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.