1. Computing Green's functions. Obtain the Green's function for the following operators using the delta-function construction:
(a) $ℒy=-y'', 0<x<1, y(0)-y'(1)=0, y(0)+y(1)=0$
(b) $ℒy=y''-y, 0<x<2π, y(0)-y(2π)=0, y'(0)-y'(2π)=0$
In (b), what goes wrong if we change the operator to $ℒy=y''+y$ (for the same boundary conditions)?
- The Green's function satisfies \(-g_{xx}=δ(x-ξ )\) with $g(0)-g'(1)=0,g(0)+g(1)=0$. So $g(x,ξ)=\begin{cases}Ax+B&0<x<ξ<1\\Bx-2B&0<ξ<x<1\end{cases}$ \begin{array}l[g]_{ξ-}^{ξ+}=0⇒Aξ+B=Bξ-2B\\ -[g_x]_{ξ-}^{ξ+}=1⇒B-A=-1\end{array}Hence $A=\frac{3-ξ}3,B=-\fracξ3$. So $g(x,ξ)=\begin{cases}\frac{3-ξ}3x-\fracξ3&0<x<ξ<1\\-\fracξ3x+\frac{2ξ}3&0<ξ<x<1\end{cases}$
- The Green's function satisfies \(g_{xx}-g=δ(x-ξ )\) with $g(0)=g(2π),g'(0)=g'(2π)$. So $g(x,ξ)=\begin{cases}A\mathrm e^x+B\mathrm e^{-x}&0<x<ξ<2π\\A\mathrm e^{x-2π}+B\mathrm e^{2π-x}&0<ξ<x<2π\end{cases}$ \begin{array}l[g]_{ξ-}^{ξ+}=0⇒A(\mathrm e^ξ-\mathrm e^{ξ-2π})+B(\mathrm e^{-ξ}-\mathrm e^{2π-ξ})=0\\ [g_x]_{ξ-}^{ξ+}=1⇒A(\mathrm e^ξ-\mathrm e^{ξ-2π})-B(\mathrm e^{-ξ}-\mathrm e^{2π-ξ})=-1\end{array}Hence $A=-\frac{\mathrm e^{2π-ξ}}{2\left(\mathrm e^{2π}-1\right)},B=-\frac{\mathrm e^ξ}{2\left(\mathrm e^{2π}-1\right)}$. So $g(x,ξ)=\begin{cases}\frac{\mathrm e^{x+2π-ξ}+\mathrm e^{ξ-x}}{2\left(1-\mathrm e^{2π}\right)}&0<x<ξ<2π\\\frac{\mathrm e^{ξ+2π-x}+\mathrm e^{x-ξ}}{2\left(1-\mathrm e^{2π}\right)}&0<ξ<x<2π\end{cases}$ If we change the operator to $ℒy=y''+y$, then $g$ satisfies \(g_{xx}+g=δ(x-ξ )\) with $g(0)=g(2π),g'(0)=g'(2π)$. So $g(x,ξ)=\begin{cases}A\cos x+B\sin x&0<x<ξ<2π\\A\cos x+B\sin x&0<ξ<x<2π\end{cases}$ So $[g_x]_{ξ-}^{ξ+}=0≠1$.
2. Green's function for an Initial Value Problem. Reconsider the IVP from Sheet 1 Q3,\[ℒy(x)≡P_2(x)y''(x)+P_1(x)y'(x)+P_0(x)y(x)=f(x)\]for $x>0$, subject to initial conditions $y(0)=y'(0)=0$. Recall that $y_1$ and $y_2$ are linearly independent solutions to the homogeneous problem $ℒy=0$ satisfying $y_1(0)=0$ and $y'_2(0)=0$.
State the ODE and initial conditions satisfied by the Green's function $g(x,ξ)$ in terms of a delta function, solve this problem for $g$, and show that this approach reproduces the expression found by variation of parameters on Sheet 1.
3. Eigenfunction expansion.
(a) Find the general solution of the Cauchy–Euler equation\[x^2y''(x)+3xy'(x)+(1+α)y(x)=0\]where α is a given positive constant.
(b) Use (a) to determine the eigenvalues $λ_j$ and eigenfunctions $y_j$ of the self-adjoint problem\[-\left(x^3 y'(x)\right)'=λ xy, y(1)=0, y(\mathrm{e})=0\]
(c) Obtain the eigenfunction expansion for the solution of the inhomogeneous problem \[\left(x^3y'(x)\right)'=x, y(1)=0, y(\mathrm{e})=0\]giving the coefficients explicitly (i.e. compute the integrals).
- Solving through change of variables By substitution $x=\mathrm e^t,y(x)=ϕ(t)⇒y'(x)=\frac1xϕ'(t),y''(x)=\frac1{x^2}\left(ϕ''(t)-ϕ'(t)\right)$ the equation becomes\[ϕ''(t)+2ϕ'(t)+(1+α)ϕ(t)=0\]The characteristic polynomial\[λ^2+2λ+1+α=0⇒λ=-1±i\sqrtα\]So the general solution is\[ϕ(t)=\mathrm e^{-t}\left(c_1\cos(\sqrtαt)+c_2\sin(\sqrtαt)\right)\]Set $t=\ln x$\[y(x)=\frac1x\left(c_1\cos(\sqrtα\ln x)+c_2\sin(\sqrtα\ln x)\right)\]Solving through trial solution We assume a trial solution $y=x^m$. The indicial equation $(m+1)^2+α=0$ has complex roots $m=-1±i\sqrtα$. So the general solution is\[y(x)=\frac1x\left(C_1\exp(i\sqrtα\ln x)+C_2\exp(-i\sqrtα\ln x)\right)\]Using Euler's formula\[y(x)=\frac1x\left(c_1\cos(\sqrtα\ln x)+c_2\sin(\sqrtα\ln x)\right)\]
- Multiply equation in (a) by $x$\[x^3y''+3x^2y'+(1+α)xy=0⇒-(x^3y')'=(1+α)xy\]y(1) = y(e) = 0$⇒c_1=0,\sin\sqrt{α}=0⇒α_j=j^2π^2⇒λ_j=1+α_j=1+j^2π^2,y_j=\frac1x\sin(jπ\ln x)$
- $ℒy(x)=-\left(x^3y'(x)\right)',f(x)=-x,r(x)=x$ Eigenfunction expansion $y(x)=\sum_{j∈ℕ}c_j y_j(x)$ By 4.31$$c_j=\frac{⟨f,y_j⟩}{λ_j⟨y_j,ry_j⟩}=\frac{-\int_1^{\rm e}\sin(jπ\ln x)\d x}{λ_j\int_1^{\rm e}\frac1x\sin^2(jπ\ln x)\d x}$$Compute the integrals\begin{align*}\int_1^{\rm e}\sin(jπ\ln x)\d x&=\int_0^1\mathrm e^x\sin(jπx)\d x\\&=\int_0^∞\mathrm e^x\sin(jπx)\d x-\int_0^∞\mathrm e^{x+1}\sin(jπ(x+1))\d x\\&=\int_0^∞\mathrm e^x\sin(jπx)\d x-(-1)^j\mathrm e\int_0^∞\mathrm e^x\sin(jπx)\d x\\&=(1-(-1)^j\mathrm e)\int_0^∞\mathrm e^x\sin(jπx)\d x\\&=(1-(-1)^j\mathrm e)L\{\sin(jπx)\}(-1)\qquad L\{\sin(ax)\}(s)=\frac{a}{a^2+s^2}\\&=(1-(-1)^j\mathrm e)\frac{jπ}{j^2π^2+1}\end{align*}\begin{align*}\int_1^{\rm e}\frac1x\sin^2(jπ\ln x)\d x&=\frac12\int_1^{\rm e}\frac1x(1-\cos(2jπ\ln x))\d x\\&=\frac12\int_0^11-\cos(2jπx)\d x\\&=\frac12\end{align*}Substitute in $c_j$$$c_j=\frac{-(1-(-1)^j\mathrm e)\frac{jπ}{j^2π^2+1}}{(1+j^2π^2)\frac12}=\frac{((-1)^j\mathrm e-1)2jπ}{(j^2π^2+1)^2}$$Substitute in $y_j$\begin{align*}y(x)&=\sum_{j∈ℕ}c_j y_j(x)\\&=\sum_{j∈ℕ}\frac{((-1)^j\mathrm e-1)2jπ}{(j^2π^2+1)^2}\frac{\sin(jπ\ln x)}x\\&=\frac{2π}x\sum_{j=1}^∞\frac{((-1)^j\mathrm e-1)j}{(j^2π^2+1)^2}\sin(jπ\ln x)\\\small\text{Fourier series}&=\frac12\left(\ln x-\frac{1-x^{-2}}{1-\mathrm e^{-2}}\right)\end{align*}
4. Eigenvalue expansion – two routes.
(a) Consider the following eigenvalue problem on $0≤x≤1$:$$ℒy≡y''+2 y'+y=λy, y'(0)+y(0)=0, y'(1)+y(1)=0$$Compute the eigenvalues $λ_k$, eigenfunctions $y_k$, and the adjoint eigenfunctions $w_k$.
(b) Under what condition on $f$ does a solution exist for the inhomogeneous problem\[ℒy(x)=f(x) \quad(0<x<1), y'(0)+y(0)=0, y'(1)+y(1)=0\]
(c) Assuming that the condition in (b) is satisfied, obtain the coefficients in an eigenfunction $y(x)=\sum_k^∞c_k y_k(x)$.
(d) Convert the problem in (b) to the equivalent Sturm-Liouville problem and show that the eigenfunction expansion of the solution to that problem matches what you found in part (c).
- Characteristic equation $m^2+2m+1=λ$ If $λ>0$, $m=-1±\sqrtλ$. General solution is\[y(x)=\mathrm e^{-x}\left(c_1\mathrm e^{\sqrtλx}+c_2\mathrm e^{-\sqrtλx}\right)\]$0=y'(0)+y(0)=y'(1)+y(1)⇒c_1=c_2=0$, no non-trivial solution. If $λ≤0$, $m=-1±i\sqrt{-λ}$. General solution is\begin{array}l y(x)=\mathrm e^{-x}\left(c_1\cos(\sqrt{-λ}x)+c_2\sin(\sqrt{-λ}x)\right)\\ y(x)+y'(x)=\sqrt{-λ}\mathrm e^{-x}\left(-c_1\sin(\sqrt{-λ}x)+c_2\cos(\sqrt{-λ}x)\right) \end{array}$0=y'(0)+y(0)=\sqrt{-λ}c_2⇒c_2=0$. For non-trivial solution, $c_1≠0$. $0=y'(1)+y(1)=-\sqrt{-λ}\mathrm e^{-1}c_1\sin\sqrt{-λ}⇒\sin\sqrt{-λ}=0⇒λ_k=-k^2π^2,y_k(x)=\mathrm e^{-x}\cos(kπx)$\begin{align*}⟨ℒy,w⟩&=\int_0^1y''w+2y'w+yw\d x=[y'w-yw'+2yw]_0^1+\int_0^1yw''-2yw'+yw\d x \end{align*}$[y'w-yw'+2yw]_0^1=[y(w-w')]_0^1⇒$adjoint boundary condition is $w'(1)=w(1),w'(0)=w(0)$ $ℒ^*w=yw''-2yw'+yw=λw⇒w_k(x)=\mathrm e^x\left(c_1\cos(kπx)+c_2\sin(kπx)\right)$ $w'(1)=w(1)⇒c_2=0$ then $w_k(x)=\mathrm e^x\cos(kπx)$. [can verify $w'(0)=w(0)$]
- $λ_0=0⇒⟨f,w_0⟩=0$. Since $w_0=\mathrm e^x$, the condition is $\int_0^1f(x)\mathrm e^x\d x=0$
- Compute the integrals$$⟨y_k,w_k⟩=\int_0^1\cos(kπx)^2\d x=\frac12$$By 4.6$$c_k=\frac{⟨f,w_k⟩}{λ_k⟨y_k,w_k⟩}=\frac{\int_0^1f(x)\mathrm e^x\cos(kπx)\d x}{-k^2π^2⋅\frac12}$$
- Sturm-Liouville problem$$\hatℒy≡(\mathrm e^{2x}y')'+\mathrm e^{2x}y=\mathrm e^{2x}f(x)\;(0<x<1), y'(0)+y(0)=0, y'(1)+y(1)=0$$By 4.31$$c_k=\frac{⟨\mathrm e^{2x}f(x),y_k⟩}{λ_k⟨y_k,\mathrm e^{2x}y_k⟩}=\frac{\int_0^1f(x)\mathrm e^x\cos(kπx)\d x}{-k^2π^2⋅\frac12}$$
5. Green's function for Sturm-Liouville. Consider the Sturm-Liouville problem$$ℒy≡-\left(p y'\right)'+q y=f, a<x<b$$where $p(x)≠0$ on $a<x<b$, with the boundary conditions\[ℬ_ℓ y≡y(a)=0, ℬ_r y≡y(b)=0\]Let $y_ℓ,y_r$ satisfy $ℒy_ℓ=ℬ_ℓy_ℓ=0,ℒy_r=ℬ_ry_r=0$, and eigenfunctions $y_k$ satisfy $ℒy_k=λ_ky_k,ℬ_ℓ y_k=ℬ_r y_k=0$.
(a) Use variation of parameters to derive the following expression for the Green's function:\[\tag{*}g(x, ξ)=\begin{cases}\frac{-y_ℓ(x) y_r(ξ)}{W(ξ)p(ξ)}&a<x<ξ<b\\\frac{-y_ℓ(ξ) y_r(x)}{W(ξ)p(ξ)}&a<ξ<x<b\end{cases}\]where $W=y_ℓ y'_r-y_ℓ'y_r$ is the Wronskian.
(b) Re-derive equation (*) by constructing the Green's function satisfying $ℒ_x g(x, ξ)=δ(x-ξ)$.
(c) Write down the eigenfunction expansion of the solution to $ℒy=f$, and hence obtain an alternative expression for the Green's function in terms of an eigenfunction expansion $g(x, ξ)=\sum_k c_k(ξ) y_k(x)$.
(d) Show that the two expressions agree by expanding (*) directly in an eigenfunction expansion and showing that the coefficients match, i.e. write $g(x, ξ)=\sum_k d_k(ξ) y_k(x)$ and show that $d_k=c_k$.
- Suppose $y(x)=c_1(x)y_ℓ(x)+c_2(x)y_r(x)$. Impose the condition $c'_1y_ℓ+c'_2y_r=0$. Differentiate, $y'=c_1y_ℓ'+c_2y_r'$. Differentiate again, $y''=c_1'y_ℓ'+c_2'y_r'+c_1y_ℓ''+c_2y_r''$. $$ℒy=-p(c_1'y_ℓ'+c_2'y_r'+c_1y_ℓ''+c_2y_r'')-p'(c_1y_ℓ'+c_2y_r')+q(c_1y_ℓ+c_2y_r)$$Substitute $ℒy_ℓ=ℒy_r=0$$$ℒy=-p(c_1y_ℓ''+c_2y_r'')$$ODE becomes$$c_1'y_ℓ'+c_2'y_r'=-\frac fp$$We have two simultaneous linear equations for $c'_1$ and $c'_2$$$\pmatrix{y_ℓ&y_r\\y_ℓ'&y_r'}\pmatrix{c_1'\\c_2'}=\pmatrix{0 \\-\frac fp}$$Note that the determinant of the matrix is $W$. Invert to get$$\pmatrix{c_1'\\c_2'}=\frac1W\pmatrix{y_r'&-y_r\\-y_ℓ'&y_ℓ}\pmatrix{0 \\-\frac fp}=-\frac f{pW}\pmatrix{-y_r\\y_ℓ}$$Integrate$$c_1=\int^x\frac{y_r(ξ)f(ξ)}{p(ξ)W(ξ)}\dξ c_2=-\int^x\frac{y_ℓ(ξ)f(ξ)}{p(ξ)W(ξ)}\dξ$$The boundary conditions lead to\begin{align*} 0=y(a)&=c_1(a)y_ℓ(a)+c_2(a)y_r(a)=c_2(a)y_r(a)\\ 0=y(b)&=c_1(b)y_ℓ(b)+c_2(b)y_r(b)=c_1(b)y_ℓ(b)\end{align*}This requires that we take $c_2(a)=c_1(b)=0$$$c_1=-\int_x^b\frac{y_r(ξ)f(ξ)}{p(ξ)W(ξ)}\dξ c_2=-\int_a^x\frac{y_ℓ(ξ)f(ξ)}{p(ξ)W(ξ)}\dξ$$We get the expression for the Green's function:\[g(x, ξ)=\begin{cases}\frac{-y_ℓ(x) y_r(ξ)}{W(ξ)p(ξ)}& a<x<ξ<b\\\frac{-y_ℓ(ξ) y_r(x)}{W(ξ)p(ξ)}& a<ξ<x<b\end{cases}\]
- For $x≠ξ$, $ℒ_xg(x,ξ)=0$, also $g(a)=y_ℓ(a)=0,g(b)=y_r(b)=0$, so $g=\begin{cases}Ay_ℓ&a<x<ξ<b\\By_r&a<ξ<x<b\end{cases}$ Continuity of $g$ at $ξ⇒Ay_ℓ(ξ)=By_r(ξ)$ Since $\int_{ξ-}^{ξ+}δ(x-ξ)=1$ we need $-p(ξ)[g_x]_{ξ-}^{ξ+}=1⇒Ay_ℓ'(ξ)-By_r'(ξ)=\frac1{p(ξ)}$$$\pmatrix{y_ℓ(ξ)&-y_r(ξ)\\y_ℓ'(ξ)&-y_r'(ξ)}\pmatrix{A\\B}=\pmatrix{0\\\frac1{p(ξ)}}$$Solve for $A,B$\[\begin{cases}A=\frac{-y_r(ξ)}{W(ξ)p(ξ)}\\B=\frac{-y_ℓ(ξ)}{W(ξ)p(ξ)}\end{cases}\]We get the expression for the Green's function:\[g(x, ξ)=\begin{cases}\frac{-y_ℓ(x) y_r(ξ)}{W(ξ)p(ξ)}& a<x<ξ<b\\\frac{-y_ℓ(ξ) y_r(x)}{W(ξ)p(ξ)}& a<ξ<x<b\end{cases}\]
- Eigenfunction expansion$$y(x)=\sum_k\frac{⟨f,y_k⟩}{λ_k⟨y_k,y_k⟩}y_k(x)=\sum_k\frac{\int_a^bf(ξ)y_k(ξ)\dξ}{λ_k⟨y_k,y_k⟩}y_k(x)$$Hence$$c_k(ξ)={y_k(ξ)\over λ_k⟨y_k,y_k⟩}$$assuming convergence.
- The coefficient of $y_k(x)$ in eigenfunction expansion of $g(x,ξ)$ is\begin{align*}d_k(ξ)&=\frac{\int_a^ξ\frac{-y_ℓ(x) y_r(ξ)}{W(ξ)p(ξ)}y_k(x)\d x+\int_ξ^b\frac{-y_ℓ(ξ) y_r(x)}{W(ξ)p(ξ)}y_k(x)\d x}{⟨y_k,y_k⟩}\\&=-\frac{y_r(ξ){\color{#f00}\int_a^ξy_ℓ(x) y_k(x)\d x}+y_ℓ(ξ){\color{#f00}\int_ξ^by_r(x)y_k(x)\d x}}{W(ξ)p(ξ)⟨y_k,y_k⟩}\end{align*}Using $ℒy_ℓ=ℒy_k=ℬ_ℓy_ℓ=ℬ_ℓy_k=0$ and integrating by parts to compute the first red expression\begin{align*}\int_a^ξy_ℓ(x)y_k(x)\d x&=\frac1{λ_k}\int_a^ξy_ℓ(x)ℒy_k(x)\d x\\&=\frac1{λ_k}\left(\left[-p(x)y_k'(x)y_ℓ(x)+p(x)y_ℓ'(x)y_k(x)\right]_a^ξ+\int_a^ξ\bcancel{ℒy_ℓ(x)}y_k(x)\d x\right)\\&=\frac{p(ξ)}{λ_k}\left(-y_k'(ξ)y_ℓ(ξ)+y_ℓ'(ξ)y_k(ξ)\right) \end{align*}Similarly\[\int_ξ^by_r(x)y_k(x)\d x=\frac{p(ξ)}{λ_k}\left(y_k'(ξ)y_r(ξ)-y_r'(ξ)y_k(ξ)\right)\]Plug in red expressions\[d_k(ξ)={y_k(ξ)\left(y_ℓ(ξ)y_r'(ξ)-y_ℓ'(ξ)y_r(ξ)\right)\over λ_kW(ξ)⟨y_k,y_k⟩}={y_k(ξ)\over λ_k⟨y_k,y_k⟩}\]
6. Legendre's equation and the Fredholm Alternative. Consider bounded solutions of the eigenvalue problem\[\tag{*}ℒy(x)≡\left(1-x^2\right) y''(x)-2 x y'(x)=λy(x), \quad-1<x<1\]
(a) Use the inner product relation to compute $ℒ^*$ and show that the boundary terms vanish identically. Why are no boundary conditions given for (*)?
(b) Convert (*) to Sturm–Liouville form. What orthogonality relation do the eigenfunctions satisfy?
(c) Verify that $y_0(x)=1$ is an eigenfunction for $λ_0=0$. For the inhomogeneous problem $ℒy(x)=f(x)$ to be solvable for $y$, what condition must $f$ satisfy?
(d) Consider the equation $ℒy(x)=-2x$. Explain via the Fredholm Alternative why this problem should have a non-unique solution. Show that\[y=x+A \log \left(\frac{1+x}{1-x}\right)+B\]is a solution for any values of $A$ and $B$. What can you conclude about the constant $A$?
(e) Find the general solution of $ℒy=1$. Does this match your reasoning in (c)?
- $\begin{aligned}[t]⟨ℒy,w⟩-⟨y,ℒw⟩&=\int_{-1}^1(1-x^2)(y''w-yw'')-2x(y'w-yw')\d x\\&=\left.(1-x^2)(y'w-yw')\right|_{-1}^1=0 \text{Since the question assumes $y$ bounded, $y'w-yw'$ is bounded.}\end{aligned}$ So ℒ is self-adjoint and the boundary terms vanish identically (See §4.7.3 Singular SL problems)
- Sturm–Liouville form $\left[\left(1-x^2\right) y'(x)\right]'=λy(x)$ The eigenfunctions satisfy the orthogonality relation (4.23) $\int_{-1}^1y_j(x)y_k(x)\d x=0,j≠k$
- The general solution to $\left[\left(1-x^2\right) y'(x)\right]'=0$ is $y(x)=c_1\log\left(1+x\over1-x\right)+c_2$ $f$ must satisfy $\int_{-1}^1f(x)\d x=\int_{-1}^1f(x)\log\left(1+x\over1-x\right)\d x=0$
- The homogeneous problem has non-trivial solution (case 2 of FAT) By (c), $\int_{-1}^1(-2x)1\d x=\int_{-1}^1(-2x)\log\left(1+x\over1-x\right)\d x=0⇒$problem has a non-unique solution. From boundary conditions $y'(1)=y'(-1)=1$ we have $A=0$
- $\left[\left(1-x^2\right) y'(x)\right]'=1⇒\left(1-x^2\right)y'(x)=x+c_1⇒y(x)=\int\frac x{1-x^2}+\frac{c_1}{1-x^2}\d x$ $⇒y(x)=-\log\sqrt{1-x^2}+c_1\log\left(1+x\over1-x\right)+c_2$ where $-\log\sqrt{1-x^2}$ is unbounded, so the problem has no bounded solution, and doesn't satisfy condition in (c): $\int_{-1}^11⋅1\d x=2≠0$.