Advanced Engineering Mathematics IISolved Sample Problems

1 Series Solution of Ordinary Differential Equations

1.1

Obtain the solution of the following differential equation in terms of Maclaurin series

d2y

dx2+ x

dy

dx− y = 0

Answer:Let f(x) =

∑∞n=0 Anxn to obtain

(2A2 − A0)x0 +

∞∑n=1

[(n + 2)(n + 1)An+2 + nAn − An]xn = 0

From the first coefficient A0 is an arbitrary constant; then for n = 1 with A1 arbitraryconstant, A3 = 0. For n ≥ 1

An+2 = − (n − 1)

(n + 2)(n + 1)An

Must now consider two cases: n even and n odd. The general solution is

f(x) = A1x + A0[1 + x2/2 − x4/24 + x6/240 + ...]

1.2

Apply Frobenius method to find the general solution of the equation:

Ly = 2x2 d2y

dx2− x

dy

dx+ (1 + x)y = 2x2y′′ − xy′ + (1 + x)y = 0

1

Since x = 0 is a regular singular point, we expect solutions of the form

y(x) = xs∞∑

k=0

Akxk =

∞∑k=0

Akxk+s

Here

y(x)′ =∞∑

k=0

(k + s)Akxk+s−1

and

y(x)′′ =∞∑

k=0

(k + s − 1)(k + s)Akxk+s−2

Substituting into the given ODE yields

∞∑k=0

2(k + s − 1)(k + s)Akxk+s −

∞∑k=0

(k + s)Akxk+s +

∞∑k=0

Akxk+s +

∞∑k=0

Akxk+s+1 = 0

and a change in the dummy index of the fourth sum readily makes the x in all sums havethe same power

∞∑k=0

2(k + s − 1)(k + s)Akxk+s −

∞∑k=0

(k + s)Akxk+s +

∞∑k=0

Akxk+s +

∞∑k=1

Ak−1xk+s = 0

The above can also be written as

[2(s − 1)s − s + 1]A0xs +

∞∑k=1

2(k + s − 1)(k + s)Akxk+s

−∞∑

k=1

(k + s)Akxk+s +

∞∑k=1

Akxk+s +

∞∑k=1

Ak−1xk+s = 0

I.e.

[2(s − 1)s − s + 1]A0xs +

∞∑k=1

[{2(k + s − 1)(k + s) − (k + s) + 1}Ak + Ak−1]xk+s = 0

2

Collecting all terms who have as common factor the lowest power of x (i.e. xs) yields theindicial equation

2s2 − 3s + 1 = 0

with roots s1 = 1 and s2 = 1/2.And the recurrence relation is

Ak = − 1

2(s + k)2 − 3(s + k) + 1Ak−1

There will be one such recurrence relationship for each root of the indicial equation, i.e.,for s = s1 = 1,

Ak = − 1

(2k + 1)kAk−1 =

(−1)k

(2k + 1)k(2k − 1)(k − 1)...(5 × 2)(3 × 1)A0

and for s = s2 = 1/2,

Ak = − 1

2k(k − 12)Ak−1 =

(−1)k

[(2k − 1)(2k − 3)...3 × 1]k!A0

where, of course, to each root, there corresponds a specific set of Ak’s.Finally the corresponding solutions are

y1(x) = x[1 +∞∑

k=1

(−1)kxk

[(2k + 1)(2k − 1)...5 × 3]k!]

and

y2(x) = x1/2[1 +∞∑

k=1

(−1)kxk

[(2k − 1)(2k − 3)...3 × 1]k!]

1.3

Use the method of Frobenius to obtain a general solution of the following differential equationvalid near x = 0

xd2y

dx2+ 2

dy

dx+ xy = 0

Hints* In the result of introducing the series representation for y into the xy term, make a

dummy index change from n to n − 1.* In the result of introducing the series representation for y into the 2y′ and into the xy′′

terms, make a dummy index change from n to n + 1.

3

* Investigate the behavior of the recurrence relationship for the smaller root s2 first.Answer:Let y = f(x) =

∑∞n=0 Anxn+s. Thus

xf(x) = x∞∑

n=0

Anxn+s =∞∑

n=0

Anxn+s+1 =∞∑

n=1

An−1xn+s

where instead of the dummy index n, the dummy n − 1 was used.Similarly

2dy

dx= 2f(x)′ = 2

∞∑n=0

(n + s)Anxn+s−1 = 2∞∑

n=−1

(n + 1 + s)An+1xn+s

Where the dummy index has been changed from n to n + 1 in order to have x raised to thepower n + s in the sum.

Finally,

xd2y

dx2= xf(x)′′ =

∞∑n=0

(n + s − 1)(n + s)Anxn+s−1 =∞∑

n=−1

(n + s)(n + 1 + s)An+1xn+s

Where the dummy index has also been changed from n to n + 1.Now, substituting the above into the original ODE yields

∞∑n=−1

(n + s)(n + 1 + s)An+1xn+s +

∞∑n=−1

2(n + 1 + s)An+1xn+s +

∞∑n=1

An−1xn+s = 0

And division by xn+s yields the general recurrence relationship

∞∑n=−1

(n + 1 + s)(n + s)An+1 +∞∑

n=−1

2(n + 1 + s)An+1 +∞∑

n=1

An−1 = 0

The indicial equation is now obtained from the above with n = −1 and it is

s(s − 1)A0 + 2sA0 = 0

Assuming A0 is arbitrary but 6= 0 one obtains the roots s1 = 0 and s2 = −1. Since s1−s2 = 1we must expect either no solution or a complete one for s2 = −1 and of course, one solutionfor s1 = 0.

Now, when n = 0 the recurrence relation becomes

(1 + s)sA1 + 2(1 + s)A1 = A1(1 + s)(s + 2) = 0

which is always true when s = −1, regardless of the value of A1, therefore A1 is also arbitraryand we necessarily have a complete solution.

4

With n ≥ 1

An+1 = (−1)1

(n + 1 + s)(n + 2 + s)An−1

which yields, for s = −1,

An+1 = (−1)1

n(n + 1)An−1

Specifically, for n = 1,

A2 = (−1)1

2A0

for n = 2

A3 = (−1)1

6A1

for n = 3

A4 = (−1)1

12A2 = (−1)

1

12(−1)

1

2A0

for n = 4

A5 = (−1)1

20A3 = (−1)

1

20(−1)

1

6A1

The recurrence relationship can finally be represented by the following two equations inmore compact form

A2l = (−1)l 1

(2l)!A0, l ≥ 0

and

A2l+1 = (−1)l 1

(2l + 1)!A1, l ≥ 0

Therefore, the general solution (corresponding to s = −1) is

y(x) = x−1∞∑

n=0

Anxn = x−1[A0 + A1x + A2x2 + ...] =

= x−1[A0 + A1x − 1

2A0x

2 − 1

6A1x

3 + ..] =

= A0[x−1 − 1

2x +

1

24x3 + ...] + A1[1 − 1

6x2 +

1

120x4 + ...] =

= A0cos(x)

x+ A1

sin(x)

x

5

1.4

Apply Frobenius method to find the general solution of the equation:

Ly = x2 d2y

dx2+ (x2 + x)

dy

dx− y = 0

Here the indicial equation is s2−1 = 0 with solutions s1 = +1, s2 = −1 and a series solutionis assured for s1. The recurrence formula is

(s + k − 1)[(s + k + 1)Ak + Ak−1] = 0

and for s = s1 = 1 this is

Ak = −Ak−1

k + 2

so that the associated solution is

y1 = xA0[1 − 1

3x +

1

3 × 4x2 − 1

3 × 4 × 5x3 + ...] =

= 2A0e−x − 1 + x

x

Now for s = s2 = −1 the recurrence formula becomes

(k − 2)(kAk + Ak−1) = 0

which yields A1 = −A0 when k = 1, 0 = 0 when k = 2 (i.e. A2 can be anything we want,including zero) and Ak = −Ak−1/k for k ≥ 3. With this, the solution associated with s = −1becomes

y2 = A01 − x

x

2 Special Functions

2.1

Obtain the general solution of the following equation in terms of Bessel functions or, ifpossible, in terms of elementary functions

xd2y

dx2− 3

dy

dx+ xy = 0

Answer:

6

As described in the class notes, the most general ODE satisfied by Bessel functions hasthe form

x2 d2y

dx2+ x(a + 2bxr)

dy

dx+ [c + dx2s − b(1 − a − r)xr + b2x2r]y = 0

The given equation can be put in this form simply multiplying by x yielding the following

x2 d2y

dx2− 3x

dy

dx+ x2y = 0

where a = −3, b = 0, c = 0, d = 1 and s = 1.The solution now can be written down directly as

y(x) = x2Zp(x) = x2(c1Jp(x) + c2Yp(x))

with

p =1

s

√(1 − a

2)2 − c = 2

Alternatively, the compatible form above can be obtained directly from Bessel’s equation

X2 d2Y

dX2+ X

dY

dX+ (X2 − P 2)Y = 0

by using the transformed variables X = d1/2xs/s = x, and Y = y/x2 and P = 2.The solution to the above Bessel’s equation is

Y (X) = c1JP (X) + c2YP (X)

and using the original variables the desired solution is

y(x) = c1x2J2(x) + c2x

2Y2(x)

3 Characteristic Functions

3.1

A uniform string is unrestrained from transverse motion at x = 0 while at x = L is attachedto a yielding support of modulus k, the ends of the string being constrained against appre-ciable movement parallel to the axis of rotation. Show that the nth critical speed ωn is givenby

ωn =µn

L

√T

ρ

7

where cot(µn) = αµn and α = T/kL.Answer: Solve

d2y

dx2+ ω2 ρ

Ty = 0

subject to the free end condition at x = 0

y′(0) = 0

and, since a yielding support is nothing but an elastic spring which exerts a restoring forceproportional to its stretch, (i.e. Ty′ = −ky),

T

kLLy′(L) = αLy′(L) = −y(L)

at x = L.Inspection shows that the functions

y1(x) = cos(

√ρ

Tωx)

and

y2(x) = sin(

√ρ

Tωx)

both satisfy the ODE, therefore, the general solution is

y(x) = c1y1(x) + c2y2(x) = c1 cos(

√ρ

Tωx) + c2 sin(

√ρ

Tωx)

The stated boundary conditions require that, c2 = 0 and that

µ =

√ρ

TωL

be the roots of

cot(µ) = αµ

3.2

Determine the Fourier coefficients of the following expansion for 0 < x < π,

f(x) = x =∞∑0

An sin(nx)

Answer:Here

An =2

π

∫ π

0x sin(nx)dx =

{ − 2n

n = even2n

n = odd

8

3.3

Use the Maple system to investigate the problem of characteristic function representations.

4 Numerical Solution of Boundary Value Problems for

Ordinary Differential Equations

4.1

Use the linear shooting method with h = 0.1 to find the solution to the following boundaryvalue problem in x ∈ [1, 2]:

Find y satisfying

y′′ = −2

xy′ +

2

x2y +

sin(ln x)

x2

and subject to

y(1) = 1

y(2) = 2

Answer:The exact solution is (check!)

y(x) = 1.1392070132x− 0.03920701320

x2− 0.3 sin(ln x) − 0.1 cos(lnx)

Numerical and analytical solutions are compared in the following table.

xi unum uex

1.5 1.48115939 1.48115942

4.2

Use the linear finite difference method with h = 0.1 to find the solution to the followingboundary value problem in x ∈ [1, 2]:

Find y satisfying

y′′ = −2

xy′ +

2

x2y +

sin(ln x)

x2

and subject to

y(1) = 1

9

y(2) = 2

Answer:The exact solution is again

y(x) = 1.1392070132x− 0.03920701320

x2− 0.3 sin(ln x) − 0.1 cos(lnx)

Numerical and analytical solutions are compared in the following table.

xi unum uex

1.5 1.48112026 1.48115942

5 Vector Analysis and Differential Geometry

5.1

For each of the following vector functions F, determine whether ∇φ = F has a solution anddetermine it if it exists.

a) F = 2xyz3i − (x2z3 + 2y)j + 3x2yz2kb) F = 2xyi + (x2 + 2yz)j + (y2 + 1)kAnswer:a) Here ∇φ = F requires ∇× F = 0 which is not the case here, so no solution.b) Here ∇× F = 0 so that

φ(x, y, z) = x2y + y2z + z + c

5.2

Let F = yzi + xyj + xzk.a) Compute the line integral of F from (0, 0, 0) to (1, 1, 1) along the path C consisting

first of the curve C1 defined by x = y2, z = 0in the x − y plane from (0, 0, 0) to (1, 1, 0) andthen the line C2, defined by x = 1, y = 1 perpendicular to the x − y plane from (1, 1, 0) to(1, 1, 1).

b) Repeat the calculation but now with C being the straight line x = y = z.Answer:a) Consider the first part of C, here dx = 2ydy and dz = 0 so that

∫C1

F · dr =∫ 1

0y3dy =

1

4

Now, for C2, dx = dy = 0 and

∫C2

F · dr =∫ 1

0zdz =

1

2

10

Thus ∫C

F · dr =∫

C1

F · dr +∫

C2

F · dr =1

4+

1

2=

3

4

b) In this case dx = dy = dz and

∫C

F · dr =∫ 1

03z2dz = 1

In this case the line integral is path dependent.

5.3

Let F = xi + yj and compute the surface integral of F over the surface S of the hemispherex2 + y2 + z2 = 1 with z ≥ 0.

Answer:Here n = xi + yj + zk and ds = dxdy/z therefore

∫ ∫SF · nds =

∫ ∫D(x2 + y2)

dxdy

z

where D is the projection of S on the x − y plane (i.e. the unit circle). Therefore

∫ ∫SF · nds = 4

∫ 1

0

∫ √1−x2

0

x2 + y2√1 − (x2 + y2)

dxdy

The final result∫ ∫

S F·nds = 43π is most easily obtained by switching to polar coordinates.

5.4

For the system of coordinates u, φ, θ defined by

x = au sin φ cos θ

y = bu sin φ sin θ

z = cu cos φ

show that the element of volume is of the form

dτ = abcu2 sin φdudφdθ

Answer:

11

The element of volume in the u1, u2, u3 coordinate system is given by

dτ = (U1 · U2 × U3)du1du2du3

Introducing u1 = u, u2 = φ, and u3 = θ and recalling that for i = 1, 2, 3

Ui =∂r

∂ui

one has

U1 = i∂x

∂u1

+ j∂y

∂u1

+ k∂z

∂u1

=

= ia sin φ cos θ + jb sin φ sin θ + kc cos φ

U2 = i∂x

∂u2

+ j∂y

∂u2

+ k∂z

∂u2

=

= iau cos φ cos θ + jbu cos φ sin θ − kcu sin φ

U3 = i∂x

∂u3

+ j∂y

∂u3

+ k∂z

∂u3

=

= −iau sin φ sin θ + jbu sin φ cos θ + k0

Therefore,

U1 · U2 × U3 =

∣∣∣∣∣∣∣∣∂x∂u1

∂y∂u1

∂z∂u1

∂x∂u2

∂y∂u2

∂z∂u2

∂x∂u3

∂y∂u3

∂z∂u3

∣∣∣∣∣∣∣∣ = |∂(x, y, z)

∂(u, φ, θ)| =

∣∣∣∣∣∣∣∣∂x∂u1

∂x∂u2

∂x∂u3

∂y∂u1

∂y∂u2

∂y∂u3

∂z∂u1

∂z∂u2

∂z∂u3

∣∣∣∣∣∣∣∣ =

=

∣∣∣∣∣∣∣a sin φ cos θ au cos φ cos θ − au sin φ sin θb sin φ sin θ bu cos φ sin θ bu sin φ cos θc cosφ − cu sin φ 0

∣∣∣∣∣∣∣ = abcu2 sin φ

so that

dτ = abcu2 sin φdudφdθ

6 Extrema; Systems of Nonlinear Equations; Calculus

of Variations

6.1

Determine extrema for the function f(x, y) = 1 − x2 − y2 at (x, y) = (0, 0).

12

Answer:Here

fx = −2x = 0

and

fy = −2y = 0

Also

fxx = −2 = fyy < 0

and since fxy = 0,

fxxfyy = 4 > f2xy

the point (0, 0) is a relative maximum.

6.2

Determine the shortest distance from the point (1, 0) to the parabola y2 = 4x.Answer:The function to be minimized is

f(x, y) = (x − 1)2 + y2

while the constraint is

g(x, y) = y2 − 4x = 0

The auxiliary function becomes

φ(x, y) = f(x, y) + λg(x, y) = (x − 1)2 + y2 + λ(y2 − 4x)

The necessary conditions for a minimum are then

φx = 2(x − 1) − 4λ = 0

φy = 2y + 2λy = 0

φλ = g(x, y) = y2 − 4x = 0

The only real solution to the above system is x = 0, y = 0 and λ = 1/2.

13

6.3

Consider the system of nonlinear equations

f1(x1, x2) = x1 + x2 − 3 = 0

f2(x1, x2) = x21 + x2

2 − 9 = 0

Let x0 = (1, 5) and use Newton’s method to find the solution.Answer:Here

J =

(∂f1

∂x1

∂f1

∂x2∂f2

∂x1

∂f2

∂x2

)=

(1 1

2x1 2x2

)

From this, the linear system for the first correction is

(1 12 10

)(s01

s02

)= −

(317

)

with solution s01 = −13

8, s0

2 = −118, which gives x1 = (−0.625, 3.625). Repeating the process

yields x2 = (−0.092, 3.092). The actual solution is x∗ = (0, 3).

6.4

Use Maple to find the solution of the following system of equations assuming the initial guessx1 = 0.1, x2 = 0.1, x3 = −0.1.

f1(x) = 3x1 − cos(x2x3) − 1

2= 0

f2(x) = x21 − 81(x2 + 0.1)2 + sin(x3) + 1.06 = 0

f3(x) = exp(−x1x2) + 20x3 +10π − 3

3= 0

Answer:The solution after 5 Newton iterations is x1 = 0.5, x2 = 0.00000002, x3 = −0.52359877.

6.5

Apply the calculus of variations to determine the function u(x) which minimizes the followingfunctional

I(u) =∫ π/2

0[(

du

dx)2 − u2 + +2xu]dx

14

with

u(0) = u(π/2) = 0

Answer:Here the Euler equation is

d2u

dx2+ u = x

the solution of which is

u = c1 cos(x) + c2 sin(x) + x

Incorporating the boundary conditions finally yields

u = x − (π

2) sin(x)

and the corresponding minimum value of I is

Imin = −(π

2)(1 − π2

12)

6.6

Find approximate solutions to the BVP consisting of finding u(x) that satisfies

u′′ + u + x = 0

subject to u(0) = u(1) = 0.Use the following coordinate functions

φi = xi(1 − x)

for i = 1, 2, ..., N , to construct by linear combination the approximation

uN =N∑

i=1

aiφi = a1φ1 + a2φ2 + ... + aNφN = a1x(1 − x) + a2x2(1 − x) + ... + aNxN(1 − x)

Then apply the Ritz method to construct approximate solutions of the equivalent variationalstatement of the problem involving only one and two terms in the above series.

Answer:The functional associated with the above problem and which is to be minimized is

I(u) =∫ 1

0[1

2(u′)2 − 1

2u2 − xu]dx

15

introducing the approximation u1(x) ≈ u(x) given by

u1 = a1x(1 − x)

into the functional, carrying out the integration and then determining the value of a1 thatmakes dI/da1 = 0 yields a1 = 5/18.

Introducing now the approximation u2(x) ≈ u(x) given by

u2 = a1x(1 − x) + a2x2(1 − x)

into the functional, carrying out the integration and then determining the values of a1 anda2 that makes ∂I/∂a1 = ∂I/∂a2 = 0 yields a1 = 71/369 and a2 = 7/41.

7 Analytical Solutions of Partial Differential Equations

7.1 Elliptic Equation - Separation of Variables Method

Find the solution u(x, y) of the following problem in 0 ≤ x ≤ 1, and 0 ≤ y ≤ 2 using themethod of separation of variables. Write down the first three terms of the solution obtained.

∂2u

∂x2+

∂2u

∂y2= 0

subject to

u(x, 0) = x

at y = 0 for 0 ≤ x ≤ 1,

u(x, 2) = 0

at y = 2 for 0 ≤ x ≤ 1,

u(0, y) = 0

at x = 0 for 0 ≤ y ≤ 2, and

u(1, y) = 0

at x = 1 for 0 ≤ y ≤ 2.Answer:Assume u(x, y) = X(x)Y (y). Substituting into the PDE yields

X ′′

X= −Y ′′

Y= −λ2

16

The associated BVP for X is

X ′′ + λ2X = 0

subject to

X(0) = X(1) = 0

The general solution

X(x) = A cos(λx) + B sin(λx)

and introducing the boundary conditions gives

Xn(x) = Bn sin(nπx)

for n = 1, 2, ..... Now the associated BVP for Y is

Y ′′ − λ2Y = 0

Since Y (0) 6= 0, the general solution is

Y (y) = C cosh(λ[y − 2]) + D sinh(λ[y − 2])

but since Y (2) = 0

Yn(y) = Dn sinh(nπ[y − 2])

The desired solution then has the form

u(x, y) =∞∑

n=1

Xn(x)Yn(y) =∞∑

n=1

an sin(nπx) sinh(nπ[y − 2])

Finally the fourth boundary condition requires that

u(x, 0) = x =∞∑

n=1

[−an sinh(2nπ)] sin(nπx)

which is the Fourier sine series representation of x with Fourier coefficients

−an sinh(2nπ) = 2∫ 1

0x sin(nπx)dx = 2

(−1)n+1

nπ

Therefore the required solution is

u(x, y) =∞∑

n=1

(−1)n 2

nπ sinh(2nπ)sin(nπx) sinh(nπ[y − 2])

17

7.2 Parabolic Equation - Separation of Variables Method

Find the temperature distribution inside a slab of thickness L and thermal diffusivity αundergoing transient heat conduction. The initial temperature distribution of the slab isT (x, 0) = f(x). The slab is quenched by forcing the temperature at its two surfaces x = 0and x = L to become equal to zero (i.e. (T (0, t) = T (L, t) = 0; Dirichlet homogeneousconditions) for t > 0.

Answer:The mathematical statement of the heat equation for this problem is:

∂T (x, t)

∂t= α

∂2T (x, t)

∂x2

subject to

T (0, t) = T (L, t) = 0

and

T (x, 0) = f(x)

for all x when t = 0.Assume now a solution exists with the form

T (x, t) = X(x)Γ(t)

Introducing the above assumption into the heat equation and rearranging yields

1

X

d2X

dx2=

1

αΓ

dΓ

dt

However since X(x) and Γ(t), the left hand side of this equation is only a function of xwhile the right hand side is a function only of t. For this to avoid being a contradiction (forarbitrary values of x and t) both sides must be equal to a constant. For physical reasons therequired constant must be negative; let us call it −ω2.

Therefore, the original PDE is transformed into the following two ODE’s

1

X

d2X

dx2= −ω2

and

1

αΓ

dΓ

dt= −ω2

General solutions to these equations are readily obtained and are

X(x) = A′ cos(ωx) + B′ sin(ωx)

18

and

Γ(t) = C exp(−ω2αt)

Substituting back into our original assumption yields

T (x, t) = X(x)Γ(t) = [A cos(ωx) + B sin(ωx)] exp(−ω2αt)

where the constant C has been combined with A′ and B′ to give A and B without losingany generality.

Now we introduce the boundary conditions. Since T (0, t) = 0, necessarily A = 0. Fur-thermore, since also T (L, t) = 0, then sin(ωL) = 0 (since B = 0 is an uninteresting trivialsolution.) and there is an infinite number of values of ω which satisfy this conditions, i.e.

ωn =nπ

L

with n = 1, 2, 3, ....Each value of ω yields an independent solution satisfying the heat equation as well as

the two boundary conditions. Therefore we have now an infinite number of independentsolutions Tn(x, t) for n = 1, 2, 3, ... given by

Tn(x, t) = [Bn sin(ωnx)] exp(−ω2nαt)

Applying the principle of superposition yields a more general solution by linear combi-nation, i.e.

T (x, t) =∞∑

n=1

Tn(x, t) =∞∑

n=1

[Bn sin(ωnx)] exp(−ω2nαt) =

∞∑n=1

[Bn sin(nπx

L)] exp(−(

nπ

L)2αt)

The last step is to ensure the values of the constants Bn are chosen so as to satisfy theinitial condition, i.e.

T (x, 0) = f(x) =∞∑

n=1

Bn sin(nπx

L) =

∞∑n=1

Bn sin(nπx

L)

Note that this is the Fourier sine series representation of the function f(x).A key property of the eigenfunctions is the orthonormality property expressed in the

case of the sin(ωnx) functions as

∫ L

0sin(

nπx

L) sin(

mπx

L)dx =

{0 , n 6= mL/2 , n = m

Using the orthonormality property one can multiply the Fourier sine series representationof f(x) by sin(mπx

L) and integrate from x = 0 to x = L to produce the result

Bn =2

L

∫ L

0f(x) sin(

nπx

L)dx

19

for n = 1, 2, 3, ....Explicit expressions for the Bn’s can be obtained for simple f(x)’s, for instance if

f(x) =

{x , 0 ≤ x ≤ L

2

L − x , L2≤ x ≤ L

then

Bn =

4Ln2π2 , n = 1, 5, 9, ...− 4L

n2π2 , n = 3, 7, 11, ...0 , n = 2, 4, 6, ...

Finally, the resulting Bn’s can be substituted into the general solution above to give

T (x, t) =∞∑

n=1

Tn(x, t) =∞∑

n=1

[2

L

∫ L

0f(x′) sin(

nπx′

L)dx′] sin(

nπx

L) exp(−n2π2αt

L2)

and for the specific f(x) given above

T (x, t) =4L

π2[exp(−π2αt

L2) sin(

πx

L) − 1

9exp(−9π2αt

L2) sin(

3πx

L) + ...]

Another important special case is when the initial temperature f(x) = Ti = constant.The result in this case is

T (x, t) =4Ti

π

∞∑n=0

1

(2n + 1)sin(

(2n + 1)πx

L) exp(−n2π2αt

L2)

7.3 Parabolic Equation - Laplace Transform Method

Find the function T (x, t) in 0 ≤ x < ∞ satisfying

∂2T

∂x2=

1

α

∂T

∂t

and subject to

T (0, t) = f(t)

T (x → ∞, t) = 0

and

T (x, 0) = 0

20

using the Laplace Transform method.

Answer:Taking Laplace transforms one gets

d2T (x, s)

dx2− s

αT (x, s) = 0

T (0, s) = f(s)

T (x → ∞, s) = 0

The solution of this problem is

T (x, s) = f(s)e−x√

s/α = f(s)g(x, s) = L[f(t) ∗ g(x, t)]

Inversion then produces

T (x, t) = f(t) ∗ g(x, t) =∫ t

0f(τ)g(x, t − τ)dτ

Inversion of g(x, s) to get g(x, t) finally gives

T (x, t) =x√4πα

∫ t

τ=0

f(τ)

(t − τ)3/2exp[− x2

4α(t − τ)]dτ

If f(t) = T0 = constant, the solution is

T (x, s) =T0

sexp(−x

√s/α)

and inverting the transform yields

T (x, t) = T0erfc(x

2√

αt)

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7.4 Parabolic Equation - Laplace Transform Method

Find a bounded solution u(r, t) of the following problem using the Laplace transform method.

∂2u

∂r2+

1

r

∂u

∂r=

∂u

∂t

subject to

u(r, 0) = 0

and

u(a, t) = u0

where t > 0, 0 < r < a with a = constant.

Answer:Taking Laplace transforms of the terms in the given PDE yields

d2u(r, s)

dr2+

1

r

du(r, s)

dr= su(r, s)

This is a Bessel-type equation with general solution given by

u(r, s) = AJ0(ir√

s) + BY0(ir√

s)

Since we look for a bounded solution, necessarily B = 0. Now transformation of the boundarycondition at r = a yields

u(a, s) =u0

s

and combining with the above gives

A =u0

s

1

J0(ia√

s)

so that the solution for u(r, s) becomes

u(r, s) = u0J0(ir

√s)

sJ0(ia√

s)

Finally, the inverse transformation gives

u(r, t) = u0[1 − 2∞∑

n=1

e−λ2nt/a2

J0(λnr/a)

λnJ1(λn)]

where λn are the roots of

J0(λn) = 0

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7.5 Hyperbolic Equation - Separation of Variables Method

For a freely vibrating square membrane of side l , supported along the boundary x = 0,x = l, y = 0, y = l, obtain the deflection w(x, y, t) in the form

w =∞∑

m=1

∞∑n=1

sin(mπx

l) sin(

nπy

l)[amn cos(ωmnt) + bmn sin(ωmnt)]

where

ωmn = π√

m2 + n2

√T

ρl2

Answer: The function w(x, y, t) must satisfy the wave equation

∂2w

∂x2+

∂2w

∂y2=

1

α2

∂2w

∂t2

where α2 = T/ρ.Assume that wp = X(x)Y (y)Θ(t) and substitute to get

1

XX ′′ +

1

YY ′′ =

1

α2

1

ΘΘ′′ = −c2

and

1

XX ′′ = −k2

introducing the boundary conditions gives

Xm(x) = bm sin(mπ

lx)

and

Yn(x) = en sin(nπ

ly)

and

c =π

l

√m2 + n2

so that

1

ΘΘ′′ = −Tπ2

ρl2(m2 + n2) = −ω2

mn

with solution

Θ(t) = f cosωmnt + g sin ωmnt

combining the above and relabeling the constants leads to the required expression.

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