Partial Differential Equations Questions and Answers

Answer: A

The value of \(m - n\) is 2.

The order is the highest derivative present, \(\partial^{3}\varphi/\partial x^{3}\), so \(m = 3\). The degree is the power of that highest-order derivative once the equation is polynomial in its derivatives; \(\partial^{3}\varphi/\partial x^{3}\) appears to the first power, so \(n = 1\).

Hence \(m - n = 2\). The square on \(\partial^{2}\varphi/\partial x^{2}\) is a trap: it is not the highest-order term, so it does not set the degree. Taking the degree as 2 gives 1, and taking the order as 2 gives 0 or 1.

Answer: A

The solution is \(y(x,t) = 10e^{-2x-t}\).

Try \(y = 10e^{-2x + kt}\), which matches the initial condition at \(t = 0\). Substituting:

\[ \begin{aligned} -2y &= 3ky + y \\ 3k &= -3,\quad k = -1 \end{aligned} \]

So \(y = 10e^{-2x-t}\). Check: \(\partial y/\partial x = -2y\), and \(3\,\partial y/\partial t + y = -3y + y = -2y\). With \(k = +1\) the right side is \(4y\), and with \(k = \pm 2\) it is \(-5y\) or \(7y\); none equals \(-2y\).

Answer: C

The true statement is that the solution cannot be obtained using separation of variables because not all the boundaries are along constant coordinate lines.

Separation of variables in Cartesian coordinates needs a domain whose sides are lines of constant x or constant y, so that each boundary condition can be imposed on one factor of the product solution. Here the left side is a slanted line through the origin, which is neither, so the method fails for this domain.

Laplace's equation itself is separable, so the first option is wrong. Non-zero boundary values are handled by superposing several separated solutions, so the second is wrong. And the slanted edge means the fourth option is wrong as posed.

Answer: B

\(a = -1,\ b = 2,\ r = 1\).

A second-order equation \(Au_{xx} + 2Bu_{xy} + Cu_{yy} + \ldots = 0\) is parabolic where \(B^{2} - AC = 0\). Here \(A = x + 2\), \(2B = 2(x + y)\) so \(B = x + y\), and \(C = 2(y - 1)\):

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

The circle has centre \((-1, 2)\) and radius 1. The other options flip the sign of a or b, which is what happens if the completed squares are read as \((x - 1)\) or \((y + 2)\).

Answer: A

The correct statement is The PDE is hyperbolic in the region -1 < x < 0, y < 0.

For \(Au_{xx} + Bu_{xy} + Cu_{yy} + \dots = 0\) the type is set by the discriminant \(B^{2} - 4AC\). Here A = 4x², B = 4(x + y), C = x² + y²:

\[ B^{2}-4AC = 16\left[(x+y)^{2} - x^{2}(x^{2}+y^{2})\right] \]

In the region -1 < x < 0, y < 0, both x and y are negative, so 2xy > 0 and \((x+y)^{2} \gt x^{2}+y^{2}\), while \(x^{2} \lt 1\) gives \(x^{2}+y^{2} \ge x^{2}(x^{2}+y^{2})\). The discriminant is positive everywhere there: hyperbolic.

For x > 1 and y < 0 the bracket is negative at, for example, x = 2, y = -0.1 (3.61 - 16.04), so the PDE is not hyperbolic throughout that region, and it is not parabolic there either, which would need the discriminant to be zero. For 0 < x < 1, y > 0 the discriminant is positive, so the PDE is hyperbolic there, not elliptic.

Answer: A

Partial derivatives exist only when there is more than one independent variable to differentiate with respect to, so the distinction is the number of independent variables and nothing else. The power of the unknown decides the degree, arbitrary constants arise in both kinds, and both demand the derivatives that appear to exist

Answer: D

Differentiating partially gives \( p = \partial z/\partial x = a \) and \( q = \partial z/\partial y = b \), so substituting back into the original relation removes both constants:

\[ z \;=\; px + qy. \]

Two constants are removed by two differentiations, and the result is of the first order

Answer: C

Each arbitrary function eliminated costs one differentiation, so one function gives a first-order equation and two give a second-order one. The wave equation, for instance, comes of eliminating the two functions in \( y = f(x - ct) + g(x + ct) \). Arbitrary functions rather than constants are what make the general solution of a partial equation so much larger than an ordinary one's

Answer: B

With \( u = x^{2} + y^{2} \) the chain rule gives \( p = 2x f'(u) \) and \( q = 2y f'(u) \), so dividing removes the unknown derivative of \( f \):

\[ \frac{p}{q} \;=\; \frac{x}{y} \quad \Longrightarrow \quad yp - xq \;=\; 0. \]

The relation says the surface is unchanged along circles about the axis, which is exactly what dependence on \( x^{2} + y^{2} \) means

Answer: A

The complete integral \( F(x,y,z,a,b) = 0 \) gives the singular solution when \( a \) and \( b \) are eliminated between it and \( \partial F/\partial a = 0 \), \( \partial F/\partial b = 0 \). Geometrically it is the envelope of the family. Setting the constants to zero merely picks one member, and replacing them by functions leads towards the general solution