What does the Fourier transform do to the operation ∂²u/∂x²?
AReplaces it with -ξ²û(ξ)
BReplaces it with iξû(ξ)
CReplaces it with ξ²û(ξ)
DReplaces it with -iξ²û(ξ)
Each spatial derivative ∂/∂x becomes multiplication by iξ under the Fourier transform. Two derivatives give (iξ)² = -ξ². This is the key property that converts differential equations into algebraic ones.
Question 2 True / False
The Fourier transform method works directly for PDEs with variable coefficients.
TTrue
FFalse
Answer: False
The Fourier transform converts constant-coefficient differential operators into polynomial multiplication. Variable coefficients become convolution operators in frequency space, which are generally no simpler than the original PDE. Other methods (such as pseudodifferential operators) are needed for variable-coefficient problems.
Question 3 Short Answer
Using the Fourier transform, what is the solution to the heat equation u_t = ku_xx on ℝ with initial data u(x,0) = f(x)?
Think about your answer, then reveal below.
Model answer: Convolution of f with the heat kernel: u(x,t) = (4πkt)^(-1/2) ∫ f(y)e^(-(x-y)²/(4kt)) dy
In frequency space, û(ξ,t) = f̂(ξ)e^(-kξ²t). Inverting gives a convolution of f with the inverse transform of e^(-kξ²t), which is the Gaussian heat kernel (4πkt)^(-1/2)e^(-x²/(4kt)).
Question 4 Multiple Choice
For the wave equation u_tt = c²u_xx on ℝ, the Fourier transform in x yields what ODE in t?
Aû_tt = -c²ξ²û
Bû_tt = c²ξ²û
Cû_t = -c²ξ²û
Dû_t = icξû
Transforming u_xx gives -ξ²û, so the wave equation becomes û_tt = -c²ξ²û. This is a simple harmonic oscillator in t for each frequency ξ, with solution û(ξ,t) = A(ξ)cos(cξt) + B(ξ)sin(cξt).