Advanced Engineering Mathematics II. Solved Sample Problems. 1 Series
Solution of Ordinary Differential Equations. 1.1. Obtain the solution of the
following ...
Advanced Engineering Mathematics II Solved Sample Problems
1
Series Solution of Ordinary Differential Equations
1.1 Obtain the solution of the following differential equation in terms of Maclaurin series dy d2 y +x −y =0 2 dx dx Answer: P n Let f (x) = ∞ n=0 An x to obtain (2A2 − A0 )x0 +
∞ X
[(n + 2)(n + 1)An+2 + nAn − An ]xn = 0
n=1
From the first coefficient A0 is an arbitrary constant; then for n = 1 with A1 arbitrary constant, A3 = 0. For n ≥ 1 An+2 = −
(n − 1) An (n + 2)(n + 1)
Must now consider two cases: n even and n odd. The general solution is f (x) = A1 x + A0 [1 + x2 /2 − x4 /24 + x6 /240 + ...]
1.2 Use the method of Frobenius to obtain a general solution of the following differential equation valid near x = 0 x
dy d2 y + 2 + xy = 0 2 dx dx 1
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 0 and into the xy 00 terms, make a dummy index change from n to n + 1. * Investigate the behavior of the recurrence relationship for the smaller root s2 first. Answer: P n+s Let y = f (x) = ∞ . Thus n=0 An x xf (x) = x
∞ X n=0
An x
n+s
∞ X
=
n=0
An x
n+s+1
=
∞ X n=1
An−1 xn+s
where instead of the dummy index n, the dummy n − 1 was used. Similarly 2
∞ ∞ X X dy = 2f (x)0 = 2 (n + s)An xn+s−1 = 2 (n + 1 + s)An+1 xn+s dx n=0 n=−1
Where the dummy index has been changed from n to n + 1 in order to have x raised to the power n + s in the sum. Finally, x
∞ ∞ X X d2 y 00 n+s−1 = xf (x) = (n + s − 1)(n + s)A x = (n + s)(n + 1 + s)An+1 xn+s n 2 dx n=0 n=−1
Where the dummy index has also been changed from n to n + 1. Now, substituting the above into the original ODE yields ∞ X
(n + s)(n + 1 + s)An+1 xn+s +
n=−1
∞ X n=−1
2(n + 1 + s)An+1 xn+s +
∞ X n=1
An−1 xn+s = 0
And division by xn+s yields the general recurrence relationship ∞ X
(n + 1 + s)(n + s)An+1 +
n=−1
∞ X n=−1
2(n + 1 + s)An+1 +
∞ X 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 = 1 we must expect either no solution or a complete one for s2 = −1 and of course, one solution for s1 = 0. Now, when n = 0 the recurrence relation becomes (1 + s)sA1 + 2(1 + s)A1 = A1 (1 + s)(s + 2) = 0 2
which is always true when s = −1, regardless of the value of A1 , therefore A1 is also arbitrary and we necessarily have a complete solution. With n ≥ 1 1 An−1 An+1 = (−1) (n + 1 + s)(n + 2 + s) which yields, for s = −1, An+1 = (−1)
1 An−1 n(n + 1)
Specifically, for n = 1, 1 A2 = (−1) A0 2 for n = 2 1 A3 = (−1) A1 6 for n = 3 A4 = (−1)
1 1 1 A2 = (−1) (−1) A0 12 12 2
for n = 4 1 1 1 A3 = (−1) (−1) A1 20 20 6 The recurrence relationship can finally be represented by the following two equations in more compact form A5 = (−1)
A2l = (−1)l
1 A0 , (2l)!
l≥0
and A2l+1 = (−1)l
1 A1 , (2l + 1)!
l≥0
Therefore, the general solution (corresponding to s = −1) is y(x) = x−1
∞ X n=0
An xn = x−1 [A0 + A1 x + A2 x2 + ...] =
1 1 = x−1 [A0 + A1 x − A0 x2 − A1 x3 + ..] = 2 6 1 1 1 1 4 x + ...] = = A0 [x−1 − x + x3 + ...] + A1 [1 − x2 + 2 24 6 120 cos(x) sin(x) + A1 = A0 x x 3
1.3 Apply Frobenius method to find the general solution of the equation: Ly = x2
dy d2 y + (x2 + x) − y = 0 2 dx dx
Here the indicial equation is s2 − 1 = 0 with solutions s1 = +1, s2 = −1 and a series solution is 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 1 1 1 x2 − x3 + ...] = y1 = xA0 [1 − x + 3 3×4 3×4×5 e−x − 1 + x = 2A0 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 = −1 becomes y2 = A0
2
1−x x
Special Functions
2.1 Obtain the general solution of the following equation in terms of Bessel functions or, if possible, in terms of elementary functions x
d2 y dy − 3 + xy = 0 2 dx dx
Answer: 4
As described in the class notes, the most general ODE satisfied by Bessel functions has the form x2
dy d2 y + x(a + 2bxr ) + [c + dx2s − b(1 − a − r)xr + b2 x2r ]y = 0 2 dx dx
The given equation can be put in this form simply multiplying by x yielding the following x2
dy d2 y − 3x + x2 y = 0 2 dx dx
where a = −3, b = 0, c = 0, d = 1 and s = 1. The solution now can be written down directly as y(x) = x2 Zp (x) = x2 (c1 Jp (x) + c2 Yp (x)) with s
1 1−a 2 ) −c=2 ( p= s 2 Alternatively, the compatible form above can be obtained directly from Bessel’s equation X2
dY d2 Y + (X 2 − P 2 )Y = 0 +X 2 dX dX
by using the transformed variables X = d1/2 xs /s = x, and Y = y/x2 and P = 2. The solution to the above Bessel’s equation is Y (X) = c1 JP (X) + c2 YP (X) and using the original variables the desired solution is y(x) = c1 x2 J2 (x) + c2 x2 Y2 (x)
3
Characteristic Functions
3.1 A uniform string is unrestrained from transverse motion at x = 0 while at x = L is attached to a yielding support of modulus k, the ends of the string being constrained against appreciable movement parallel to the axis of rotation. Show that the nth critical speed ωn is given by µn ωn = L
s
5
T ρ
where cot(µn ) = αµn and α = T /kL. Answer: Solve ρ d2 y + ω2 y = 0 2 dx T subject to the free end condition at x = 0 y 0 (0) = 0 and, since a yielding support is nothing but an elastic spring which exerts a restoring force proportional to its stretch, (i.e. T y 0 = −ky), T Ly 0 (L) = αLy 0 (L) = −y(L) kL at x = L. Inspection shows that the functions
r
y1 (x) = cos( and
ρ ωx) T
r
ρ ωx) T both satisfy the ODE, therefore, the general solution is y2 (x) = sin(
r
r
ρ ρ ωx) + c2 sin( ωx) y(x) = c1 y1 (x) + c2 y2 (x) = c1 cos( T T The stated boundary conditions require that, c2 = 0 and that r
µ=
ρ ωL T
be the roots of cot(µ) = αµ
3.2 Determine the Fourier coefficients of the following expansion for 0 < x < π, f (x) = x =
∞ X 0
Answer: Here 2 An = π
Z 0
π
An sin(nx)
(
x sin(nx)dx =
6
− n2 2 n
n = even n = odd
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 boundary value problem in x ∈ [1, 2]: Find y satisfying 2 2 sin(ln x) y 00 = − y 0 + 2 y + x x x2 and subject to y(1) = 1 y(2) = 2 Answer: The exact solution is (check!) y(x) = 1.1392070132x −
0.03920701320 − 0.3 sin(ln x) − 0.1 cos(ln x) x2
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 following boundary value problem in x ∈ [1, 2]: Find y satisfying 2 2 sin(ln x) y 00 = − y 0 + 2 y + x x x2 and subject to y(1) = 1 7
y(2) = 2 Answer: The exact solution is again y(x) = 1.1392070132x −
0.03920701320 − 0.3 sin(ln x) − 0.1 cos(ln x) x2
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 and determine it if it exists. a) F = 2xyz 3 i − (x2 z 3 + 2y)j + 3x2 yz 2 k b) F = 2xyi + (x2 + 2yz)j + (y 2 + 1)k Answer: 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) = x2 y + y 2 z + 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 = y 2 , z = 0in the x − y plane from (0, 0, 0) to (1, 1, 0) and then 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 Z C1
F · dr =
Z
1
0
y 3 dy =
1 4
Now, for C2 , dx = dy = 0 and Z C2
F · dr =
Z
1
0
8
zdz =
1 2
Thus Z
F · dr =
C
Z C1
F · dr +
Z C2
3 1 1 + = 4 2 4
F · dr =
b) In this case dx = dy = dz and Z C
F · dr =
Z
1
0
3z 2 dz = 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 hemisphere x2 + y 2 + z 2 = 1 with z ≥ 0. Answer: Here n = xi + yj + zk and ds = dxdy/z therefore Z Z S
F · nds =
Z Z D
(x2 + y 2 )
dxdy z
where D is the projection of S on the x − y plane (i.e. the unit circle). Therefore Z Z S
The final result
RR S
F · nds = 4
Z
1
0
Z
√ 1−x2
q
x2 + y 2
1 − (x2 + y 2 )
0
dxdy
F·nds = 43 π 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: 9
The element of volume in the u1 , u2 , u3 coordinate system is given by dτ = (U1 · U2 × U3 )du1 du2 du3 Introducing u1 = u, u2 = φ, and u3 = θ and recalling that for i = 1, 2, 3 Ui =
∂r ∂ui
one has ∂x ∂y ∂z +j +k = ∂u1 ∂u1 ∂u1 = ia sin φ cos θ + jb sin φ sin θ + kc cos φ U1 = i
∂x ∂y ∂z +j +k = ∂u2 ∂u2 ∂u2 = iau cos φ cos θ + jbu cos φ sin θ − kcu sin φ U2 = i
∂x ∂y ∂z +j +k = ∂u3 ∂u3 ∂u3 = −iau sin φ sin θ + jbu sin φ cos θ + k0 U3 = i
Therefore, ∂x ∂y ∂z ∂x ∂x ∂x ∂u1 ∂u1 ∂u1 ∂u1 ∂u2 ∂u3 ∂x ∂y ∂z ∂y ∂y ∂y ∂(x, y, z) =| | = U1 · U2 × U3 = ∂u ∂u1 ∂u2 ∂u3 = ∂u ∂u ∂(u, φ, θ) ∂x2 ∂y2 ∂z2 ∂z ∂z ∂z ∂u ∂u ∂u ∂u ∂u ∂u 1 2 3 3 3 3 a sin φ cos θ au cos φ cos θ − au sin φ sin θ = abcu2 sin φ = b sin φ sin θ bu cos φ sin θ bu sin φ cos θ c cos φ − cu sin φ 0
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 − y 2 at (x, y) = (0, 0). 10
Answer: Here fx = −2x = 0 and fy = −2y = 0 Also fxx = −2 = fyy < 0 and since fxy = 0, 2 fxx fyy = 4 > fxy
the point (0, 0) is a relative maximum.
6.2 Determine the shortest distance from the point (1, 0) to the parabola y 2 = 4x. Answer: The function to be minimized is f (x, y) = (x − 1)2 + y 2 while the constraint is g(x, y) = y 2 − 4x = 0 The auxiliary function becomes φ(x, y) = f (x, y) + λg(x, y) = (x − 1)2 + y 2 + λ(y 2 − 4x) The necessary conditions for a minimum are then φx = 2(x − 1) − 4λ = 0 φy = 2y + 2λy = 0 φλ = g(x, y) = y 2 − 4x = 0 The only real solution to the above system is x = 0, y = 0 and λ = 1/2. 11
6.3 Consider the system of nonlinear equations f1 (x1 , x2 ) = x1 + x2 − 3 = 0 f2 (x1 , x2 ) = x21 + x22 − 9 = 0 Let x0 = (1, 5) and use Newton’s method to find the solution. Answer: Here ∂f1 ∂x1 ∂f2 ∂x1
J=
∂f1 ∂x2 ∂f2 ∂x2
!
=
1 1 2x1 2x2
!
From this, the linear system for the first correction is 1 1 2 10
!
s01 s02
!
=−
3 17
!
with solution s01 = − 13 , s02 = − 11 , which gives x1 = (−0.625, 3.625). Repeating the process 8 8 2 yields x = (−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 guess x1 = 0.1, x2 = 0.1, x3 = −0.1. 1 =0 2 f2 (x) = x21 − 81(x2 + 0.1)2 + sin(x3 ) + 1.06 = 0 10π − 3 =0 f3 (x) = exp(−x1 x2 ) + 20x3 + 3 f1 (x) = 3x1 − cos(x2 x3 ) −
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 following functional Z
I(u) =
0
π/2
[(
du 2 ) − u2 + +2xu]dx dx 12
with u(0) = u(π/2) = 0 Answer: Here the Euler equation is d2 u +u=x dx2 the solution of which is u = c1 cos(x) + c2 sin(x) + x Incorporating the boundary conditions finally yields π u = x − ( ) sin(x) 2 and the corresponding minimum value of I is π2 π Imin = −( )(1 − ) 2 12
6.6 Find approximate solutions to the BVP consisting of finding u(x) that satisfyies u00 + 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 X i=1
ai φi = a1 φ1 + a2 φ2 + ... + aN φN = a1 x(1 − x) + a2 x2 (1 − x) + ... + aN xN (1 − x)
Then apply the Ritz method to construct approximate solutions of the equivalent variational statement 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 Z
I(u) =
1 0
1 1 [ (u0 )2 − u2 − xu]dx 2 2 13
introducing the approximation u1 (x) ≈ u(x) given by u1 = a1 x(1 − x) into the functional, carrying out the integration and then determining the value of a1 that makes dI/da1 = 0 yields a1 = 5/18. Introducing now the approximation u2 (x) ≈ u(x) given by u2 = a1 x(1 − x) + a2 x2 (1 − x) into the functional, carrying out the integration and then determining the values of a1 and a2 that makes ∂I/∂a1 = ∂I/∂a2 = 0 yields a1 = 71/369 and a2 = 7/41.
7 7.1
Analytical Solutions of Partial Differential Equations 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 the method of separation of variables. Write down the first three terms of the solution obtained. ∂ 2u ∂ 2u + =0 ∂x2 ∂y 2 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 Y 00 X 00 =− = −λ2 X Y 14
The associated BVP for X is X 00 + λ2 X = 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 00 − λ2 Y = 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 ∞ X
u(x, y) =
n=1
Xn (x)Yn (y) =
∞ X n=1
an sin(nπx) sinh(nπ[y − 2])
Finally the fourth boundary condition requires that u(x, 0) = x =
∞ X
[−an sinh(2nπ)] sin(nπx)
n=1
which is the Fourier sine series representation of x with Fourier coefficients −an sinh(2nπ) = 2
Z 0
1
x sin(nπx)dx = 2
(−1)n+1 nπ
Therefore the required solution is u(x, y) =
∞ X
(−1)n
n=1
2 sin(nπx) sinh(nπ[y − 2]) nπ sinh(2nπ) 15
7.2
Parabolic Equation - Separation of Variables Method
Find the temperature distribution inside a slab of thickness L and thermal diffusivity α undergoing transient heat conduction. The initial temperature distribution of the slab is T (x, 0) = f (x). The slab is quenched by forcing the temperature at its two surfaces x = 0 and x = L to become equal to zero (i.e. (T (0, t) = T (L, t) = 0; Dirichlet homogeneous conditions) for t > 0. Answer: The mathematical statement of the heat equation for this problem is: ∂ 2 T (x, t) ∂T (x, t) =α ∂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 dΓ 1 d2 X = 2 X dx αΓ dt However since X(x) and Γ(t), the left hand side of this equation is only a function of x while the right hand side is a function only of t. For this to avoid being a contradiction (for arbitrary values of x and t) both sides must be equal to a constant. For physical reasons the required constant must be negative; let us call it −ω 2 . Therefore, the original PDE is transformed into the following two ODE’s 1 d2 X = −ω 2 2 X dx and 1 dΓ = −ω 2 αΓ dt General solutions to these equations are readily obtained and are X(x) = A0 cos(ωx) + B 0 sin(ωx) 16
and Γ(t) = C exp(−ω 2 αt) Substituting back into our original assumption yields T (x, t) = X(x)Γ(t) = [A cos(ωx) + B sin(ωx)] exp(−ω 2 αt) where the constant C has been combined with A0 and B 0 to give A and B without lossing any generality. Now we introduce the boundary conditions. Since T (0, t) = 0, necessarily A = 0. Furthermore, since also T (L, t) = 0, then sin(ωL) = 0 (since B = 0 is an uninteresting trivial solution.) 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 independent solutions Tn (x, t) for n = 1, 2, 3, ... given by Tn (x, t) = [Bn sin(ωn x)] exp(−ωn2 αt) Applying the principle of superposition yields a more general solution by linear combination, i.e. T (x, t) =
∞ X n=1
Tn (x, t) =
∞ X
[Bn sin(ωn x)] exp(−ωn2 αt) =
n=1
∞ X
[Bn sin(
n=1
nπ nπx )] exp(−( )2 αt) L L
The last step is to ensure the values of the constants Bn are chosen so as to satisfy the initial condition, i.e. T (x, 0) = f (x) =
∞ X n=1
Bn sin(
∞ X nπx nπx )= ) Bn sin( L L n=1
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(ωn x) functions as Z 0
L
mπx nπx ) sin( )dx = sin( L L
(
0 , n 6= m L/2 , n = m
Using the orthonormality property one can multiply the Fourier sine series representation ) and integrate from x = 0 to x = L to produce the result of f (x) by sin( mπx L 2ZL nπx )dx f (x) sin( Bn = L 0 L 17
for n = 1, 2, 3, .... Explicit expressions for the Bn ’s can be obtained for simple f (x)’s, for instance if (
x , 0 ≤ x ≤ L2 L − x , L2 ≤ x ≤ L
f (x) = then
Bn =
4L n2 π 2 − n4L 2 π2
0
, n = 1, 5, 9, ... , n = 3, 7, 11, ... , n = 2, 4, 6, ...
Finally, the resulting Bn ’s can be substituted into the general solution above to give
T (x, t) =
∞ X n=1
Tn (x, t) =
Z ∞ X 2
[
n=1
L
L
0
f (x0 ) sin(
n2 π 2 αt nπx0 nπx )dx0 ] sin( ) exp(− ) L L L2
and for the specific f (x) given above
T (x, t) =
1 9π 2 αt π 2 αt πx 3πx 4L ) − exp(− ) + ...] [exp(− ) sin( ) sin( 2 2 2 π L L 9 L L
Another important special case is when the initial temperature f (x) = Ti = constant. The result in this case is T (x, t) =
7.3
∞ (2n + 1)πx n2 π 2 αt 4Ti X 1 sin( ) exp(− ) π n=0 (2n + 1) L L2
Parabolic Equation - Laplace Transform Method
Find the function T (x, t) in 0 ≤ x < ∞ satisfying 1 ∂T ∂ 2T = 2 ∂x α ∂t and subject to T (0, t) = f (t) T (x → ∞, t) = 0 and T (x, 0) = 0 18
using the Laplace Transform method. Answer: Taking Laplace transforms one gets d2 T¯(x, s) s − T¯(x, s) = 0 2 dx α 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) =
Z 0
t
f (τ )g(x, t − τ )dτ
Inversion of g¯(x, s) to get g(x, t) finally gives T (x, t) = √
x 4πα
Z
t
τ =0
x2 f (τ ) ]dτ exp[− (t − τ )3/2 4α(t − τ )
If f (t) = T0 = constant, the solution is q T0 exp(−x s/α) T¯ (x, s) = s
and inverting the transform yields x T (x, t) = T0 erf c( √ ) 2 αt
19
7.4
Parabolic Equation - Laplace Transform Method
Find a bounded solution u(r, t) of the following problem using the Laplace transform method. ∂u ∂ 2 u 1 ∂u = + ∂r2 r ∂r ∂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 d2 u¯(r, s) 1 d¯ u(r, s) = s¯ u(r, s) + dr2 r dr 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 boundary condition at r = a yields u0 u¯(a, s) = s and combining with the above gives u0 1 √ A= s J0 (ia s) so that the solution for u¯(r, s) becomes
√ J0 (ir s) √ u¯(r, s) = u0 sJ0 (ia s)
Finally, the inverse transformation gives 2
u(r, t) = u0 [1 − 2
2
∞ X e−λn t/a J0 (λn r/a) n=1
where λn are the roots of J0 (λn ) = 0
20
λn J1 (λn )
]
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=
∞ ∞ X X m=1 n=1
sin(
nπy mπx ) sin( )[amn cos(ωmn t) + bmn sin(ωmn t)] l l
where
s
√
ωmn = π m2 + n2
T ρl2
Answer: The function w(x, y, t) must satisfy the wave equation 1 ∂ 2w ∂ 2w ∂ 2w + = 2 2 ∂x2 ∂y 2 α ∂t where α2 = T /ρ. Assume that wp = X(x)Y (y)Θ(t) and substitute to get 1 1 1 00 1 00 X + Y = 2 Θ00 = −c2 X Y α Θ and 1 00 X = −k 2 X introducing the boundary conditions gives mπ x) Xm (x) = bm sin( l and nπ Yn (x) = en sin( y) l and π√ 2 c= m + n2 l so that T π2 1 00 2 Θ = − 2 (m2 + n2 ) = −ωmn Θ ρl with solution Θ(t) = f cos ωmn t + g sin ωmn t combining the above and relabeling the constants leads to the required expression. 21
