Problem 157 - Calculus 3

Question

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

Step-by-Step Solution

Answer

Yes, \(f(x, y, t)=e^{-2t} \sin x \sin y\) is a solution to the heat equation.

Step 1: Compute the Time Derivative

First, we need to find the partial derivative of the function \(f(x, y, t) = e^{-2t} \sin x \sin y\) with respect to time \(t\). This is given by:\[\frac{\partial f}{\partial t} = \frac{\partial}{\partial t}(e^{-2t} \sin x \sin y) = -2e^{-2t} \sin x \sin y.\]

Step 2: Compute the First Spatial Derivative

Next, compute the partial derivative of \(f(x, y, t)\) with respect to \(x\):\[\frac{\partial f}{\partial x} = e^{-2t} \cos x \sin y.\]

Step 3: Compute the Second Spatial Derivative with respect to x

Now, compute the second derivative of \(f(x, y, t)\) with respect to \(x\):\[\frac{\partial^2 f}{\partial x^2} = e^{-2t}(-\sin x)\sin y = -e^{-2t} \sin x \sin y.\]

Step 4: Compute the First Spatial Derivative with respect to y

Next, compute the partial derivative of \(f(x, y, t)\) with respect to \(y\):\[\frac{\partial f}{\partial y} = e^{-2t} \sin x \cos y.\]

Step 5: Compute the Second Spatial Derivative with respect to y

Now, compute the second derivative of \(f(x, y, t)\) with respect to \(y\):\[\frac{\partial^2 f}{\partial y^2} = e^{-2t} \sin x (-\sin y) = -e^{-2t} \sin x \sin y.\]

Step 6: Verify the Heat Equation

Substitute the computed derivatives into the heat flow equation \[\frac{\partial f}{\partial t} = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}\]: \[-2e^{-2t} \sin x \sin y = -e^{-2t} \sin x \sin y - e^{-2t} \sin x \sin y\] which simplifies to: \[-2e^{-2t} \sin x \sin y = -2e^{-2t} \sin x \sin y.\] This confirms that both sides are equal, so \(f(x, y, t)\) satisfies the heat equation.

Key Concepts

Partial Derivatives

Understanding partial derivatives is crucial when working with multivariable functions like the one presented in the heat equation. Partial derivatives involve differentiating a function with respect to one variable while keeping the others constant. This is like a focused zooming-in technique to see how the function changes as you tweak just one variable.
In our problem, the function we have is given by:
\[f(x, y, t) = e^{-2t} \sin x \sin y\]
To find the partial derivative with respect to time \(t\), we adjust only \(t\) and treat \(x\) and \(y\) as constants. Doing so, we find:
\[\frac{\partial f}{\partial t} = -2e^{-2t} \sin x \sin y\]
Then, we do the same with spatial variables \(x\) and \(y\), resulting in first and second partial derivatives respectively. The computations show us how each variable individually impacts the entire function, which is crucial for solving partial differential equations like the heat equation.

Heat Flow

The heat flow describes how the temperature distribution changes over time in a given space, which in this case is the \(xy\)-plane. The heat equation used here is a type of partial differential equation that models this phenomenon. It ensures that the rate of change in temperature is proportional to the curvature or the diffusion of heat across the plane.
For the given function,

• The change over time is captured by \(\frac{\partial f}{\partial t}\).
• The diffusion or spread of heat along the \(x\)-direction is represented by \(\frac{\partial^2 f}{\partial x^2}\).
• The diffusion along the \(y\)-direction is captured by \(\frac{\partial^2 f}{\partial y^2}\).

By understanding these components, we can see how an initial temperature configuration evolves. This model can also represent various other materials and conditions beyond just temperature, making it incredibly versatile.

Solution Verification

The solution verification is the process where you ensure the proposed solution actually satisfies the given equation under certain conditions. Here, we verify that:
\[f(x, y, t) = e^{-2t} \sin x \sin y\]
is indeed a valid solution to the heat equation. Verification involves plugging the partial derivatives back into the heat equation:
\[\frac{\partial f}{\partial t} = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}\]
After calculating, both sides of the equation need to match. With our derivations:

• \(-2e^{-2t} \sin x \sin y\) matches \(-e^{-2t} \sin x \sin y - e^{-2t} \sin x \sin y\)

This equality confirms that the function appropriately represents a heat flow according to the defined initial conditions, which is why solution verification is an essential step in solving differential equations.