A second-order equation \( Ar + Bs + Ct + \cdots = 0 \) is classified by the sign of
Answer: C
The discriminant \( B^{2} - 4AC \) decides the type, exactly as it does for a conic: negative gives elliptic, zero parabolic and positive hyperbolic. The classification governs what kind of data the problem needs and how disturbances travel, so it is settled before any method of solution is chosen
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Laplace's equation \( z_{xx} + z_{yy} = 0 \) is of which type?
Answer: B
Here \( A = 1 \), \( B = 0 \) and \( C = 1 \), so
\[ B^{2} - 4AC \;=\; 0 - 4 \;=\; -4 \;<\; 0, \]
which is the elliptic case. Elliptic problems carry data on the whole boundary of a closed region and have smooth solutions throughout, with no direction of propagation
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The heat equation \( z_{t} = \alpha z_{xx} \) is of which type?
Answer: A
Written in the standard form the only second derivative is \( z_{xx} \), so \( A = \alpha \) with \( B = C = 0 \) and
\[ B^{2} - 4AC \;=\; 0, \]
the parabolic case. Parabolic problems march forward in time from an initial state, and their solutions smooth out as they go, which is why heat conduction has no reverse
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The wave equation \( z_{tt} = c^{2} z_{xx} \) is of which type?
Answer: D
Rearranged as \( z_{tt} - c^{2}z_{xx} = 0 \) we have \( A = 1 \), \( B = 0 \), \( C = -c^{2} \), so
\[ B^{2} - 4AC \;=\; 4c^{2} \;>\; 0 \]
for any non-zero \( c \), the hyperbolic case. Hyperbolic equations carry disturbances along characteristics at finite speed, which is what makes a wave a wave
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D'Alembert's solution of the one-dimensional wave equation has the form
Answer: C
The two arbitrary functions represent disturbances travelling in opposite directions without change of shape:
\[ y(x,t) \;=\; f(x - ct) + g(x + ct). \]
A single function would describe a wave running one way only, which cannot meet two initial conditions, and a product would not satisfy the equation at all
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In \( y = f(x - ct) \), the constant \( c \) is
Answer: B
The shape \( f \) is unchanged if \( x \) and \( t \) increase together so as to keep \( x - ct \) fixed, which requires \( dx/dt = c \). So the profile moves bodily at speed \( c \). Amplitude belongs to \( f \) itself, and frequency and wavelength appear only once a particular periodic \( f \) is chosen
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A string of length \( L \) fixed at both ends has, for its fundamental mode, a wavelength of
Answer: A
Fixed ends force nodes at both, and the fundamental has no interior node, so the length holds half a wave:
\[ L \;=\; \frac{\lambda}{2} \quad \Longrightarrow \quad \lambda \;=\; 2L. \]
The frequency is then \( c/2L \). A quarter wave in the length would need one free end, as in a closed organ pipe
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A string of length 2 m carries waves at 40 m/s. Its fundamental frequency is
Answer: D
The fundamental wavelength is twice the length, so \( \lambda = 4 \) m, and
\[ f \;=\; \frac{c}{\lambda} \;=\; \frac{40}{4} \;=\; 10\ \mathrm{Hz}. \]
Equivalently \( f = c/2L = 40/(2 \times 2) \). Taking the wavelength as the length itself rather than twice it would double the answer to 20 Hz
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In the heat equation \( \partial u/\partial t = \alpha\,\partial^{2}u/\partial x^{2} \), the constant \( \alpha \) has dimensions of
Answer: C
For the two sides to agree dimensionally, \( \alpha \) must carry the dimensions of the left side divided by those of \( u/x^{2} \), that is
\[ \left[\alpha\right] \;=\; \frac{\Theta/T}{\Theta/L^{2}} \;=\; \frac{L^{2}}{T}, \]
units of square metres per second. This is why a diffusion length grows as the square root of time rather than in proportion to it
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Three of these four describe the one-dimensional heat equation. The exception is
Answer: B
Finite-speed propagation along characteristics belongs to the hyperbolic wave equation. The classical heat equation is parabolic, its solutions smooth with time, and it does require both an initial distribution and conditions at the ends. Its disturbances spread instantly in principle, which is a known shortcoming of the model
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