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Asymptotic expansions

  1. Aug 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Find two term asymptotic expansions for the roots of the following equations for ε → 0.
    $$x^2+x-4ε = 0, \quad x^2+cos(εx)=5 \quad \text{and} \quad εx^3-x+9=0$$



    2. Relevant equations
    The 2 term asymptotic expansion is $$x\approx x_0+εx_1.$$


    3. The attempt at a solution
    Plugging the asymptotic expansion into the first equation gives $$(x_0+εx_1)^2+x_0+εx_1-4ε=0$$
    $$x_0^2+2εx_0x_1+ε^2x_1^2+x_0+εx_1-4ε=0.$$
    Equation constant and order ε terms one gets $$x_0^2+x_0=0 \quad \text{and} \quad 2x_0x_1+x_1-4=0.$$
    $$ x_0 = 0,-1 \Rightarrow x_1=4,-4.$$
    Thus our 2 term expansions are
    $$x \approx 4ε \quad \text{and} \quad x \approx -1-4ε.$$
    First we expand the cos term as a power series
    $$\cos(εx) = 1 - \frac{(εx)^2}{2}+\cdots \approx 1.$$
    Plugging the asymptotic expansion into the second equation gives $$(x_0+εx_1)^2+1=5$$
    $$x_0^2+2εx_0x_1+ε^2x_1^2-4=0.$$
    Equation constant and order ε terms one gets $$x_0^2-4=0 \quad \text{and} \quad 2x_0x_1=0.$$
    $$ x_0 = \pm 2 \Rightarrow x_1=0.$$
    Thus our 2 term expansions are
    $$x \approx 2 \quad \text{and} \quad x \approx -2.$$
    Plugging the asymptotic expansion into the third equation gives $$ε(x_0+εx_1)^3-x_0-εx_1+9=0$$
    $$ε(x_0^3+2εx_0x_1+\cdots+εx_0^2x_1+\cdots)-x_0-εx_1+9=0.$$
    Are the first 2 correct and not too sure on how to continue with the final one.
    Please help!!!!!!
     
  2. jcsd
  3. Aug 10, 2013 #2

    mfb

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    You can determine x0 if you set ε=0 and solve the equation.
    Your expansion of (x0+εx1)3 is wrong. The right expansion will make the following steps easier.
     
  4. Aug 10, 2013 #3
    Thanks.
    So I take it the first 2 are good.

    Cheers.
     
  5. Aug 11, 2013 #4

    pasmith

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    I'm not sure that "[itex]x \sim \pm 2[/itex]" really qualifies as a "two-term expansion".

    Here the appropriate expansion is in powers of [itex]\epsilon^2[/itex], since changing the sign of [itex]\epsilon[/itex] does not change [itex]\cos(\epsilon x)[/itex], and so should not change the values of [itex]x[/itex] which satisfy the equation [itex]x^2 + \cos(\epsilon x) = 5[/itex].

    Try again with [itex]x \sim x_0 + x_2 \epsilon^2[/itex].
     
  6. Aug 12, 2013 #5
    Thanks for that.

    Using the new asymptotic expansion i get
    $$x \approx 2+\frac{\epsilon^2}{2} \quad \text{and} \quad x \approx -2-\frac{\epsilon^2}{2}.$$

    Can someone confirm this.

    Cheers.
     
  7. Aug 12, 2013 #6

    pasmith

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    That is correct.
     
  8. Aug 12, 2013 #7
    Does that remain true for singular perturbation problems such as the third problem here?
     
  9. Aug 13, 2013 #8

    mfb

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    x0 is by definition the solution for x at ε=0.
    If you do not get a converging taylor expansion around ε=0, the whole method is not feasible anyway.
     
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