Question

The ends of a thin, laterally insulated bar of length 2 are maintained at a temperature of zero degrees. At time \(t=0\), the temperature profile is $$ u(x, 0)=100 \sin \left(\frac{\pi x}{2}\right) \text {. } $$ At time \(t=1\), the temperature at the center of the bar has decreased to a value of 70 ; that is, \(u(1,1)=70\). (a) What is the thermal diffusivity \(\kappa\) of the bar? (b) What is \(u_{x}(2,1)\) ?

Short answer

Answer: (a) The thermal diffusivity (𝜅) is given by: $$\kappa = -\frac{1}{\pi^2} \ln{\left(\frac{7}{10}\right)}$$ (b) The spatial derivative of the temperature at x=2 and t=1, 4u_x(2,1) is given by: $$u_x(2,1) = -100 e^{-\ln{\left(\frac{7}{10}\right)}} \frac{\pi}{2}$$

Step-by-step solution

Write down the heat equation

The heat equation for an insulated bar in one spatial dimension and time is: $$ \frac{\partial u}{\partial t} = \kappa \frac{\partial^2 u}{\partial x^2} $$ where \(u(x,t)\) is the temperature distribution in the bar, \(\kappa\) is the thermal diffusivity, and \(t\) and \(x\) represent time and position along the bar.

Write the general solution to the heat equation

For this problem, we observe that the given initial temperature profile has a sine function which suggests that we can use separation of variables to obtain the general solution. The solution will take the form: $$ u(x,t) = \sum_{n=1}^\infty A_n e^{-\kappa n^2 \pi^2 t} \sin{\left(\frac{n\pi x}{2}\right)} $$ To determine the coefficients \(A_n\), we apply the initial condition \(u(x,0) = 100 \sin{\left(\frac{\pi x}{2}\right)}\).

Apply the initial condition to find the coefficients \(A_n\)

Plugging the initial condition into the general solution: $$ 100 \sin{\left(\frac{\pi x}{2}\right)} = \sum_{n=1}^\infty A_n e^{-\kappa n^2 \pi^2 \cdot 0} \sin{\left(\frac{n\pi x}{2}\right)} $$ Since \(e^0 = 1\), this simplifies to: $$ 100 \sin{\left(\frac{\pi x}{2}\right)} = \sum_{n=1}^\infty A_n \sin{\left(\frac{n\pi x}{2}\right)} $$ It's clear that only \(A_1\) is nonzero, and \(A_1 = 100\). So the solution becomes: $$ u(x,t) = 100 e^{-\kappa \pi^2 t} \sin{\left(\frac{\pi x}{2}\right)} $$

Determine the thermal diffusivity \(\kappa\) using the condition \(u(1,1) = 70\)

Plugging in the given values, we have: $$ 70 = 100 e^{-\kappa \pi^2 \cdot 1} \sin{\left(\frac{\pi}{2}\right)} $$ Simplifying and solving for \(\kappa\): $$ \frac{7}{10} = e^{-\kappa \pi^2} $$ Taking the natural logarithm of both sides: $$ \ln{\left(\frac{7}{10}\right)} = -\kappa \pi^2 $$ Then, we get the value of \(\kappa\): $$ \kappa = -\frac{1}{\pi^2} \ln{\left(\frac{7}{10}\right)} $$

Evaluate \(u_x(2,1)\)

To find \(u_x\), we need to take the derivative of \(u(x,t)\) with respect to \(x\): $$ u_x(x,t) = 100 e^{-\kappa \pi^2 t} \frac{\pi}{2} \cos{\left(\frac{\pi x}{2}\right)} $$ Now, we can plug in the given values to find \(u_x(2,1)\): $$ u_x(2,1) = 100 e^{-\left(-\frac{1}{\pi^2} \ln{\left(\frac{7}{10}\right)}\right) \pi^2\cdot 1} \frac{\pi}{2} \cos{\left(\frac{\pi \cdot 2}{2}\right)} $$ This simplifies to: $$ u_x(2,1) = 100 e^{-\ln{\left(\frac{7}{10}\right)}} \frac{\pi}{2} \cos{(\pi)} $$ Finally, we use the property that \(\cos{(\pi)}=-1\) and obtain the result: $$ u_x(2,1) = -100 e^{-\ln{\left(\frac{7}{10}\right)}} \frac{\pi}{2} $$ So, the expressions for the thermal diffusivity \(\kappa\) and \(u_x(2,1)\) are: (a) \(\kappa = -\frac{1}{\pi^2} \ln{\left(\frac{7}{10}\right)}\) (b) \(u_x(2,1) = -100 e^{-\ln{\left(\frac{7}{10}\right)}} \frac{\pi}{2}\)

Key concepts

Thermal Diffusivity

Thermal diffusivity, denoted as \( \kappa \), is a property that measures a material's ability to conduct heat relative to its ability to store heat. It tells us how quickly heat spreads through a material. The heat equation, \( \frac{\partial u}{\partial t} = \kappa \frac{\partial^2 u}{\partial x^2} \), incorporates \( \kappa \) to relate changes in temperature \( u(x,t) \) over time \( t \) to changes in temperature over the spatial dimension \( x \).
- A higher \( \kappa \) means heat propagates more quickly through the material.
- Conversely, a lower \( \kappa \) implies slower heat conduction.
In context, we found the thermal diffusivity \( \kappa \) for the insulated bar by applying boundary conditions and initial conditions. Given \( u(1,1) = 70 \) and an initial temperature profile expressed as a sine function, \( \kappa \) was determined to be \( -\frac{1}{\pi^2} \ln{\left(\frac{7}{10}\right)} \). This value characterizes how heat dissipates through this specific bar.

Separation of Variables

The method of separation of variables is a mathematical technique used to solve partial differential equations, like the heat equation. This method breaks down a complex equation into simpler, separate parts, simplifying the process of finding a solution. In our context, the solution to the heat equation was expressed as
\[ u(x,t) = \sum_{n=1}^\infty A_n e^{-\kappa n^2 \pi^2 t} \sin{\left(\frac{n\pi x}{2}\right)} \]
Separation of variables allowed us to separate \( u(x,t) \) into functions of \( x \) and \( t \), effectively simplifying the problem by dealing with one variable at a time.
- First, identify sinusoidal functions in the initial term, which indicates potential use of sine and cosine functions.- Use these in the solution to address spatial and temporal changes separately.
This technique is particularly useful when both initial values and boundary conditions are harmonic, as is evidenced by the sine function present in the initial condition of this problem.

Initial Temperature Profile

The initial temperature profile describes the temperature distribution along the bar at time \( t = 0 \). It is expressed as \( u(x, 0) = 100 \sin \left(\frac{\pi x}{2}\right) \), a sine function that indicates how temperature varies spatially but not yet temporally.
- This sine profile indicates a peak temperature at the center of the bar, with zero at the ends due to boundary conditions.
- It's essential when using separation of variables to define the coefficients \( A_n \) in the general solution.
Applying this profile as an initial condition allowed us to determine remaining components of the heat distribution solution, giving us insight into how the temperature evolves with time. The profile's sinusoidal nature matched the general solution's form, ensuring continuity from initial conditions to the solution of the heat equation.