Numerical Schemes for Evolution Equations · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Turning change into steps

Mathematics · Differential Equations · ages 23-24
Name ______________________   Date ____________

  1. For k > 0, one-dimensional FTCS heat stepping must satisfy dt <= dx^2/(2k) to meet its all-mode stability bound.

    Circle one:   True   False

  2. In the centred approximation to u_xx, which coefficients multiply the old left, centre and right values before division by dx^2?

    • (1, -2, 1)
    • (1, -1, 1)
    • (-1, 2, -1)
    • (1, 2, 1)

  3. For FTCS with r = 0.3, if L = C = R = 7, the updated centre is 7.3.

    Circle one:   True   False

  4. FTCS heat stepping uses dt = 0.01, dx = 0.1 and diffusivity k = 1. What is the mesh ratio r?

    Answer: ______________

  5. For the periodic one-dimensional heat equation with k > 0, backward Euler with a centred space difference has |G| <= 1 for every positive dt and dx.

    Circle one:   True   False

  6. For FTCS heat stepping with dx = 0.1 and k = 1, what is the largest dt allowed by the all-mode bound r <= 0.5?

    • 0.050
    • 0.005
    • 0.500
    • 0.100

  7. One FTCS heat step has r = 0.5, old centre C = 4, old left value L = 1 and old right value R = 2. What is C_new?

    Answer: ______________

  8. For a smooth solution, what are the consistency orders of FTCS heat stepping, using a forward time quotient and a centred second space quotient?

    • Second order in time, first order in space
    • First order in time, second order in space
    • Second order in time, second order in space

  9. Lee starts FTCS heat stepping with r = 0.2. He halves dx, keeps dt and k fixed, and claims r is unchanged. What happened to r?

    • Only round-off error changes with the grid
    • The mesh ratio falls from 0.2 to 0.05
    • The mesh ratio rises from 0.2 to 0.8

  10. Keep k fixed in r = k dt/dx^2. Doubling dx and quartering dt leaves r unchanged.

    Circle one:   True   False

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Answer key

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

Turning change into steps W1-mt_RuDbOapexB-s1

  1. True · The bound r = k dt/dx^2 <= 0.5 gives dt <= dx^2/(2k).
  2. (1, -2, 1) · The numerator L - 2C + R has coefficients 1, -2 and 1 in left, centre, right order.
  3. False · The spatial difference is 7 - 14 + 7 = 0, so the update leaves the centre at 7.
  4. 1 · r = k dt/dx^2 = 1(0.01)/(0.1^2) = 0.01/0.01 = 1.
  5. True · G = 1/(1 + 4r sin^2(theta/2)). For r >= 0 its denominator is at least 1, so 0 < G <= 1.
  6. 0.005 · The cap is dx^2/(2k) = 0.01/2 = 0.005.
  7. 1.5 · C_new = 4 + 0.5(1 - 8 + 2) = 4 - 2.5 = 1.5.
  8. First order in time, second order in space · The forward time quotient has leading error proportional to dt. The centred second space quotient has leading error proportional to dx^2.
  9. The mesh ratio rises from 0.2 to 0.8 · Halving dx quarters the denominator of r = k dt/dx^2. The unchanged numerator makes r four times as large: 0.8.
  10. False · The numerator becomes one quarter as large and the denominator becomes four times as large. The new ratio is r/16.
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