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Chapter 9: Problem 1

For any constants \(k, \omega, c_{1}\), and \(c_{2}\), verify that the function $$ u(x, t)=e^{-k \omega^{2} t}\left(c_{1} \cos \omega x+c_{2} \sin \omega x\right) $$ satisfies the differential equation $$ u_{t}=k u_{x x} $$

Short Answer

Expert verified
The function satisfies the differential equation because the computed partial derivatives confirm the given equation.

Step by step solution

01

Compute the partial derivative of u with respect to t

The function given is \[ u(x, t) = e^{-k \omega^2 t} \left( c_{1} \cos \omega x + c_{2} \sin \omega x \right) \] To find \( u_t \), we need to differentiate \( u(x, t) \) with respect to t. Using the chain rule: \[ u_t = \frac{\partial}{\partial t} \left( e^{-k \omega^2 t} \right) \left( c_{1} \cos \omega x + c_{2} \sin \omega x \right) = -k \omega^2 e^{-k \omega^2 t} \left( c_{1} \cos \omega x + c_{2} \sin \omega x \right) = -k \omega^2 u(x,t) \]
02

Compute the second partial derivative of u with respect to x

Next, we determine \( u_{xx} \). First, find the first partial derivative of u with respect to x: \[ u_{x} = \frac{\partial}{\partial x} \left( e^{-k \omega^2 t} \right) \left( c_{1} \cos \omega x + c_{2} \sin \omega x \right) = e^{-k \omega^2 t} \left( -c_{1} \omega \sin \omega x + c_{2} \omega \cos \omega x \right) \] Now, take the second derivative: \[ u_{xx} = \frac{\partial}{\partial x} \left( e^{-k \omega^2 t} \right) \left( -c_{1} \omega \sin \omega x + c_{2} \omega \cos \omega x \right) = e^{-k \omega^2 t} \left( -c_{1} \omega^2 \cos \omega x - c_{2} \omega^2 \sin \omega x \right) \] \[ u_{xx} = -\omega^2 e^{-k \omega^2 t} \left( c_{1} \cos \omega x + c_{2} \sin \omega x \right) = -\omega^2 u(x,t) \]
03

Verify the differential equation

Now substitute \( u_t \) and \( u_{xx} \) into the differential equation \( u_{t} = k u_{xx} \): We have \( u_t = -k \omega^2 u(x,t) \) and \( u_{xx} = -\omega^2 u(x,t) \). Aggregate: \[ -k \omega^2 u(x,t) = k (-\omega^2 u(x,t)) \] Simplifying: \[ -k \omega^2 u = -k \omega^2 u \] Therefore, the function \( u(x, t) \) satisfies the differential equation \( u_{t} = k u_{xx} \).

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Key Concepts

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

Chain rule
The Chain Rule is an essential tool in calculus. It allows us to compute the derivative of a function that is composed of other functions. For functions involving multiple variables, the chain rule helps in determining partial derivatives. Imagine you have a composite function like the one in this problem: $$u(x, t) = e^{-k \theta^{2} t}\big(c_{1} \text{cos}(\theta x) + c_{2} \text{sin}(\theta x)\big)$$. To find the time derivative \(u_t\), apply the chain rule. Here, the term \(e^{-k \theta^2 t}\) depends on t, while \(c_1 \text{cos}(\theta x) + c_2 \text{sin}(\theta x)\) remains constant with respect to t. So, by chain rule: $$u_t = \frac{\text{d}}{\text{d} t} \big(e^{-k \theta^2 t}\big) \big(c_1 \text{cos}(\theta x) + c_2 \text{sin}(\theta x)\big) = -k \theta^2 e^{-k \theta^2 t} \big(c_1 \text{cos}(\theta x) + c_2 \text{sin}(\theta x)\big) = -k (\theta^2) u(x, t)$$. This shows the power of the chain rule in simplifying and verifying complex functions.
Second partial derivative
A second partial derivative is the partial derivative of a function taken twice with respect to one or more variables. Given $$u(x, t) = e^{-k \theta^2 t} (c_1 \text{cos}(\theta x) + c_2 \text{sin}(\theta x))$$, first find \(u_x\) by differentiating with respect to x: $$u_x = -c_1 \theta e^{-k \theta^2 t} \text{sin}(\theta x) + c_2 \theta e^{-k \theta^2 t} \text{cos}(\theta x)$$. Then, we take the derivative of this result to find the second partial derivative \(u_{xx}\): $$u_{xx} = \frac{\text{d}}{\text{x}} \big(-c_1 \theta e^{-k \theta^2 t} \text{sin}(\theta x) + c_2 \theta e^{-k \theta^2 t} \text{cos}(\theta x)\big) = -c_1 \theta^2 e^{-k \theta^2 t} \text{cos}(\theta x) - c_2 \theta^2 e^{-k \theta^2 t} \text{sin}(\theta x)$$. Simplifying, we see: $$u_{xx} = -\theta^2 u(x, t)$$. This demonstrates how second partial derivatives are approached and why they are crucial for verifying differential equations.
Verification of differential equations
Verification involves proving that a given function satisfies a differential equation. Here, we need to show that the function $$u(x, t) = e^{-k \theta^2 t} (c_1 \text{cos}(\theta x) + c_2 \text{sin}(\theta x))$$ satisfies the equation \(u_t = k u_{xx}\). From our computations: $$u_t = -k (\theta^2) u$$ and $$u_{xx} = -(\theta^2) u$$. To verify, substitute these results into the original differential equation: $$-k \theta^2 u = k (-\theta^2) u$$. Both sides are equal, confirming that our function \(u(x, t)\) indeed satisfies the differential equation \(u_t = k u_{xx}\). This verification step is crucial in ensuring the function is a proper solution to the differential equation provided.

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Most popular questions from this chapter

Verify that the function \(u(x, t)=e^{-t} \cos x\) solves the initial boundary value problem $$ \left\\{\begin{array}{l} u_{t}=u_{x x}, \quad 00 \\ u_{x}(0, t)=u_{x}(\pi, t)=0, \quad t \geq 0 \\ u(x, 0)=\cos x, \quad 0 \leq x \leq \pi \end{array}\right. $$ Note that here the boundary conditions are specified on the values of the spatial derivative. Such boundary conditions are examples of Neumann ones.

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