Question

In each exercise, the solution of a partial differential equation is given. Determine the unspecified coefficient function. \(x u_{x}+b(x, t) u_{t}=0 ; \quad u(x, t)=x e^{-t}\)

Short answer

Question: Determine the unknown coefficient function b(x, t) for the given PDE \(x u_{x}+b(x, t) u_{t}=0\) and its solution \(u(x, t) = x e^{-t}\). Answer: The unknown coefficient function b(x, t) is 1.

Step-by-step solution

Differentiate u(x, t) with respect to x and t

We need the partial derivatives \(u_x\) and \(u_t\) to substitute in the given PDE. Differentiating u(x, t) with respect to x and t, we have: $$ u_x = \frac{\partial}{\partial x}(x e^{-t})=e^{-t} $$ and $$ u_t = \frac{\partial}{\partial t}(x e^{-t})=-x e^{-t} $$

Plug the partial derivatives and solution into the PDE

Now, we plug our solution, \(u(x, t)\), and the partial derivatives, \(u_x\) and \(u_t\), into the PDE: $$ xu_{x}+b(x, t)u_{t}=0 \Rightarrow x(e^{-t})+b(x, t)(-x e^{-t})=0 $$

Solve for the coefficient function b(x, t)

Now we need to solve for the coefficient function \(b(x, t)\). By simplifying the equation, we can isolate b(x, t): $$ x e^{-t} - b(x, t) x e^{-t} = 0 $$ Factor out the common term \(x e^{-t}\): $$ x e^{-t}(1- b(x, t)) = 0 $$ Since \(x e^{-t}\) is a non-zero term, we can divide both sides of the equation by this term: $$ 1- b(x, t) = 0 $$ Finally, solve for \(b(x, t)\): $$ b(x, t) = 1 $$ The unknown coefficient function is \(b(x, t) = 1\).

Key concepts

Coefficient Function

In partial differential equations, finding the coefficient function is crucial for solving the equation completely. Essentially, the coefficient function acts like a multiplier for one of the derivative terms in the equation.

In our original exercise, the task was to find such an unspecified function, denoted as \(b(x, t)\), which is the coefficient of \(u_t\) in the PDE. The importance of determining \(b(x, t)\) lies in enabling us to fully understand the relationship between the variables within the PDE.

To find \(b(x, t)\), we needed to manipulate the given PDE equation, \(xu_x + b(x, t)u_t = 0\), by plugging in the known solution \(u(x, t) = xe^{-t}\) and its derivatives. After simplification, this allowed us to isolate and find the value of \(b(x, t)\), which turned out to be 1. Knowing this coefficient function now enables us to effectively solve and interpret the behavior of the PDE under specified conditions.

Partial Derivatives

Partial derivatives play a fundamental role in partial differential equations. They allow us to measure how a function changes as its variables change independently. In our exercise, we were dealing with the function \(u(x, t) = xe^{-t}\).

We calculated its partial derivatives with respect to \(x\) and \(t\):

• The partial derivative of \(u\) with respect to \(x\), denoted \(u_x\), was found to be \(e^{-t}\). This represents how the function \(u\) changes as the variable \(x\) varies, while keeping \(t\) constant.
• Similarly, the partial derivative of \(u\) with respect to \(t\), denoted \(u_t\), was calculated as \(-xe^{-t}\). This indicates how \(u\) changes when \(t\) varies, with \(x\) held constant.

These derivatives are essential when plugging into the given PDE to analyze and solve it further. Understanding how to calculate and interpret partial derivatives allows us to tackle more complex functions and PDEs by breaking them down into more manageable parts.

Solution Verification

Solution verification is the process used to confirm that the derived solution satisfies the original equation or system in question. After determining the partial derivatives and substituting them, we need to ensure they adhere to the structure of the given PDE.

In our exercise, after determining \(u_x = e^{-t}\) and \(u_t = -xe^{-t}\), we substituted these values back into the equation \(xu_x + b(x, t)u_t = 0\). Upon substitution, the equation became \(xe^{-t} - b(x, t)xe^{-t} = 0\).

To verify our solution, we simplified this to \(xe^{-t}(1 - b(x, t)) = 0\). Since \(xe^{-t}\) is non-zero, the only way the product equals zero is if the inner factor \(1 - b(x, t)\) becomes zero, leading us to conclude that \(b(x, t) = 1\).

This step reinforced our derived coefficient indicating that solving mathematical equations is not just about finding answers, but also ensuring that those answers meet the original conditions and parameters given by the equation.