Second Order Partial Differential Equations Questions and Answers

Answer: A

The value is \(\dfrac{e^{-49t}}{2\sqrt{2}}(e^{48t} - 1)\).

Rewrite the initial data as sine modes, which satisfy the boundary conditions:

\[ \begin{aligned} \sin 4x\cos 3x &= \tfrac{1}{2}(\sin 7x + \sin x) \\ u(x,t) &= \tfrac{1}{2}(e^{-49t}\sin 7x + e^{-t}\sin x) \end{aligned} \]

since the mode \(\sin nx\) decays as \(e^{-n^{2}t}\). At \(x = \pi/4\), \(\sin(7\pi/4) = -\tfrac{1}{\sqrt{2}}\) and \(\sin(\pi/4) = \tfrac{1}{\sqrt{2}}\):

\[ \begin{aligned} u &= \frac{e^{-t} - e^{-49t}}{2\sqrt{2}} \\ &= \frac{e^{-49t}}{2\sqrt{2}}(e^{48t} - 1) \end{aligned} \]

The (1 - e^48t) option has the sign of \(\sin(7\pi/4)\) wrong, and the \(4\sqrt{2}\) option halves the correct coefficient.

Answer: C

The value is \(\dfrac{1}{2}\left(e^{\pi} + \dfrac{5}{3}\right)\).

d'Alembert's formula with wave speed c = 3:

\[ \begin{aligned} u &= \tfrac{1}{2}[f(x + 3t) + f(x - 3t)] \\ &\quad + \tfrac{1}{6}\int_{x - 3t}^{x + 3t}\sin s\,ds \end{aligned} \]

At \(x = \pi/2\), \(t = \pi/6\), the end points are \(\pi\) and 0:

\[ \begin{aligned} u &= \tfrac{1}{2}(e^{\pi} + 1) + \tfrac{1}{6}(-\cos\pi + \cos 0) \\ &= \tfrac{1}{2}(e^{\pi} + 1) + \tfrac{1}{3} = \tfrac{1}{2}\left(e^{\pi} + \tfrac{5}{3}\right) \end{aligned} \]

The 1/3 options drop the \(e^{0} = 1\) term, and the minus signs reverse the integral of the initial velocity.

Answer: A

The answer is \(\hat{u}(w, t) = \frac{2a}{a^2+w^2} e^{-kw^2 t}\).

Transforming the equation in x turns the second derivative into multiplication by \(-w^2\), leaving an ordinary differential equation in t:

\[ \begin{aligned} \frac{\partial \hat{u}}{\partial t} &= -k w^2 \hat{u} \\ \hat{u}(w, t) &= \hat{u}(w, 0)\, e^{-k w^2 t} \end{aligned} \]

The transform of the initial condition is

\[ \begin{aligned} \hat{u}(w, 0) &= \int_{-\infty}^{\infty} e^{-a|x|} e^{iwx} dx \\ &= \frac{2a}{a^2 + w^2}. \end{aligned} \]

The option with a in the numerator has lost the factor 2 from the two halves of the line; the one with \(e^{-ka^2 t}\) decays at the wrong rate; and the Gaussian is the solution kernel in x, not this transform.

Answer: A

The maximum value is 1.

With zero initial velocity, D'Alembert's formula reduces to the average of two travelling copies of the initial shape:

\[ \begin{aligned} u(x, t) &= \tfrac{1}{2}[f(x - ct) + f(x + ct)] \\ u(0, t) &= \tfrac{1}{2}[f(-ct) + f(ct)] \end{aligned} \]

For \(0 \lt t \lt 1/c\) both \(|ct|\) and \(|-ct|\) are below 1, so both terms equal 1 and \(u(0, t) = 1\). For \(t \gt 1/c\) both copies have moved past the origin and \(u(0, t) = 0\).

So the largest value is 1. The value \(\frac{1}{2}\) is what a point outside the initial pulse sees when one half-pulse passes it, which is the usual trap; \(\frac{1}{4}\) and 0 are not the maximum at the origin.

Answer: B

The value is \(T(\frac{1}{2}, \frac{1}{3}) = \frac{5}{6}\).

Try \(T(x, y) = x + y\). It is harmonic, since both second derivatives are 0, and it meets every boundary condition: \(T(x, 0) = x\), \(T(0, y) = y\), \(T(x, 1) = x + 1\) and \(T(1, y) = 1 + y\). By uniqueness of the solution of the Dirichlet problem, this is the solution.

\[ \begin{aligned} T\left(\frac{1}{2}, \frac{1}{3}\right) &= \frac{1}{2} + \frac{1}{3} \\ &= \frac{5}{6} \end{aligned} \]

\(\frac{7}{6}\) adds an extra \(\frac{1}{3}\), \(\frac{1}{6}\) is the difference \(\frac{1}{2} - \frac{1}{3}\), and \(\frac{1}{2}\) is only the x-coordinate.

Answer: D

The equation is parabolic on \(\mathbf{R}^{2}\).

For \(Au_{xx} + 2Bu_{xy} + Cu_{yy} + \ldots = 0\) the type is decided by the sign of \(B^{2} - AC\). Here \(A = x^{2}\), \(2B = -2xy\) so \(B = -xy\), and \(C = y^{2}\):

\[ B^{2} - AC = x^{2}y^{2} - x^{2}y^{2} = 0 \]

The discriminant vanishes at every point, so the equation is parabolic throughout the plane. (Indeed the principal part is \((x\partial_{x} - y\partial_{y})^{2}\) up to lower-order terms, a perfect square.)

Hyperbolic would need \(B^{2} - AC \gt 0\) and elliptic \(B^{2} - AC \lt 0\); neither happens anywhere, including the first quadrant singled out in the first option.

Answer: B

The problem is \(X'' + \lambda X = 0\), \(X'(0) = X'(c) = 0\).

Substituting \(u = X(x)T(t)\) into \(u_{t} = u_{xx}\) gives \(XT' = X''T\), so \(T'/T = X''/X = -\lambda\), a constant. This yields \(X'' + \lambda X = 0\).

The boundary conditions are on \(u_{x}\): \(u_{x}(0, t) = X'(0)T(t) = 0\) for all \(t\), and similarly at \(x = c\). For a non-trivial \(T\) this forces \(X'(0) = 0\) and \(X'(c) = 0\), Neumann conditions (insulated ends).

Option A has Dirichlet conditions, which would come from \(u = 0\) at the ends. Options C and D mix the two kinds or bring in the value 1, which no condition gives.

Answer: C

The solution is \(z = \phi_{1}(x + y) + \phi_{2}(x - y)\).

This is the wave equation \(z_{xx} = z_{yy}\). Its auxiliary equation \(m^{2} - 1 = 0\) has distinct roots \(m = \pm 1\), so the general solution is d'Alembert's: an arbitrary function of \(x + y\) plus an arbitrary function of \(x - y\).

Check: for \(z = \phi_{1}(x + y)\), \(z_{xx} = \phi_{1}''\) and \(z_{yy} = \phi_{1}''\), so they cancel; likewise for \(\phi_{2}(x - y)\). Factors of \(x\) or \(y\) in front belong to repeated roots and do not satisfy this equation.

Answer: D

The equation is elliptic.

Expand \(3\Delta u = 3u_{xx} + 3u_{yy}\), so the principal part is \(3u_{xx} + 4u_{xy} + 3u_{yy}\). For \(Au_{xx} + 2Bu_{xy} + Cu_{yy}\), the type depends on \(B^{2} - AC\):

\[ B^{2} - AC = 2^{2} - 3 \cdot 3 = -5 \lt 0 \]

A negative discriminant means elliptic. The term \(-u^{2}\) makes the equation non-linear (so 'Linear' is wrong), but it is a lower-order term and does not change the type. Hyperbolic would need a positive and parabolic a zero discriminant.

Answer: C

The two solutions satisfy \(u_{1} - u_{2} \equiv 0\).

The difference \(w = u_{1} - u_{2}\) solves the homogeneous problem \(w_{tt} - w_{xx} = 0\) with \(w(x, 0) = 0\) and \(w_{t}(x, 0) = 0\). By d'Alembert's formula (or the energy method), the only solution is \(w \equiv 0\). So the Cauchy problem for the wave equation has a unique solution and \(u_{1} = u_{2}\).

With \(u_{1} = u_{2} = u\), \(u_{1} = 2u_{2}\) and \(u_{1} + u_{2} = 0\) would force \(u \equiv 0\), which does not meet \(u(x, 0) = \cos x\), and \(u_{1}u_{2} = u^{2} = 1\) fails too.