(a) Reduce the partial differential equation y ∂²z/∂x² + (x + y) ∂²z/∂x∂y + x ∂²z/∂y² = 0 to canonical form and hence solve it…

(a) Canonical form and solution The given partial differential equation is y (∂² z)/(∂ x²) + (x + y) (∂² z)/(∂ x ∂ y) + x (∂² z)/(∂ y²) = 0. Comparing this with the standard form A zₓₓ + B z_xy + C z_yy = 0, we identify the coefficients as A = y, B = x + y, and C = x. The discriminant is calculated as Δ = B² - 4AC = (x + y)² - 4xy = x² + 2xy + y² - 4xy = (x - y)². Since Δ > 0 for x ≠ y, the equation is of the hyperbolic type. The characteristic equation is A (dy)² - B dx dy + C (dx)² = 0, which becomes y (dy)² - (x + y) dx dy + x (dx)² = 0. Factoring this expression yields (dy - dx)(y dy - x dx) = 0. This provides two families of characteristic curves: dy - dx = 0 implies y - x = c₁ and y dy - x dx = 0 implies y² - x² = c₂. We define the new variables ξ = y - x and η = y² - x².

With ξₓ = −1, ξ_y = 1, ηₓ = −2x, η_y = 2y the chain rule gives zₓ = −z_ξ − 2x z_η, z_y = z_ξ + 2y z_η and zₓₓ = z_ξξ + 4x z_ξη + 4x² z_ηη − 2z_η, z_xy = −z_ξξ − 2(x + y) z_ξη − 4xy z_ηη, z_yy = z_ξξ + 4y z_ξη + 4y² z_ηη + 2z_η. Substituting, the z_ξξ and z_ηη terms cancel and y zₓₓ + (x + y) z_xy + x z_yy = −2(x − y)² z_ξη − 2(y − x) z_η = −2ξ² z_ξη − 2ξ z_η = 0, so the canonical form is z_ξη + (1/ξ) z_η = 0 (ξ ≠ 0). To solve this, we set u = z_η, which simplifies the equation to u_ξ + 1/(ξ) u = 0. This is a first-order linear differential equation in ξ. The integrating factor is e^∫ 1/(ξ) dξ = ξ. Multiplying the equation by ξ gives ξ u_ξ + u = 0, or d/(dξ)(ξ u) = 0. Integrating with respect to ξ yields ξ u = F(η), where F is an arbitrary function. Substituting back u = z_η, we get z_η = (F(η))/(ξ). Integrating with respect to η gives the general solution z = 1/(ξ) ∫ F(η) dη + H(ξ). Letting G(η) = ∫ F(η) dη, the final solution is z = (G(η))/(ξ) + H(ξ). Substituting the original variables, the general solution is z = (G(y² - x²))/(y - x) + H(y - x).

(b)(i) Fixed-point binary to decimal The problem specifies 10-bit floating point numbers with 5 bits as the fractional part. This implies the binary point is located after the fifth bit, creating a 5-bit integer part and a 5-bit fractional part.

For the number 1010111011, the binary point is placed as 10101.11011. The integer part 10101₂ = 16 + 4 + 1 = 21. The fractional part .11011₂ is calculated as 2⁻¹ + 2⁻² + 2⁻⁴ + 2⁻⁵ = 0.5 + 0.25 + 0.0625 + 0.03125 = 0.84375. Thus, the decimal equivalent is 21.84375.

For the number 0100101010, the binary point is placed as 01001.01010. The integer part 01001₂ = 8 + 1 = 9. The fractional part .01010₂ is calculated as 2⁻² + 2⁻⁴ = 0.25 + 0.0625 = 0.3125. Thus, the decimal equivalent is 9.3125.

(b)(ii) Mixed-base sum in hexadecimal To evaluate the sum, we convert all terms to the decimal system. The first term is printed with subscript 10, but its digits are only 0 and 1 and it is added to octal and binary numbers, so it is taken as the binary number (11001.1011)₂. (Read literally in base 10, the sum would be 11073.4136, whose hexadecimal form does not terminate.) Converting (11001.1011)₂ to decimal: 16 + 8 + 1 + 0.5 + 0.125 + 0.0625 = 25.6875. Converting the octal term (33.24)₈ to decimal: 3 × 8 + 3 + 2 × 8⁻¹ + 4 × 8⁻² = 24 + 3 + 0.25 + 0.0625 = 27.3125. Converting the binary term (101101)₂ to decimal: 32 + 8 + 4 + 1 = 45. The sum is 25.6875 + 27.3125 + 45 = 98. To convert 98 to hexadecimal: 98 = 6 × 16 + 2. Therefore, the final result is (62)₁₆.

(c) Cylinder in a stream with circulation The complex velocity potential for a uniform stream of velocity V in the negative x-direction is f(z) = -Vz. According to Milne-Thomson's circle theorem, the complex potential w(z) for flow around a cylinder of radius a is given by w(z) = f(z) + f̄(a²/z), where f̄ is the conjugate function. Since f(z) = −Vz has the real coefficient −V, f̄(a²/z) = −V(a²/z). Thus, the potential for the cylinder is w(z) = -V(z + (a²)/z).

When a circulation k is added, the potential becomes w(z) = -V(z + (a²)/z) - ik/(2π) ln z. To find the stagnation points, we solve dw/dz = 0: -V(1 - (a²)/(z²)) - ik/(2π z) = 0. Multiplying by z²: -V(z² - a²) - ikz/(2π) = 0, which simplifies to Vz² + ik/(2π)z - Va² = 0. Using the quadratic formula, the stagnation points are z = (-ik/(2π) ± √((ik/(2π))² + 4V² a²))/2V = -ik/(4π V) ± √(a² - (k²)/(16π² V²)). For k < 4πVa, |z| = a, so both stagnation points lie on the cylinder at z = a eⁱθ with sin θ = −k/(4πVa): they are below the horizontal diameter, symmetric about the y-axis. For k = 4πVa they coincide at z = −ia; for k > 4πVa both lie on the imaginary axis and only the one with |z| > a is in the fluid.

Yes, a lifting tendency in the vertical direction is produced. By Blasius' theorem, Fₓ − iF_y = (iρ/2)∮(dw/dz)² dz; the coefficient of 1/z in (dw/dz)² is 2(−V)(−ik/2π) = iVk/π, so Fₓ − iF_y = (iρ/2)(2πi)(iVk/π) = −iρVk. Hence Fₓ = 0 and F_y = ρVk: an upward lift ρVk per unit length (Kutta–Joukowski) with zero drag. The circulation creates an asymmetry in the flow velocity around the cylinder; the velocity is higher on the upper side (where the anticlockwise circulation adds to the leftward stream) and lower on the lower side, which is why the stagnation points move below the horizontal. By Bernoulli's theorem, this velocity difference results in a pressure difference, generating a net force perpendicular to the free stream. This is the Magnus effect, which results in lift while the drag remains zero (d'Alembert's paradox).