Question

In a study of heat transfer, we find that for a solid rod. there is a relationship between the second derivative of the temperature with respect to position along the rod and the first with respect to time. (A linear temperature change with position would imply as much heat flowing into a region as out. so the temperature there would not change with time.) $$ \frac{\partial^{2} T(x, t)}{\partial x^{2}}=b \frac{\partial T(x, t)}{\partial t} $$ (a) Separate variables. That is, assume a solution that is a product of a function of \(x\) and a function of \(1 .\) plug it in. then divide by it. Obtain two ordinary differential equations. (b) Consider a fairly simple, if somewlat unrealistic. case. Suppose the temperature is 0 at \(x=0\) and \(x=L\), and positive in betwaen. Write down the simplest function of.r that (1) fits these conditions and \((2)\) obey s the differential equation involving x. Does your choice determine the value. including sign. of some constant? (c) Obtain the full \(T(x, t)\) for this case.

Short answer

The general solution to the problem is \(T(x, t) = (A\sin(n\pi x/L))Ce^{-bn^2\pi^2t/L^2}\), where \(n\) is an integer, \(A\) and \(C\) are constants, and \(b\), \(L\) are given constants.

Step-by-step solution

Separation of Variables: Step 1

Assume a solution that is a product of a function of \(x\) and a function of \(t\). Let \(T(x, t) = X(x)T(t)\).

Separation of Variables: Step 2

Substitute this into the given equation: \(X''(x)T(t) = bX(x)T'(t)\).

Separation of Variables: Step 3

Now, divide by \(X(x)T(t)\) to get: \(\frac{X''(x)}{X(x)} = b \frac{T'(t)}{T(t)}\).

Obtain two ordinary differential equations

Since the left side is a function of \(x\) and the right side is a function of \(t\), both sides must be equal to a constant. Let this constant be -\(k^2\). Therefore, we get two ordinary differential equations: \(X'' + k^2X = 0\) and \(T' + bk^2T = 0\).

Determine the simplest function that fits the conditions

From the first ordinary differential equation, it can be seen that a solution of the form \(X(x) = A\sin(kx) + B\cos(kx)\) will satisfy it. With the given boundary conditions that the temperature is \(0\) at \(x=0\) and \(x=L\), we can solve for the constants \(A\) and \(B\). From \(X(0) = 0\), we get \(B = 0\). And from \(X(L) = 0\), we get \(k = n\pi /L\).

Obtain the full \(T(x, t)\) for this case

Substitute \(k = n\pi /L\) into the second ordinary differential equation gives \(T'(t) + b(n \pi/L)^2T = 0\), whose solution is of the form \(T(t) = Ce^{-bn^2\pi^2t/L^2}\). Therefore, the general solution is \(T(x, t) = (A\sin(n\pi x/L))Ce^{-bn^2\pi^2t/L^2}\).

Key concepts

Separation of Variables

When dealing with the complex world of heat transfer in solid rods, the separation of variables technique provides a silver bullet. It’s a mathematical method that simplifies the intricate dance of temperature variations over time and through space. By assuming the temperature, represented by the symbol \( T(x, t) \), can be broken down into separate functions of space (\( X(x) \)) and time (\( T(t) \)), we can transform a daunting partial differential equation (PDE) into more manageable, individual ordinary differential equations (ODEs).

Recall our original PDE: \[ \frac{\partial^{2} T(x, t)}{\partial x^{2}}=b \frac{\partial T(x, t)}{\partial t} \]. Through the magic of separation of variables, we express \( T(x, t) \) as the product of \( X(x) \) and \( T(t) \), and voilà!—the PDE no longer seems so intimidating. Instead, we now face two ODEs that are closer to terra firma where tried and true methods can lead to a solution.

Partial Differential Equations

Partial differential equations (PDEs) are like the weather patterns of mathematics—complicated, influenced by multiple factors, and challenging to predict. Specifically, in the context of heat transfer, PDEs describe how heat dissipates or accumulates across various points in a rod and across time. The initial daunting equation is our forecast model: \[ \frac{\partial^{2} T(x, t)}{\partial x^{2}}=b \frac{\partial T(x, t)}{\partial t} \], where \( T(x, t) \) represents the temperature at a point \( x \) along the rod at time \( t \).

The equation relates the second spatial derivative (a measure of concavity in the temperature profile) and the first time derivative (a measure of how quickly the temperature changes over time). To tackle this PDE, we use separation of variables, giving us a way to forecast temperature changes like meteorologists do with the weather, using separate models for different atmospheric layers.

Boundary Conditions

Boundary conditions are the stage directions for a play— they tell the characters, namely the temperature in our scenario, where to start and where to end their performance. In the case of heat transfer within a rod, these conditions stipulate what the temperature must be at the edges (boundaries), and they can dramatically affect the entire solution within the domain.

For our rod, it's dictated that the temperature is zero at both ends: \( T(0, t) = 0 \) and \( T(L, t) = 0 \). In our simplification process, the boundary conditions help us pinpoint the exact functions to use. In this case, we determined that \( X(x) = A\sin(kx) + B\cos(kx) \) must satisfy both our ODE and the condition that at \( x=0 \) and \( x=L \), the temperature is zero. This leads to the revelation that \( B = 0 \) and \( k = n\pi /L \), shaping the choreography our temperature function must follow through space.

Ordinary Differential Equations

Ordinary differential equations (ODEs) are like the building blocks of the mathematical universe. They describe the relationship between a function and its derivatives—relatively simpler compared to PDEs. After using separation of variables on our PDE, we're left with two ODEs to solve: \[ X'' + k^2X = 0 \] and \[ T' + bk^2T = 0 \].

These equations show a clearer path; one tells us how the shape of the temperature distribution (\( X(x) \)) changes with position along the rod, while the other reveals how temperature at a point (\( T(t) \)) evolves over time. We handle each ODE with familiar techniques from calculus and apply our boundary conditions to uncover the full temperature story, represented by the harmonious merging of space and time in our final solution \( T(x, t) \). The end result—\( T(x, t) = (A\sin(n\pi x/L))Ce^{-bn^2\pi^2t/L^2} \)—is not just an equation but the elegant narrative of heat traveling through the rod.