Let's go through the problem step by step: ### **Step 1: Given** - \( u(x, y) \) is harmonic in the rectangle \( x_1 \leq x \leq x_2, \ y_1 \leq y \leq y_2 \). - Want to show there exists a conjugate harmonic function \( v(x, y) \) so that \( f = u + iv \) is analytic in \( D \), and that \( v(x, y) \) is uniquely determined up to an additive constant. ### **Step 2: Cauchy-Riemann Equations** The Cauchy-Riemann equations are: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \] \[ \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \] Given that \( u \) is harmonic: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \] ### **Step 3: Integral Representation for \( v(x, y) \)** Given the hint, we define: \[ v(x, y) = \int_{y_0}^y \frac{\partial u}{\partial x}(x, y') \, dy' + A(x) \] ### **Step 4: Determine \( A(x) \)** Differentiate \( v(x, y) \) with respect to \( x \): \[ \frac{\partial v}{\partial x}(x, y) = \int_{y_0}^y \frac{\partial^2 u}{\partial x^2}(x, y') \, dy' + A'(x) \] From the Cauchy-Riemann equations: \[ \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} \] Set \( y' = y \) at the upper limit: \[ \frac{\partial v}{\partial x}(x, y) = \frac{\partial u}{\partial x}(x, y) \cdot 0 + A'(x) \] But to get the correct match, take derivative inside the integral: \[ \frac{\partial v}{\partial x}(x, y) = \int_{y_0}^y \frac{\partial^2 u}{\partial x^2}(x, y') \, dy' + A'(x) \] But since \( u \) is harmonic: \[ \frac{\partial^2 u}{\partial x^2} = -\frac{\partial^2 u}{\partial y^2} \] So, \[ \int_{y_0}^y \frac{\partial^2 u}{\partial x^2}(x, y') \, dy' = -\int_{y_0}^y \frac{\partial^2 u}{\partial y^2}(x, y') \, dy' \] \[ = -\left[\frac{\partial u}{\partial y}(x, y) - \frac{\partial u}{\partial y}(x, y_0)\right] \] Thus, \[ \frac{\partial v}{\partial x}(x, y) = -\left[\frac{\partial u}{\partial y}(x, y) - \frac{\partial u}{\partial y}(x, y_0)\right] + A'(x) \] But we want \( \frac{\partial v}{\partial x}(x, y) = -\frac{\partial u}{\partial y}(x, y) \). So set: \[ A'(x) = \frac{\partial u}{\partial y}(x, y_0) \] \[ A(x) = \int_{x_0}^x \frac{\partial u}{\partial y}(x', y_0) dx' + C \] But from the text, \[ A(x) = - \int_{x_0}^x \frac{\partial u}{\partial y}(x', y_0) dx' \] (They included the minus to match the negative sign.) ### **Step 5: Final Formula for \( v(x, y) \)** Combine everything: \[ v(x, y) = \int_{y_0}^y \frac{\partial u}{\partial x}(x, y') dy' - \int_{x_0}^x \frac{\partial u}{\partial y}(x', y_0) dx' \] This is equation (17.1). ### **Step 6: Verify Cauchy-Riemann Equations** Compute \( \frac{\partial v}{\partial y} \): \[ \frac{\partial v}{\partial y} = \frac{\partial u}{\partial x}(x, y) \] Compute \( \frac{\partial v}{\partial x} \): \[ \frac{\partial v}{\partial x} = \int_{y_0}^y \frac{\partial^2 u}{\partial x^2}(x, y') dy' - \frac{\partial u}{\partial y}(x, y_0) \] But since \( u \) is harmonic, \[ \int_{y_0}^y \frac{\partial^2 u}{\partial x^2}(x, y') dy' = -\left[\frac{\partial u}{\partial y}(x, y) - \frac{\partial u}{\partial y}(x, y_0)\right] \] So, \[ \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}(x, y) \] So the Cauchy-Riemann equations are satisfied. ### **Step 7: Line Integral Form** From (17.2), for any path: \[ v(x, y) = \int_{(x_0, y_0)}^{(x, y)} \left[-\frac{\partial u}{\partial y}(x', y') dx' + \frac{\partial u}{\partial x}(x', y') dy'\right] \] This is an exact differential because \( u \) is harmonic, so the integral is path-independent up to a constant. ### **Step 8: Uniqueness up to a Constant** If you change the base point \( (x_0, y_0) \), the only difference is an additive constant, so \( v(x, y) \) is unique up to a constant. --- ## **Final Answer (Summary)** **Given a harmonic function \( u(x, y) \) in a rectangle, there exists a conjugate harmonic function \( v(x, y) \), unique up to an additive constant, such that \( f = u + iv \) is analytic in \( D \). One explicit formula for \( v(x, y) \) is:** \[ v(x, y) = \int_{y_0}^y \frac{\partial u}{\partial x}(x, y') dy' - \int_{x_0}^x \frac{\partial u}{\partial y}(x', y_0) dx' \] **or, equivalently,** \[ v(x, y) = \int_{(x_0, y_0)}^{(x, y)} \left[-\frac{\partial u}{\partial y}(x', y') dx' + \frac{\partial u}{\partial x}(x', y') dy'\right] \] **where the value is determined up to an arbitrary additive constant.** --- **If you want any step further explained, let me know!**