Second Order Partial Differential Equations Questions and Answers

Q: Let \(u_{1}\) and \(u_{2}\) be solution of the equation :\(\Delta u = 1\) in D\(u = 0\) on \(\partial D\)where \(D = \{(x, y) \in \mathbf{R}^{2}/(x - 2)^{2} + (y - 1)^{2} \le 1/4\}\) and \(\partial D = \{(x, y) \in \mathbf{R}^{2}/(x - 2)^{2} + (y - 1)^{2} = 1/4\}\) is the boundary of D.If \(u_{1}(2, 1) = u_{2}(2, 1)\); then :

The answer is \(u_{1} \equiv u_{2}\) in D.The Dirichlet problem for Poisson's equation on a bounded domain has at most one solution: \(w = u_{1} - u_{2}\) satisfies \(\Delta w = 0\) in D and \(w = 0\) on \(\partial D\), so by the maximum principle \(w\) attains its maximum and minimum on the boundary, where it is 0. Hence \(w \equiv 0\). The condition \(u_{1}(2, 1) = u_{2}(2, 1)\) is automatically true.Neither solution is constant: a constant has \(\Delta u = 0\), not 1. (Explicitly, \(u = \tfrac{1}{4}((x - 2)^{2} + (y - 1)^{2} - \tfrac{1}{4})\).)

Q: The equation \(3u_{xx} - x^{2}u_{y} + 5u = 0\) transforms under the change of variables to polar coordinates is :

The equation remains a parabolic equation.The type of a second-order linear PDE \(Au_{xx} + 2Bu_{xy} + Cu_{yy} + \cdots = 0\) is decided by the sign of \(B^{2} - AC\). Here \(A = 3\) and \(B = C = 0\) (\(u_{y}\) is first order), so \(B^{2} - AC = 0\): the equation is parabolic.A smooth, invertible change of variables such as polar coordinates multiplies the discriminant by the square of the Jacobian, so its sign, and hence the type, does not change. The equation is also linear in \(u\), so it does not become non-linear, and it cannot become hyperbolic or elliptic.

Q: Let \(u_{1}\) and \(u_{2}\) be solutions of the equation :\(\Delta u = 0\) in D\(\dfrac{\partial u}{\partial n} = g(s)\) on \(\partial D\)where \(D = \{(x, y) \in \mathbf{R}^{2} / \frac{x^{2}}{4} + \frac{y^{2}}{9} \le 1\}\) is an ellipse with boundary \(\partial D = \{(x, y) \in \mathbf{R}^{2} /\) \(\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1\}\). If \(u_{1}(0, \frac{1}{2}) - u_{2}(0, \frac{1}{2}) = 1\). Then :

The answer is \(u_{1} \equiv u_{2} + 1\) in D.Let \(w = u_{1} - u_{2}\). Then \(\Delta w = 0\) in D and \(\partial w / \partial n = 0\) on the boundary. Green's identity gives\[ \begin{aligned} \int_{D} |\nabla w|^{2} &= \int_{\partial D} w \frac{\partial w}{\partial n} - \int_{D} w\, \Delta w \\ &= 0 \end{aligned} \]so \(\nabla w = 0\) and \(w\) is constant: solutions of the Neumann problem are unique up to an additive constant. The given value \(w(0, \frac{1}{2}) = 1\) fixes the constant as 1.So \(u_{1} - u_{2}\) does not change sign (C), the two are not identical (D), and they are not 'dependent' in any multiplicative sense (A).

Q: The equation\[ xu_{xx} - yu_{yy} + \frac{1}{2}(u_{x} - u_{y}) = 0 \]is hyperbolic in the maximal domain :

The equation is hyperbolic in \(\{(x, y) : xy \gt 0\}\).For \(Au_{xx} + Bu_{xy} + Cu_{yy} + \ldots = 0\), the type is set by \(B^{2} - 4AC\). Here \(A = x\), \(B = 0\), \(C = -y\):\[ B^{2} - 4AC = 4xy \]The equation is hyperbolic where \(4xy \gt 0\), elliptic where \(xy \lt 0\) and parabolic on the axes. The maximal hyperbolic domain is the whole region \(xy \gt 0\), the first and third quadrants together.Options A and D give only one of the two quadrants, so they are not maximal, and option B, \(x \lt 0, y \gt 0\), is where the equation is elliptic.

Q: The partial differential equation\[ \begin{aligned} &x^{2} \frac{\partial^{2} z}{\partial x^{2}} - y^{2} \frac{\partial^{2} z}{\partial y^{2}} \\ &= xyz \end{aligned} \]is :

The equation is hyperbolic.For \(A z_{xx} + 2B z_{xy} + C z_{yy} + \ldots = 0\) the type is set by the discriminant \(B^{2} - AC\). Here\[ \begin{aligned} A &= x^{2},\quad B = 0,\quad C = -y^{2} \\ B^{2} - AC &= x^{2}y^{2} \end{aligned} \]which is positive whenever \(xy \neq 0\), so the equation is hyperbolic away from the coordinate axes (where it degenerates to parabolic). The right-hand side \(xyz\) involves no second derivatives and does not affect the type.It would be elliptic only if \(A\) and \(C\) had the same sign, and parabolic only if the discriminant vanished identically. 'Laplacian' is not a type; the Laplace equation is the model elliptic equation.

Q: The region in which the partial differential equation\[ (1 - x^{2})u_{xx} - u_{yy} = 0 \]is an ellipse if :

The equation is elliptic when |x| > 1.For \(Au_{xx} + Bu_{xy} + Cu_{yy} = 0\), the type depends on \(B^{2} - 4AC\): negative means elliptic. Here \(A = 1 - x^{2}\), \(B = 0\), \(C = -1\):\[ \begin{aligned} B^{2} - 4AC &= 0 - 4(1 - x^{2})(-1) \\ &= 4(1 - x^{2}) \end{aligned} \]This is negative when \(x^{2} \gt 1\), that is \(|x| \gt 1\). For \(|x| \lt 1\) it is positive and the equation is hyperbolic, and at \(|x| = 1\) it is parabolic. 'x > 1' and 'x < 1' each describe only part of, or the wrong part of, the line.

Q: The region in which the partial differential equation,\[ yu_{xx} + 2xyu_{xy} + xu_{yy} = u_{x} + u_{y} \]is hyperbolic if :

The equation is hyperbolic where \(xy \lt 0\).For \(A u_{xx} + 2B u_{xy} + C u_{yy} + \ldots = 0\), the type depends on \(B^{2} - AC\). Here \(A = y\), \(2B = 2xy\) so \(B = xy\), and \(C = x\):\[ B^{2} - AC = x^{2}y^{2} - xy = xy(xy - 1) \]This is positive (hyperbolic) when \(xy \lt 0\) or \(xy \gt 1\). Of the options, only \(xy \lt 0\) describes a region where the equation is hyperbolic throughout.\(xy \lt 1\) includes \(0 \lt xy \lt 1\), where \(B^{2} - AC \lt 0\) and the equation is elliptic. \(xy \neq 0\) and \(xy \neq 1\) also include that elliptic band.

Q: Consider the following statements :(I) Wave equation is of parabolic type.(II) Heat equation is of elliptic type.Then :

Both (I) and (II) are false.Classify by \(B^{2} - AC\). The wave equation \(u_{tt} = c^{2}u_{xx}\) has \(A = c^{2}\), \(C = -1\), \(B = 0\), so \(B^{2} - AC = c^{2} \gt 0\): it is hyperbolic, not parabolic.The heat equation \(u_{t} = k u_{xx}\) has only one second-order term, so \(B^{2} - AC = 0\): it is parabolic, not elliptic.Laplace's equation \(u_{xx} + u_{yy} = 0\) is the standard elliptic example. The statements have the types of the wave and heat equations mixed up.

Q: The partial differential equation\[ \begin{aligned} &\frac{\partial^{2} z}{\partial x^{2}} + y\frac{\partial^{2} z}{\partial y^{2}} = 0, \\ &y \gt 0 \end{aligned} \]is :

The equation is elliptic.Compare with \(Az_{xx} + Bz_{xy} + Cz_{yy} = 0\): here \(A = 1\), \(B = 0\), \(C = y\). Then\[ B^{2} - 4AC = -4y \lt 0 \quad (y \gt 0) \]A negative discriminant means elliptic. It would be hyperbolic for \(y \lt 0\) and parabolic on \(y = 0\), but the question restricts to \(y \gt 0\). 'Laplacian' is not a type; Laplace's equation is itself the model elliptic equation.

21.

Let \(u_{1}\) and \(u_{2}\) be solution of the equation :

\(\Delta u = 1\) in D
\(u = 0\) on \(\partial D\)

where \(D = \{(x, y) \in \mathbf{R}^{2}/(x - 2)^{2} + (y - 1)^{2} \le 1/4\}\) and \(\partial D = \{(x, y) \in \mathbf{R}^{2}/(x - 2)^{2} + (y - 1)^{2} = 1/4\}\) is the boundary of D.

If \(u_{1}(2, 1) = u_{2}(2, 1)\); then :

A.\(u_{1}\) is a constant function
B.\(u_{2}\) is a constant function
C.\(u_{1} \equiv u_{2} \equiv\) a constant function
D.\(u_{1} \equiv u_{2}\) in D

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The answer is \(u_{1} \equiv u_{2}\) in D.

The Dirichlet problem for Poisson's equation on a bounded domain has at most one solution: \(w = u_{1} - u_{2}\) satisfies \(\Delta w = 0\) in D and \(w = 0\) on \(\partial D\), so by the maximum principle \(w\) attains its maximum and minimum on the boundary, where it is 0. Hence \(w \equiv 0\). The condition \(u_{1}(2, 1) = u_{2}(2, 1)\) is automatically true.

Neither solution is constant: a constant has \(\Delta u = 0\), not 1. (Explicitly, \(u = \tfrac{1}{4}((x - 2)^{2} + (y - 1)^{2} - \tfrac{1}{4})\).)

Discuss in Forum

22.

The equation \(3u_{xx} - x^{2}u_{y} + 5u = 0\) transforms under the change of variables to polar coordinates is :

A.a non-linear equation
B.a hyperbolic equation
C.an elliptic equation
D.a parabolic equation

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The equation remains a parabolic equation.

The type of a second-order linear PDE \(Au_{xx} + 2Bu_{xy} + Cu_{yy} + \cdots = 0\) is decided by the sign of \(B^{2} - AC\). Here \(A = 3\) and \(B = C = 0\) (\(u_{y}\) is first order), so \(B^{2} - AC = 0\): the equation is parabolic.

A smooth, invertible change of variables such as polar coordinates multiplies the discriminant by the square of the Jacobian, so its sign, and hence the type, does not change. The equation is also linear in \(u\), so it does not become non-linear, and it cannot become hyperbolic or elliptic.

Discuss in Forum

23.

Let \(u_{1}\) and \(u_{2}\) be solutions of the equation :

\(\Delta u = 0\) in D
\(\dfrac{\partial u}{\partial n} = g(s)\) on \(\partial D\)

where \(D = \{(x, y) \in \mathbf{R}^{2} / \frac{x^{2}}{4} + \frac{y^{2}}{9} \le 1\}\) is an ellipse with boundary \(\partial D = \{(x, y) \in \mathbf{R}^{2} /\) \(\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1\}\). If \(u_{1}(0, \frac{1}{2}) - u_{2}(0, \frac{1}{2}) = 1\). Then :

A.\(u_{1}\) and \(u_{2}\) are dependent solutions
B.\(u_{1} \equiv u_{2} + 1\) in D
C.\(u_{1} - u_{2}\) changes sign in D
D.\(u_{1} \equiv u_{2}\) in D

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The answer is \(u_{1} \equiv u_{2} + 1\) in D.

Let \(w = u_{1} - u_{2}\). Then \(\Delta w = 0\) in D and \(\partial w / \partial n = 0\) on the boundary. Green's identity gives

\[ \begin{aligned} \int_{D} |\nabla w|^{2} &= \int_{\partial D} w \frac{\partial w}{\partial n} - \int_{D} w\, \Delta w \\ &= 0 \end{aligned} \]

so \(\nabla w = 0\) and \(w\) is constant: solutions of the Neumann problem are unique up to an additive constant. The given value \(w(0, \frac{1}{2}) = 1\) fixes the constant as 1.

So \(u_{1} - u_{2}\) does not change sign (C), the two are not identical (D), and they are not 'dependent' in any multiplicative sense (A).

Discuss in Forum

24.

The equation

\[ xu_{xx} - yu_{yy} + \frac{1}{2}(u_{x} - u_{y}) = 0 \]

is hyperbolic in the maximal domain :

A.\(\{(x, y) \in \mathbf{R}^{2} / x \gt 0, y \gt 0\}\)
B.\(\{(x, y) \in \mathbf{R}^{2} / x \lt 0, y \gt 0\}\)
C.\(\{(x, y) \in \mathbf{R}^{2} / xy \gt 0\}\)
D.\(\{(x, y) \in \mathbf{R}^{2} / x \lt 0, y \lt 0\}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The equation is hyperbolic in \(\{(x, y) : xy \gt 0\}\).

For \(Au_{xx} + Bu_{xy} + Cu_{yy} + \ldots = 0\), the type is set by \(B^{2} - 4AC\). Here \(A = x\), \(B = 0\), \(C = -y\):

\[ B^{2} - 4AC = 4xy \]

The equation is hyperbolic where \(4xy \gt 0\), elliptic where \(xy \lt 0\) and parabolic on the axes. The maximal hyperbolic domain is the whole region \(xy \gt 0\), the first and third quadrants together.

Options A and D give only one of the two quadrants, so they are not maximal, and option B, \(x \lt 0, y \gt 0\), is where the equation is elliptic.

Discuss in Forum

25.

The partial differential equation

\[ \begin{aligned} &x^{2} \frac{\partial^{2} z}{\partial x^{2}} - y^{2} \frac{\partial^{2} z}{\partial y^{2}} \\ &= xyz \end{aligned} \]

is :

A.hyperbolic
B.parabolic
C.elliptic
D.Laplacian

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The equation is hyperbolic.

For \(A z_{xx} + 2B z_{xy} + C z_{yy} + \ldots = 0\) the type is set by the discriminant \(B^{2} - AC\). Here

\[ \begin{aligned} A &= x^{2},\quad B = 0,\quad C = -y^{2} \\ B^{2} - AC &= x^{2}y^{2} \end{aligned} \]

which is positive whenever \(xy \neq 0\), so the equation is hyperbolic away from the coordinate axes (where it degenerates to parabolic). The right-hand side \(xyz\) involves no second derivatives and does not affect the type.

It would be elliptic only if \(A\) and \(C\) had the same sign, and parabolic only if the discriminant vanished identically. 'Laplacian' is not a type; the Laplace equation is the model elliptic equation.

Discuss in Forum

26.

The region in which the partial differential equation

\[ (1 - x^{2})u_{xx} - u_{yy} = 0 \]

is an ellipse if :

A.\(x \gt 1\)
B.\(x \lt 1\)
C.\(|x| \gt 1\)
D.\(|x| \lt 1\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The equation is elliptic when |x| > 1.

For \(Au_{xx} + Bu_{xy} + Cu_{yy} = 0\), the type depends on \(B^{2} - 4AC\): negative means elliptic. Here \(A = 1 - x^{2}\), \(B = 0\), \(C = -1\):

\[ \begin{aligned} B^{2} - 4AC &= 0 - 4(1 - x^{2})(-1) \\ &= 4(1 - x^{2}) \end{aligned} \]

This is negative when \(x^{2} \gt 1\), that is \(|x| \gt 1\). For \(|x| \lt 1\) it is positive and the equation is hyperbolic, and at \(|x| = 1\) it is parabolic. 'x > 1' and 'x < 1' each describe only part of, or the wrong part of, the line.

Discuss in Forum

27.

The region in which the partial differential equation,

\[ yu_{xx} + 2xyu_{xy} + xu_{yy} = u_{x} + u_{y} \]

is hyperbolic if :

A.\(xy \neq 1\)
B.\(xy \neq 0\)
C.\(xy \lt 0\)
D.\(xy \lt 1\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The equation is hyperbolic where \(xy \lt 0\).

For \(A u_{xx} + 2B u_{xy} + C u_{yy} + \ldots = 0\), the type depends on \(B^{2} - AC\). Here \(A = y\), \(2B = 2xy\) so \(B = xy\), and \(C = x\):

\[ B^{2} - AC = x^{2}y^{2} - xy = xy(xy - 1) \]

This is positive (hyperbolic) when \(xy \lt 0\) or \(xy \gt 1\). Of the options, only \(xy \lt 0\) describes a region where the equation is hyperbolic throughout.

\(xy \lt 1\) includes \(0 \lt xy \lt 1\), where \(B^{2} - AC \lt 0\) and the equation is elliptic. \(xy \neq 0\) and \(xy \neq 1\) also include that elliptic band.

Discuss in Forum

28.

Consider the following statements :

(I) Wave equation is of parabolic type.
(II) Heat equation is of elliptic type.

Then :

A.Only (I) is true
B.Only (II) is true
C.Both (I) and (II) are true
D.Both (I) and (II) are false

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

Both (I) and (II) are false.

Classify by \(B^{2} - AC\). The wave equation \(u_{tt} = c^{2}u_{xx}\) has \(A = c^{2}\), \(C = -1\), \(B = 0\), so \(B^{2} - AC = c^{2} \gt 0\): it is hyperbolic, not parabolic.

The heat equation \(u_{t} = k u_{xx}\) has only one second-order term, so \(B^{2} - AC = 0\): it is parabolic, not elliptic.

Laplace's equation \(u_{xx} + u_{yy} = 0\) is the standard elliptic example. The statements have the types of the wave and heat equations mixed up.

Discuss in Forum

29.

The partial differential equation

\[ \begin{aligned} &\frac{\partial^{2} z}{\partial x^{2}} + y\frac{\partial^{2} z}{\partial y^{2}} = 0, \\ &y \gt 0 \end{aligned} \]

is :

A.hyperbolic
B.parabolic
C.elliptic
D.Laplacian

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The equation is elliptic.

Compare with \(Az_{xx} + Bz_{xy} + Cz_{yy} = 0\): here \(A = 1\), \(B = 0\), \(C = y\). Then

\[ B^{2} - 4AC = -4y \lt 0 \quad (y \gt 0) \]

A negative discriminant means elliptic. It would be hyperbolic for \(y \lt 0\) and parabolic on \(y = 0\), but the question restricts to \(y \gt 0\). 'Laplacian' is not a type; Laplace's equation is itself the model elliptic equation.

Second Order Partial Differential Equations Questions and Answers

Topics

Other Subjects

Tags: Second Order Partial Differential Equations Questions and Answers

About Us

Second Order Partial Differential Equations Questions and Answers

Topics

Other Subjects

Tags: Second Order Partial Differential Equations Questions and Answers

About Us

Stay informed about all important Dates & Notifications