Differential equations 2 problem sheet 4

Order relations.
1. Determine the order as ϵ → 0 of $\frac{\sqrtϵ}{1-\cosϵ}$
2. Obtain an asymptotic expansion of exp(tan ϵ) in integer powers of ϵ up to O(ϵ⁴).
3. Show that log ϵ = o(ϵ^−p) as ϵ → 0⁺ for all p > 0.
1. $\frac{\sqrt{ϵ}}{1-\cos ϵ}=\frac{ϵ^{1/2}}{\frac12ϵ^2+O(ϵ^4)}∼2ϵ^{-3/2}$
2. Substitute $\tanϵ=ϵ+\frac{ϵ^3}3+O\left(ϵ^5\right)$ into $\exp(x)=1+x+\frac{x^2}2+\frac{x^3}6+\frac{x^4}{24}+O\left(x^5\right)$ we get\begin{align*}\exp(\tanϵ)&=1+ϵ+\frac{ϵ^3}3+O\left(ϵ^5\right)+\frac12\left(ϵ+\frac{ϵ^3}3+O\left(ϵ^5\right)\right)^2+\frac16\left(ϵ+O\left(ϵ^3\right)\right)^3+\frac1{24}\left(ϵ+O\left(ϵ^3\right)\right)^4+O\left(ϵ^5\right)\\&=1+ϵ+\frac{ϵ^3}3+\frac{ϵ^2}2+\frac{ϵ^4}3+\frac{ϵ^3}6+O\left(ϵ^5\right)\\&=1+ϵ+\frac{ϵ^2}2+\frac{ϵ^3}2+\frac{ϵ^4}3+O\left(ϵ^5\right)\end{align*}
3. Using $\lim_{x→0^+}\frac{\log x}{x^{-1}}=0$, let $x=ϵ^p$, $\lim_{ϵ→0^+}\frac{p\log ϵ}{ϵ^{-p}}=0$, factor out constant $p$, $\lim_{ϵ→0^+}\frac{\log ϵ}{ϵ^{-p}}=0$
Approximating roots. Find expansions for all roots of the equations below as ϵ → 0 with two nonzero terms in each expansion:
1. $ϵx^3-x+1=0$
2. $\tan x=\fracϵx$ [Hint: The roots are near nπ for integer n, but the root near zero must be treated separately]
1. Setting $ϵ=0$ we get $x=1$. The root near 1 is $x=\sum_{n≥0}\frac{\binom{3n}{n}}{2n+1}⋅ϵ^n$ To find the far-off roots: Dominant balance: $x=O(ϵ^{-1/2})$ Set $x=Xϵ^{-1/2},X=X_0+ϵ^{1/2}X_1+O(ϵ)$. Plugging in the equation$$X^3-X+ϵ^{1/2}=0$$$O(ϵ^0):X_0^3-X_0=0⇒X_0=0,±1⇒±1$ are new solutions $O(ϵ^{1/2}):3X_0^2X_1-X_1+1=0⇒X_1=-1/2$ $X=±1-\frac12ϵ^{1/2}+O(ϵ)$ $x=±ϵ^{-1/2}-\frac12+O(ϵ^{1/2})$
2. Putting $x∼x_0+ϵx_1+O(ϵ^2)$ into the equation, expanding $\tan x$ at $x_0$,$$\left(\tan x_0+(1+\tan^2 x_0)ϵx_1+O(ϵ^2)\right)\left(x_0+ϵx_1+O(ϵ^2)\right)=ϵ$$Equating the constant terms\[x_0\tan x_0=0⇒x_0=kπ,k∈\Bbb Z\]The equation becomes$$\left(ϵx_1+O(ϵ^2)\right)\left(kπ+ϵx_1+O(ϵ^2)\right)=ϵ$$For $k≠0$, equating the coefficients of $ϵ$\[x_1=\frac1{kπ}\]we obtain the asymptotic expansion$$x\sim kπ+\fracϵ{kπ}+O(ϵ^2)$$
  The root near zero must be treated separately because leading order is different: For $k=0$, equating the coefficient of $ϵ^1$ we get $0=1$😯 When $x\to0$, $\tan x=x+O(x^3)$, so $x(x+O(x^3))=ϵ$, we must have $x=O(ϵ^{1/2})$. Let$$x∼ϵ^{1/2}(x_0+ϵx_1+O(ϵ^2))$$Plugging into the equation, if expanding $\tan x$ at $x_0$,\[\left(\tan(ϵ^{1/2}x_0)+(1+\tan^2(ϵ^{1/2}x_0))ϵ^{3/2}x_1+O(ϵ^{5/2})\right)ϵ^{1/2}\left(x_0+ϵx_1+O(ϵ^2)\right)=ϵ\]which is cumbersome. To get rid of $\tan()$ let us expand at 0 instead\[\left(ϵ^{1/2}x_0+ϵ^{3/2}x_1+O(ϵ^{5/2})+\frac13\left(ϵ^{1/2}x_0+O(ϵ^{3/2})\right)^3+O(ϵ^{5/2})\right)ϵ^{1/2}\left(x_0+ϵx_1+O(ϵ^2)\right)=ϵ\]Equating the coefficient of $ϵ^1$ we get$$x_0^2=1⇒x_0=±1$$Equating the coefficient of $ϵ^2$, using $x_0^2=1$ we get$$2x_0x_1+\frac{x_0^4}3=0⇒x_1=-\frac16x_0$$we obtain the asymptotic expansions $±ϵ^{1/2}(1-\fracϵ6+O(ϵ^2))$
Regular perturbation. Find the first two terms in an asymptotic expansion in powers of the small parameter ϵ of the solution to$$xy'(x)+y(x)=ϵy(x)^{1/2}, x>1, y(1)=1$$Explain why the expansion is not valid as $x→∞$. What form of rescaling would be necessary to examine the behaviour for large x? [You do not need to carry out the resulting analysis.]
Letting ϵ = 0, we obtain $$ xy'(x) + y(x) = 0⇒xy(x)=\text{const}$$Using $y(1)=1$, we get $y(x)=\frac1x$. Assume a better approximation of the form$$ y(x) = \frac 1x + ϵy_0(x)+ O(ϵ^2)$$Differentiate this to get$$ y'(x) = -\frac 1{x^2} + ϵy_0'(x) + O(ϵ^2)$$The ODE becomes\begin{align*}ϵy_0'(x)x+ ϵy_0(x)+ O(ϵ^2)&=ϵ\left(\frac 1x + ϵy_0(x)+ O(ϵ^2)\right)^{1/2}\\&=\fracϵ{x^{1/2}}\left(1+\fracϵ2y_0(x)x+O(ϵ^2)\right) \end{align*}Equate the coefficient of ϵ$$y_0'(x)x+y_0(x)=x^{-1/2}⇒y_0(x)=\frac2{x^{1/2}}+\frac{c}{x}$$$$y_0(1)=0⇒c=-2⇒y(x)=\frac1x+ϵ\frac{2(x^{1/2}-1)}x$$The expansion becomes nonuniform when $\frac{1}{x}∼ϵ{x^{1/2}\over x}⇔x∼ϵ^{-2}$. Rescale $X=ϵ^2x$ with $X=O(1)$.
Inner and outer expansions. Find inner and outer expansions, correct up to and including terms of order ϵ, for the function$$f(x;ϵ)=\frac{\mathrm{e}^{-x/ϵ}}x+\frac{\sin x}x-\operatorname{coth}x$$for x > 0 and 0 < ϵ ≪ 1.
Compare the inner and outer approximations with the exact function f by plotting all three on the same graph for various small values of ϵ.
For $x=O(1)$, the first term is exponentially small, so the outer expansion is$$f(x;ϵ)∼\frac{\sin x}x-\operatorname{coth}x$$To find the inner expansion, rescale $x$ using $x=ϵy$, with $y=O(1)$, and set $f(x;ϵ)=F(y;ϵ)$. Plugging this into the original equation, $$ F(y;ϵ)=\frac{e^{-y}}{ϵy}+\frac{\sin(ϵy)}{ϵy}-\coth(ϵy)$$Using Taylor expansions\begin{array}l\frac{\sin(ϵy)}{ϵy}=1+O(ϵ^2)\\\coth(ϵy)=\frac{1}{ϵy}+\frac{ϵy}{3}+O(ϵ^3)\end{array}Plugging into $F(y;ϵ)$$$F(y;ϵ)=1+\frac{e^{-y}-1}{ϵy}-\frac{ϵy}{3}+O(ϵ^2)$$Plot for $ϵ=0.1$
Singular perturbation. Use matched asymptotic expansions to find leading-order outer and inner solutions to the boundary value problem$$ϵy''(x)+xy'(x)+y(x)=0, 1<x<2, y(1)=0, y(2)=1$$where 0 < ϵ ≪ 1. Construct the leading-order composite solution.
How would the analysis change if ϵ were small and negative? [You do not need to find the solution in this case.]
The boundary layer is at $x=1$. To find inner expansion, let $x=1+ϵX,X=O(1),y(x)=Y(X)$ Then $y'(x)=\frac{Y'(X)}ϵ,y''(x)=\frac{Y''(X)}{ϵ^2}$. Plugging in the equation\[\frac{Y''(X)}ϵ+Y'(X)(\frac1ϵ+X)+Y(X)=0⇒Y''_0(X)+Y'_0(X)=0\]Using $Y_0(0)=0$, we obtain $Y_0(X)=C\left(1-e^{-X}\right)$ To find outer expansion, use regular expansion $y∼y_0(x)+ϵy_1(x)+⋯$\[xy_0'(x)+y_0(x)=0,y_0(2)=1⇒y_0(x)=\frac2x\]By the matching principle\[\lim_{x→1^+}y_0(x)=2=\lim_{X→∞}Y_0(X)=C\]hence $C=2$. The common limit is 2 so the leading-order composite solution is\[y_0(x)+Y_0(X)-2=\frac2x-2e^{-(x-1)/ϵ}\]If $0<-ϵ≪1$, the above $X=\frac{x-1}ϵ<0$, $\lim_{X→-∞}Y_0(X)=∞$❌so the boundary layer must be $x=2$.
Boundary layer. Construct leading-order inner and outer solutions to$$ϵu''(x)+u'(x)=\frac{u(x)+u(x)^3}{1+3 u(x)^2}, 0<x<1, u(0)=0, u(1)=1$$where 0 < ϵ ≪ 1. [You will only be able to determine the outer solution implicitly.]
Sketch a graph of the leading-order composite solution.
Boundary layer is at $x=0$. To find outer expansion $u_0(x)$ satisfies$$u_0'(x)=\frac{u_0(x)+u_0(x)^3}{1 + 3u_0(x)^2}⇔\frac{d}{dx}\ln(u_0(x)+u_0(x)^3)=1$$Integrating gives$$\ln\Big(u_0(x)+u_0(x)^3\Big)=x+C$$Plugging in boundary condition $u_0(1)=1$ gives $C=\ln2-1$, hence our outer expansion satisfies$$u_0(x)+u_0(x)^3=2e^{x-1}⇒u_0(0)≈0.560074$$By monotonicity, $u_0(x)$ is unique.
To find inner expansion, let $x=ϵX,u(x)=U(X)⇒u'(x)=\frac{U'(X)}ϵ,u''(x)=\frac{U''(X)}{ϵ^2}$
$$U_0''(X)+U_0'(X)=0,U_0(0)=0⇒U_0(X)=C(1-e^{-X})$$Using matching principle$$\lim_{X→∞}U_0(X)=C=u_0(0)$$The leading-order composite solution is$$u_0(x)+U_0(X)-\text{common limit}=u_0(x)-u_0(0)e^{-x/ϵ}$$
Outer expansion
Inner expansion
Composite expansion

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