1. Consider the boundary value problem\begin{equation} \begin{aligned} L y &≡ y''(x)+p_1(x) y'(x)+p_0(x) y(x)=f(x), \\ y(-π)&=0, \\ y(π)&=0, \end{aligned}\end{equation} where $p_1, p_0$ and $f$ are given smooth functions and prime denotes differentiation with respect to $x$.
   1. Determine the adjoint operator and adjoint boundary conditions.
   2. Assuming that there is no zero eigenfunction,
      1. Derive a formula for the coefficients of an eigenfunction expansion for the solution to (1), stating what the eigenfunctions satisfy. You may use without proof orthogonality relations between eigenfunctions if stated clearly.
      2. Express the solution in terms of a Green's function, stating clearly what problem the Green's function satisfies.
   3. Show the equivalence of the two solution forms in part (b) by expanding the Green's function as a series in the eigenfunctions.
   4. Suppose now that $p_1(x) ≡ 0$ and $p_0(x)=δ(x)$, the delta function. Using without proof that eigenvalues $λ_k$ satisfying $L y_k=λ_k y_k$ are strictly negative, obtain an algebraic relation satisfied by eigenvalues and show that the eigenvalues form a discrete infinite set.
2. 1. Consider Hermite's equation \begin{equation} y''-2 x y'+2 n y=0 \end{equation} where $n$ is a non-negative integer and prime denotes differentiation with respect to $x$.
      1. Classify $x=∞$ as an ordinary, regular singular, or irregular singular point of the ODE (2), giving reasons.
      2. The Rodrigues' formula for the $n$th order Hermite polynomials is \begin{equation} H_n(x)=(-1)^n e^{x^2} \frac{d^n}{d x^n}\left(e^{-x^2}\right) \end{equation} Prove that $H_n$ given by (3) is a solution of Hermite's equation (2). [Hint: define $P_n=\frac{d^n}{d x^n}\left(e^{-x^2}\right)$ and show first that $P_{n+1}=-2 x P_n-2 n P_{n-1}$.]
   2. Consider now the hypergeometric equation $$ x(1-x) y''(x)+\left[\frac{3}{2}-(a+b+1) x\right] y'(x)-a b y(x)=0, $$ where $a$ and $b$ are real numbers.
      1. Determine the indicial exponents for a series solution about $x=0$ and give the form of two linearly independent solutions. [You do not need to explicitly compute the solutions.]
      2. By considering a series solution about $x=0$, find all conditions on $a, b$ under which:
         (I) One solution is a polynomial.
         (II) One solution is equal to $y(x)=\frac{1}{1-x}$ for ${|x|}<1$.
3. 1. 1. If $f(x) ∼ g(x)$ as $x → x_0$ and $g(x)=\mathrm{o}(h(x))$ as $x → x_0$, show that $f(x)=\mathrm{o}(h(x))$ as $x → x_0$.
      2. What does it mean to say that $f(ϵ)$ has a valid asymptotic expansion $$ \sum_{k=0}^∞ a_k ϵ^k $$ as $ϵ → 0$? Explain why $$ e^{-1 / ϵ}+\sum_{k=0}^∞ a_k ϵ^k $$ would also be a valid expansion.
   2. Consider solutions to the following differential equation for $y(x)$ $$ y''+y+ϵ y^3=0 $$ as an asymptotic expansion $y ∼ y_0+ϵ y_1+…$, where $0<ϵ ≪ 1$ is a small parameter and prime denotes differentiation with respect to $x$.
      1. For initial conditions $y(0)=1, y'(0)=0$, show that the asymptotic solution breaks down at $x=O\left(ϵ^α\right)$, for $α<0$ which you should determine. [See hint at the bottom of the page.]
      2. For periodic boundary conditions $y(0)=y(2 π),   y'(0)=y'(2 π)$, use the Fredholm Alternative Theorem to show that the asymptotic expansion yields only the trivial solution.
   3. Consider now the following equation for $Y(x, s)$ : $$ Y_{x x}+2 ϵ Y_s+ϵ Y^3+Y=0, $$ where subscript denotes partial differentiation. We seek solutions that are $2 π$ periodic in $x$, and that satisfy $Y(0,0)=1$, as an asymptotic expansion $Y ∼ Y_0+ϵ Y_1+…$, with $Y_0=A(s) \cos x$. Obtain the function $A(s)$ such that a solution for $Y_1$ is ensured to exist. You do not need to determine $Y_1$.
      [Hint: throughout this problem you may use without proof the identities \begin{aligned} \cos ^3 x&=\frac{1}{4}(3 \cos x+\cos 3 x) \\ \sin ^3 x&=\frac{1}{4}(3 \sin x-\sin 3 x) \\ \cos ^2 x \sin x&=\frac{1}{4}(\sin x+\sin 3 x) \\ \sin ^2 x \cos x&=\frac{1}{4}(\cos x-\cos 3 x) . \end{aligned}]

Solution

1. 1. adjoint operator $ℒ^*$ defined by $⟨ℒy,w⟩=⟨y,ℒ^*w⟩$ where $⟨u,v⟩=∫_{-π}^π uv\,dx$
      \begin{align*} ⟨ℒy,w⟩&=[y'w-yw'+p_1yw]_{-π}^π+∫_{-π}^π(w''-(p_1w)'+p_0w)y\,dx\\ &=y'(π)w(π)-y'(-π)w(-π)+∫_{-π}^π(w''-(p_1w)'+p_0w)y\,dx\\ ⇒ℒ^*w&≡ w''-(p_1w)'+p_0w\text{ with }w(-π)=w(π)=0 \end{align*}
   2. 1. Multiply both sides of $ℒy=f$ by $w_k$—adjoint eigenfunctions satisfying \begin{cases} ℒ^*w_k=λ_kw_k\\w_k(±π)=0 \end{cases}Then $⟨ℒy,w_k⟩=⟨y,ℒ^*w_k⟩=λ_k⟨y,w_k⟩$. Now write $y=\sum_jc_jy_j$ where \begin{cases} ℒy_j=λ_jy_j\\y_j(±π)=0 \end{cases}and use orthogonality $∫_{-π}^πy_jw_k\,dx=0$ if $k≠j$.
         $⇒λ_kc_kn_k=⟨f,w_k⟩$ where $n_k=∫_{-π}^πy_k(x)w_k(x)dx$
         $⇒c_k={⟨f,w_k⟩\overλ_kn_k}$
      2. Green’s function $g(x,ξ)$ satisfies \begin{cases} ℒg(x,ξ)=δ(x-ξ)\\ g(±π,0)=0 \end{cases}Then the solution is $y(x)=∫_{-π}^πf(ξ)g(x,ξ)dξ$
   3. Rewriting the solution in (b):\begin{align*} y(x)&=\sum_k{∫_{-π}^πf(ξ)w_k(ξ)dξ\overλ_kn_k}⋅y_k(x)\\ &=∫_{-π}^π\sum_k{y_k(x)w_k(ξ)\overλ_kn_k}⋅f(ξ)\,dξ\\ &=∫_{-π}^π\tilde g(x,ξ)f(ξ)\,dξ\text{ with }\tilde g=\sum_k{w_k(ξ)\overλ_kn_k}y_k(x) \end{align*}Now write $g$ from (ii) as $g(x,ξ)=\sum_kd_ky_k(x)$ with $d_k=d_k(ξ)$ and consider $ℒg=δ(x-ξ)$. Multiply by $w_k(x)$ and integrate from $-π$ to $π$: $⟨ℒg,w_k⟩=⟨δ(x-ξ),w_k⟩=w_k(ξ)$ and $⟨ℒg,w_k⟩=⟨g,ℒ^*w_k⟩=λ_k⟨g,w_k⟩$
      Insert $g=\sum_kd_ky_k(x)$ and use orthogonality: $λ_kd_kn_k=w_k(ξ)⇒d_k={w_k(ξ)\overλ_kn_k}⇒g(x,ξ)=\sum_k{w_k(ξ)\overλ_kn_k}y_k(x)=\tilde g$, ∴ the solutions agree.
   4. $y''+δ(x)y=λy,y(±π)=0⇒y=\begin{cases} A\cosμx+B\sinμx&x<0\\ C\cosμx+D\sinμx&x>0 \end{cases}$ where $μ=\sqrt{-λ}$
      $y(±π)=0⇒\color{red}\begin{cases} A\cosμπ-B\sinμπ=0\\ C\cosμπ+D\sinμπ=0 \end{cases}$
      Jump conditions $\left.y\right|_{0-}^{0+}=0⇒A=C$
      $\left.y'\right|_{0-}^{0+}+y(0)=0⇒\color{red}μ(D-B)+A=0$
      Red equations give system $\begin{pmatrix}\cos μ π & -\sin μ π & 0 \\ \cos μ π & 0 & \sin μ π \\ 1 & -μ & μ\end{pmatrix}\begin{pmatrix}A \\ B \\ D\end{pmatrix}=0$ which only has non-trivial solutions if $2μ\cosμπ\sinμπ-\sin^2μπ=0$
      • $\sinμπ=0⇒λ_k=-k^2,k∈ℕ$
      • or $2μ=\tanμπ$ which also has a discrete infinite set of solutions, as can be seen by plotting $\tanμπ$ and $2μ$
        
3. 1. 1. $f(x)∼g(x)$ as $x→x_0⇒\lim_{x→x_0}\frac fg=1$
         $g(x)=o(h(x))$ as $x→x_0⇒\lim_{x→x_0}\frac gh=0$
         Then $\lim_{x→x_0}\frac fh=\lim_{x→x_0}\frac fg\lim_{x→x_0}\frac gh=0⇒f(x)=o(h(x))$ as $x→x_0$
      2. $f$ has asymptotic expansion $\sum_{k=0}^∞ a_k ϵ^k$, then $f=\sum_{k=0}^N a_k ϵ^k+o(ϵ^N)∀N$ as $ϵ → 0$, but $e^{-1/ϵ}=o(ϵ^N)∀N$ it is transcendentally small (can show with limit, but not necessary)
         $∴\sum_{k=0}^∞a_kϵ^k+e^{-1/ϵ}$ is also valid [can stop series at any $N$, the remainder will be $o(ϵ^N)$]
   2. 1. Let $y∼y_0+ϵy_1+⋯$
         $O(1)$: $\begin{cases}y_0''+y_0=0\\y_0(0)=1\\y_0'(0)=0\end{cases}$ has solution $y_0=\cos x$
         $O(ϵ)$: $y_1''+y_1=-y_0^3=-\frac14(3\cos x+\cos3x)$, $y_1(0)=y_1'(0)=0$
         Due to presence of complementary function $\cos x$ on right side, $y_1$ will have terms of form $x\cos x,x\sin x$
         Hence in expansion $y∼y_0+ϵy_1$ will have $ϵy_1∼y_0$ when $ϵx\cos x=O(\cos x)$, ie. $x=O(ϵ^{-1})$, the asymptotic expansion breaks down.
      2. $y∼y_0+ϵy_1+⋯$ gives $y_0=A\cos x+B\sin x$
         At $O(ϵ)$: $y_1''+y_1=-y_0^3$
         $y_1(2π)=y_1(0),y_1'(2π)=y_1'(0)$
         For this BVP we can apply Fredholm: since $\cos x,\sin x$ both satisfy the homogeneous problem, for there to exist a solution, it must hold that $\color{red}∫_0^{2π}y_0^3\cos x\,dx=∫_0^{2π}y_0^3\sin x\,dx=0$
         But $y_0^3=\frac34(A^3+AB^2)\cos x+\frac34(B^3+A^2B)\sin x+⋯(\text{terms orthogonal to both }\sin x,\cos x)$
         Hence the red equation is only satisfied by trivial solution $A=B=0$
   3. Taking $Y∼Y_0+ϵY_1+⋯$ gives $\begin{cases}(Y_0)_{xx}+Y_0=0\\Y_0(0,0)=1\end{cases}$ and we are given the solution $Y_0=A(s)\cos x$
      At $O(ϵ)$: $(Y_1)_{xx}+Y_1=-2(Y_0)_s-Y_0^3=-2A'(s)\cos x-\frac14A(s)^3(3\cos x+\cos3x)$ with periodic boundary conditions $Y_1(2π,s)=Y_1(0,s),Y_1'(2π,s)=Y_1'(0,s)$
      We can view this as an ODE BVP in $x$ and again apply Fredholm Alternative Theorem. The solvability condition is\[∫_0^{2π}[2A'(s)\cos x+\frac14A(s)^3(3\cos x+\cos3x)]\cos x\,dx=0\]which holds if $2A'(s)+\frac34A(s)^3=0⇒\frac1{A^3}dA=-\frac38ds⇒-\frac1{2A^2}=-\frac38s+C$, and $Y_0(0,0)=1⇒A(0)=1⇒C=-\frac12$, so a solution to $Y_1$ exists if $A(s)=±\frac2{\sqrt{3s+4}}$