Second Order Partial Differential Equations Questions and Answers

Q: The function \(u(r, \theta)\) satisfying the Laplace equation\[ \begin{aligned} &\frac{\partial^{2} u}{\partial r^{2}} + \frac{1}{r}\frac{\partial u}{\partial r} \\ &+ \frac{1}{r^{2}}\frac{\partial^{2} u}{\partial \theta^{2}} = 0, \end{aligned} \]\(e \lt r \lt e^{2}\), subject to the conditions \(u(e, \theta) = 1,\ u(e^{2}, \theta) = 0\) is

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.

Q: The diffusion equation\(\frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial u}{\partial t},\)\(u = u(x, t),\ u(0, t) = 0 = u(\pi, t),\ u(x, 0) = \cos x \sin 5x\)admits the solution

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 \(\cos x\sin 5x\).

Q: Let \(\Omega = \{(x,y) \in \mathbb{R}^{2} \mid x^{2} + y^{2} \lt 1\}\) be the open unit disc in \(\mathbb{R}^{2}\) with boundary \(\partial\Omega\). If \(u(x,y)\) is the solution of the Dirichlet problem\[ \begin{aligned} u_{xx} + u_{yy} &= 0 \quad \text{in } \Omega \\ u(x,y) &= 1 - 2y^{2} \quad \text{on } \partial\Omega, \end{aligned} \]then \(u\left(\frac{1}{2}, 0\right)\) is equal to

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

Q: The steady state solution for the heat equation\[ \begin{aligned} &\frac{\partial u}{\partial t} - \frac{\partial^{2} u}{\partial x^{2}} = 0, \\ &0 \lt x \lt 2,\ t \gt 0, \end{aligned} \]with the initial condition u(x, 0) = 0, \(0 \lt x \lt 2\) and the boundary conditions u(0, t) = 1 and u(2, t) = 3, \(t \gt 0\), at x = 1 is

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.

Q: Let u(x, t) be the solution of the wave equation\[ \begin{aligned} &\frac{\partial^{2} u}{\partial t^{2}} - \frac{\partial^{2} u}{\partial x^{2}} = 0, \\ &0 \lt x \lt \pi,\ t \gt 0, \end{aligned} \]with the initial conditions\[ u(x,0) = \sin x + \sin 2x + \sin 3x, \]\[ \frac{\partial u}{\partial t}(x,0) = 0, \; 0 \lt x \lt \pi \]and the boundary conditions u(0, t) = u(\(\pi\), t) = 0, \(t \geq 0\). Then, the value of \(u\left(\frac{\pi}{2}, \pi\right)\) is

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.

Q: Let \(D = \{(x, y) \in \mathbb{R}^{2} : x \gt 0\) and \(y \gt 0\}\). If the following second-order linear partial differential equation\[ \begin{aligned} y^{2}\frac{\partial^{2}u}{\partial x^{2}} - x^{2}\frac{\partial^{2}u}{\partial y^{2}} \\ + y\frac{\partial u}{\partial y} = 0 \text{ on } D \end{aligned} \]is transformed to\[ \begin{aligned} &\left(\frac{\partial^{2}u}{\partial \eta^{2}} - \frac{\partial^{2}u}{\partial \xi^{2}}\right) + \left(\frac{\partial u}{\partial \eta} + \frac{\partial u}{\partial \xi}\right)\frac{1}{2\eta} \\ &+ \left(a\frac{\partial u}{\partial \eta} + b\frac{\partial u}{\partial \xi}\right)\frac{1}{2\xi} = 0 \text{ on } D \end{aligned} \]for some \(a, b \in \mathbb{R}\), via the co-ordinate transform \(\eta = \frac{x^{2}}{2}\) and \(\xi = \frac{y^{2}}{2}\), then which one of the following is correct?

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.

Q: The partial differential equation\[ \begin{aligned} &(1 + x^{2})\frac{\partial^{2} u}{\partial x^{2}} + 2x(1 - y^{2})\frac{\partial^{2} u}{\partial y \partial x} \\ &+ (1 - y^{2})\frac{\partial^{2} u}{\partial y^{2}} + x\frac{\partial u}{\partial x} + (1 - y^{2})\frac{\partial u}{\partial y} = 0 \end{aligned} \]is

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.

Q: Let \(u(x,t)\) be the solution of the following initial-boundary value problem\[ \begin{aligned} &\frac{\partial u}{\partial t} - \frac{\partial^{2} u}{\partial x^{2}} = 0, \quad x \in (0,\pi), \quad t \gt 0, \\ &u(0,t) = u(\pi,t) = 0, \quad u(x,0) = \sin 4x \cos 3x. \end{aligned} \]Then, for each \(t \gt 0\), the value of \(u\left(\frac{\pi}{4}, t\right)\) is

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 - e48t) option has the sign of \(\sin(7\pi/4)\) wrong, and the \(4\sqrt{2}\) option halves the correct coefficient.

The function \(u(r, \theta)\) satisfying the Laplace equation

\[ \begin{aligned} &\frac{\partial^{2} u}{\partial r^{2}} + \frac{1}{r}\frac{\partial u}{\partial r} \\ &+ \frac{1}{r^{2}}\frac{\partial^{2} u}{\partial \theta^{2}} = 0, \end{aligned} \]

\(e \lt r \lt e^{2}\), subject to the conditions \(u(e, \theta) = 1,\ u(e^{2}, \theta) = 0\) is

A.\(\ln(e/r)\)
B.\(\ln(e/r^{2})\)
C.\(\ln(e^{2}/r)\)
D.\(\sum_{n=1}^{\infty} \left(\frac{r - e^{2}}{e - e^{2}}\right) \sin n\theta\)

Exam Year(s)

GATE Mathematics - 2012,

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.

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The diffusion equation

\(\frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial u}{\partial t},\)

\(u = u(x, t),\ u(0, t) = 0 = u(\pi, t),\ u(x, 0) = \cos x \sin 5x\)

admits the solution

A.\(\frac{e^{-36t}}{2}\left[\sin 6x + e^{20t}\sin 4x\right]\)
B.\(\frac{e^{-36t}}{2}\left[\sin 4x + e^{20t}\sin 6x\right]\)
C.\(\frac{e^{-20t}}{2}\left[\sin 3x + e^{15t}\sin 5x\right]\)
D.\(\frac{e^{-36t}}{2}\left[\sin 5x + e^{20t}\sin x\right]\)

Exam Year(s)

GATE Mathematics - 2012,

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 \(\cos x\sin 5x\).

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Let \(\Omega = \{(x,y) \in \mathbb{R}^{2} \mid x^{2} + y^{2} \lt 1\}\) be the open unit disc in \(\mathbb{R}^{2}\) with boundary \(\partial\Omega\). If \(u(x,y)\) is the solution of the Dirichlet problem

\[ \begin{aligned} u_{xx} + u_{yy} &= 0 \quad \text{in } \Omega \\ u(x,y) &= 1 - 2y^{2} \quad \text{on } \partial\Omega, \end{aligned} \]

then \(u\left(\frac{1}{2}, 0\right)\) is equal to

A.\(-1\)
B.\(\frac{-1}{4}\)
C.\(\frac{1}{4}\)
D.1

Exam Year(s)

GATE Mathematics - 2015,

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.

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If the characteristic curves of the partial differential equation \(xu_{xx} + 2x^{2}u_{xy} = u_{x} - 1\) are \(\mu(x, y) = c_{1}\) and \(\nu(x, y) = c_{2}\), where \(c_{1}\) and \(c_{2}\) are constants, then

A.\(\mu(x, y) = x^{2} - y,\ \nu(x, y) = y\)
B.\(\mu(x, y) = x^{2} + y,\ \nu(x, y) = y\)
C.\(\mu(x, y) = x^{2} + y,\ \nu(x, y) = x^{2}\)
D.\(\mu(x, y) = x^{2} - y,\ \nu(x, y) = x^{2}\)

Exam Year(s)

GATE Mathematics - 2018,

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.

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The steady state solution for the heat equation

\[ \begin{aligned} &\frac{\partial u}{\partial t} - \frac{\partial^{2} u}{\partial x^{2}} = 0, \\ &0 \lt x \lt 2,\ t \gt 0, \end{aligned} \]

with the initial condition u(x, 0) = 0, \(0 \lt x \lt 2\) and the boundary conditions u(0, t) = 1 and u(2, t) = 3, \(t \gt 0\), at x = 1 is

Exam Year(s)

GATE Mathematics - 2022,

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.

Discuss in Forum

Let u(x, t) be the solution of the wave equation

\[ \begin{aligned} &\frac{\partial^{2} u}{\partial t^{2}} - \frac{\partial^{2} u}{\partial x^{2}} = 0, \\ &0 \lt x \lt \pi,\ t \gt 0, \end{aligned} \]

with the initial conditions

\[ u(x,0) = \sin x + \sin 2x + \sin 3x, \]

\[ \frac{\partial u}{\partial t}(x,0) = 0, \; 0 \lt x \lt \pi \]

and the boundary conditions u(0, t) = u(\(\pi\), t) = 0, \(t \geq 0\). Then, the value of \(u\left(\frac{\pi}{2}, \pi\right)\) is

A.-1/2
B.0
C.1/2
D.1

Exam Year(s)

GATE Mathematics - 2022,

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.

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Let \(u = u(x, t)\) be the solution of

\(\frac{\partial u}{\partial t} - 4\frac{\partial^{2} u}{\partial x^{2}} = 0,\) \(0 \lt x \lt 1,\ t \gt 0,\)
\(u(0, t) = u(1, t) = 0,\ t \ge 0,\)
\(u(x, 0) = \sin(\pi x),\ 0 \le x \le 1.\)

Define \(g(t) = \int_{0}^{1}(u(x, t))^{2}dx\), for \(t \gt 0\). Which one of the following is correct?

A.\(g\) is decreasing on \((0, \infty)\) and \(\lim_{t \to \infty} g(t) = 0\)
B.\(g\) is decreasing on \((0, \infty)\) and \(\lim_{t \to \infty} g(t) = 1/4\)
C.\(g\) is increasing on \((0, \infty)\) and \(\lim_{t \to \infty} g(t)\) does not exist
D.\(g\) is increasing on \((0, \infty)\) and \(\lim_{t \to \infty} g(t) = 3\)

Exam Year(s)

GATE Mathematics - 2024,

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.

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Let \(D = \{(x, y) \in \mathbb{R}^{2} : x \gt 0\) and \(y \gt 0\}\). If the following second-order linear partial differential equation

\[ \begin{aligned} y^{2}\frac{\partial^{2}u}{\partial x^{2}} - x^{2}\frac{\partial^{2}u}{\partial y^{2}} \\ + y\frac{\partial u}{\partial y} = 0 \text{ on } D \end{aligned} \]

is transformed to

\[ \begin{aligned} &\left(\frac{\partial^{2}u}{\partial \eta^{2}} - \frac{\partial^{2}u}{\partial \xi^{2}}\right) + \left(\frac{\partial u}{\partial \eta} + \frac{\partial u}{\partial \xi}\right)\frac{1}{2\eta} \\ &+ \left(a\frac{\partial u}{\partial \eta} + b\frac{\partial u}{\partial \xi}\right)\frac{1}{2\xi} = 0 \text{ on } D \end{aligned} \]

for some \(a, b \in \mathbb{R}\), via the co-ordinate transform \(\eta = \frac{x^{2}}{2}\) and \(\xi = \frac{y^{2}}{2}\), then which one of the following is correct?

A.\(a = 2, b = 0\)
B.\(a = 0, b = -1\)
C.\(a = 1, b = -1\)
D.\(a = 1, b = 0\)

Exam Year(s)

GATE Mathematics - 2024,

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.

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The partial differential equation

\[ \begin{aligned} &(1 + x^{2})\frac{\partial^{2} u}{\partial x^{2}} + 2x(1 - y^{2})\frac{\partial^{2} u}{\partial y \partial x} \\ &+ (1 - y^{2})\frac{\partial^{2} u}{\partial y^{2}} + x\frac{\partial u}{\partial x} + (1 - y^{2})\frac{\partial u}{\partial y} = 0 \end{aligned} \]

A.elliptic in the region \(\{(x,y) \in \mathbb{R}^{2} : |y| \le 1\}\)
B.hyperbolic in the region \(\{(x,y) \in \mathbb{R}^{2} : |y| \gt 1\}\)
C.elliptic in the region \(\{(x,y) \in \mathbb{R}^{2} : |y| \gt 1\}\)
D.hyperbolic in the region \(\{(x,y) \in \mathbb{R}^{2} : |y| \lt 1\}\)

Exam Year(s)

GATE Mathematics - 2025,

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.

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10.

Let \(u(x,t)\) be the solution of the following initial-boundary value problem

\[ \begin{aligned} &\frac{\partial u}{\partial t} - \frac{\partial^{2} u}{\partial x^{2}} = 0, \quad x \in (0,\pi), \quad t \gt 0, \\ &u(0,t) = u(\pi,t) = 0, \quad u(x,0) = \sin 4x \cos 3x. \end{aligned} \]

Then, for each \(t \gt 0\), the value of \(u\left(\frac{\pi}{4}, t\right)\) is

A.\(\dfrac{e^{-49t}}{2\sqrt{2}}(e^{48t} - 1)\)
B.\(\dfrac{e^{-49t}}{2\sqrt{2}}(1 - e^{48t})\)
C.\(\dfrac{e^{-49t}}{2\sqrt{2}}(1 + e^{48t})\)
D.\(\dfrac{e^{-49t}}{4\sqrt{2}}(1 - e^{48t})\)

Exam Year(s)

GATE Mathematics - 2025,

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.

Second Order Partial Differential Equations Questions and Answers

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