Asymptotic Expansions for Root Equations | Two-Term Expansions for ε → 0

[squenshl]

Homework Statement

Find two term asymptotic expansions for the roots of the following equations for ε → 0.
$$x^2+x-4ε = 0, \quad x^2+cos(εx)=5 \quad \text{and} \quad εx^3-x+9=0$$

Homework Equations

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

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!

 

[mfb]

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.

 

[squenshl]

Thanks.
So I take it the first 2 are good.

Cheers.

 

[pasmith]

[x^2 + \cos(\epsilon x) = 5]
Thus our 2 term expansions are
$$x \approx 2 \quad \text{and} \quad x \approx -2.$$

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

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

Try again with x \sim x_0 + x_2 \epsilon^2.

 

[squenshl]

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.

 

[pasmith]

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.

That is correct.

 

[Jorriss]

You can determine x₀ if you set ε=0 and solve the equation.
.

Does that remain true for singular perturbation problems such as the third problem here?

 

[mfb]

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.