(a) State uniqueness theorem for the existence of unique solution of the initial value problem dy/dx = f(x, y), y(x₀) = y₀ in the…

(a) Picard–Lindelöf theorem: Let f(x,y) be continuous in the rectangle R: |x−x₀|≤a, |y−y₀|≤b, and satisfy the Lipschitz condition in y:

|f(x,y₁)−f(x,y₂)| ≤ K|y₁−y₂| for all (x,y₁),(x,y₂)∈R. Let M=max|f| on R and h=min(a,b/M). Then the initial value problem dy/dx=f(x,y), y(x₀)=y₀ has a unique solution for |x−x₀|≤h.

For dy/dx=2√y, y(1)=0. Since the right side is real only for y≥0, take R: |x−1|≤a, 0≤y≤b. Here f=2√y is continuous on R, so existence is assured by the Cauchy–Peano theorem. But uniqueness fails because f is not Lipschitz at y=0:

|2√y−0|/|y−0| = 2/√y → ∞ as y→0⁺. Hence no Lipschitz constant K exists on any rectangle reaching y=0.

To find all solutions, for y>0 write z=√y. Then dz/dx = (dy/dx)/(2√y)=1, so z=x−c, giving y=(x−c)². At y=0 the solution may remain zero. Therefore all solutions on an interval containing x=1 are:

• y(x)=0 for all x;
• for each c≥1, y(x)=0 for x≤c and y(x)=(x−c)² for x≥c. Thus y(1)=0 is satisfied, but the solution is not unique.

(b) Let θ be the angle of the string from the downward vertical. The radial equation is T − mg cosθ = m v²/l. Energy from the lowest point, where the speed is u, gives v² = u² − 2gl(1−cosθ). Hence T(θ)=mg cosθ + (m/l)[u² −2gl(1−cosθ)] = (m u²/l) −2mg +3mg cosθ. For the opposite end of the same diameter, the angle is θ+π. Therefore T(θ+π)= (m u²/l) −2mg +3mg cos(θ+π) = (m u²/l) −2mg −3mg cosθ. Adding, T(θ)+T(θ+π)= 2m u²/l −4mg, which is independent of θ. Thus the sum of tensions at the ends of any diameter is constant.

For a complete revolution, the string remains taut if T(π)≥0, i.e. (m u²/l)−5mg≥0, so u²≥5gl.

(c) Stokes’ theorem: If S is an oriented surface bounded by a closed curve C, with orientation induced positively on C, then ∮_C F·dr = ∬_S (∇×F)·n dS.

Here F = xy î + yz ĵ + zx k̂. Thus ∇×F = (∂(zx)/∂y − ∂(yz)/∂z) î + (∂(xy)/∂z − ∂(zx)/∂x) ĵ + (∂(yz)/∂x − ∂(xy)/∂y) k̂ = (−y) î + (−z) ĵ + (−x) k̂.

The surface is r(x,y)= x î + y ĵ + (1−x²) k̂, 0≤x≤1, −2≤y≤2. Then r_x × r_y = (1,0,−2x) × (0,1,0) = (2x,0,1), which points upward. Hence n dS = (2x,0,1) dx dy. On S, z=1−x², so (∇×F)·(r_x×r_y) = (−y)(2x)+(−z)(0)+(−x)(1) = −2xy−x. Therefore ∬_S (∇×F)·n dS = ∫_x=0^1 ∫_y=−2^2 (−2xy−x) dy dx = ∫_0^1 (−4x) dx = −2.

Now compute the line integral around the boundary. With upward orientation, C is traversed counterclockwise in the xy-domain.

C₁: y=−2, x from 0 to 1, z=1−x². F·dr = −2x dx +(1−x²)x(−2x dx) = (−2x−2x²+2x⁴) dx. So ∫_C₁ F·dr = −19/15.

C₂: x=1, z=0, y from −2 to 2. F·dr=0, so integral =0.

C₃: y=2, x from 1 to 0, z=1−x². F·dr = 2x dx +(1−x²)x(−2x dx) = (2x−2x²+2x⁴) dx. So ∫_C₃ F·dr = −11/15.

C₄: x=0, z=1, y from 2 to −2. F·dr = y dy, and ∫_2^−2 y dy=0.

Thus ∮_C F·dr = −19/15 + 0 −11/15 + 0 = −2. Both sides equal −2, so Stokes’ theorem is verified.