Mathematics 373 – Applied Mathematics (PDE) Selected Solutions – Midterm Problem Set March 12, 2001

I. B. As always, the value of a solution of a first order linear PDE with prescribed values along some curve in the plane is only uniquely determined on the subset of the plane consisting of all points on characteristics that intersect the curve where the values are prescribed. Here the characteristics are the curves y = arcsin(x) + c for all real c. With the condition u(x, 0) = \sqrt{1 − x^{2}}, we have to determine which characteristics intersect the x-axis (the given curve). By considering the graph of the arcsin function, we see that only the characteristics with −π/2 ≤ c ≤ π/2 intersect the x-axis here. (If c < −π/2, the graph is shifted “too far down” along the y-axis; if c > π/2 it is shifted “too far up”.) Hence the region where the solution is uniquely determined is

{(x, y) : −1 ≤ x ≤ 1, arcsin(x) − π/2 ≤ y ≤ arcsin(x) + π/2}

II. A. Using D’Alembert’s solution of the wave equation:

u(x, t) = \frac{1}{2}(f(x − ct) + f(x + ct)) + \frac{1}{2c} ∫_{x−ct}^{x+ct} g(s) ds

we assume that if f(x), g(x) = 0 for all |x| > B. Then |x| > B+ct is the same as x < −(B+ ct) or x > B +ct. From this we get x+ct < −B (which implies x−ct = (x+ct)−2ct < −B too since we take t ≥ 0) or x − ct > B (which implies x + ct = (x − ct) + 2ct > B too). Therefore f(x − ct) = f(x + ct) = 0. Similarly the interval [x − ct, x + ct] ⊂ (−∞, −B] in the first case, and [x − ct, x + ct] ⊂ [B, ∞) in the second case. Therefore the interval of integration is contained in the range where g is known to be zero. Hence ∫_{x−ct}^{x+ct} g(s) ds = 0.

II. B. We have

KE + P E = ∫_{−∞}^{∞} ρu_{t}^{2} + T u_{x}^{2} dx

Hence by the hint and the wave equation:

\frac{d}{dt}(KE + P E) = \frac{d}{dt} ∫_{−∞}^{∞} ρu_{t}^{2} + T u_{x}^{2} dx
= ∫_{−∞}^{∞} 2ρu_{t}u_{tt} + 2T u_{x}u_{xt} dx
= 2T ∫_{−∞}^{∞} u_{t}u_{xx} + u_{x}u_{xt} dx

Integrating by parts in the first term (U = u_{t}, dV = u_{x} dx), we get

\begin{matrix}= 2T u_{t}u_{x}|_{−∞}^{∞} + ∫_{−∞}^{∞} −u_{xt}u_{x} + u_{x}u_{xt} dx \\ = 2T u_{t}u_{x}|_{−∞}^{∞} + 0 \end{matrix}

Now, by part A of the problem, if |x| > B + ct, we have u(x, t) = 0. Hence the limit as x → ±∞ of u or any of its derivatives must be zero. Hence

\frac{d}{dt}(KE + P E) = 0

which is what we wanted to show.

IV. B. The eigenvalues and eigenfunctions of the boundary value problem

\{ \begin{matrix}X′′(x) = −λX(x) \\ X(0) = 0 \\ X′(L) = 0 \end{matrix}

come only from the case λ > 0, so X = A cos(\sqrt{λ}x)+B sin(\sqrt{λ}x). The condition X(0) = 0 implies A = 0. Then the other boundary condition says B\sqrt{λ} cos(\sqrt{λ}L) = 0. So \sqrt{λ}L = \frac{(2n−1)π}{2} for some n ≥ 1 (that is, \sqrt{λ}L has the form (odd integer)·π/2).

eigenvalues: λ = (\frac{(2n−1)π}{2L} )^{2}.

eigenfunctions: ϕ_{n}(x) = sin (\frac{(2n−1)πx}{2L}).

IV B. To derive the orthogonality relation

0 = ∫_{0}^{L} sin (\frac{(2n − 1)πx}{2L}) sin (\frac{(2m − 1)πx}{2L}) dx

when m ≠ n, use the trig identity

sin(A) sin(B) = \frac{1}{2}(cos(A − B) − cos(A + B))

then integrate.

V. A. The Fourier sine coefficients of the given initial string shape are

B_{n} = \frac{2}{L} ∫_{0}^{L} f(x) sin (\frac{nπx}{L}) dx

= \frac{2}{L} [∫_{0}^{a} \frac{hx}{a} sin (\frac{nπx}{L}) dx + ∫_{a}^{L} \frac{h(L − x)}{(L − a)} sin (\frac{nπx}{L}) dx]

= \frac{2hL^{2}}{a(L − a)} · sin (\frac{nπa}{L}) · \frac{1}{(nπ)^{2}}

The solution of the wave equation is

u(x, t) = ∑_{n=1}^{∞} B_{n} sin (\frac{nπx}{L}) cos (\frac{nπct}{L})