Solution for PDE ut - e^-tux + u = 0

We are given a partial differential equation (PDE) and an initial condition. We need to find which of the given expressions for \(u(x, t)\) is a solution that satisfies both the PDE and the initial condition.

The given PDE is:

\(\frac{\partial u}{\partial t} - e^{-t} \frac{\partial u}{\partial x} + u = 0\)

This can be written as:

\(u_t - e^{-t}u_x + u = 0\)

The initial condition is:

\(u(x, 0) = x\)

We will check each option by calculating the partial derivatives \(u_t\) and \(u_x\) for each given expression of \(u(x, t)\), substituting them into the PDE, and verifying if the equation holds true. We will also check if the initial condition is satisfied by each expression.

Checking Option 1: \(u(x, t) = e^t(x + e^t - 1)\)

First, let's rewrite the expression: \(u(x, t) = xe^t + e^{2t} - e^t\)

• Calculate \(u_t\):
  \(u_t = \frac{\partial}{\partial t}(xe^t + e^{2t} - e^t) = xe^t + 2e^{2t} - e^t\)
• Calculate \(u_x\):
  \(u_x = \frac{\partial}{\partial x}(xe^t + e^{2t} - e^t) = e^t\)
• Substitute \(u_t\), \(u_x\), and \(u\) into the PDE \(u_t - e^{-t}u_x + u = 0\):
  \((xe^t + 2e^{2t} - e^t) - e^{-t}(e^t) + (xe^t + e^{2t} - e^t)\)
  \(= xe^t + 2e^{2t} - e^t - 1 + xe^t + e^{2t} - e^t\)
  \(= 2xe^t + 3e^{2t} - 2e^t - 1\)

This expression \(2xe^t + 3e^{2t} - 2e^t - 1\) is not equal to 0 for all \(x\) and \(t\). So, this option is not a solution to the PDE.

• Check initial condition \(u(x, 0) = x\):
  \(u(x, 0) = e^0(x + e^0 - 1) = 1(x + 1 - 1) = x\)

The initial condition is satisfied, but the PDE is not.

Checking Option 2: \(u(x, t) = e^{-t}(x - e^{-t} + 1)\)

First, let's rewrite the expression: \(u(x, t) = xe^{-t} - e^{-2t} + e^{-t}\)

• Calculate \(u_t\):
  \(u_t = \frac{\partial}{\partial t}(xe^{-t} - e^{-2t} + e^{-t}) = -xe^{-t} - (-2)e^{-2t} + (-1)e^{-t} = -xe^{-t} + 2e^{-2t} - e^{-t}\)
• Calculate \(u_x\):
  \(u_x = \frac{\partial}{\partial x}(xe^{-t} - e^{-2t} + e^{-t}) = e^{-t}\)
• Substitute \(u_t\), \(u_x\), and \(u\) into the PDE \(u_t - e^{-t}u_x + u = 0\):
  \((-xe^{-t} + 2e^{-2t} - e^{-t}) - e^{-t}(e^{-t}) + (xe^{-t} - e^{-2t} + e^{-t})\)
  \(= -xe^{-t} + 2e^{-2t} - e^{-t} - e^{-2t} + xe^{-t} - e^{-2t} + e^{-t}\)
  \(= (-xe^{-t} + xe^{-t}) + (2e^{-2t} - e^{-2t} - e^{-2t}) + (-e^{-t} + e^{-t})\)
  \(= 0 + 0 + 0 = 0\)

The expression simplifies to 0, which means this option is a solution to the PDE.

• Check initial condition \(u(x, 0) = x\):
  \(u(x, 0) = e^{-0}(x - e^{-0} + 1) = 1(x - 1 + 1) = x\)

The initial condition is also satisfied.

Since Option 2 satisfies both the PDE and the initial condition, it is the correct solution.

Checking Option 3: \(u(x, t) = x - e^t + 1\)

• Calculate \(u_t\):
  \(u_t = \frac{\partial}{\partial t}(x - e^t + 1) = -e^t\)
• Calculate \(u_x\):
  \(u_x = \frac{\partial}{\partial x}(x - e^t + 1) = 1\)
• Substitute \(u_t\), \(u_x\), and \(u\) into the PDE \(u_t - e^{-t}u_x + u = 0\):
  \((-e^t) - e^{-t}(1) + (x - e^t + 1)\)
  \(= -e^t - e^{-t} + x - e^t + 1\)
  \(= x - 2e^t - e^{-t} + 1\)

This expression \(x - 2e^t - e^{-t} + 1\) is not equal to 0 for all \(x\) and \(t\). So, this option is not a solution to the PDE.

• Check initial condition \(u(x, 0) = x\):
  \(u(x, 0) = x - e^0 + 1 = x - 1 + 1 = x\)

The initial condition is satisfied, but the PDE is not.

Checking Option 4: \(u(x, t) = xe^t\)

• Calculate \(u_t\):
  \(u_t = \frac{\partial}{\partial t}(xe^t) = xe^t\)
• Calculate \(u_x\):
  \(u_x = \frac{\partial}{\partial x}(xe^t) = e^t\)
• Substitute \(u_t\), \(u_x\), and \(u\) into the PDE \(u_t - e^{-t}u_x + u = 0\):
  \((xe^t) - e^{-t}(e^t) + (xe^t)\)
  \(= xe^t - 1 + xe^t\)
  \(= 2xe^t - 1\)

This expression \(2xe^t - 1\) is not equal to 0 for all \(x\) and \(t\). So, this option is not a solution to the PDE.

• Check initial condition \(u(x, 0) = x\):
  \(u(x, 0) = xe^0 = x(1) = x\)

The initial condition is satisfied, but the PDE is not.

Based on the verification of all options, only the expression in Option 2 satisfies both the given PDE and the initial condition.