Mathematics 173 – Applied Mathematics 1 Exam 1 Solutions – October 14, 1999

I. A) The boundary of D consists of two pieces. The first, α_{1}(t), is the arc of the unit circle

(cos t, sin t) for t ∈ [π/4, 5π/4] traced counterclockwise. The second, α_{2}(t) = (t, t), the segment of the line y = x for t ∈ [−\sqrt{2}/2, \sqrt{2}/2]. The line integral is:

∫_{α} F · T ds = ∫_{α_{1}} F · T ds + ∫_{α_{2}} F · T ds

= ∫_{π/4}^{5π/4}(cos^{2} t − sin t, cos t + sin t) · (− sin t, cos t) dt

+ ∫_{−\sqrt{2}/2}^{\sqrt{2}/2}(t^{2} − t, 2t) · (1, 1) dt

= ∫_{π/4}^{5π/4} − cos^{2} t sin t + 1 + sin t cos t dt + ∫_{−\sqrt{2}/2}^{\sqrt{2}/2} t^{2} + t dt

= [(cos^{3} t)/3 + t − (cos^{2} t)/2]_{π/4}^{5π/4} + [t^{3}/3 + t^{2}/2]_{−\sqrt{2}/2}^{\sqrt{2}/2}

= \frac{1}{3}(−\sqrt{2}/2)^{3} + 5π/4 − \frac{1}{2}(−\sqrt{2}/2)^{2} + \frac{1}{3}(\sqrt{2}/2)^{3} + \frac{1}{2}(\sqrt{2}/2)^{2}

− \frac{1}{3}(\sqrt{2}/2)^{3} − π/4 + \frac{1}{2}(\sqrt{2}/2)^{2} − \frac{1}{3}(−\sqrt{2}/2)^{3} − \frac{1}{2}(−\sqrt{2}/2)^{2}

= π

B) First, since F = (x^{2} − y, x + y) = (P, Q), we have

\frac{∂Q}{∂x} − \frac{∂P}{∂y} = 1 − (−1) = 2

For the double integral over D, if rectangular coordinates are used, then the region must be split into two portions, since the equations of the upper and lower boundaries of the region are different for x ∈ [−1, −\sqrt{2}/2] and for x ∈ [−\sqrt{2}/2, \sqrt{2}/2]. In the first range, the circle forms the upper and lower boundary; in the second, the line is the lower boundary. The double integral is the following sum of iterated integrals:

∫_{−1}^{−\sqrt{2}/2} ∫_{−\sqrt{1−x^{2}}}^{√\overline{1−x^{2}}} 2 dy dx + ∫_{−\sqrt{2}/2}^{\sqrt{2}/2} ∫_{x}^{√\overline{1−x^{2}}} 2 dy dx

It is simpler to use polar coordinates, since the region of integration is D = {(r, θ) : 0 ≤ r ≤ 1, π/4 ≤ θ ≤ 5π/4}. We have

∫∫_{D} 2 dA = ∫_{π/4}^{5π/4} ∫_{0}^{1} 2r dr dθ = π.

(This can also be seen since ∫∫_{D} 2 dA equals twice the area of D, which is 2 · π/2 = π.)

III. A) To verify that u(x, t) = e^{−kt} sin x + cos(x/\sqrt{k}) is a solution of this boundary value

problem, we may simply compute as follows:

u_{t} = −ke^{−kt} sin x u_{xx} = −e^{−kt} sin x − \frac{1}{k} cos(x/\sqrt{k}) .

Hence,

ku_{xx} + cos(x/\sqrt{k}) = k[−e^{−kt} sin x − \frac{1}{k} cos(x/\sqrt{k})] + cos(x/\sqrt{k}) = −ke^{−kt} sin x − cos(x/\sqrt{k}) + cos(x/\sqrt{k}) = −ke^{−kt} = u_{t} .

Then

u(x, 0) = e^{0} sin(x) + cos(x/\sqrt{k}) = sin(x) + cos(x/\sqrt{k})
u(0, t) = e^{−kt} sin(0) + cos(0) = 1
u(π, t) = e^{−kt} sin(π) + cos(π/\sqrt{k}) = cos(π/\sqrt{k})

B) An equilibrium solution is one where u_{t} ≡ 0, so u = u(x). In this case, the PDE u_{t} = ku_{xx} + cos(x/\sqrt{k}) becomes the ordinary differential equation:

\frac{d^{2}u}{dx^{2}} = −\frac{1}{k} cos(x/\sqrt{k}).

This can be solved for u by integrating twice with respect to x:

\frac{du}{dx} = ∫ \frac{d^{2}u}{dx^{2}} dx

= ∫ −\frac{1}{k} cos(x/\sqrt{k}) dx

= −\frac{\sqrt{k}}{k} sin(x/\sqrt{k}) + c_{1}

Then,

\begin{aligned}u & = ∫ −\frac{\sqrt{k}}{k} sin(x/\sqrt{k}) + c_{1} dx \\ & = cos(x/\sqrt{k}) + c_{1}x + c_{2} \end{aligned}

where c_{1}, c_{2} are constants. Using the boundary conditions, at x = 0, u(0) = 0 ⇒ 0 = cos(0)+c_{2}. Hence c_{2} = −1. Then at x = π, u(π) = 0 implies 0 = cos(π/\sqrt{k})+c_{1}π −1. Solving for c_{1}, we get

c_{1} = \frac{1 − cos(π/\sqrt{k})}{π}.

IV. A) By the linearity of differentiation and the assumption that (u_{n})_{t} = k(u_{n})_{xx} for all n,

if u(x, t) = ∑_{n} c_{n}u_{n}(x, t), then

u_{t} = \frac{∂}{∂t} (∑_{n} c_{n}u_{n}(x, t))

= ∑_{n} c_{n}(u_{n})_{t}

= ∑_{n} c_{n}k(u_{n})_{xx}

= k \frac{∂^{2}}{∂x^{2}} (∑_{n} c_{n}u_{n})

= ku_{xx}

Similarly, if each u_{n} satisfies u_{n}(0, t) = 0 = u_{n}(L, t), then u(0, t) = ∑_{n} c_{n}u_{n}(0, t) = ∑_{n} c_{n} · 0 = 0 and u(L, t) = ∑_{n} c_{n}u_{n}(L, t) = ∑_{n} c_{n} · 0 = 0.

B) By “separation of variables plus superposition” recipe, if

u(x, 0) = sin (\frac{3πx}{L}) − 5 sin (\frac{7πx}{L})

= ∑_{n=1}^{∞} d_{n} sin (\frac{nπx}{L})

then the solution of this boundary value problem is

u(x, t) = ∑_{n=1}^{∞} d_{n}e^{−k(\frac{nπ}{L})^{2}t} sin (\frac{nπx}{L})

The d_{n} are found by the formula

d_{n} = \frac{2}{L} ∫_{0}^{L} (sin (\frac{3πx}{L}) − 5 sin (\frac{7πx}{L})) sin (\frac{nπx}{L}) dx

= \frac{2}{L} ∫_{0}^{L} sin (\frac{3πx}{L}) sin (\frac{nπx}{L}) dx

− 5\frac{2}{L} ∫_{0}^{L} sin (\frac{7πx}{L}) sin (\frac{nπx}{L}) dx

= \{ \begin{matrix}1 & if n = 3 \\ −5 & if n = 7 \\ 0 & otherwise \end{matrix} by the orthogonality relation. This shows

u(x, t) = u_{3}(x, t) − 5u_{7}(x, t) = e^{−k(\frac{3π}{L})^{2}t} sin (\frac{3πx}{L}) − 5e^{−k(\frac{7π}{L})^{2}t} sin (\frac{7πx}{L})