Green's Functions for Boundary Value Problems · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Build a Green's function for fixed boundaries

Mathematics · Differential Equations · ages 21-22
Name ______________________   Date ____________
  1. At s=1/4, the pieces Ax and B(1-x) meet continuously. With B nonzero, what is A/B?

    Answer: ______________

  2. For -u'' on [0,1] with zero endpoint values, what is G(1/4,1/2)?

    • 1/8
    • 3/8
    • 1/4
    • 1/2
  3. For -u''=f on [0,1] with u(0)=u(1)=0, which formula is the Green's function?

    • G=s(1-x) for x<=s; G=x(1-s) for x>=s
    • G=x(1-s) for x<=s; G=s(1-x) for x>=s
    • G=x(1-s) for x<=s; G=-s(1-x) for x>=s
  4. For the taught separated boundary problem, how are the two homogeneous pieces used to construct G?

    • Meet each endpoint condition, match values, then set the slope jump.
    • Meet each endpoint condition, match values, then make both slopes equal.
    • Meet only the left endpoint condition, then keep one line across the interval.
  5. For the two-zero-endpoint kernel G(x,s)=s(1-x) on x>s, fix s=1/5. What is its right-side x derivative? Give a decimal.

    Answer: ______________

  6. For -2u''=f on [0,1] with zero endpoint values, put a unit source at s=1/4. The left x slope of G is 3/8. What is the right slope? Give a decimal.

    Answer: ______________

  7. For -u''=f on [0,1] with u(0)=0 and u'(1)=0, which Green's function meets the conditions and unit-source jump?

    • G=s for x<=s; G=x for x>=s
    • G=-x for x<=s; G=-s for x>=s
    • G=x for x<=s; G=s for x>=s
    • G=x for x<=s; G=s(1-x) for x>=s
  8. The taught Green's function for -u'' on [0,1] with u(0)=u(1)=0 satisfies G(x,s)=G(s,x).

    Circle one:   True   False

  9. For -u'', replacing u(1)=0 by u'(1)=0 while retaining u(0)=0 changes the Green's function.

    Circle one:   True   False

  10. For -u''=1 on [0,1] with u'(0)=u'(1)=0, why is there no solution?

    • A positive forcing requires positive derivatives at every point of the interval.
    • A second-order equation cannot use derivative conditions at its endpoints.
    • Constant homogeneous solutions make every forcing incompatible with these conditions.
    • Integration gives -u'(1)+u'(0)=1, but the conditions make that left side zero.
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Build a Green's function for fixed boundaries W1-mt_2ctFrfKIgW-s1

  1. 3 · Continuity gives A/4=3B/4. Thus A=3B and A/B=3.
  2. 1/8 · Since x<s, use G=x(1-s)=(1/4)(1/2)=1/8.
  3. G=x(1-s) for x<=s; G=s(1-x) for x>=s · These branches vanish at their respective endpoints, agree at s and have right slope minus left slope equal to -1.
  4. Meet each endpoint condition, match values, then set the slope jump. · Each piece meets its own endpoint condition. Continuity and the unit-source jump determine their scaling.
  5. -0.2 · Hold s fixed. The x derivative of s(1-x) is -s=-1/5=-0.2.
  6. -0.125 · The operator coefficient makes the jump -1/2. The right slope is 3/8-1/2=-1/8=-0.125.
  7. G=x for x<=s; G=s for x>=s · The left piece vanishes at zero and the right piece has zero x derivative. Both equal s at the join, and the slope jump is 0-1=-1.
  8. True · For x<=s, both expressions equal x(1-s). For x>=s, both equal s(1-x).
  9. True · A zero right derivative requires a constant right piece. A zero right value uses a piece proportional to 1-x.
  10. Integration gives -u'(1)+u'(0)=1, but the conditions make that left side zero. · Integrating over [0,1] gives -u'(1)+u'(0)=1. The two zero endpoint derivatives would require 0=1.
Worksheet · LightMySky