Let's solve the problem step by step: --- ## **Part (a): Show the Approximate Momentum Equation and Deduce the Integral** ### **Step 1: The Momentum Equation** Given: - Thin wake: \( u \) varies rapidly with \( y \), slowly with \( x \) - \( u = U + u_1 \), where \( u_1 \) is small and negative - \( U \) is the uniform flow, \( u_1 \) is the disturbance #### **Linearized x-momentum equation (incompressible, viscous, steady):** \[ u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} = \nu \frac{\partial^2 u}{\partial y^2} \] - \( u = U + u_1 \) - \( v \) is small for thin wakes (streamwise is dominant) Neglecting nonlinear terms (since \( u_1 \ll U \)), and neglecting \( v \frac{\partial u}{\partial y} \): \[ (U + u_1) \frac{\partial (U + u_1)}{\partial x} = \nu \frac{\partial^2 (U + u_1)}{\partial y^2} \] Since \( U \) is constant and \( u_1 \) is small: \[ U \frac{\partial u_1}{\partial x} = \nu \frac{\partial^2 u_1}{\partial y^2} \] This matches the required result. --- ### **Step 2: Deduce the Integral Conservation** The equation is: \[ U \frac{\partial u_1}{\partial x} = \nu \frac{\partial^2 u_1}{\partial y^2} \] Integrate both sides over \( y \) from \( -\infty \) to \( +\infty \): \[ U \frac{\partial}{\partial x} \left( \int_{-\infty}^{\infty} u_1\, dy \right) = \nu \int_{-\infty}^{\infty} \frac{\partial^2 u_1}{\partial y^2} dy \] But \[ \int_{-\infty}^{\infty} \frac{\partial^2 u_1}{\partial y^2} dy = \left[ \frac{\partial u_1}{\partial y} \right]_{-\infty}^{+\infty} = 0 \] (because disturbances die out far from the wake) So, \[ U \frac{\partial}{\partial x} \left( \int_{-\infty}^{\infty} u_1\, dy \right) = 0 \] \[ \implies \int_{-\infty}^{\infty} u_1\, dy = \text{constant} \] --- ## **Part (b): Similarity Solution** ### **Step 1: Assume Similarity Form** Let \[ u_1 = F(x)\, f(\eta), \qquad \eta = \frac{y}{g(x)} \] Compute derivatives: - \(\frac{\partial u_1}{\partial x} = F'(x)f(\eta) + F(x)f'(\eta)\frac{d\eta}{dx}\) - \(\frac{\partial^2 u_1}{\partial y^2} = F(x)f''(\eta)\left(\frac{1}{g(x)^2}\right)\) But \(\frac{d\eta}{dx} = -\frac{y}{g(x)^2}g'(x) = -\eta \frac{g'(x)}{g(x)}\). So \[ \frac{\partial u_1}{\partial x} = F'(x)f(\eta) - F(x)\eta f'(\eta)\frac{g'(x)}{g(x)} \] \[ \frac{\partial^2 u_1}{\partial y^2} = F(x)f''(\eta)\frac{1}{g(x)^2} \] Plug into the PDE: \[ U\left[ F'(x)f(\eta) - F(x)\eta f'(\eta)\frac{g'(x)}{g(x)} \right] = \nu F(x)f''(\eta)\frac{1}{g(x)^2} \] Divide both sides by \(F(x)\): \[ U\left( \frac{F'(x)}{F(x)} f(\eta) - \eta f'(\eta) \frac{g'(x)}{g(x)} \right) = \nu f''(\eta) \frac{1}{g(x)^2} \] ### **Step 2: Seek Separation of Variables** For similarity, coefficients of \(f\) and its derivatives must be constant, independent of \(x\): Let \(g(x) = a x^{1/2}\), so \(\frac{g'(x)}{g(x)} = \frac{1}{2x}\). Let \(F(x) = \frac{A}{x^{1/2}}\), so \(\frac{F'(x)}{F(x)} = -\frac{1}{2x}\). Substitute: \[ U\left( -\frac{1}{2x} f(\eta) - \eta f'(\eta) \frac{1}{2x} \right) = \nu f''(\eta) \frac{1}{a^2 x} \] Multiply both sides by \(x\): \[ U\left( -\frac{1}{2} f(\eta) - \frac{1}{2} \eta f'(\eta) \right) = \frac{\nu}{a^2} f''(\eta) \] Rearrange: \[ f''(\eta) + \frac{U a^2}{2\nu} \left[ f(\eta) + \eta f'(\eta) \right] = 0 \] Let \(k = \frac{U a^2}{2\nu}\): \[ f''(\eta) + k \left[ f + \eta f' \right] = 0 \] Try a Gaussian solution: \(f(\eta) = e^{-c \eta^2}\) - \(f'(\eta) = -2c\eta e^{-c\eta^2}\) - \(f''(\eta) = (-2c + 4c^2\eta^2) e^{-c\eta^2}\) Plug in: \[ (-2c + 4c^2\eta^2) e^{-c\eta^2} + k (1 - 2c\eta^2) e^{-c\eta^2} = 0 \] \[ \left[ -2c + 4c^2\eta^2 + k - 2k c \eta^2 \right] e^{-c\eta^2} = 0 \] \[ -2c + k + (4c^2 - 2k c) \eta^2 = 0 \] Set coefficients of \(\eta^2\) and constant to zero: - Constant: \(-2c + k = 0 \implies c = k/2\) - \(\eta^2\): \(4c^2 - 2k c = 0 \implies 4c^2 = 2k c \implies 2c = k \implies c = k/2\) So it is self-consistent. Recall \(k = \frac{U a^2}{2\nu}\) and \(a^2 = 4\nu/U\) (so that \(k = 2\)), thus: So, \(c = 1\). Thus, the solution is: \[ u_1 = \frac{A}{x^{1/2}} e^{-U y^2/(4\nu x)} \] --- ## **Final Solution** ### **(a)** - **Momentum equation:**                                                   \[ U \frac{\partial u_1}{\partial x} = \nu \frac{\partial^2 u_1}{\partial y^2} \] - **Integral conservation:**                                                     \[ \int_{-\infty}^\infty u_1\,dy = \text{constant} \] ### **(b)** - **Similarity solution:** \[ u_1 = \frac{A}{x^{1/2}} e^{-U y^2/(4\nu x)} \] where \(A\) is constant. --- ### **Sketch: Velocity Profiles at Two Different \(x\)** - For small \(x\): The Gaussian is tall and narrow. - For larger \(x\): The Gaussian is shorter and broader (since denominator in exponent is larger, and prefactor \(\frac{1}{\sqrt{x}}\) is smaller). So as you move downstream, the wake spreads out and the disturbance becomes weaker and wider. --- **If you need an explicit plot or further clarification, let me know!**