(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 rectangular region R: |x - x₀| ≤ a, |y - y₀| ≤ b. Test the existence and uniqueness of the solution of the initial value

Question

(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 rectangular region R: |x - x₀| ≤ a, |y - y₀| ≤ b. Test the existence and uniqueness of the solution of the initial value problem dy/dx = 2√y, y(1) = 0, in a suitable rectangle R. If more than one solution exist, then find all the solutions. (15 marks)

(b) A heavy particle hanging vertically from a fixed point by a light inextensible string of length l starts to move with initial velocity u in a circle so as to make a complete revolution in a vertical plane. Show that the sum of tensions at the ends of any diameter is constant. (15 marks)

(c) State Stokes' theorem and verify it for the vector field F⃗ = xyî + yzĵ + zxk̂ over the surface S, which is the upwardly oriented part of the cylinder z = 1 - x², for 0 ≤ x ≤ 1, -2 ≤ y ≤ 2. (20 marks)

UPSC Answer Check · Accepted Answer

(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.

(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…

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