Second Order Partial Differential Equations Questions and Answers

Answer: A

The particular integral is \(-\frac{x}{2}\cos(x - y)\).

For \(\sin(x - y)\), replacing \(D_{x}^{2}\) by \(-1\) and \(D_{y}^{2}\) by \(-1\) makes the denominator \(-1 - (-1) = 0\), the case of failure. The rule is then to multiply by \(x\) and differentiate the denominator with respect to \(D_{x}\):

\[ \begin{aligned} \frac{1}{D_{x}^{2} - D_{y}^{2}}\sin(x - y) &= x\cdot\frac{1}{2D_{x}}\sin(x - y) \\ &= \frac{x}{2}\int \sin(x - y)\,dx \\ &= -\frac{x}{2}\cos(x - y). \end{aligned} \]

Check: with \(u = -\frac{x}{2}\cos(x - y)\), \(u_{xx} - u_{yy} = \sin(x - y)\). The option that adds \(\sin(x - y)\) differs only by a term the operator annihilates, so it is not what the inverse operator produces; the particular integral carries no complementary-function part. \(-\frac{x}{2}\sin(x - y) + \cos(x - y)\) gives \(-\cos(x - y)\) under the operator, and \(\frac{3x}{2}\sin(x - y)\) also produces a cosine term, so neither satisfies the equation.

Answer: C

The solution is \(\ln(e^{2}/r)\).

The boundary values do not depend on \(\theta\), so look for a radial solution. With \(u_{\theta\theta} = 0\) the equation becomes \((ru')' = 0\), whose solutions are \(u = A + B\ln r\). Applying the conditions:

\[ \begin{aligned} A + B &= 1 \quad (r = e) \\ A + 2B &= 0 \quad (r = e^{2}) \\ B &= -1,\ A = 2 \end{aligned} \]

So \(u = 2 - \ln r = \ln(e^{2}/r)\); check: \(u(e) = 2 - 1 = 1\) and \(u(e^{2}) = 0\). \(\ln(e/r)\) gives 0 and -1 at the two radii, and \(\ln(e/r^{2})\) gives -1 and -3. The series vanishes at \(r = e^{2}\), but at \(r = e\) it becomes \(\sum \sin n\theta\), not the constant 1, and its terms \((r - e^{2})\sin n\theta\) do not satisfy Laplace's equation.

Answer: A

The solution is \(\frac{e^{-36t}}{2}\left[\sin 6x + e^{20t}\sin 4x\right]\).

Write the initial data as a sine series using \(\cos A\sin B = \frac{1}{2}[\sin(A + B) - \sin(A - B)]\):

\[ \cos x\sin 5x = \tfrac{1}{2}(\sin 6x + \sin 4x). \]

Each mode \(\sin nx\) satisfies the boundary conditions and decays as \(e^{-n^{2}t}\) under \(u_{xx} = u_{t}\). So

\[ \begin{aligned} u &= \tfrac{1}{2}\left(e^{-36t}\sin 6x + e^{-16t}\sin 4x\right) \\ &= \tfrac{e^{-36t}}{2}\left(\sin 6x + e^{20t}\sin 4x\right) \end{aligned} \]

The option with \(\sin 4x\) carrying \(e^{-36t}\) attaches the decay rates to the wrong modes. The others use \(\sin 3x, \sin 5x\) or \(\sin 5x, \sin x\), which are not the modes of \(\cos x\sin 5x\).

Answer: C

\(u\left(\frac{1}{2}, 0\right) = \) \(\frac{1}{4}\).

On the unit circle \(x^{2} + y^{2} = 1\), so the boundary data can be rewritten:

\[ 1 - 2y^{2} = (x^{2} + y^{2}) - 2y^{2} = x^{2} - y^{2} \]

The function \(x^{2} - y^{2}\) is harmonic (\(u_{xx} + u_{yy} = 2 - 2 = 0\)) and matches the boundary values, so by uniqueness of the Dirichlet problem it is the solution: \(u(x, y) = x^{2} - y^{2}\).

Then \(u\left(\frac{1}{2}, 0\right) = \frac{1}{4}\). Taking \(1 - 2y^{2}\) itself as the solution would give 1, but that function is not harmonic, so 1 is wrong; \(-1\) and \(-\frac{1}{4}\) have the wrong sign.

Answer: A

The characteristics are \(\mu = x^{2} - y\) and \(\nu = y\).

For \(Au_{xx} + 2Bu_{xy} + Cu_{yy} + \cdots = 0\) the characteristic directions satisfy \(A\,dy^{2} - 2B\,dx\,dy + C\,dx^{2} = 0\). Here \(A = x\), \(2B = 2x^{2}\) and \(C = 0\):

\[ \begin{aligned} x\,dy^{2} - 2x^{2}\,dx\,dy &= 0 \\ x\,dy\,(dy - 2x\,dx) &= 0 \end{aligned} \]

So \(dy = 0\), giving \(y = c\), or \(dy = 2x\,dx\), giving \(y - x^{2} = c\), equivalently \(x^{2} - y = c\). The equation is hyperbolic (\(B^{2} - AC = x^{4} \gt 0\) for \(x \neq 0\)), so there are two real families.

\(x^{2} + y\) comes from a sign error in the second family. \(\nu = x^{2}\) would mean \(x\) is constant, \(dx = 0\); putting \(dx = 0\) into the characteristic equation leaves \(x\,dy^{2} = 0\), which forces \(dy = 0\) too, so the lines \(x = \) constant are not characteristics.

Answer: B

The steady-state value at x = 1 is 2.

At steady state u no longer depends on t, so the equation reduces to \(u'' = 0\), and u is linear in x. The initial condition does not affect the steady state; only the boundary values do:

\[ \begin{aligned} u(x) &= a + bx,\ u(0) = 1,\ u(2) = 3 \\ u(x) &= 1 + x,\ \ u(1) = 2 \end{aligned} \]

The steady temperature is the straight line between the end values, so at the midpoint it is their average, (1 + 3)/2 = 2. 1 and 3 are the end values themselves, and 4 is their sum.

Answer: B

The value is 0.

With zero initial velocity and fixed ends on (0, \(\pi\)), each mode sin kx oscillates as cos kt, so by superposition

\[ u(x,t) = \sin x\cos t + \sin 2x\cos 2t + \sin 3x\cos 3t \]

At x = \(\pi/2\), t = \(\pi\):

\[ \begin{aligned} u &= (1)(-1) + (0)(1) + (-1)(-1) \\ &= -1 + 0 + 1 = 0 \end{aligned} \]

The answer 1 comes from forgetting that \(\sin(3\pi/2) = -1\), and \(\pm 1/2\) has no source in this solution. D'Alembert's formula, u = [g(x + t) + g(x - t)]/2 with g extended as an odd \(2\pi\)-periodic function, gives the same 0.

Answer: A

\(g\) is decreasing on \((0, \infty)\) and \(\lim_{t \to \infty} g(t) = 0\).

The initial data is a single Fourier sine mode, so the solution is \(u(x, t) = e^{-4\pi^{2}t}\sin(\pi x)\). Then

\[ \begin{aligned} g(t) &= e^{-8\pi^{2}t}\int_{0}^{1}\sin^{2}(\pi x)\,dx \\ &= \tfrac{1}{2}e^{-8\pi^{2}t} \end{aligned} \]

This is strictly decreasing and tends to 0. The heat equation with zero boundary values always dissipates energy, so the increasing options are wrong, and 1/4 would need the solution to settle to a non-zero state, which zero boundary conditions forbid.

Answer: B

The coefficients are \(a = 0, b = -1\).

With \(\eta = x^{2}/2\), \(\xi = y^{2}/2\): \(u_{x} = xu_{\eta}\), \(u_{xx} = u_{\eta} + x^{2}u_{\eta\eta}\), \(u_{y} = yu_{\xi}\), \(u_{yy} = u_{\xi} + y^{2}u_{\xi\xi}\). Substituting:

\[ \begin{aligned} & x^{2}y^{2}(u_{\eta\eta} - u_{\xi\xi}) + y^{2}u_{\eta} \\ & - x^{2}u_{\xi} + y^{2}u_{\xi} = 0 \end{aligned} \]

Divide by \(x^{2}y^{2}\) and use \(x^{2} = 2\eta\), \(y^{2} = 2\xi\):

\[ \begin{aligned} &(u_{\eta\eta} - u_{\xi\xi}) + (u_{\eta} + u_{\xi})\frac{1}{2\eta} \\ &- u_{\xi}\frac{1}{2\xi} = 0. \end{aligned} \]

Comparing with the given form, \(a = 0\) and \(b = -1\). Any non-zero \(a\) would need a \(u_{\eta}/\xi\) term, which the computation does not produce.

Answer: B

The equation is hyperbolic in the region |y| > 1.

For \(Au_{xx} + 2Bu_{xy} + Cu_{yy} + \ldots = 0\), the type is fixed by the sign of \(B^{2} - AC\). Here \(A = 1 + x^{2}\), \(2B = 2x(1 - y^{2})\), \(C = 1 - y^{2}\):

\[ \begin{aligned} B^{2} - AC &= x^{2}(1 - y^{2})^{2} - (1 + x^{2})(1 - y^{2}) \\ &= (1 - y^{2})(x^{2} - x^{2}y^{2} - 1 - x^{2}) \\ &= (y^{2} - 1)(1 + x^{2}y^{2}) \end{aligned} \]

Since \(1 + x^{2}y^{2} \gt 0\), the sign is that of \(y^{2} - 1\): positive (hyperbolic) for |y| > 1, negative (elliptic) for |y| < 1, and zero (parabolic) on |y| = 1. So 'elliptic for |y| ≤ 1' wrongly includes the parabolic lines, and the other two options reverse the types.