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Tutorial Sheet 5

This document is a tutorial sheet for a Complex Analysis and Differential Equations-II course at the Indian Institute of Technology Indore, led by Dr. Santanu Manna. It includes various problems related to partial differential equations, boundary value problems, and heat equations, requiring students to classify, solve, and analyze different equations under specified conditions. The tutorial aims to enhance understanding of mathematical concepts and their applications in engineering and physics.

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16 views2 pages

Tutorial Sheet 5

This document is a tutorial sheet for a Complex Analysis and Differential Equations-II course at the Indian Institute of Technology Indore, led by Dr. Santanu Manna. It includes various problems related to partial differential equations, boundary value problems, and heat equations, requiring students to classify, solve, and analyze different equations under specified conditions. The tutorial aims to enhance understanding of mathematical concepts and their applications in engineering and physics.

Uploaded by

charannelavalli
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Indian Institute of Technology Indore

MA203 Complex Analysis and Differential Equations-II


(Autumn Semester 2024)
Instructor: Dr. Santanu Manna
Tutorial Sheet 5

1. Classify the following partial differential equations :

(a) 5uxx − 3uyy + (cos x)ux + ey uy + u = 0,

(b) ex uxx + ey uyy u = 0,

(c) sin2 xuxx + sin 2xuxy + cos2 xuyy = x,

2. Solve the following PDEs by the method of separation of variables :

(c) uxx = uy + 2u, with u(0, y) = 0 and ux (0, y) = 1 + e−3y ,

(d) zxx − 2zx + zy = 0,

(e) ux = 2ut + u where u(x, 0) = 6e−3x .

3. Find the solution u(x, t) of heat equation

∂u ∂2u
= k 2 , 0 < x < 1, t > 0,
∂t ∂x
∂u(1,t)
with initial and boundary conditions u(x, 0) = 1 − x and u(0, t) = 10, ∂x = 0.
2
4. Solve the differential equation ∂u 2∂ u
∂t = α ∂x2 for the condition of heat along a rod without
radiation, subject to the following conditions:

i. u ̸= ∞ as t → ∞ ,
∂u
ii. ∂x = 0 for x = 0 and x = l ,

iii. u = lx − x2 for t = 0, between x = 0 and x = l.

5. The ends A and B of a rod 20 cm long have the temperature at 30o C and 80o C until steady-
state prevails. The temperature of the ends are changed to 40o C and 60o C respectively.
Find the temperature distribution in the rod at time t.

6. Find the solution of the Laplace equation

∂ 2 u 1 ∂u 1 ∂2u
+ + = 0,
∂r2 r ∂r r2 ∂θ2
outside the circle r = a, that satisfies the boundary condition

u(a, θ) = f (θ), 0 ≤ θ < 2π,

on the circle.

1
7. Let u(x, t) be a continuously differential function and satisfying the equation

∂2u ∂2u
= k , 0 < x < l, t > 0,
∂t2 ∂x2
subject to the initial conditions
∂u(x, 0)
u(x, 0) = C sin3 (πx/l) and = 0,
∂t
and boundary conditions
u(0, t) = u(l, t) = 0, ∀ t.
Then find u(x, t) (C and k are constants).

8. A bar AB of length 10 cm has its ends A and B kept at 30◦ and 100◦ temperatures re-
spectively, until steady-state condition is reached. Then the temperature at A is lowered
to 20◦ and that at B to 40◦ and these temperatures are maintained. Find the subsequent
temperature distribution in the bar.

9. Solve ut = c2 uxx when

i. u ̸= ∞ as t → ∞ ,

ii. ux = 0 when x = 0 for all t ,

iii. u = 0 when x = L for all t ,

iv. u = u0 = constant when t = 0 for 0 < x < L .

10. Find the displacement of a string stretched between two fixed points at a distance 2c apart
when the string is initially at rest in equilibrium position and points of the string are given
initial
( velocities v where
x
c, if 0 < x ≤ c
v = 2c−x , and x being the distance measured from one end.
c , if c < x < 2c.

11. Find the displacement of a string stretched between the fixed points (0, 0) and (1, 0) and
released from rest from the position A sin πx + B sin 2πx.

12. If u(r, θ) satisfying the equation

∂ 2 u 1 ∂u 1 ∂2u
+ + = 0,
∂r2 r ∂r r2 ∂θ2
π
within the region of the plane bounded by r = a, r = b, θ = 0 and θ = 2. Its value along
the other boundary is zero (see, Fig). Then prove that
∞
2 X (r/b)4m−2 − (b/r)4m−2 sin(4m − 2)θ
u(r, θ) = .
π (a/b)4m−2 − (b/a)4m−2 (2m − 1)3
m=1

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