Partial Differential Equations Questions and Answers

Answer: B

It is the heat equation.

The one-dimensional heat (diffusion) equation is \(u_t=\alpha\,u_{xx}\) (subscripts denoting partial derivatives): the first time derivative equals a constant times the second space derivative. Here the constant is 1.

The wave equation has the second time derivative, \(u_{tt}=c^2u_{xx}\). Laplace's equation has no time derivative at all, \(\nabla^{2}u = 0\). There is no standard 'elasticity equation' of this form.

Answer: B

The solution is \(u = e^{x}\sin y\).

Check Laplace's equation term by term: \(u_{xx} = e^{x}\sin y\) and \(u_{yy} = -e^{x}\sin y\), so \(u_{xx} + u_{yy} = 0\) everywhere.

For \(e^{x+y}\) the sum is \(2e^{x+y}\); for \(\sin x\sin y\) it is \(-2\sin x\sin y\); for \(\cos x\cos y\) it is \(-2\cos x\cos y\). None of these vanishes for all \(x\) and \(y\), so none is a solution.

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.