Question

In each exercise, the solution of a partial differential equation is given. Determine the unspecified coefficient function. \(u_{x}+b(x, t) u_{t}=0 ; \quad u(x, t)=f\left(x^{3}-t\right)\), where \(f(z)\) is continuously differentiable

Short answer

Answer: The function \(b(x, t)\) is \(3x^2\).

Step-by-step solution

Find partial derivatives

We are given that \(u(x, t) = f(x^3 - t)\). To find \(u_x\) and \(u_t\), we need to use the chain rule. For \(u_x\), we have: \(u_x = \frac{\partial u}{\partial x} = \frac{\partial f(x^3 - t)}{\partial x} = f'(x^3 - t) \cdot \frac{\partial (x^3 - t)}{\partial x} = 3x^2 f'(x^3 - t)\). For \(u_t\), we have: \(u_t = \frac{\partial u}{\partial t} = \frac{\partial f(x^3 - t)}{\partial t} = f'(x^3 - t) \cdot \frac{\partial (x^3 - t)}{\partial t} = -f'(x^3 - t)\).

Substitute the partial derivatives into the given partial differential equation

Now, we substitute the values of \(u_x\) and \(u_t\) into the given partial differential equation, which is \(u_x + b(x, t)u_t = 0\). So, we have: \(3x^2 f'(x^3 - t) + b(x, t)(-f'(x^3 - t)) = 0\).

Solve for the function \(b(x, t)\)

To find \(b(x, t)\), we rearrange the equation obtained in step 2: \(b(x,t)f'(x^3-t) = 3x^2 f'(x^3 - t)\). Now, we can divide both sides by \(f'(x^3 - t)\), assuming that \(f'(x^3 - t)\) is not equal to 0: \(b(x, t) = 3x^2\). Thus, the function \(b(x, t)\) is: \(b(x, t) = 3x^2\).

Key concepts

Chain Rule

The Chain Rule is a crucial concept in calculus, particularly useful when dealing with composite functions. It allows us to differentiate a composite function by forming a connection between the derivatives of its inner and outer functions. When you have a function such as \( u(x, t) = f(x^3 - t) \), it is composed of an outer function \( f \) and an inner function \( x^3 - t \).

To apply the Chain Rule, we identify these components and calculate the derivative by multiplying the derivative of the outer function by the derivative of the inner function. In our example exercise, when finding \( u_x \) (the partial derivative with respect to \( x \)), we treat \( x^3 - t \) as \( z \), allowing us to express \( f(x^3 - t) \) as \( f(z) \).

• First, differentiate the outer function: \( f'(z) \).
• Then, differentiate the inner function: \( \frac{d(x^3 - t)}{dx} = 3x^2 \).

By multiplying these derivatives, we get \( u_x = 3x^2 f'(x^3 - t) \). Similarly, when finding \( u_t \), since the function is independent of \( t \) and reducing by \( 1 \) in the inner function derivative, we obtain \( u_t = -f'(x^3 - t) \). This understanding of the Chain Rule is powerful for solving complex problems in calculus and differential equations.

Partial Derivatives

Partial Derivatives extend the concept of a derivative from single-variable calculus to multi-variable functions. They measure how a multivariable function changes as one of its variables changes, while keeping other variables constant. In our exercise, we deal with a function \( u(x, t) = f(x^3 - t) \) that depends on two variables, \( x \) and \( t \).

To compute partial derivatives using the Chain Rule, we do the following:

• For \( \frac{\partial u}{\partial x} \) or \( u_x \), consider \( t \) constant and then differentiate with respect to \( x \).
• For \( \frac{\partial u}{\partial t} \) or \( u_t \), think of \( x \) as constant and differentiate with respect to \( t \).

For \( u = f(x^3 - t) \), by treating \( x^3 - t \) as a single variable \( z \), we can easily differentiate:
\( u_x = 3x^2 f'(x^3 - t) \) shows how \( u \) changes as \( x \) changes.
\( u_t = - f'(x^3 - t) \) shows how \( u \) changes as \( t \) changes.
This technique simplifies the process of resolving partial derivatives and is a fundamental tool in understanding the behavior of multivariable functions.

Coefficient Function

The Coefficient Function plays a significant role in solving partial differential equations (PDEs). It is an unknown function that multiplies another function, affecting the behavior of the system described by the PDE. In our specific exercise, the goal was to determine the coefficient function \( b(x, t) \) in the equation \[ u_x + b(x, t) u_t = 0 \].

Here's how we found the coefficient function:

• First, we substituted the expressions for \( u_x \) and \( u_t \) into the given PDE to form a new equation containing \( b(x, t) \).
• We had \( 3x^2 f'(x^3 - t) + b(x, t)(-f'(x^3 - t)) = 0 \).
• Re-arranging terms, we isolated \( b(x, t) \): \( b(x,t)f'(x^3-t) = 3x^2 f'(x^3 - t) \).

Then, assuming \( f'(x^3 - t) \) is non-zero, dividing both sides by it allowed us to simplify and solve for \( b(x, t) \):\[ b(x, t) = 3x^2 \].
This process highlights how coefficient functions can sometimes depend on only one of the variables of the PDE, connecting solutions to physical phenomena prescribed by the equation.