Graduate · Heat semigroup modes
Heat semigroup modes: Comparable broad and fine scales
Heat semigroup modes: investigate comparable broad and fine scales with higher-mode amplitude a = 1.4; higher frequency k = 4.
Predict what will happen, press Play, then pause and explain what changed. Every control also works without playback.
Watch the relationship
PausedHD animation studio
From experiment to screen.
Present a crisp Canvas scene, save a full-HD image, or capture your model as a silent video.
Understand what you are seeing
The idea behind the motion.
Fourier sine modes are eigenfunctions of the periodic Laplacian, so heat evolution damps each mode at a rate proportional to frequency squared. A short-wavelength component can disappear much faster than a broad component. Orthogonality also gives an exact expression for the spatially averaged quadratic energy. This investigation starts with Higher-mode amplitude a = 1.4; Higher frequency k = 4. Predict the result before playing, then change one parameter while keeping the other fixed to test the reason for the change.
A relationship to keep
u=e^(−0.2t)sin x+a e^(−0.2k²t)sin(kx)
Read the symbols alongside the explanation. A diagram shows the relationship; the assumptions tell you when it applies.
- STEP 1
Set up the mathematical model
Inspect the initial combination sin x+a sin(kx) on a full 2π-periodic interval. The starting case is “Comparable broad and fine scales.”
- STEP 2
Follow the changing quantity
Evolve time from zero to three with diffusivity 0.2. Track the two amplitudes separately while their sum changes shape.
- STEP 3
Explain and test the result
Compare decay rates 0.2 and 0.2k². Check that mean half-square energy equals one quarter of the sum of the squared amplitudes.
Your turn to explain
Make a prediction. Test your reasoning.
Keep Higher-mode amplitude a = 1.4; Higher frequency k = 4. Pause the timeline at 100%. Given time = 3, calculate fundamental amplitude, higher-mode amplitude, mean half-square energy. Show the substitution into the displayed formula.
Use the values specified in the question. Reset restores the initial values for this investigation.
Compare your explanation
At t=3, the amplitudes are exp(−0.2(3))=0.548812 and 1.4exp(−0.2(4)²(3))=0.000095. Orthogonality gives mean half-square energy [0.548812²+0.000095²]/4=0.075299. Results: Fundamental amplitude: 0.549; Higher-mode amplitude: 0; Mean half-square energy: 0.075. Decimal values are rounded; retain the original parameters when checking.
Work through a full lesson
Connect the animation to a worked example and practice questions.