Let's solve each part step by step with explanations: --- ### **Part (a):** #### **Step 1: Show the momentum equation** Given: - The flow velocity has nearly returned to \( U \), so \( u = U + u_1 \), where \( u_1 \) is small and negative. - The wake is thin: variation in \( y \) is much faster than in \( x \). The 2D steady Navier-Stokes equation for \( u \) in the \( x \)-direction (neglecting pressure gradients and with high Reynolds number): \[ u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} = \nu \frac{\partial^2 u}{\partial y^2} \] Since \( u_1 \) is small compared to \( U \), and changes in \( y \) are much more rapid, we can approximate: - \( u \approx U \) - \( v \) is very small (mass conservation in thin wake) - Thus, neglect \( v \frac{\partial u}{\partial y} \) So the equation reduces to: \[ U \frac{\partial u_1}{\partial x} = \nu \frac{\partial^2 u_1}{\partial y^2} \] **Explanation:** We neglected nonlinear and cross terms because \( u_1 \) is small and changes in \( y \) dominate. --- #### **Step 2: Deduce \(\int_{-\infty}^{\infty} u_1 \, dy = \text{constant}\)** Integrate both sides of the simplified equation over \( y \): \[ U \frac{\partial}{\partial x} \int_{-\infty}^{\infty} u_1 \, dy = \nu \int_{-\infty}^{\infty} \frac{\partial^2 u_1}{\partial y^2} \, dy \] The right side is: \[ \nu \frac{\partial u_1}{\partial y} \Big|_{-\infty}^{\infty} \] Since \( u_1 \to 0 \) as \( |y| \to \infty \), this term vanishes. Therefore: \[ \frac{\partial}{\partial x} \int_{-\infty}^{\infty} u_1 \, dy = 0 \] So, \[ \int_{-\infty}^{\infty} u_1 \, dy = \text{constant} \] **Explanation:** This shows conservation of momentum deficit in the wake as we move downstream. --- ## **Part (b):** #### **Step 1: Similarity Solution Ansatz** Assume: \[ u_1 = F(x) f(\eta), \quad \eta = \frac{y}{g(x)} \] Substitute into the momentum equation: \[ U \frac{\partial u_1}{\partial x} = \nu \frac{\partial^2 u_1}{\partial y^2} \] Calculate derivatives: - \(\frac{\partial u_1}{\partial x} = F'(x) f(\eta) + F(x) f'(\eta) \frac{\partial \eta}{\partial x}\) - \(\frac{\partial \eta}{\partial x} = -\frac{y}{g(x)^2} g'(x) = -\eta \frac{g'(x)}{g(x)}\) - \(\frac{\partial^2 u_1}{\partial y^2} = F(x) f''(\eta) \frac{1}{g(x)^2}\) So the equation becomes: \[ U \left[ F'(x) f(\eta) - F(x) f'(\eta) \eta \frac{g'(x)}{g(x)} \right] = \nu F(x) f''(\eta) \frac{1}{g(x)^2} \] Divide both sides by \( F(x) \): \[ U \left[ \frac{F'(x)}{F(x)} f(\eta) - f'(\eta) \eta \frac{g'(x)}{g(x)} \right] = \nu f''(\eta) \frac{1}{g(x)^2} \] For a similarity solution, all \( x \)-dependence must combine such that the ODE in \( \eta \) is independent of \( x \). This is achieved if: - \( \frac{F'(x)}{F(x)} \sim \frac{1}{x} \) - \( \frac{g'(x)}{g(x)} \sim \frac{1}{x} \) - \( \frac{1}{g(x)^2} \sim \frac{1}{x} \) Let \( g(x) = \sqrt{C x} \), \( F(x) = \frac{A}{x^{1/2}} \). **Explanation:** We set up the similarity solution and balanced the \( x \)-dependence for self-consistency. --- #### **Step 2: Obtain the Solution and Final Form** Plug \( g(x) = \sqrt{4 \nu x / U} \) into the equation, and after simplification, the ODE for \( f(\eta) \) is: \[ f''(\eta) + \frac{1}{2} \eta f'(\eta) = 0 \] Integrate: - First integral: Let \( f'(\eta) = h(\eta) \) - \( h' + \frac{1}{2} \eta h = 0 \implies \frac{d h}{h} = -\frac{1}{2} \eta d\eta \implies \ln h = -\frac{1}{4} \eta^2 + \text{const} \) - \( h(\eta) = C e^{-\eta^2/4} \) - Integrate again: \( f(\eta) = C \int e^{-\eta^2/4} d\eta \), but for a wake, the Gaussian is the main form. So, \[ u_1 = \frac{A}{x^{1/2}} e^{-U y^2/(4 \nu x)} \] **Explanation:** We derived the similarity solution and found the explicit functional form, matching the required result. --- ### **Final Answer:** #### **(a)** - The **momentum equation** simplifies to: \[ U \frac{\partial u_1}{\partial x} = \nu \frac{\partial^2 u_1}{\partial y^2} \] - The **integral** of \( u_1 \) over \( y \) is **constant**: \[ \int_{-\infty}^{\infty} u_1 \, dy = \text{constant} \] #### **(b)** - The **similarity solution** is: \[ u_1 = \frac{A}{x^{1/2}} e^{-U y^2 / (4 \nu x)} \] where \( A \) is a constant. --- **Sketch:** - At two different downstream \( x \) values, the Gaussian profile widens and its peak lowers as \( x \) increases (since the denominator in the exponent increases and the prefactor decreases as \( 1/\sqrt{x} \)). - The wake "spreads out" and "weakens" as you go further downstream.