(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 for Pp+Qq=R, dx/(y-z)=dy/(z-x)=dz/(x-y). Let ds be the common parameter. Then dx/ds=y-z, dy/ds=z-x, dz/ds=x-y. Adding the three differentials gives d(x+y+z)/ds=(y-z)+(z-x)+(x-y)=0, so c1=x+y+z. Multiplying the three equations by x, y, z and adding gives d(x²+y²+z²)/ds=2[x(y-z)+y(z-x)+z(x-y)]=0, so c2=x²+y²+z². The gradients of c1 and c2 are (1,1,1) and (2x,2y,2z); they are not parallel except on x=y=z, so the general integral is F(c1,c2)=0.

On φ=0, xy=1, write x=t, y=1/t, t≠0. Then c1=t+1/t=s and c2=t²+1/t²=s²-2. Also s²-4=(t-1/t)²≥0, so |s|≥2. Thus the first curve only forces F(s,s²-2)=0 on the two intervals |s|≥2; it does not force c2=c1²-2 for all c1. On the circle x+y+φ=0, x²+y²+φ²=a², we have c1=0 and c2=a².

We need a C¹ function g such that g(s)=s²-2 for |s|≥2 and g(0)=a². To make the implicit surface a graph φ(x,y) on the circle, also require g'(0)>2|a|. One explicit choice is g(s)=s²-2+((a²+2)/16)(s²-4)²(1+2s), |s|≤2, g(s)=s²-2, |s|≥2. At s=±2, (s²-4)² and its derivative vanish, so g is C¹. At s=0, g(0)=-2+((a²+2)/16)(16)=a². The derivative of (s²-4)²(1+2s) at 0 is 32, so g'(0)=((a²+2)/16)(32)=2(a²+2)>2|a|. Take F(c1,c2)=c2-g(c1). Then an integral surface is x²+y²+φ²-g(x+y+φ)=0. It contains the first curve because there c2=c1²-2 and |c1|≥2. It contains the circle because there c1=0 and c2=a².

For H=x²+y²+φ²-g(x+y+φ), H_x=2x-g', H_y=2y-g', H_z=2φ-g'. Where H_z≠0, ∂φ/∂x=-(2x-g')/(2φ-g') and ∂φ/∂y=-(2y-g')/(2φ-g'). Then (y-φ)∂φ/∂x+(φ-x)∂φ/∂y = -[(y-φ)(2x-g')+(φ-x)(2y-g')]/(2φ-g') = -[(y-x)(2φ-g')]/(2φ-g')=x-y. On the first curve, φ=0 and g'(c1)=2c1, so H_z=-2c1≠0. On the circle, |φ|≤|a|, so H_z=2φ-g'(0)≤2|a|-2(a²+2)<0. Hence the PDE is satisfied on the required curves. The surface is not unique.

(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.