Graduate · Variational energy · 364 of 650
Variational energy · Initial sine amplitude a=1.5; Deviation penalty b=2.5
Use the remaining amplitude A=0.6, two fifths of the initial amplitude. Calculate E[u_A], its derivative with respect to A, and the minimum within this trial family. Givens: Initial sine amplitude a=1.5; Deviation penalty b=2.5.
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01 · Make a prediction
Your practice question
Use E[u]=½∫₀¹u′²dx+2.5∫₀¹(u−x)²dx with u_A(x)=x+A sin(πx). The initial amplitude is 1.5. Use the remaining amplitude A=0.6, two fifths of the initial amplitude. Calculate E[u_A], its derivative with respect to A, and the minimum within this trial family.
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A starting point
The endpoints are fixed for every A. Integrate the squared derivative and squared sine term to obtain E=½+A²(π²/4+b/2).
Work through the reasoning
Step 1
Identify the model and target
Use E[u]=½∫₀¹u′²dx+2.5∫₀¹(u−x)²dx with u_A(x)=x+A sin(πx). The initial amplitude is 1.5. Use the remaining amplitude A=0.6, two fifths of the initial amplitude. The governing relation is E[u]=½∫₀¹u′²dx+b∫₀¹(u−x)²dx; u=x+A sin(πx). The endpoints are fixed for every A. Integrate the squared derivative and squared sine term to obtain E=½+A²(π²/4+b/2).
Step 2
Substitute and calculate
The remaining amplitude is A=(1.5)(1−0.6)=0.6. Exact integration gives E=1/2+A²(π²/4+2.5/2)=1.838264. Differentiating that expression gives A(π²/2+2.5)=4.460881.
Step 3
Check the mathematical meaning
The coefficient π²/4+2.5/2 is positive, so the reduced energy has unique minimum 1/2 at A=0. The current energy exceeds that minimum by 1.338264. Remaining sine amplitude: 0.6; Functional energy: 1.838; Derivative with respect to amplitude: 4.461; Minimum energy: 0.5. Decimal values are rounded, so use unrounded intermediate values.
The answer
The remaining amplitude is A=(1.5)(1−0.6)=0.6. Exact integration gives E=1/2+A²(π²/4+2.5/2)=1.838264. Differentiating that expression gives A(π²/2+2.5)=4.460881. The coefficient π²/4+2.5/2 is positive, so the reduced energy has unique minimum 1/2 at A=0. The current energy exceeds that minimum by 1.338264. Animation check: Remaining sine amplitude: 0.6; Functional energy: 1.838; Derivative with respect to amplitude: 4.461; Minimum energy: 0.5. Decimal displays are rounded; retain the original parameters and exact π until the final step.
Compare the method as well as the result. A different valid method may reach the same answer. Keep exact values until the last step when the question asks for rounding.
02 · Explore the model
See the mathematical relationship move.
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Watch the relationship
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The mathematical idea
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. Use E[u]=½∫₀¹u′²dx+2.5∫₀¹(u−x)²dx with u_A(x)=x+A sin(πx). The initial amplitude is 1.5. Use the remaining amplitude A=0.6, two fifths of the initial amplitude. The requested state occurs at 60% playback; use the exact target stated in the question for your calculation.
E[u]=½∫₀¹u′²dx+b∫₀¹(u−x)²dx; u=x+A sin(πx)
03 · Reflect and transfer
Explain what changes and why.
Why does checking a one-parameter trial family alone not prove that a candidate minimizes a functional over every admissible function?
This is a distinct guided scenario using a reusable mathematical model. Similar-looking diagrams can represent different given values and conclusions; they are not different mathematical theories.