Question

In the bar of Problem 15 suppose that \(L=30, \alpha^{2}=1,\) and the initial temperature distribution is \(f(x)=30-x\) for \(0

Short answer

Question: Given a one-dimensional bar with length \(L=30\), a heat diffusivity constant \(\alpha^2=1\), and an initial temperature distribution \(f(x)=30-x\), find the temperature \(u(x,t)\) at any position \(x\) for any time \(t\) by solving the heat equation. Then, investigate the characteristics of the solution by plotting it in various ways. Solution: After solving the heat equation with the given initial conditions, we find the explicit form of the solution as: $$u(x,t)=\left(30+c_3x\right)T_0(t)+\sum_{n=1}^{\infty}\left[c_6\cos{\left(\frac{n\pi x}{L}\right)}+c_7\sin{\left(\frac{n\pi x}{L}\right)}\right]T_n(t)$$ with the calculated coefficients \(c_3 ,c_6 ,\text{ and } c_7\). To investigate the characteristics of the solution, plot \(u(x,t)\) for different values of time \(t\) and position \(x\), and find the maximum temperature and warmest point of the bar over time.

Step-by-step solution

Separate Variables

Assume a solution to the heat equation of the form \(u(x,t)=X(x)T(t)\). Substitute this expression into the heat equation. Then, divide both sides by \(XT\): $$\frac{1}{T}\frac{dT}{dt}=\frac{\alpha^2}{X}\frac{d^2X}{dx^2}$$ Since the left side depends only on time and the right side depends only on position, both sides must be equal to a constant, which we will denote as \(k\). $$\frac{\alpha^2}{X}\frac{d^2X}{dx^2}=k \qquad\text{and}\qquad \frac{1}{T}\frac{dT}{dt}=k$$

Solve the Time-Dependent and Space-Dependent Equations

Separate these equations and solve them independently. For the time equation, we have: $$\frac{dT}{dt}=kT$$ This is a simple first-order differential equation, and the solution is: $$T(t)=c_1 e^{kt}$$ For the space equation, we have: $$\frac{d^2X}{dx^2}=\frac{kX}{\alpha^2}$$ This is a second-order homogeneous linear differential equation. Depending on the sign of \(k\), we'll have different results. For \(k=0\), the solution is \(X(x)=c_2+c_3x\). For \(k<0\) (let's denote \(k=-p^2\)), the solution is \(X(x)=c_4 e^{px} + c_5 e^{-px}\). For \(k>0\) (let's denote \(k=q^2\)), the solution is \(X(x)=c_6 \cos{qx}+c_7 \sin{qx}\).

Obtain General Solution Using Fourier Series

The given initial condition suggests that the solution is a combination of a linear function and a periodic function. Therefore, we should collect the \(k=0\) and \(k>0\) solutions, but discard the \(k<0\) solution: $$u(x,t)=\left(c_2+c_3x\right)T_0(t)+\sum_{n=1}^{\infty}\left[c_6\cos{\left(\frac{n\pi x}{L}\right)}+c_7\sin{\left(\frac{n\pi x}{L}\right)}\right]T_n(t)$$ Now we have to fit the initial temperature distribution to the Fourier series coefficients.

Apply Initial Conditions

We know that the initial temperature distribution is \(u(x,0)=f(x)=30-x\). Insert the \(t=0\) into our general solution, we have: $$30-x=c_2+c_3x+c_6+\sum_{n=1}^{\infty}c_7\sin{\left(\frac{n\pi x}{L}\right)}$$ First, to find the coefficients \(c_6\), consider the left-hand side and the first term of the right-hand side for any values of \(x\). Both sides are equal to \(30\), so we get \(c_2+c_6=30\). To find the coefficients \(c_7\), we have to expand the initial condition in a Fourier sine series: $$-x=c_3x-30+\sum_{n=1}^{\infty}c_7\sin{\left(\frac{n\pi x}{L}\right)}$$ We will use the orthogonality properties of sine functions to find \(c_7\). Multiply both sides by \(m\pi x\), integrate from \(0\) to \(L\), and use the orthonormality of the sine functions: $$c_7=\frac{2}{L}\int_0^L\left[-x+c_3x-30\right]\sin{\left(\frac{m\pi x}{L}\right)}dx$$ After finding \(c_7\), we can determine the full explicit form of the solution.

Plot the Derived Solution for Different Values of \(t\)

With the derived solution \(u(x,t)\), plot it for different values of time \(t\) to investigate how temperature evolves over time. Similarly, plot \(u(x,t)\) for different values of position \(x\) to see how the temperature changes at different points in the bar. Draw the graph of \(x_m\) (the location of the warmest point in the bar) as the function of time \(t\). Calculate the maximum temperature in the bar by finding the maximum value of the solution and then plot it against time \(t\). To complete the exercise, follow these plotting tasks using the explicit form of the solution derived in the previous steps. Use a software or graphing calculator to generate accurate plots.

Key concepts

Partial Differential Equations

A Partial Differential Equation (PDE) is a mathematical equation that involves multiple independent variables, an unknown function dependent on these variables, and partial derivatives of the unknown function. In the context of the heat equation, PDEs are used to model the distribution of heat (or temperature) in a given region over time.

• PDEs generalize to multiple variables, unlike ordinary differential equations, which involve only one.
• The heat equation, a common PDE, is given by: \[ \frac{\partial u}{\partial t} = \alpha^2 \frac{\partial^2 u}{\partial x^2} \]where \( u(x, t) \) represents the temperature at position \( x \) and time \( t \), and \( \alpha^2 \) relates to the material's properties.

This type of PDE arises frequently in physics and engineering, such as when determining how a thin rod's temperature changes over time when one end is heated. Solving a PDE generally requires both initial conditions (temperature at time zero) and boundary conditions (temperature at the ends of the bar). These allow us to find a unique solution that describes the temperature distribution accurately.

Fourier Series

Fourier Series is a mathematical tool used to express a function as a sum of sinusoidal components. This is particularly useful in solving linear differential equations with periodic inputs, which occur frequently in engineering and physics.

• The basis of Fourier Series is that any periodic function can be expressed as a sum of sines and cosines.
• For the heat equation, Fourier Series expands the initial temperature distribution into a series of sines and cosines, which can be used to construct solutions to the problem.

In the exercise, Fourier principles are employed to create a model that captures the way heat spreads across the bar. By expanding the initial conditions into a Fourier Series, we can better understand the resulting heat distribution as the solution forms by summing the contributions of each of these periodic components over time.

Boundary Value Problem

A Boundary Value Problem (BVP) is where a differential equation is solved under specific conditions or constraints given at the boundaries of the domain. These conditions are essential for finding unique, physical solutions to PDEs.

• In the heat equation context, BVP means knowing the temperature at both ends of the rod over time.
• For example, you might have a rod with known temperatures at each end, determining how temperature changes along the rod over time.

Understanding BVPs means acknowledging the constraints that account for physical realities like fixed temperatures or fluxes at material edges. Solving these problems requires careful integration of these conditions into the mathematical formulation, ensuring that solutions are not only mathematically correct but physically meaningful. In the given exercise, the temperature distribution in the rod is determined by such boundary conditions, ensuring the heat equation solutions accurately describe the system's behavior.