MATH 543: METHODS OF APPLIED MATHEMATICS I

Final Exam: Solutions 04 January 2002 Friday 15.30-17.30 SAZ:18

QUESTIONS: (25)1. Show that

| sin αx| = \frac{2}{π} − \frac{4}{π} ∑_{n=1}^{∞} \frac{cos 2nαx}{4n^{2} − 1}

where α is any real number.

Solution: Since the function is even we shall use the even part of the orthonormal set \{\sqrt{\frac{α}{2π}}, \sqrt{\frac{α}{π}} cos nαx\} for n = 1, 2, ⋯ over the interval [−\frac{π}{α}, \frac{π}{α}]. Then

| sin αx| = a_{0} + ∑_{k=1}^{∞} a_{k} cos kαx

where

a_{0} = \frac{α}{π} ∫_{0}^{π/α} sin αxdx = \frac{2}{pi},

a_{k} = \frac{2α}{π} ∫_{0}^{π/α} sin αx cos kαxdx = \frac{2}{π} \frac{1 + (−1)^{k}}{1 − k^{2}} ,

From the above expression we have k = 2n, n = 1, 2 ⋯ and the result follows.

(25)2. Find a two term perturbation of

u′ + u = \frac{1}{1 + εu}, t > 0, 0 < ε << 1

with u(0) = 0. (a) use regular perturbation (b) use the Poincare-Lindstedt method. Compare your results and discuss the validity of the approximations.

Solution: (a) Regular perturbation.

u(t) = u_{0}(t) + εu_{1}(t) + O(ε^{2})

where

u′_{0} + u_{0} = 1, u_{0}(0) = 0,
u′_{1} + u_{1} + u_{0} (u′_{0} + u_{0}) = 0, u_{1}(0) = 0,

Solving u_{0} and u_{1} we obtain that

u(t) = 1 − e^{−t} + ε(te^{−t} + e^{−t} − 1) + O(ε^{2})

(b) Poincare’-Lindstaedt Method.

t = \frac{τ}{w},

w = 1 + εw_{1} + O(ε^{2}), u(τ) = u_{0} + εu_{1} + O(ε^{2})

with the following perturbation equations

u_{0, τ} + u_{0} = 1, u_{0}(0) = 0, u_{0}(τ) = 1 − e^{−τ}
u_{1, τ} + u_{1} + 1 + w_{1} e^{−τ} − e^{−τ} = 0,

(1)

We choose w_{1} = 1 and hence we get u_{1} = −1 + e^{−τ} hence

u(τ) = 1 − e^{−tau} − ε(1 − e^{−τ}) + O(ε^{2})

and τ = (1 + ε) t + O(ε^{2}) Both perturbation method gives uniformly valid approximation but the PL method gives more better approximation (25)3. (a) Let

I(λ) = ∫_{a}^{b} f(t) e^{λ g(t)} dt

where f is continuous and and g is smooth enough in the interval [a, b]. Furthermore the maximum of g occurs at t = a (one of the end points of the integration interval) with g′(t) < 0 for all t ∈ [a, b]. Prove that the dominant term is

I(λ) ∼ −\frac{f(a) e^{λg(a)}}{λg′(a)}

for λ >> 1

(b) Verify the following approximation for large λ

∫_{1}^{2} \sqrt{3 + t} e^{\frac{λ}{1+t}} dt ∼ \frac{8}{λ} e^{λ/2}

Solution: The dominant term will at the point where the function g has maximium hence

I(λ) ∼ f(a) e^{λg(a)} ∫_{a}^{b} e^{λ g′(a)(t−a)}dt ∼ \frac{f(a)e^{λg(a)}}{λg′(a)} ∫_{0}^{λg′(b−a)} e^{u}du

since λ >> 1 the the integral above gives −1 hence the result follows.

(25)4. Solve the following boundary value problem:

u′′ + w^{2}u = f(x), u(−π) = u′(π) = 0

where w is a positive constant.

Solution: Using the Green’s function technique we find that

G(x, y) = \{ \begin{matrix}U_{<}, x < y \\ U_{>}, x > y \end{matrix}

where

U_{<} = A cos wx + B sin wx, U_{>} = C cos wx + D sin wx

Here A, B, C, D are constants (wrt x) to be determined. Using the boundary conditions we obtain

U_{<} = \frac{A}{sin wπ} sin w(x + π), U_{>} = \frac{D}{sin wπ} cos w(x − π)

Here there is a condition sin wπ ≠ 0 but this not so important. The result will not be effected by this constraint. Using the Continuity and jump conditions at x = y we obtain

U_{<} = −\frac{1}{w cos 2wπ} cos w(y − π) sin w(x + π),

U_{>} = −\frac{1}{w cos 2wπ} cos w(x − π) sin w(y + π)

then the solution is

u(x) = −\frac{cos w(x − π)}{w cos 2wπ} ∫_{−π}^{x} f(y) sin w(y + π) dy

−\frac{sin w(x + π)}{w cos 2wπ} ∫_{x}^{π} f(y) cos w(y − π) dy

Here we must assume that cos 2wπ ≠ 0. If cos2wπ = 0 it means that there exists nontrivial solution of the homogenous equation satisfying both boundary conditions (Prove this statement). In that case the above Green’s function technique fails. We must modify the definition of the Green’s function as we have done it in SET 5 page 8.