Perturbation Methods-Multiscale expansions

Question

I was looking at some of the problems from the chapter Multiple-scale expansion in Introduction to Perturbation Methods by Mark H. Holmes. I came across this question to find the first term expansion:

$\epsilon y''+\epsilon y'+y=cos(t)$, where $y(0)=0=y'(0)$. I took my slow and fast scale variables as $t_1=t$ and $t_2=\sqrt{\epsilon}t$. My question arises when we have to find the $O(1)$ expansion of the term.I'm aware that I need to use Taylors expansion, but something doesn't seem to fit quiet properly. It would be great if someone could help me out with this.

Answer 1

What you should get an approximation for is depicted in the following plot of numerical solutions for some perturbation parameters. This picture corresponds to some extended understanding of "boundary layer", as there is indeed some rapid compensation for the gap between the value $1$ at $t=0$ of the outer solution $y_{\rm outer}(t)=\cos(t)$ and the initial condition $y(0)=0$. This first segment has indeed also a width of $\sim\sqrt{ϵ}$, but then overshoots the outer solution and oscillates with frequency $1/\sqrt{ϵ}$, dampened by the friction term containing the first derivative giving a uniform hull curve for the amplitude of the oscillation of $e^{-t/2}$. Taking these building blocks to first satisfy $y(0)=0$ and then also $y'(0)=0$ gives the first order approximation $$ y(t)\approx \cos(t)-e^{-t/2}\left(\cos\left(\frac{t}{\sqrt{ϵ}}\right)+\frac{\sqrt{ϵ}}2\sin\left(\frac{t}{\sqrt{ϵ}}\right)\right) $$

Let's be a bit more ingeniuous about it. We know that largely $y(t)=\cos(t)+u(t)$ where $u$ will be small in later times. But then $$ ϵu''+ϵu'+u=ϵ\cos(t)+ϵ\sin(t) $$ Adjust the decomposition of $y$ for the new right side, now $y(t)=(1+ϵ)\cos(t)+ϵ\sin(t)+u(t)$, then the differential equation reads $$ \begin{split} ϵ[-(1+ϵ)\cos(t)-ϵ\sin(t)+u''(t)]&+ϵ[-(1+ϵ)\sin(t)+ϵ\cos(t)+u(t)]\\&+[(1+ϵ)\cos(t)+ϵ\sin(t)+u(t)]=\cos(t) \end{split} \\\iff\\ ϵu''+ϵu'+u=2ϵ^2\sin(t)\iff u''+u'+\frac1ϵu=2ϵ\sin(t) $$ which now is essentially a homogeneous equation.

Approach via parametrized sinusoid

As this is a harmonic oscillator with damping, use the form proposed in the hint at the end of the section, set $u(t)=A(t)\cos(ωt+θ(t))$ with $ω=\sqrt{\frac1ϵ}$ and $A(t),θ(t)$ "slow" moving, in the sense that their derivatives are small against $ω$. Then \begin{align} u'(t)&=A'(t)\cos(ωt+θ(t))-A(t)\sin(ωt+θ(t))(ω+θ'(t)) \\ u''(t)&=A''(t)\cos(ωt+θ(t))-2A'(t)\sin(ωt+θ(t))(ω+θ'(t)) \\ & ~ ~ ~ ~- A(t)\cos(ωt+θ(t))(ω+θ'(t))^2-A(t)\sin(ωt+θ(t))θ''(t) \\ \end{align} Comparing coefficients in $u''(t)+u'(t)+ω^2u(t)$ results in the identies \begin{align} 0&=A''(t)+A'(t)- A(t)(2ω+θ'(t))θ'(t)\\ 0&=-A(t)θ''(t)-2A'(t)(ω+θ'(t)) -A(t)(ω+θ'(t))\\[.7em] \implies 2A(t)θ'(t)&=\sqrt{ϵ}[A''(t)+A'(t)-2A(t)θ'(t)^2]\\ 2A'(t)+A(t)&=-\sqrt{ϵ}[A(t)θ''(t)+ (2A'(t)+A(t))θ'(t)] \end{align} so that in first approximation $A(t)=A_0e^{-t/2}$ and $θ(t)=θ_0$. Inserting into the right side gives the next order of approximation as \begin{align} 2θ'(t)&=\sqrt{ϵ}[\tfrac14-\tfrac12]&&\implies& θ(t)=θ_0-\frac{\sqrt{ϵ}}8t\\ 2A'(t)+A(t)&=0&&\implies& \text{unchanged} \end{align}

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Now adjust the coefficients for the initial conditions in \begin{align} y(t)&=(1+ϵ)\cos(t)+ϵ\sin(t)+A_0e^{-t/2}\cos\left((ω-\tfrac18\sqrt{ϵ})t+θ_0\right)\\ y'(t)&=-(1+ϵ)\sin(t)+ϵ\cos(t)-\tfrac12 A_0e^{-t/2}\cos\left((ω-\tfrac18\sqrt{ϵ})t+θ_0\right) \\&~~~~-(ω-\tfrac18\sqrt{ϵ})A_0e^{-t/2}\sin\left((ω-\tfrac18\sqrt{ϵ})t+θ_0\right) \\\hline 0=y(0)&=(1+ϵ)+A_0\cos\left(θ_0\right)\\ 0=y'(0)&=ϵ-\tfrac12 A_0\cos\left(θ_0\right)-(ω-\tfrac18\sqrt{ϵ})A_0\sin\left(θ_0\right)\\ &=\tfrac12(1+3ϵ)-ω(1-\tfrac18ϵ)A_0\sin\left(θ_0\right) \end{align} It remains to solve the polar decomposition $A_0(\cos(θ_0),\sin(θ_0))=(-(1+ϵ), \tfrac12\sqrtϵ+O(ϵ^{3/2}))$ giving $θ_0=\pi-\tfrac12\sqrtϵ$ and $A_0=1+\frac98ϵ$.

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Approach via WKB approximation

This all follows much more easily if the basis functions are computed via a WKB expansion $u(t)=e^{S/\delta}$, allowing for complex exponents, and then the real combinations are taken. The equation $ϵu''+ϵu'+u=0$ then leads to $$ ϵ(δS''+S'^2)+ϵδS'+δ^2=0 $$ Also here one finds balancing for $δ^2=ϵ$ so that $$ S'^2+1=-δ(S''+S') $$ Using an expansion in powers of $δ$ (and setting $s_k=S_k'$) gives $s_0=\pm i$, $2s_0s_1=-s_0\implies s_1=-\frac12$, $s_1^2+2s_0s_2=-s_1\implies s_2=\mp\frac{i}8$ so that the basis solutions up to that approximation order are in their real form $$ y_1(t)=e^{-t/2}\cos\left(\left(\frac1δ-\fracδ8\right)t\right),~~ y_2(t)=e^{-t/2}\sin\left(\left(\frac1δ-\fracδ8\right)t\right). $$ Now use shortly $ω=\left(\frac1δ-\fracδ8\right)=\sqrt{\frac1ϵ}(1-\fracϵ8)$ and match the initial conditions for \begin{align} y(t)&=(1+ϵ)\cos(t)+ϵ\sin(t)+e^{-t/2}[A_0\cos(ωt)+B_0\sin(ωt)]\\ y'(t)&=-(1+ϵ)\sin(t)+ϵ\cos(t)-\tfrac12 e^{-t/2}[A_0\cos(ωt)+B_0\sin(ωt)]-ωe^{-t/2}[A_0\sin(ωt)-B_0\cos(ωt)] \\\hline 0=y(0)&=(1+ϵ)+A_0\\ 0=y'(0)&=ϵ-\tfrac12 A_0+ωB_0=\tfrac12(1+3ϵ)+ωB_0 \end{align} so that more directly $A_0=-(1+ϵ)$ and $B_0=-\tfrac12\sqrtϵ(1+O(ϵ))$.

With some slightly more computation one can get also the next terms in the expansions of $A_0,B_0$

Answer 2

Lutzl's answers is excellent, but I wanted to add the direct multiple-scales version.

Let $t_1=t$ and $t_2=\epsilon^\alpha t$, and $y(t)=Y(t_1,t_2)$. Substituting into the differential equation gives

$$ \epsilon Y_{t_1t_2}+2\epsilon^{1+\alpha}Y_{t_1t_2}+\epsilon^{1+2\alpha}Y_{t_2t_2}+\epsilon Y_{t_1}+\epsilon^{1+\alpha}Y_{t_2}+Y=\cos(t_1). $$

If we pick $\alpha>0$ then the leading order equation will be $Y_0=\cos(t_1)$ which doesn't satisfy the initial conditions, so we must have $\alpha<0$ and hence $\alpha=-1/2$ by dominant balance (this was your error, you had $\alpha=1/2$). So the system to solve is $$ \epsilon Y_{t_1t_2}+2\epsilon^{1/2}Y_{t_1t_2}+Y_{t_2t_2}+\epsilon Y_{t_1}+\epsilon^{1/2}Y_{t_2}+Y=\cos(t_1)\quad Y(0,0)=0,\quad \epsilon^{1/2}Y_{t_1}(0,0)+Y_{t_2}(0,0)=0. $$

Expand $Y$ as $Y_0+\epsilon^{1/2}Y_1+\ldots$ to give, at leading order,

$$ Y_{0,t_2t_2}+Y_{0}=\cos(t_1),\quad Y_0(0,0)=0,\quad Y_{0,t_2}(0,0)=0, $$ with solution $Y_0=A(t_1)\cos(t_2)+B(t_1)\sin(t_2)+\cos(t_1)$, where $A(0)+1=0$ and $B(0)=0$ to satisfy the initial conditions.

The $O(\epsilon^{1/2})$ equation is

$$ Y_{1,t_2t_2}+Y_{1}=-2Y_{0,t_1t_2}-Y_{0,t_2},\quad Y_1(0,0)=0,\quad Y_{0,t_1}+Y_{1,t_2}(0,0)=0, $$ and substituting $Y_0$, $$ Y_{1,t_2t_2}+Y_{1}=(2A'(t_1)+A(t_1))\sin(t_2)-(2B'(t_1)+B(t_1))\cos(t_2). $$ The $\sin(t_2)$ and $\cos(t_2)$ terms are secular, so we must remove them by setting $$ 2A'+A=0,\quad 2B'+B=0 $$ which, coupled with the initial conditions $A(0)=-1$ and $B(0)=0$, give $A(t_1)=-e^{-t_1/2}$ and $B(t_1)=0$.

So we finally get $$ Y_0(t_1,t_2) = -e^{-t_1/2}\cos(t_2)+\cos(t_1) $$ or, in terms of $y$ and $t$, $$y(t)=-e^{-t/2}\cos\left(\frac{t}{\sqrt{\epsilon}}\right)+\cos(t)+O(\sqrt{\epsilon}).$$