/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 71 Verify that the function \(u=e^{... [FREE SOLUTION] | 91Ó°ÊÓ

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Verify that the function \(u=e^{-u^{2} k^{2} t} \sin k x\) is a solution of the heat conduction equation \(u_{t}=\alpha^{2} u_{x x}\)

Short Answer

Expert verified
The function satisfies the heat conduction equation, verifying it is a solution.

Step by step solution

01

Identify the Given Function and Equation

The function provided is \( u = e^{-u^2 k^2 t} \sin k x \), and we need to determine if it satisfies the heat conduction equation \( u_{t} = \alpha^2 u_{xx} \). The given parameters include time \( t \), space \( x \), and constants \( k \) and \( \alpha \).
02

Compute the Partial Derivative with Respect to Time

Differentiate the function \( u(x, t) = e^{-u^2 k^2 t} \sin k x \) with respect to \( t \). Using the chain rule, we have:\[ u_t = \frac{\partial}{\partial t} \left(e^{-u^2 k^2 t} \sin k x \right) = -u^2 k^2 e^{-u^2 k^2 t} \sin k x \]
03

Compute the Partial Derivative with Respect to Space Once

Differentiate the function \( u = e^{-u^2 k^2 t} \sin k x \) once with respect to \( x \):\[ u_x = \frac{\partial}{\partial x} \left( e^{-u^2 k^2 t} \sin k x \right) = k e^{-u^2 k^2 t} \cos k x \]
04

Compute the Second Partial Derivative with Respect to Space

Differentiate \( u_x = k e^{-u^2 k^2 t} \cos k x \) again with respect to \( x \):\[ u_{xx} = \frac{\partial}{\partial x} \left( k e^{-u^2 k^2 t} \cos k x \right) = -k^2 e^{-u^2 k^2 t} \sin k x \]
05

Compare Expressions of \( u_t \) and \( \alpha^2 u_{xx} \)

Substitute the expressions for \( u_t \) and \( u_{xx} \) into the heat conduction equation. We need to validate the equation:\[ -u^2 k^2 e^{-u^2 k^2 t} \sin k x = \alpha^2 (-k^2 e^{-u^2 k^2 t} \sin k x) \]Simplifying the right side, we get:\[ -\alpha^2 k^2 e^{-u^2 k^2 t} \sin k x \]Since both sides are equal, the function satisfies the heat equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Differential Equations
Partial Differential Equations (PDEs) are equations that involve rates of change with respect to continuous variables. In physics, they are often used to describe phenomena such as heat, sound, fluid dynamics, elasticity, and quantum mechanics.
For example, the heat conduction equation is a PDE, which in this case is written as:
  • \( u_{t} = \alpha^2 u_{xx} \)
Here, \( u_t \) represents the rate of change of the temperature \( u \) with respect to time \( t \), and \( u_{xx} \) is the second derivative of \( u \) with respect to the spatial variable \( x \).
This equation essentially states that the rate of change of temperature over time at any point in a medium is proportional to the spatial curvature of the temperature at that point. The constant \( \alpha^2 \) is the thermal diffusivity, which dictates how quickly heat spreads through the medium. PDEs can be quite complex, requiring specific mathematical techniques, such as the separation of variables, Fourier series, and others, to find solutions.
Partial Derivatives
Partial derivatives involve the derivative of a function with respect to one variable while keeping the other variables constant. Consider a function \( u(x, t) \), depending on time \( t \) and space \( x \).
To compute the partial derivative of \( u \) with respect to \( t \), denoted as \( u_t \), treat \( x \) as a constant and differentiate the rest.
  • In the given function: \( u(t, x) = e^{-u^2 k^2 t} \sin k x \)
  • The partial derivative with respect to \( t \) is:\[ u_t = -u^2 k^2 e^{-u^2 k^2 t} \sin k x \]
This shows how the function changes as time progresses, assuming the position is fixed.
Similarly, for the partial derivative with respect to \( x \), denoted \( u_x \), treat \( t \) as constant:
  • \( u_x = k e^{-u^2 k^2 t} \cos k x \)
In practice, computing partial derivatives is crucial for understanding how multi-variable functions behave and interact in various dimensions.
Chain Rule
The chain rule is a fundamental tool in calculus used for computing the derivative of a composite function. When dealing with partial derivatives, it tells us how to differentiate functions that are compositions of other functions.
In the context of our problem, the function \( u(t, x) = e^{-u^2 k^2 t} \sin k x \) can be viewed as a composition of simpler functions.
Applying the chain rule to find \( u_t \), recognize that \( e^{-u^2 k^2 t} \) is dependent on \( t \), hence:
  • Find the derivative of the exponent with respect to \( t \): \[ \frac{d}{dt}(-u^2 k^2 t) = -u^2 k^2 \]
  • Multiply this result by the derivative of the exponential part:\[ u_t = -u^2 k^2 e^{-u^2 k^2 t} \sin k x \]
For spatial partial derivatives, the chain rule helps in determining how to differentiate the function elements effectively, ensuring a comprehensive approach in solving complex dynamics as seen with PDEs.
Mastering the chain rule allows for handling diverse scenarios where function compositions appear, making it a reliable method in calculus toolkit.

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