1. We consider a second-order, linear differential operator $𝔏 y(x)$ for $a<x<b$ supplemented by linear homogeneous boundary conditions $𝔅_1 y, 𝔅_2 y$.
   1. Explain how the adjoint operator $𝔏^*$ and the corresponding adjoint boundary conditions are defined.
      State the three possibilities for the existence and uniqueness of the solutions of the inhomogeneous problem $𝔏 y=f$, forming the Fredholm Alternative.
   2. Assuming that the solution $y(x)$ of the inhomogeneous ODE $𝔏 y=f$ takes the form $$ y(x)=\sum_i c_i y_i(x) $$ where $y_i$ are the eigenfunctions of the differential operator, provide an expression for the coefficients $c_i$ and determine under what condition all the coefficients are well defined. Use this expression to express the Green's function of the problem in terms of these eigenfunctions.
   3. Let 𝔏 be given by $𝔏 y ≡-x y''(x)$, $a=1$ and $b=2$, and let the boundary conditions be $𝔅_1 y ≡ y'(1)=0, 𝔅_2 y ≡ y'(2)=0$. Determine the condition on $f$ for the existence of solutions for the inhomogeneous ODE $𝔏 y=f$.
   4. Consider now the case when $𝔏 y ≡ y''(x)+4 y(x)$, with $a=0$ and $b=1$, and the general boundary conditions $$ \left.\begin{aligned} 𝔅_1 y &≡ α_1 y(a)+α_2 y'(a)+β_1 y(b)+β_2 y'(b)=0 \\ 𝔅_2 y &≡ α_3 y(a)+α_4 y'(a)+β_3 y(b)+β_4 y'(b)=0 \end{aligned}\right\} $$ where $\left(α_1, α_2, β_1, β_2\right)$ and $\left(α_3, α_4, β_3, β_4\right)$ are linearly independent.
      1. Give an example of values for $\left(α_1, α_2, β_1, β_2\right)$ and $\left(α_3, α_4, β_3, β_4\right)$ for which the solution of the inhomogeneous problem $𝔏 y=f$ is either undefined or not uniquely defined. Justify your answer.
      2. Taking now $𝔅_1 y ≡ y(0)=0, 𝔅_2 y ≡ y(1)=0$, determine an eigenfunction decomposition for the Green's function.
2. 1. Show that any second order linear operator $$ 𝔏 y ≡ P_2(x) y''+P_1(x) y'+P_0(x) y $$ with $P_2(x) ≠ 0$ can be converted to a Sturm-Liouville operator.
      Assuming that the boundary conditions are self-adjoint, show that the eigenvalue problem $$ 𝔏 y=λ y $$ has real eigenvalues, and that the eigenfunctions are orthogonal with respect to a modified inner product.
      [You may use results from Sturm-Liouville theory if clearly stated.]
   2. For each $n ∈ ℕ_0$, we define the function $H_n: ℝ → ℝ$ by $$ H_n(x)=(-1)^n e^{x^2} \frac{d^n}{d x^n} e^{-x^2} $$ Determine the explicit expressions for $H_1(x), H_2(x), H_3(x)$ and $H_4(x)$, and then show that, in general, $H_n(x)$ is a polynomial of degree $n$.
   3. Using $∫_{-∞}^∞ e^{-x^2} d x=\sqrt{π}$, show the orthogonality conditions $$ ∫_{-∞}^∞ H_m(x) H_n(x) e^{-x^2} d x=2^n n ! \sqrt{π} δ_{n m} $$
   4. By considering a Taylor expansion in the $t$ variable, verify that $$ e^{2 x t-t^2}=\sum_{n=0}^∞ \frac{1}{n !} H_n(x) t^n . $$ Using this result, or otherwise, show that $y=H_n(x)$ satisfies the ODE $$ y''-2 x y'+2 n y=0 $$ and determine its Sturm-Liouville form.
      [You may find it helpful to note that $2 x t-t^2=x^2-(x-t)^2$.]
3. 1. For a function $f(ϵ)$ to have an asymptotic expansion of the form $$ f(ϵ) ∼ \sum_k a_k ϕ_k(ϵ)   \text { as } ϵ → 0 $$ what are the conditions that should be satisfied by the gauge functions $ϕ_k$ and by the series?
      Explain the difference between an asymptotic expansion and a convergent series.
      Give an example of two functions that have the same asymptotic expansion but are different.
   2. Consider the differential equation $$ y''(x)+y(x)=ϵ y^n(x) $$ for $0<x<2 π, 0<ϵ ≪ 1$ and $n ∈ ℕ$.
      1. Take the boundary conditions $y'(0)=0$ and $y'(2 π)=0$. Using the Fredholm Alternative Theorem, give the values of $n$ when the asymptotic expansion $$ y ∼ y_0+ϵ y_1+… $$ yields only the trivial solution.
      2. In the case $n=3$, now considering the initial conditions $y(0)=0$ and $y'(0)=1$, show that the asymptotic expansion breaks down for $x$ of a certain order that you should determine.
         [Hint: You may use the identity $\sin ^3 x=\frac{1}{4}(3 \sin x-\sin 3 x)$.]
   3. Consider the boundary value problem $$ ϵ y''(x)+p(x) y'(x)+y(x)=0 $$ subject to the conditions $y(0)=1$ and $y(1)=0$, where $0<ϵ ≪ 1$ is a small parameter. We further assume that $p(x)$ is strictly negative and that $p(0)=p(1)=-1$. Determine the location of the boundary layer and construct leading order inner and outer solutions.

Solution

1. 1. $𝔏^*$ is defined by $⟨𝔏y,w⟩=⟨y,𝔏^*w⟩$ where $⟨y,w⟩=\int_a^by(x)\overline{w(x)}dx$ \begin{split} 𝔏y&=P_2 y''+P_1 y'+P_0 y \\ 𝔏^*y&=\left(P_2 w\right)''-\left(P_1 w\right)'+P_0 w\\ ⟨𝔏 y, w⟩-\left\langle y, 𝔏^* w\right\rangle&=\left[P_2 w y'-\left(P_2 w\right)' y+P_1 w y\right]_{a}^{b} \\&=\require{mathtools}\mathmakebox[.9width]{\left(K_1^* w\right)\left(𝔅_1 y\right)+\left(K_2^* w\right)\left(𝔅_2 y\right)+\left(K_1 y\right)\left(𝔅_1^* w\right)+\left(K_2 y\right)\left(𝔅_2^* w\right)}\end{split}where $K_1y,K_2y$ are linearly independent of $B_1y,B_2y$, and $K_1^*w,K_2^*w$ are linearly independent of $B_1^*w,B_2^*w$.
      $𝔅_1^* w,𝔅_2^* w$ is the corresponding adjoint boundary conditions.
      
      Theorem 2.2. Fredholm Alternative (linear ODE version)
      Exactly one of the following possibilities occurs.
      1. The homogeneous problem (H+BC) has only the zero solution. In this case the solution of (N+BC) is unique.
      2. The homogeneous problem (H+BC) admits non-trivial solutions, and so does (H*+BC*). In this case there are two sub-possibilities:
      2(a) if $⟨f,w⟩=0$ for all $w$ satisfying (H*+BC*), then (N+BC) has a non-unique solution;
      2(b) otherwise, (N+BC) has no solution.
   2. Let $𝔏y_i=λ_iy_i,𝔏^*w_i=λ_iw_i$. Interchange the inner product with ∑ and use orthogonality \[⟨f,w_k⟩=⟨y,𝔏^*w_k⟩=\left<\sum_ic_iy_i,λ_kw_k\right>=\sum_ic_iλ_k⟨y_i,w_k⟩=c_kλ_k⟨y_k,w_k⟩\] If $λ_k≠0$, we can solve for $c_k$\[c_k={⟨f,w_k⟩\overλ_k⟨y_k,w_k⟩}\] If $λ_k=0$, either
      • $⟨f,w_k⟩=0$, then $c_k$ is arbitrary: the solution is non-unique; or
      • $⟨f,w_k⟩≠0$, it is inconsistent: the solution does not exist.
      \[y(x)=\sum_ic_iy_i(x)=\sum_i{\int_a^bf(ξ)w_i(ξ)y_i(x)dξ\overλ_i⟨y_i,w_i⟩}=\int_a^bf(ξ)\sum_i{w_i(ξ)y_i(x)\overλ_i⟨y_i,w_i⟩}dξ\]So the Green's function is $\sum_i{w_i(ξ)y_i(x)\overλ_i⟨y_i,w_i⟩}$
   3. $𝔏^* w ≡-(x w(x))'',⟨𝔏y,w⟩-⟨y,𝔏^*w⟩=[(xw(x))'y(x)-xw(x)y'(x)]_1^2$
      $𝔅_1^* w = (xw(x))'|_1$, $𝔅_2^* w ≡ (xw(x))'|_2$ \begin{eqnarray*} 𝔏^* w=0&⇒&w(x)=\frac Ax+B,(xw(x))'=B\\ 𝔅_1^* w=0&⇒&B=0\\ 𝔅_2^* w=0&⇒&B=0 \end{eqnarray*} $w=\frac Ax$, so $𝔏 y=f$ exists solution iff $∫_1^2f(x)\frac1xdx=0$
   4. 1. By (a), the solution of $𝔏 y=f$ is either undefined or not uniquely defined iff $𝔏 y=0$ has a non-trivial solution.
         $𝔏 y=0⇒y(x)=A\cos(2x)+B\sin(2x)$\begin{array}ly(0)=A&y(1)=A\cos2+B\sin2\\\frac{y'(0)}2=B&\frac{y'(1)}2=-A\sin2+B\cos2\end{array}$\begin{cases}y(1)=y(0)\cos2+\frac{y'(0)}2\sin2\\\frac{y'(1)}2=-y(0)\sin2+\frac{y'(0)}2\cos2\end{cases}$ is satisfied by $y(x)=A\cos(2x)+B\sin(2x)$ for all $A,B$
      2. $𝔏w≡ w''+4 w,⟨𝔏y,w⟩-⟨y,𝔏^*w⟩=[y'(x)w(x)-y(x)w'(x)]_0^1$, so the adjoint BC\begin{array}l 𝔅_1^*w=w(0)\\𝔅_2^*w=w(1) \end{array}$𝔏w=λw⇒w''+(4-λ)w=0$
         For $λ>4$, $w(x)=Ae^{\sqrt{λ-4}x}+Be^{-\sqrt{λ-4}x}$, only $w=0$ can satisfy $w(0)=w(1)=0$.
         For $λ=4$, $w(x)=Ax+B$, only $w=0$ can satisfy $w(0)=w(1)=0$.
         For $λ<4$, $w(x)=A\cos(\sqrt{4-λ}x)+B\sin(\sqrt{4-λ}x)$ \begin{array}l w(0)=0⇒A=0\\w(1)=0⇒A\cos(\sqrt{4-λ})+B\sin(\sqrt{4-λ})=0 \end{array}$B≠0⇒\sin(\sqrt{4-λ})=0⇒\sqrt{4-λ}=kπ,k=1,2…$
         Let $B=1,y_k(x)=w_k(x)=\sin(kπx)$, then $⟨y_k,w_k⟩=\frac12$
         By (b) Green's function $\sum_{k=1}^∞{w_k(ξ)y_k(x)\overλ_k⟨y_k,w_k⟩}=\sum_{k=1}^∞{2\sin(kπξ)\sin(kπx)\over4-k^2π^2}$
2. 1. Comparing$$r(x)𝔏y ≡ r(x)P_2(x) y''+r(x)P_1(x) y'+r(x)P_0(x) y$$with$$\hat𝔏y≡-(py')'+qy$$we need $\frac{P_1(x)}{P_2(x)}=\frac{p'(x)}{p(x)}⇒p(x)=∫\exp\left(P_1(x)\over P_2(x)\right)dx⇒r(x)=-\frac{p(x)}{P_2(x)},q(x)=r(x)P_0(x)$
      $𝔏y=λy⇔\hat𝔏y=λry$
      Assuming that the boundary conditions are self-adjoint, $\hat𝔏y=λry$ is fully self-adjoint $$r(x)>0⇒⟨y_k,ry_k⟩=\int r(x)\left|y_k(x)\right|^2>0$$ $$λ_k^*⟨y_k,ry_k⟩=⟨y_k,𝔏y_k⟩=⟨𝔏y_k,y_k⟩=λ_k⟨ry_k,y_k⟩⇒λ_k∈ℝ$$and for $j≠k$,$$λ_k⟨y_j,ry_k⟩=⟨y_j,𝔏y_k⟩=⟨𝔏y_j,y_k⟩=λ_j⟨ry_j,y_k⟩⇒⟨y_j,ry_k⟩=0$$
   2. $H_0(x)=1,H_1(x)=2x,H_2(x)=4x^2-2,H_3(x)=8x^3-12x,H_4=4 \left(4 x^4-12 x^2+3\right)$.
      Suppose $H_{n-1}(x)$ is a polynomial of degree $n-1$,\begin{align*} H_n(x)&=(-1)^n e^{x^2} \frac{d^n}{d x^n} e^{-x^2} \\&=(-1)^n e^{x^2} \frac{d}{dx}\left(\frac{d^{n-1}}{d x^{n-1}} e^{-x^2}\right) \\&=-e^{x^2} \frac{d}{d x}\left(e^{-x^2}H_{n-1}(x)\right) \\&=-e^{x^2} \left(e^{-x^2}H'_{n-1}(x)-2xe^{-x^2}H_{n-1}(x)\right) \\&=-H'_{n-1}(x)+2xH_{n-1}(x) \end{align*}$\deg H'_{n-1}(x)≤n-2,\deg2xH_{n-1}(x)=n⇒H_n(x)$ is a polynomial of degree $n$.
   3. Let $I_{m,n}=∫_{-∞}^∞ H_m(x) H_n(x) e^{-x^2}\ d x$. \begin{align*} I_{m,n}&=(-1)^n∫_{-∞}^∞ H_m(x)⋅\frac{d^n}{dx^n}e^{-x^2}\ d x \\&=\left[H_m(x)⋅\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right]_{-∞}^∞+(-1)^{n-1}∫_{-∞}^∞ \frac{d}{dx}H_m(x)⋅\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\ d x \\&=\left[\frac{d}{dx}H_m(x)⋅\frac{d^{n-2}}{dx^{n-2}}e^{-x^2}\right]_{-∞}^∞+(-1)^{n-2}∫_{-∞}^∞ \frac{d^2}{dx^2}H_m(x)⋅\frac{d^{n-2}}{dx^{n-2}}e^{-x^2}\ d x \\&⋮ \\&=\left[\frac{d^{n-1}}{dx^{n-1}}H_m(x)⋅e^{-x^2}\right]_{-∞}^∞+∫_{-∞}^∞\frac{d^n}{dx^n}H_m(x)⋅e^{-x^2}\ d x \end{align*} For $m<n$, $\deg H_m(x)<n$, so $\frac{d^n}{dx^n}H_m(x)=0$, so $I(m,n)=0$.
      For $m>n$, by symmetry, $I(m,n)=0$.
      For $m=n$, $H_n(x)=-H'_{n-1}(x)+2xH_{n-1}(x),H_1'(x)=2⇒\frac{d^n}{dx^n}H_n(x)=2n\frac{d^{n-1}}{dx^{n-1}}H_{n-1}(x)=⋯=2^nn!⇒I(n,n)=∫_{-∞}^∞2^nn!⋅e^{-x^2}\ d x=2^nn!\sqrtπ$.
   4. $H_n(x)=(-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}$ Replacing $x$ with $y=t-x$
      $=e^{x^2}\left.\frac{d^n}{dy^n}e^{-y^2}\right|_{t=0}=e^{x^2}\left.\frac{d^n}{dt^n}e^{-y^2}\right|_{t=0}=\left.\frac{d^n}{dt^n}e^{2tx-t^2}\right|_{t=0}$
      So $e^{2 x t-t^2}=\sum_{n=0}^∞\frac{1}{n!} H_n(x) t^n$ \begin{align*} \sum_{n=1}^∞\frac d{dx}[e^{-x^2}H_n'(x)]\frac{t^n}{n!} &=\frac d{dx}\left(e^{-x^2}\sum_{n=0}^∞H_n'(x)\frac{t^n}{n!}\right)\\ &=\frac d{dx}\left(e^{-x^2}\frac d{dx}e^{2xt-t^2}\right)\\ &=2t\frac d{dx}e^{-(x-t)^2}\\ &=-2t\frac d{dt}e^{-(x-t)^2}\\ &=-2e^{-x^2}t\frac d{dt}e^{2xt-t^2}\\ &=-2e^{-x^2}\sum_{n=1}^∞H_n(x)\frac{nt^n}{n!}\\ \end{align*} $⇒\frac d{dx}[e^{-x^2}H_n'(x)]=-2ne^{-x^2}H_n(x)$[Sturm-Liouville]
      $⇒H_n''(x)-2 xH_n'(x)+2 nH_n(x)=0$
3. 1. $ϕ_{k+1}(ϵ)=o(ϕ_k(ϵ))$ as $ϵ→0$, i.e. each term is smaller magnitude than the previous term.
      $f(ϵ)-\sum_{k=0}^N a_kϕ_k(ϵ)=o(ϕ_N(ϵ))$, i.e. the approximation gets more accurate as $N$ increases.
      For a convergent series of the form\[f(ϵ)=\sum^∞_{k=0}a_kϕ_k(ϵ),\]we require that the partial sum\[f_N(ϵ) =\sum^N_{k=0}a_kϕ_k(ϵ),\]converges to $f(ϵ)$ as $N→∞$, with $ϵ$ held fixed. For an asymptotic expansion\[f(ϵ)∼\sum_ka_kϕ_k(ϵ),\]we instead require that the partial sum converges asymptotically to $f(ϵ)$ as $ϵ→0$, with $N$ held fixed.
      $\frac1{1-ϵ}$ and $\frac1{1-ϵ}+e^{-1/ϵ}$ has the same asympototic expansion $\sum_{k≥0}ϵ^k$ as $ϵ→0$
      $\frac1{1-ϵ^{-1}}$ and $\frac1{1-ϵ^{-1}}+\logϵ$ has the same asympototic expansion $\sum_{k≤0}ϵ^k$ as $ϵ→0$
   2. 1. At O(1) we get $y_0''(x)+y_0(x)=0,y_0'(0)=y_0'(2π)=0⇒y_0(x)=A\cos x$
         At $O(ϵ)$ we get $y_1''(x)+y_1(x)=A\cos^nx$ by FAT since $y''+y$ is self-adjoint, the condition for existence of solution is $⟨A\cos^nx,y_0(x)⟩=0⇔∫_0^{2π}\cos^{n+1}x\mathrm{~d}x=0$
         For $n$ odd, $∫_0^{2π}\cos^{n+1}x\mathrm{~d}x>0$; for $n$ even, by symmetry $∫_0^{2π}\cos^{n+1}x\mathrm{~d}x=0$.
         So when $n$ is odd, the asympototic expansion yields only the trivial solution.
      2. O(1): $y_0''(x)+y_0(x)=0,y_0(0)=0,y_0'(0)=1⇒$\[y_0(x)=\sin x\]O(ϵ): $y_1''(x)+y_1(x)=\sin^3x=\frac{3 \sin x-\sin 3 x}4,y_1(0)=y_1'(0)=0⇒$\[y_1(x)=\frac{9\sin x+\sin3x-12 x\cos x}{32}=\frac{x^5}{20}+O(x^7)\]$⇒y(x;ϵ)∼\sin x+ϵ\frac{9\sin x+\sin3x-12 x\cos x}{32}+…$
         The expansion becomes nonuniform when $\sin x∼ϵx^5⇔x=O(ϵ^{-1/4})$
   3. $p(x)$ is strictly negative, so boundary layer at right ($x=1$)
      Outer expansion: substitute $y∼y_0+O(ϵ)$ into the equation\[p(x)y'_0(x)+y_0(x)=0,y_0(0)=1⇒y_0(x)=\exp\left(-\int_0^x\frac1{p(x)}dx\right)\]Inner expansion: dominant balance $ϵy''(x)$ with $p(x)y'(x)$, let $x=1-ϵX,y(x)=Y(X),X∼X_0+O(ϵ)$, then $y'(x)=-ϵ^{-1}Y'(X),y''(x)=ϵ^{-2}Y''(X)$\[Y_0''(X)-p(1)Y_0'(X)=0,Y_0(0)=0⇒Y_0(X)=C(1-e^{-X})\]From matching $C=\lim_{X→∞}Y_0(X)=\lim_{x→1-}y_0(x)=\exp\left(-\int_0^1\frac1{p(x)}dx\right)$
      $⇒y∼\exp\left(-\int_0^1\frac1{p(x)}dx\right)(1-e^{x-1\overϵ})+O(ϵ)$