Problem 157 - Calculus Volume 3

Question

The equation for heat flow in the \(x y\) -plane is \(\frac{\partial f}{\partial t}=\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}} . \quad\) Show that \(f(x, y, t)=e^{-2 t} \sin x \sin y\) is a solution.

Step-by-Step Solution

Answer

Yes, the function satisfies the heat equation in the given form.

Step 1: Verify the Partial Derivative with Respect to Time

Start by computing the partial derivative of the function \( f(x, y, t) = e^{-2t} \sin x \sin y \) with respect to \( t \). This derivative is given by:\[\frac{\partial f}{\partial t} = \frac{\partial}{\partial t}\left(e^{-2t} \sin x \sin y\right)\]Applying the derivative, we get:\[\frac{\partial f}{\partial t} = -2e^{-2t} \sin x \sin y\]

Step 2: Compute the Partial Derivative with Respect to x

Next, compute the partial derivative of \( f(x, y, t) \) with respect to \( x \):\[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(e^{-2t} \sin x \sin y\right)\]This is:\[\frac{\partial f}{\partial x} = e^{-2t} \cos x \sin y\]Now, take the second partial derivative with respect to \( x \):\[\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} \left(e^{-2t} \cos x \sin y\right)\]Which simplifies to:\[\frac{\partial^2 f}{\partial x^2} = -e^{-2t} \sin x \sin y\]

Step 3: Compute the Partial Derivative with Respect to y

Compute the partial derivative of \( f(x, y, t) \) with respect to \( y \):\[\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\left(e^{-2t} \sin x \sin y\right)\]This is:\[\frac{\partial f}{\partial y} = e^{-2t} \sin x \cos y\]Now, compute the second partial derivative with respect to \( y \):\[\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} \left(e^{-2t} \sin x \cos y\right)\]Which simplifies to:\[\frac{\partial^2 f}{\partial y^2} = -e^{-2t} \sin x \sin y\]

Step 4: Check Heat Equation Condition

Combine the second partial derivatives to verify the heat equation:\[\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = -e^{-2t} \sin x \sin y + (-e^{-2t} \sin x \sin y)\]Simplify:\[\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = -2e^{-2t} \sin x \sin y\]Compare this with the earlier result from Step 1. Since\[\frac{\partial f}{\partial t} = -2e^{-2t} \sin x \sin y\]the condition of the heat equation \( \frac{\partial f}{\partial t} = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} \) is satisfied.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental aspect of calculus, especially in multivariable functions. Like normal derivatives, which provide the rate of change of a function with respect to one variable, partial derivatives do this in the context when you have functions depending on multiple variables. You focus on one variable at a time.
If you have a function \( f(x, y, t) \), it means the function depends on three variables. The partial derivative with respect to \( x \) is written as \( \frac{\partial f}{\partial x} \), calculated by treating \( y \) and \( t \) as constants. Similarly, you can compute \( \frac{\partial f}{\partial y} \) and \( \frac{\partial f}{\partial t} \).

• To find \( \frac{\partial f}{\partial x} \), it involves differentiating the function with respect to \( x \) while keeping \( y \) and \( t \) constant.
• \( \frac{\partial f}{\partial y} \) is found in the same way, but differentiated concerning \( y \).
• \( \frac{\partial f}{\partial t} \) targets how the function changes over time with \( x \) and \( y \) held steady.

Taking the second partial derivatives, \( \frac{\partial^2 f}{\partial x^2} \) or \( \frac{\partial^2 f}{\partial y^2} \), involves differentiating again with respect to the same variable. This helps understand the curvature of the surface defined by the function.

Heat Flow

Heat flow describes how thermal energy moves from one region of a system to another. It's an essential concept in physics and engineering, helping us understand how heat distributes over time.
The heat equation, like \( \frac{\partial f}{\partial t} = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} \), is a type of partial differential equation modeling how heat diffuses.

• In the equation, \( \frac{\partial f}{\partial t} \) indicates how temperature changes over time.
• The sum \( \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} \) models the spatial distribution of temperature.

The heat equation is powerful because it captures the essence of diffusion processes. The function \( f(x, y, t) = e^{-2t} \sin x \sin y \) in the original exercise is proposed as a solution, representing a specific scenario where heat smoothly changes over time in the \( x-y \) plane, decaying exponentially due to the \( e^{-2t} \) factor.

Differential Equations

Differential equations involve equations that contain derivatives, showing how a quantity changes with respect to another. When solving differential equations, you often find functions that satisfy these dynamic relationships.
Partial differential equations (PDEs) involve multiple variables. The heat equation from the exercise is a perfect example, with both spatial and time derivatives. They're crucial in understanding real-world phenomena.

• Solving a PDE involves finding a function that satisfies the given equation, adhering to certain conditions.
• The solution \( f(x, y, t) = e^{-2t} \sin x \sin y \) satisfies the heat equation by ensuring both sides of the equation equal as steps show.

Differential equations can be quite complex but breaking them down step by step and understanding each part like the right derivatives and setup can make them easier to handle.