Partial Differential Equations Questions and Answers

Answer: C

The complete integral of a first-order equation in two independent variables carries two arbitrary constants, one for each variable, and the general solution replaces them by an arbitrary function. Counting one constant per order is the ordinary-equation habit, and it does not carry across

Answer: B

If \( A \), \( B \) and \( C \) depend on \( x \) and \( y \), the discriminant does too, and it may change sign. The Tricomi equation \( y z_{xx} + z_{yy} = 0 \) is elliptic for \( y > 0 \) and hyperbolic for \( y < 0 \), a mixed type that arises physically at transonic speeds. Linearity is not what decides it

Answer: C

The possible case is a(x) periodic and b(y) exponential.

Substituting \(f = a(x)b(y)\) into Laplace's equation and dividing by ab separates the variables:

\[ \frac{a''(x)}{a(x)} = -\frac{b''(y)}{b(y)} = -k^{2} \]

A negative constant gives \(a'' = -k^{2}a\), so a is a sine or cosine (periodic), and \(b'' = k^{2}b\), so b is a combination of \(e^{ky}\) and \(e^{-ky}\) (exponential). This is the familiar separated solution such as \(\sin kx\, e^{ky}\).

The two functions always have opposite-signed constants, so they cannot both be exponential or both periodic. A linear b(y) needs the constant to be zero, and then a(x) is linear too, not periodic.

Answer: D

The equation \(\frac{\partial u}{\partial t} = \alpha \frac{\partial^{2} u}{\partial x^{2}}\) is parabolic.

A second-order equation \(A u_{xx} + B u_{xt} + C u_{tt} + \ldots = 0\) is classified by the discriminant \(B^{2} - 4AC\). Here \(A = \alpha\), \(B = 0\) and \(C = 0\), since there is no second derivative in \(t\), so

\[ B^{2} - 4AC = 0 - 4\alpha \cdot 0 = 0 \]

A zero discriminant means parabolic. This is the heat (diffusion) equation, the standard parabolic example. Elliptic needs a negative discriminant (Laplace's equation) and hyperbolic a positive one (the wave equation). 'Circular' is not a class of partial differential equation.

Answer: A

The solution at \(t = 1\) is \(u(x,1) = e^{-(x-1)^{2}}\).

The equation \(u_{t} + u_{x} = 0\) is the linear advection (one-way wave) equation with unit speed. Along the characteristics \(x - t = \text{constant}\), \(u\) does not change, so the general solution is \(u(x,t) = f(x - t)\): the initial profile moves to the right at speed 1 without changing shape.

\[ \begin{aligned} u(x,0) &= f(x) = e^{-x^{2}} \\ u(x,t) &= e^{-(x-t)^{2}} \\ u(x,1) &= e^{-(x-1)^{2}} \end{aligned} \]

The option with \((x+1)\) moves the profile left, which would be the solution of \(u_{t} - u_{x} = 0\). \(e^{-x^{2}}\) leaves the profile where it was, and \(e^{-1}\) is a constant, which is not a translated Gaussian at all.

Answer: B

The equation is hyperbolic.

A second-order equation \(Au_{xx} + 2Bu_{xy} + Cu_{yy} + \dots = 0\) is classified by \(B^{2} - AC\) (equivalently by \(b^{2} - 4ac\) when the mixed coefficient is written as b). Here the coefficients are 1, 4 and 2:

\[ b^{2} - 4ac = 4^{2} - 4(1)(2) = 8 \gt 0 \]

A positive discriminant means hyperbolic, zero means parabolic and negative means elliptic. The coefficients are constants, so the type is the same everywhere and the equation is not of mixed type.

Answer: A

The equation represents linear wave propagation.

\(\varphi_t + a\varphi_x = 0\) is the first-order linear advection (one-way wave) equation. Its general solution is \(\varphi(x, t) = f(x - at)\): any initial profile travels unchanged at speed \(a\), which is pure wave propagation.

Transient heat conduction is the diffusion equation, \(\varphi_t = \alpha \varphi_{xx}\), with a second derivative in space. Newton's law of cooling is an ordinary differential equation in time, \(dT/dt = -k(T - T_\infty)\), with no spatial derivative. Radiative transfer describes intensity along a ray with absorption, emission and scattering terms, which this equation lacks.

Answer: A

The order is (i) Wave equation, (ii) Heat equation, (iii) Laplace equation.

Equation (i) has a second derivative in time equal to \(c^{2}\) times a second derivative in space: \(u_{tt} = c^{2}u_{xx}\) is the one-dimensional wave equation, with \(c\) the wave speed. Equation (ii) has only a first derivative in time: \(u_{t} = c^{2}u_{xx}\) is the one-dimensional heat (diffusion) equation, with \(c^{2}\) the diffusivity. Equation (iii), \(u_{xx} + u_{yy} = 0\), has no time variable at all and is the two-dimensional Laplace equation for steady potentials.

The other options swap these labels. The quick test is the order of the time derivative: second order means waves, first order means diffusion, and none means a steady (Laplace) problem.

Answer: B

The solution is u(x, y) = f(x - cy).

Try \(u = f(\xi)\) with \(\xi = x - cy\). Then \(\partial u/\partial y = -cf'(\xi)\) and \(\partial u/\partial x = f'(\xi)\), so

\[ \begin{aligned} \frac{\partial u}{\partial y} + c\frac{\partial u}{\partial x} &= -cf' + cf' \\ &= 0 \end{aligned} \]

for any differentiable f. This is the one-way wave equation: u is constant along the lines x - cy = constant.

f(x + cy) gives \(cf' + cf' = 2cf'\), not zero. f(cx + y) gives \(f' + c^{2}f'\), and f(cx - y) gives \(-f' + c^{2}f'\), which vanishes only if c = 1, excluded because c is greater than 1.

Answer: B

The solution is \(T(x, y) = x + y\).

It satisfies Laplace's equation, since both second derivatives are zero, and it matches every boundary condition: \(T(x, 0) = x\), \(T(0, y) = y\), \(T(x, 1) = x + 1\) and \(T(1, y) = 1 + y\). The solution of Laplace's equation with given boundary values is unique, so this is the answer.

\(x - xy + y\) gives \(T(x, 1) = 1\), not \(1 + x\). \(x + xy + y\) gives \(T(x, 1) = 2x + 1\). \(-x + y\) gives \(T(x, 0) = -x\). Each fails at least one boundary.