Math 401-501/502 Extra Examples for Test 1— Spring 2013

• Singular perturbation of polynomials
  1. Find the scaling required to determine the large roots of $\varepsilon x^4 +\varepsilon^2 x - 1=0$.

     Solution. The strategy here is different from the other problems because $\varepsilon$ is in two coefficients. For this problem, we need to factor out $\varepsilon x$. Doing this yields $\varepsilon x(x^3+\varepsilon)-1=0$. Since we are looking for $x$ such that $|x|\gg 1$ and since $0<\varepsilon \ll 1$, $x^3+\varepsilon\sim x^3$. It follows that $\varepsilon x^4-1\sim 0$, and so $x=\mathcal O(\varepsilon^{-1/4})$. Thus, we take $z=\varepsilon^{1/4}x$. To actually obtain approximations for the roots, we replace $x$ in the polynomial by $z$ to get $z^2 +\varepsilon^{7/4}z -1=0$. The perturbation expansion for the solution would have the form $x=a\varepsilon^{-1/4}+b\varepsilon^{3/2} +\mathcal O(\varepsilon^{13/4})$. Not particularly nice, but not horrible either.

• Two-Times Method
  1. Consider $\ddot y +y +\varepsilon\big( \dot y +y^2\big)=0$, $y(0)=0, \dot y(0)=1$, $0 < \varepsilon \ll 1$. Use the two-time method with $\tau=\varepsilon t$ and $\sigma = t$ to find an order $\varepsilon^0$ approximation to $y(t,\varepsilon)$ without secular terms.

     Solution. We will break this problem into steps.

     1. Apply chain rule. We assume that $y=y(\sigma,\tau,\varepsilon)$. By the chain rule we have $\dot y = \frac{\partial y}{\partial \sigma}\frac{d\sigma}{dt} + \frac{\partial y}{\partial \tau}\frac{d\tau}{dt} = \frac{\partial y}{\partial \sigma}+\frac{\partial y}{\partial \tau}\varepsilon$ and $\ddot y=\frac{\partial^2\! y}{\partial \sigma^2}+2\frac{\partial^2\! y}{\partial \sigma \partial \tau}\varepsilon + \frac{\partial^2\! y}{\partial \tau^2} \varepsilon^2$. Using these equations to replace $\dot y, \ddot y$ in the differential equation yields $ \frac{\partial^2\! y}{\partial \sigma^2}+2\frac{\partial^2\! y}{\partial \sigma \partial \tau}\varepsilon + \frac{\partial^2\! y}{\partial \tau^2} \varepsilon^2 + y+ \big(\frac{\partial y}{\partial \sigma}+\frac{\partial y}{\partial \tau}\varepsilon + y^2 \big)\varepsilon=0 $. The initial conditons beome $\dot y=1 = \frac{\partial y}{\partial \sigma}+\frac{\partial y}{\partial \tau}\varepsilon$, where $\sigma=0=\tau$, and $y(0,0,\varepsilon)=0$.
     2. Obtain pertubation equationsLet the perturbation expansion for $y$ be given by $y(\sigma,\tau, \varepsilon) = u(\sigma,\tau)+v(\sigma,\tau)\varepsilon + \mathcal O(\varepsilon^2)$. Plugging this into the equation then gives, after collecting terms, the following:

        \[ \frac{\partial^2\! u}{\partial \sigma^2} + u + \bigg(\frac{\partial^2\! v}{\partial \sigma^2} + v + 2\frac{\partial^2\! u}{\partial \sigma \partial \tau} + \frac{\partial u}{\partial \sigma} + u^3 \bigg) \varepsilon + \mathcal O(\varepsilon^2) =0.\]

        In addition, the inital conditons are $u+\varepsilon v +\mathcal O(\epsilon^2) = 0$ and $ 1 = \frac{\partial u}{\partial \sigma}+ \bigg(\frac{\partial u}{\partial \tau} + \frac{\partial v}{\partial \sigma}\bigg)\varepsilon + \mathcal O(\varepsilon^2)$. Equating to the coefficients of $\varepsilon^0$ and $\varepsilon^1$ to 0 results in these initial value problems:

        \[\frac{\partial^2\! u}{\partial \sigma^2} + u =0,\ u(0,0)=0,\ \frac{\partial u}{\partial \sigma}(0,0) = 1. \] \[ \frac{\partial^2\! v}{\partial \sigma^2} + v + 2\frac{\partial^2\! u}{\partial \sigma \partial \tau} + \frac{\partial u}{\partial \sigma} + u^2 =0,\ v(0,0)=0,\ \frac{\partial v}{\partial \sigma}(0,0) + \frac{\partial u}{\partial \tau}(0,0) = 0. \]

     3. Solve the equations for $u$. The equation for $u$ is a standard linear, 2nd order, constant coefficient, homogeneous differential equation in $\sigma$. The variable $\tau$ doesn't come into play. The solution to the equation is $u(\sigma,\tau)= A(\tau)\sin(\sigma) + B(\tau)\cos(\sigma)$, where the coefficients $A$ and $B$ depend on $\tau$ and are to be determined later. The initial conditons $u(0,0)=0$ and $\frac{\partial u}{\partial \sigma}(0,0)=1$ imply that $B(0)=0$ and $A(0)=1$.
     4. Plug $u$ into the equations for $v$. Doing so gets us the equation

        \[ \frac{\partial^2\! v}{\partial \sigma^2} + v + 2A'(\tau)\cos(\sigma) - 2B'(\tau)\sin(\sigma) + A(\tau)\cos(\sigma) - B(\tau)\sin(\sigma) +(A(\tau)\sin(\tau)+B(\tau)\cos(\sigma))^2=0\] Since $(A\cos(\sigma)+B\cos(\sigma))^2= A^2\sin^2(\sigma)+2AB\sin(\sigma)\cos(\sigma) + B^2\cos^2(\sigma)= \frac12\big(A^2+B^2)+ \frac12\big(B^2-A^2)\cos(2\sigma)+ AB\sin(2\sigma)$, we see that, after a little algebra, the equation above becomes

        \[ \frac{\partial^2\! v}{\partial \sigma^2} + v +(2A'+A)\cos(\sigma)-(2B'+B)\sin(\sigma)+\frac12\big(A^2+B^2)+ \frac12\big(B^2-A^2) \cos(2\sigma)+ AB\sin(2\sigma) \]

     5. Eliminate secular terms. To avoid secular tems in $v$, we must get rid of the "resonant" terms with $\sin(\sigma)$ and $\cos(\sigma)$. To do that, we set $2A'+A=0$ and $2B'+B=0$. These are easy to solve; the solutions are $A(\tau)=A(0)e^{-\tau/2}$ and $B(\tau)=B(0)e^{-\tau/2}$. Using the values $A(0)=1$ and $B(0)=0$, we see that $A(\tau)=e^{-\tau/2}$ and $B(\tau)=0$.
     6. Find $u$. Because $u(\sigma,\tau) = A(\tau)\sin(\sigma) + B(\tau)\cos(\sigma)$, we have $u (\sigma,\tau) = e^{-\tau/2}\sin(\sigma)$. Finally, using $\tau=\varepsilon t$ and $\sigma =t$, we see that the lowest order term is $u(t,\varepsilon t) = e^{-(\varepsilon/2)t}\sin(t)$. We remark that to get $v$ and eliminate any secular terms in it, one would have to find the equations for the next higher order term in the perturbation expansion.

• Boundary Layer Problems
  1. Find the lowest-order uniform (composite) expansion for the solution of $\varepsilon y'' + y' +\frac y{x+1} =2$, $y(0)=0$, $y(1)=3$. The boundary layer is on the right and we assume that $0<\varepsilon \ll 1$,

     Solution. For $x\approx 1$, drop the $\varepsilon$ term and work with $y' +\frac y{x+1} =2$, $y(1)=3$. This is a first order linear equation with integrating factor (see an ODE book) $\mu = \exp\big(\int (x+1)^{-1}dx\big) = \exp(\log(x+1)) = x+1$. Hence $[(x+1)y]'=2(x+1)$ and so $(x+1)y=x^2+2x +C$. Since $y(1)=3$, $C=3$. The outer solution is thus $y_{out}=x+1+2(x+1)^{-1}$. To find the inner solution, we go back to the orginal equation $\varepsilon y'' + y' +\frac y{x+1} =2$ and change variables to $z=x/\varepsilon$. After simplifying, we obtain the equation $d^2y/dz^2+dy/dz+4\varepsilon y(1+\varepsilon z)^{-1}=2\varepsilon$. Dropping $\mathcal O(\varepsilon)$ terms, we arrive at $d^2y/dz^2+dy/dz=0$. The solution to this equation is $y=A+Be^{-z}$. The boundary condition is $y(x=0)=y(z=0)=0$. Using this in the previous equation we see that $B=-A$ and $y=A(1-e^{-z})$. Putting back $x$ then gives us the boundary layer solution $y_{BL}=A(1-e^{-x/\varepsilon})$. The next step is to find $A$ and $y_{overlap}$. To do this, let $w=x/\sqrt{\varepsilon}$ and then note that, with $w$, $y_{BL} =A(1-e^{-w/\sqrt{\varepsilon}})$ and $y_{out} = \sqrt{\varepsilon}w + 1+2(\sqrt{\varepsilon}w+1)^{-1}$. We remember that $\lim_{\varepsilon \downarrow 0} y_{BL} = \lim_{\varepsilon \downarrow 0}y_{out}= y_{overlap}$. Doing the limits the implies $A=3=y_{overlap}$. The uniform solution is then $y_{unif}=y_{BL}+y_{out}-y_{overlap} = x+1+2(x+1)^{-1}-3e^{-x/\varepsilon}$.

  2. Find $y_{unif}$ for the boundary layer problem $\varepsilon y'' + 4y' = 4$, $0 < \varepsilon \ll 1$, $y(0)=1,y(1)=3$, on the interval $0 < x <1$.

     Answer. $y_{unif} = 2+x-e^{-4x/\varepsilon}$.