Question 4
For , you may use the complete family Consider two endpoint conditions and , where is specified. Compare these with initial data , .
Tasks
Solve the initial-data problem for an arbitrary real . Explain why it always determines exactly one member of the family.
Determine all solutions of the endpoint problem for and arbitrary .
Determine all solutions for , treating and separately. Explain why two conditions need not select one solution.
For , , sketch the solutions with . Explain why their common endpoint values do not violate uniqueness for initial-value problems.
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Question 4 – Solution
Strategy. Translate each condition into a constraint on and check whether the second constraint supplies new information.
Step 1: Use value and slope at the same point. From and , initial data give Both constants are fixed, for every real . The coefficients of the normalized equation are continuous everywhere, consistent with initial-value uniqueness.
Step 2: Move the second datum to . The first condition gives . At the other endpoint, This particular pair of endpoint conditions also determines one solution.
Step 3: Detect a redundant or incompatible condition. At , every function has . Consequently, When , the second condition adds no restriction on . When , it contradicts the first condition and the equation. Merely counting two stated conditions misses this distinction.
See the diagram in the original worksheet below.
Step 4: Compare endpoint agreement with initial-data agreement. The four curves share their endpoint positions but have different initial slopes . At their slopes are , also different. The initial-value theorem requires agreement of both value and first derivative at the same point. Agreement of values at two separate points is a different requirement, so the sketch creates no contradiction.