Question

Determine the values of the constant \(\alpha\), if any, for which the specified function is a solution of the given partial differential equation. $$ u(x, t)=e^{-2 \alpha t} \cos \alpha x, \quad u_{t}-u_{x x}=0 $$

Short answer

Answer: The function u(x, t) satisfies the partial differential equation for α = 0 and α = 2.

Step-by-step solution

Compute partial derivatives of u(x,t)

To start, let's compute the first partial derivative with respect to \(t\): $$ u_t = \frac{\partial}{\partial t} (e^{-2 \alpha t} \cos(\alpha x)) $$ Now, apply the chain rule to find \(u_t\): $$ u_t = -2 \alpha e^{-2 \alpha t} \cos(\alpha x) $$ Next, compute the second partial derivative with respect to \(x\): $$ u_{xx} = \frac{\partial^2}{\partial x^2} (e^{-2 \alpha t} \cos(\alpha x)) $$ Applying the chain rule twice, we get: $$ u_{xx} = -\alpha^2 e^{-2 \alpha t} \cos(\alpha x) $$

Substitute partial derivatives into the partial differential equation

Now, substitute \(u_t\) and \(u_{xx}\) into the partial differential equation, \(u_t - u_{xx} = 0\): $$ -2 \alpha e^{-2 \alpha t} \cos(\alpha x) - (-\alpha^2 e^{-2 \alpha t} \cos(\alpha x)) = 0 $$

Simplify the equation

We can now simplify the equation by factoring out the common term \(e^{-2 \alpha t} \cos(\alpha x)\): $$ \left(-2 \alpha + \alpha^2\right) e^{-2 \alpha t} \cos(\alpha x) = 0 $$ For this equation to hold for all \(x\) and \(t\), the term in parentheses must be equal to 0: $$ -2 \alpha + \alpha^2 = 0 $$

Solve the equation for alpha

Finally, we solve the equation for \(\alpha\): $$ \alpha(\alpha - 2) = 0 $$ This gives us two possible solutions for \(\alpha\): $$ \alpha = 0, \quad \alpha = 2 $$ Thus, the specified function \(u(x, t) = e^{-2 \alpha t} \cos(\alpha x)\) is a solution of the given partial differential equation \(u_t - u_{xx} = 0\) for the values \(\alpha = 0\) and \(\alpha = 2\).

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate compositions of functions. It is key when dealing with partial derivatives, especially in the context of partial differential equations like we see in this exercise. Here’s a basic understanding of how it works.

In simple terms, the chain rule allows us to find the derivative of a function based on its composed parts. Suppose you have a function that is a composition of two functions: that is, a function inside another function. In such cases, the chain rule helps determine the derivative of this composite function.

In the original exercise, the function is given as a product of two functions: an exponential function and a trigonometric function. When computing the partial derivatives, applying the chain rule is essential. For example, in computing the partial derivative with respect to time, we keep the spatial variable constant and apply the chain rule to the time-dependent component, yielding \( u_t = -2 \alpha e^{-2 \alpha t} \cos (\alpha x) \). When dealing with the spatial component, the chain rule is applied twice because of the second derivative requirement, resulting in \( u_{xx} = -\alpha^2 e^{-2 \alpha t} \cos (\alpha x) \).

Understanding the application of the chain rule is crucial for solving partial differential equations effectively.

Solution Verification

Verifying a solution in mathematics, particularly in the realm of differential equations, involves checking if your proposed solution satisfies the given equation or system of equations. This is a systematic process to ensure accuracy in your results.

In the exercise, the function was proposed to solve the partial differential equation \( u_t - u_{xx} = 0 \). To verify if this function actually solves this equation, we substitute the derived formulas for \( u_t \) and \( u_{xx} \) from the chain rule application back into the original equation.

Once substituted, we ended up with the equation: \(-2 \alpha e^{-2 \alpha t} \cos (\alpha x) - (-\alpha^2 e^{-2 \alpha t} \cos (\alpha x)) = 0\). Simplifying this yields the requirement that \(-2 \alpha + \alpha^2 = 0 \). This simplification step is where the verification truly happens. If the equality holds true, it confirms that the function is indeed a solution.

In mathematical problem-solving, not just arriving at a function but verifying it as a correct solution is imperative. It aligns the function with the conditions set by the differential equation itself.

Second Order Differential Equations

Second-order differential equations involve the second derivatives of a function and are critical in describing various physical phenomena, such as wave motion, heat conduction, and vibrations. A second-order partial differential equation, like in this exercise, includes terms involving the second derivative with respect to one or more independent variables.

In tackling second-order differential equations, understanding how to compute and manipulate derivatives becomes significant. In our exercise, after applying the chain rule, we obtained the second derivative of the spatial component, \( u_{xx} = -\alpha^2 e^{-2 \alpha t} \cos(\alpha x) \). The equation \( u_t - u_{xx} = 0 \) is a simple but classic example of a second-order differential equation known as the "Telegrapher's Equation" or "Wave Equation" in specific scenarios.

Recognizing and solving these equations form a substantial part of mathematical physics and engineering, providing solutions that describe dynamic systems. Through solving them, you uncover properties like stability, oscillation, and propagation characteristics.

For students, gaining mastery in handling second-order differential equations involves practice with the mathematical tools, like the chain rule, and a clear understanding of how higher-order derivatives influence the behavior of solutions.