Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

a. Show that $$u(x, t)=20 e^{-t} \sin \pi x, \quad 0 \leq x \leq 1, \quad 0 \leq t$$ solves $$u_{t}(x, t)=\frac{1}{\pi^{2}} u_{x x}(x, t), \quad u(x, 0)=20 \sin \pi x, \quad \text { and } \quad u(0, t)=u(1, t)=0$$ b. Describe a physical problem for which this is a solution. c. What is the 'eventual' value of \(u(x, t)\) (what is \(\left.\lim _{t \rightarrow \infty} u(x, t)\right) ?\) d. At what time, \(t,\) will the maximum value of \(u(x, t)\) for \(0 \leq x \leq 1\) be \(20 ?\)

Step-by-Step Solution

Answer

a. PDE & conditions are satisfied. b. Decaying heat in a rod. c. 0. d. At \(t=0\).

Step 1: Verify Initial Condition

Check if the initial condition is satisfied by substituting \(t = 0\) into \(u(x, t)\). Compute:\[u(x, 0) = 20 e^0 \sin(\pi x) = 20 \sin(\pi x)\]This matches the initial condition \(u(x, 0) = 20 \sin \pi x\).

Step 2: Calculate Spatial Derivatives

Find the second spatial derivative of \(u(x, t)\):\[\frac{\partial u}{\partial x} = 20 e^{-t} \pi \cos(\pi x)\]Taking the second derivative:\[\frac{\partial^2 u}{\partial x^2} = -20 e^{-t} \pi^2 \sin(\pi x)\]Substitute into the PDE \(u_{t}(x, t)=\frac{1}{\pi^{2}} u_{x x}(x, t)\):\[-20 e^{-t} = \frac{1}{\pi^2} (-20 e^{-t} \pi^2 \sin(\pi x)) = -20 e^{-t}\sin(\pi x)\]These terms agree, confirming the PDE is satisfied.

Step 3: Verify Boundary Conditions

Check the boundary conditions \(u(0, t) = 0\) and \(u(1, t) = 0\):For \(x=0\), \(u(0, t) = 20 e^{-t} \sin(0)= 0\).For \(x=1\), \(u(1, t) = 20 e^{-t} \sin(\pi)= 0\).Both boundary conditions are satisfied.

Step 4: Describe Physical Problem

The solution represents the temperature distribution in a one-dimensional rod of length 1, with insulated ends, initially having a temperature distribution given by \(20 \sin \pi x\). Over time, the temperature exponentially decays due to heat loss.

Step 5: Find Eventual Temperature

Calculate the limit as \(t \rightarrow \infty\):\[\lim_{t \rightarrow \infty} u(x, t) = \lim_{t \rightarrow \infty} 20 e^{-t} \sin(\pi x) = 0\]Therefore, the eventual temperature distribution is zero.

Step 6: Determine Time for Maximum Value

Find the time when the maximum value of \(u(x, t)\) is 20:The maximum value of \(\sin(\pi x)\) is at \(\sin(\frac{\pi}{2}) = 1\), therefore:\[20 e^{-t} \times 1 = 20\]This implies \(e^{-t} = 1\), thus \(t = 0\). The maximum value is only achieved at \(t = 0\).

Key Concepts

Initial and Boundary Conditions

In the context of solving partial differential equations (PDEs), initial and boundary conditions are necessary to uniquely determine a solution. They act like guidelines or rules that the solution must adhere to at specific points or boundaries. For the heat equation example given, we have both initial and boundary conditions to consider.

**Initial Conditions** refer to the state of the system at the beginning of observation, often time zero. In this exercise, the initial condition is specified as:

• \( u(x, 0) = 20 \sin(\pi x) \)

This indicates the temperature distribution along a rod at time \( t = 0 \), forming a sine wave shape.**Boundary Conditions** ensure that the solution behaves appropriately at the edges of the domain. They can be specified at any time. For our equation, the boundary conditions are:

• \( u(0, t) = 0 \)
• \( u(1, t) = 0 \)

These conditions describe a rod where both ends are kept at zero temperature all the time, possibly representing the rod being in contact with a wall at each end.

These conditions must be checked to verify a function solves the PDE properly, essentially 'anchoring' the solution within a defined domain.

Separation of Variables

Separation of variables is a widely used mathematical technique for solving partial differential equations, especially when initial and boundary conditions are provided. It involves decomposing a PDE into simpler, solvable ordinary differential equations (ODEs).

The essence of this method lies in assuming that the solution can be represented as the product of functions, each depending on a single variable. For the heat equation, the assumption might take the form of:

• \( u(x, t) = X(x)T(t) \)

where \( X(x) \) is a function of space and \( T(t) \) is a function of time.

This assumption allows us to split the original PDE into two ODEs:

• One for the spatial component, which will involve the function \( X(x) \)
• Another for the temporal component, involving the function \( T(t) \)

Each of these ODEs can be solved independently, showing how the spatial and temporal behaviors of the solution are separate yet interconnected. By applying initial and boundary conditions to these ODEs, we can find particular solutions that, when multiplied, satisfy the original PDE under its constraints.

Heat Equation

The heat equation is a foundational concept in the study of differential equations and mathematical physics. It is a type of partial differential equation that describes how heat diffuses through a given region over time.

In one dimension, the heat equation is given by:

• \( u_t(x, t) = \frac{1}{\pi^2} u_{xx}(x, t) \)

Here, \( u(x,t) \) denotes the temperature at position \( x \) and time \( t \). The equation effectively states that the rate of change of temperature with respect to time is proportional to the curvature (second derivative) of the temperature profile along the spatial dimension.

**Physical Interpretation**The heat equation is invaluable for modeling thermal behavior. It captures how temperature gradients lead to heat flow from hotter regions to cooler ones, illustrating a diffusion process. In practical terms, the given solution represents temperature diminishing over time, corresponding to the phenomenon of heat dissipating naturally.

**Application to the Given Problem**The provided function \( u(x, t) = 20 e^{-t} \sin \pi x \) is a specific solution to the given heat equation under particular initial and boundary conditions. It mirrors a physical scenario where a rod experiences a quick temperature decline across its length, eventually stabilizing as time progresses and the system reaches thermal equilibrium.