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anonymous

  • 4 years ago

do initial conditions have to be where t=0,

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  1. anonymous
    • 4 years ago
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    No. "Initial conditions" is an arbitrary marker. So is t=0.

  2. anonymous
    • 4 years ago
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    boundary is when they give t=a, t=b. rather than y(a)=k1 , y'(a)= k2 and so on.

  3. anonymous
    • 4 years ago
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    If those are the only boundaries they give you, I can't tell you anything about "initial conditions". I would guess that they mean for t=a to be the initial condition, but it's not necessarily 0 unless there's other information.

  4. TuringTest
    • 4 years ago
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    There's not really enough info for me to understand what you're doing, but it looks to me like you have boundary values, not initial values.

  5. anonymous
    • 4 years ago
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    Show that the problem y'=2x, y(0)=0, y(1)=100, has no solution. Is this an initial value problem?

  6. TuringTest
    • 4 years ago
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    It looks like a boundary value problem, but I don't think it really is. It just demonstrates that the two initial values contradict\[y'=2x\]\[y=x^2+C\]\[y(0)=C=0\]\[y(1)=1+C=100\to C=99\]so C is trying to be two different numbers at once, which is not possible. Not sure what you call this type of 'impossible initial values' set-up...

  7. TuringTest
    • 4 years ago
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    I think BVP's are at least second order

  8. anonymous
    • 4 years ago
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    if it is an initial value problem then there is no solution since c cant be two different numbers at once, where as if this was a boundary problem we just identified the boundary values for the boundary conditions.

  9. TuringTest
    • 4 years ago
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    Yes, I think it is an IVP with two contradictory values. A first-order BVP doesn't really make any sense since you only need one condition to find C, and boundary conditions come in pairs or more.

  10. anonymous
    • 4 years ago
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    thanks

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