Question

$$\text{ If }u=f(x-ct)+g(x+ct),\text{ show that }\frac{{\partial }^{2}u}{\partial x^2}=\frac{1}{c^2}\frac{{\partial }^{2}u}{\partial t^2}$$.

Short answer

We showed that $\frac{{\text{d}}^{2}u}{{\text{dx}}^2}=\frac{1}{c^2}\frac{{\text{d}}^{2}u}{{\text{dt}}^2}$.

Step-by-step solution

Understanding the given function

The function given is a combination of two functions, where: $u=f(x-ct)+g(x+ct)$. This represents the solution as two wave functions traveling in opposite directions.

First partial derivative with respect to x

Differentiate both terms of the function with respect to x: $\frac{\text{d}}{\text{dx}}[f(x-ct)]=f^{\prime }(x-ct)$, $\frac{\text{d}}{\text{dx}}[g(x+ct)]=g^{\prime }(x+ct)$. So, $\frac{\text{d}u}{\text{dx}}=\frac{\text{d}}{\text{dx}}[f(x-ct)+g(x+ct)]=f^{\prime }(x-ct)+g^{\prime }(x+ct)$.

Second partial derivative with respect to x

Differentiate the first derivative with respect to x again: $\frac{{\text{d}}^{2}u}{{\text{dx}}^2}=\frac{\text{d}}{\text{dx}}[f^{\prime }(x-ct)+g^{\prime }(x+ct)]=f^{″}(x-ct)+g^{″}(x+ct)$.

First partial derivative with respect to t

Differentiate the given function with respect to t: $\frac{\text{d}}{\text{dt}}[f(x-ct)]=-c{f}^{\prime }(x-ct)$, $\frac{\text{d}}{\text{dt}}[g(x+ct)]=c{g}^{\prime }(x+ct)$. So, $\frac{\text{d}u}{\text{dt}}=\frac{\text{d}}{\text{dt}}[f(x-ct)+g(x+ct)]=-c{f}^{\prime }(x-ct)+c{g}^{\prime }(x+ct)$.

Second partial derivative with respect to t

Differentiate the first derivative with respect to t again: $\frac{{\text{d}}^{2}u}{{\text{dt}}^2}=\frac{\text{d}}{\text{dt}}[-c{f}^{\prime }(x-ct)+c{g}^{\prime }(x+ct)]=c^2{f}^{″}(x-ct)+c^2{g}^{″}(x+ct)$.

Relate the second derivatives of x and t

Compare the second derivatives obtained: $\frac{{\text{d}}^{2}u}{{\text{dx}}^2}=f^{″}(x-ct)+g^{″}(x+ct)$, $\frac{{\text{d}}^{2}u}{{\text{dt}}^2}=c^2[f^{″}(x-ct)+g^{″}(x+ct)]$. So, $\frac{{\text{d}}^{2}u}{{\text{dx}}^2}=\frac{1}{c^2}\frac{{\text{d}}^{2}u}{{\text{dt}}^2}$.

Key concepts

Partial Derivatives

Partial derivatives are a fundamental tool in calculus, especially when dealing with functions of multiple variables. Unlike ordinary derivatives, which are taken with respect to a single variable in a single-variable function, partial derivatives are taken with respect to one variable while keeping other variables constant. This is particularly useful in the context of wave equations, where a function might depend on both space and time.

For example, consider the function given in the exercise: $u=f(x-ct)+g(x+ct)$. When we take the partial derivative of $u$ with respect to $x$, we treat $t$ as a constant, and vice versa.

• The first partial derivative with respect to $x$ is $\frac{\text{d}u}{\text{dx}}=f^{\prime }(x-ct)+g^{\prime }(x+ct)$.
• The first partial derivative with respect to $t$ is $\frac{\text{d}u}{\text{dt}}=-c{f}^{\prime }(x-ct)+c{g}^{\prime }(x+ct)$.

Understanding these partial derivatives is crucial for tackling more complex topics like second-order differential equations in the context of wave functions.

Second-Order Differential Equations

Second-order differential equations involve second derivatives and are common in physics and engineering, especially when describing wave phenomena. In the given exercise, you're dealing with a linear homogeneous second-order differential equation.

The general form of the wave equation involves both spatial and temporal variables, and it requires second partial derivatives:

• The second partial derivative with respect to $x$ is $\frac{{\text{d}}^{2}u}{{\text{dx}}^2}=f^{″}(x-ct)+g^{″}(x+ct)$.
• The second partial derivative with respect to $t$ is $\frac{{\text{d}}^{2}u}{{\text{dt}}^2}=c^2{f}^{″}(x-ct)+c^2{g}^{″}(x+ct)$.

By comparing these second partial derivatives, you relate them in the form: $\frac{{\text{d}}^{2}u}{{\text{dx}}^2}=\frac{1}{c^2}\frac{{\text{d}}^{2}u}{{\text{dt}}^2}$. Recognizing and solving second-order differential equations is an essential skill for understanding wave behavior and other physical phenomena.

Wave Functions

Wave functions are used to describe waves and oscillations in various physical contexts, from sound waves to light waves to quantum mechanical probability waves. The exercise deals with a particular form of solution to the wave equation, given by the superposition of two wave functions: $u=f(x-ct)+g(x+ct)$.

These functions represent waves traveling in opposite directions.

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