(a) Find the integral surface of the following quasi-linear equation (y - φ) (∂ φ)/(∂ x) + (φ - x) (∂ φ)/(∂ y) = x - y, which…

(a) Put z = φ, p = ∂z/∂x, q = ∂z/∂y. The equation is (y − z)p + (z − x)q = x − y. By Lagrange's method, the auxiliary equations are

dx/(y − z) = dy/(z − x) = dz/(x − y).

Adding numerators and denominators: d(x + y + z) = 0, so c₁ = x + y + z. Using multipliers x, y, z: x dx + y dy + z dz = 0, so c₂ = x² + y² + z². The general integral is F(x + y + φ, x² + y² + φ²) = 0.

Surface through the curve φ = 0, xy = 1. On this curve c₁ = x + y and c₂ = x² + y² = (x + y)² − 2xy = c₁² − 2. Hence the required surface is c₂ = c₁² − 2:

x² + y² + φ² = (x + y + φ)² − 2, i.e. xy + yφ + φx = 1.

Check: φ = (1 − xy)/(x + y) gives ∂φ/∂x = −(1 + y²)/(x + y)² and ∂φ/∂y = −(1 + x²)/(x + y)²; substituting, (y − φ)∂φ/∂x + (φ − x)∂φ/∂y = x − y.

Surface through the circle x + y + φ = 0, x² + y² + φ² = a². On the circle c₁ = 0 and c₂ = a², both constant, so the circle is itself a characteristic curve. The Cauchy problem is therefore not unique: every surface F(x + y + φ, x² + y² + φ²) = 0 with F(0, a²) = 0 contains it, for example

x² + y² + φ² − a² = (x + y + φ)·G(x + y + φ), G arbitrary,

including the sphere x² + y² + φ² = a² and the plane x + y + φ = 0.

Remark. The surface xy + yφ + φx = 1 does not contain the circle, since on the circle xy + yφ + φx = (c₁² − c₂)/2 = −a²/2 ≠ 1. A single surface through both curves exists only for a non-analytic choice such as c₂ = g(c₁) with g(s) = s² − 2 for |s| ≥ 2 and g(0) = a².

(b) The nodes are x0=-2, x1=-1, x2=0, x3=1, x4=2, x5=3, x6=4. The ordinates are f0=-55, f1=-9, f2=1, f3=5, f4=33, f5=115, f6=281.

(i) Simpson's 3/8 rule is obtained by integrating the cubic interpolant through four equally spaced points. On [x0,x3], with t=(x-x0)/h, the integrals of the four Lagrange basis functions over t=0,1,2,3 are 3/8, 9/8, 9/8, 3/8, so ∫ from x0 to x3 f(x) dx ≈ (3h/8)(f0+3f1+3f2+f3). Since 6=2×3, apply this to [x0,x3] and [x3,x6] and add: (3h/8)(f0+3f1+3f2+f3)+(3h/8)(f3+3f4+3f5+f6) =(3h/8)[f0+3f1+3f2+2f3+3f4+3f5+f6]. With h=1, =(3/8)[-55+3(-9)+3(1)+2(5)+3(33)+3(115)+281]. Now f0+f6=226, f1+f2+f4+f5=-9+1+33+115=140, so 3×140=420, and 2f3=10. Hence the sum is 226+420+10=656, and (3/8)×656=246.

(ii) Trapezoidal rule uses linear interpolation on each subinterval, i.e. the sum of six trapezoids: ∫≈(h/2)[(f0+f1)+(f1+f2)+...+(f5+f6)] =(h/2)[f0+2(f1+f2+f3+f4+f5)+f6]. With h=1, f1+f2+f3+f4+f5=-9+1+5+33+115=145, so =(1/2)[-55+2(145)+281]=(1/2)(516)=258. As a check, the exact antiderivative is (5/4)x⁴-x³+x²+x. At x=4 it is 320-64+16+4=276; at x=-2 it is 20+8+4-2=30. The exact integral is 246, so Simpson's 3/8 is exact for this cubic. Simpson's 3/8: 246. Trapezoidal: 258.

(c) Let ρ be the constant density, and assume steady motion with no body forces. Continuity for incompressible flow is ∂u/∂x+∂v/∂y+∂w/∂z=0. With r²=x²+y² and w=0, use the product rule: ∂u/∂x=B[2x/r⁴-4x(x²-y²)/r⁶]=2Bx(3y²-x²)/r⁶, ∂v/∂y=2B[x/r⁴-4xy²/r⁶]=2Bx(x²-3y²)/r⁶. Thus ∂u/∂x+∂v/∂y=0. Also ∂v/∂x=2B[y/r⁴-4x²y/r⁶]=2By(y²-3x²)/r⁶, ∂u/∂y=B[-2y/r⁴-4y(x²-y²)/r⁶]=2By(y²-3x²)/r⁶, so ∂v/∂x=∂u/∂y; the flow is irrotational away from (0,0). A potential can be found by integrating u with respect to x: Φ=-Bx/r²+h(y). Then ∂Φ/∂y=2Bxy/r⁴+h'(y)=v, so h'(y)=0 and Φ=-Bx/r². The speed is q²=u²+v²=B²[(x²-y²)²+4x²y²]/r⁸=B²(x²+y²)²/r⁸=B²/r⁴. For steady inviscid incompressible irrotational flow, Bernoulli's theorem gives p+(1/2)ρq²=C. Equivalently, because (u·∇)u=∇(q²/2), Euler's x-equation gives p_x=2ρB²x/r⁶. Integrating with respect to x, put u=r², du=2x dx: p=2ρB²∫ x/r⁶ dx=-ρB²/(2r⁴)+C(y). Differentiating this with respect to y gives p_y=2ρB²y/r⁶+C'(y). The y-equation of Euler requires p_y=2ρB²y/r⁶, so C'(y)=0. The z-equation gives p_z=0, which is satisfied. Hence p(x,y)=C-(ρB²)/(2(x²+y²)²). The constant C is fixed by a reference pressure. The formula is valid for (x,y)≠(0,0); the velocity and pressure are singular at the origin. If ρ is in kg/m³ and B in m³/s, p is in Pa.