function spreads out over space as time progresses. For this equation, \( t \) not only scales the function but changes its shape over time. Let’s break down what happens:- As \( t \) increases, the denominator in the exponential part becomes larger, which effectively flattens and spreads the curve.
- Conversely, for small \( t \), the Gaussian is sharp and narrow.
- In physical diffusion contexts, the time growth represents how much diffusion has occurred, with the peak height decreasing as the spread increases.The parameter \( D \), which is the diffusion coefficient, also plays a crucial role in this.

• A larger \( D \) results in quicker spreading and vice versa.

Through observing time-dependent solutions, one can visually grasp how the process evolves over time, which is poignant in diffusion problems or heat equations in physics. Studying different \( t \) outcomes as done in the exercise shows the dynamic behaviour of solutions, allowing one to predict future states of the system.