Graduate · Variational energy
Variational energy: A weak penalty on deviation
Variational energy: investigate a weak penalty on deviation with initial sine amplitude a = 1; deviation penalty b = 0.5.
Predict what will happen, press Play, then pause and explain what changed. Every control also works without playback.
Watch the relationship
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Understand what you are seeing
The idea behind the motion.
An energy functional assigns a number to an entire function. The family x+A sin(πx) respects fixed endpoints for every A, making it possible to vary shape without violating boundary data. The exact energy is one half plus a positive quadratic multiple of A², so the straight function is the unique minimizer within this family. This investigation starts with Initial sine amplitude a = 1; Deviation penalty b = 0.5. 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
E[u]=½∫₀¹u′²dx+b∫₀¹(u−x)²dx; u=x+A sin(πx)
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
Verify that the endpoint values remain zero and one even when the sine amplitude changes. The starting case is “A weak penalty on deviation.”
- STEP 2
Follow the changing quantity
Reduce A from the chosen initial amplitude to zero. Compare the moving curve with the reference straight line and the energy readout.
- STEP 3
Explain and test the result
Differentiate the reduced energy with respect to A. Distinguish this finite-dimensional family calculation from a proof over every admissible function.
Your turn to explain
Make a prediction. Test your reasoning.
Keep Initial sine amplitude a = 1; Deviation penalty b = 0.5. Pause the timeline at 20%. Given remaining sine amplitude = 0.8, calculate functional energy, derivative with respect to amplitude, minimum 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
The remaining amplitude is A=(1)(1−0.2)=0.8. Exact integration gives E=1/2+A²(π²/4+0.5/2)=2.239137. Differentiating that expression gives A(π²/2+0.5)=4.347842. Results: Functional energy: 2.239; Derivative with respect to amplitude: 4.348; Minimum energy: 0.5. Decimal values are rounded; retain the original parameters when checking.
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