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- Find the general real solution of the differential equation: \begin{equation} ℒy ≡ \frac{1}{4} y''(x)+y'(x)+\left(1+\frac{π^2}{4}\right) y(x)=0, \end{equation} where prime denotes differentiation with respect to $x$. Consider the boundary value problem \begin{equation} ℒy(x)=f(x) \text { on } 0< x< 1, y'(0)+y(0)=0, y(1)=0, \end{equation} with $L y$ as in (1). State the definition of the Green's function $g(x, ξ)$ using the delta function $δ(x)$. Determine the Green's function for the problem (2) explicitly.
- Find the eigenvalues $λ_n$ and corresponding eigenfunctions $y_n(x)$ for the problem \begin{equation} M y=-\frac{λ}{x} y \end{equation} where the operator $M$ is given by \begin{equation} M y ≡ x y''+y'+\frac{4}{x} y, 1< x< e^π, y(1)=0, y\left(e^π\right)=0 \end{equation} Is the operator self-adjoint with respect to the inner product $$ ⟨f, g⟩=∫_1^{e^π} f(t) g(t) \mathrm{d} t ? $$
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- Consider the equation
\begin{equation}
x y''(x)+(2 x-1) y'(x)+\frac{1}{x} y(x)=0 .
\end{equation}
Here prime denotes differentiation with respect to $x$. Classify the points $x=0, x=∞$ as ordinary, regular, or irregular singular points, giving reasons.
For $x=0$, determine the indicial equation and indicial exponents. Find the series expansion about $x=0$ for the solution of (5) that satisfies $$ y'(0)=1 $$ and from it obtain the solution in closed form. Why is one initial condition sufficient to determine this solution uniquely? - Consider Chebyshev's equation
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d} x}\left(\left(1-x^2\right)^{1/2} \frac{\mathrm{d} y}{\mathrm{d} x}\right)+n^2\left(1-x^2\right)^{-1/2} y(x)=0, -1< x< 1
\end{equation}
for integers $n ⩾ 0$.
Find the general solution to Chebyshev's equation, using the substitution $x=\cos z$. Denote by $T_n(x)=\cos (n \arccos (x))$ the polynomial solution of Chebyshev's equation of degree $n$. Show that the $T_n(x)$ satisfy the orthogonality relation $$ ∫_{-1}^1 T_n(x) T_m(x)\left(1-x^2\right)^{-1/2} \mathrm{~d} x=0 \text { for } m ≠ n . $$ Determine the expansion of the function $g(x)=\left(1-x^2\right)^{1/2}$ in terms of the $T_n(x)$.
- Consider the equation
\begin{equation}
x y''(x)+(2 x-1) y'(x)+\frac{1}{x} y(x)=0 .
\end{equation}
Here prime denotes differentiation with respect to $x$. Classify the points $x=0, x=∞$ as ordinary, regular, or irregular singular points, giving reasons.
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- Find the first two terms in an asymptotic expansion for all 5 roots, $x$, of the polynomial $$ x^5+x^2-ϵ=0, \text { as } ϵ → 0 $$ Note: some roots will be complex valued.
- Consider the ODE problem $$ ϵ \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+(x+1) \frac{\mathrm{d} y}{\mathrm{~d} x}+y=0, y(0)=0, y(1)=1 $$ as $ϵ → 0$. Find the first terms of outer and inner solutions. You may assume that the boundary layer is near $x=0$. Match the first term of the inner solution to the one-term outer solution, and give a composite solution.
Solution
- characteristic equation $\frac14m^2+m+1+{π^2\over4}=0$ has solution $m=-2+iπ$
general solution $y=e^{-2x}(C_1\sin(πx)+C_2\cos(πx))$
Denote derivative wrt $x$ by $ℒ_x$
$ℒ_xg=δ(x-ξ)$ for $x,ξ∈(0,1),x≠ξ$
boundary condition $g_x(0,ξ)+g(0,ξ)=0,g(1,ξ)=0$
Let $g(x,ξ)=\begin{cases}e^{-2x}(A\sin(πx)+B\cos(πx))&0<x<ξ<1\\e^{-2x}(C\sin(πx)+D\cos(πx))&0<ξ<x<1\end{cases}$
$g_x(0,ξ)+g(0,ξ)=0⇒πA-2B+B=0⇒B=πA$
$g(1,ξ)=-e^{-2}D=0⇒D=0$
So $g(x,ξ)=\begin{cases}Ae^{-2x}(\sin(πx)+π\cos(πx))&0<x<ξ<1\\Ce^{-2x}\sin(πx)&0<ξ<x<1\end{cases}$
By continuity $[g]_{ξ-}^{ξ+}=0⇒A(\sin(πξ)+π\cos(πξ))=C\sin(πξ)$
Integrating $ℒ_xg=δ(x-ξ)$ across $x=ξ$ we get\begin{align*}4=[g_x]_{ξ-}^{ξ+}=&C\left(-2e^{-2ξ}\sin(πξ)+e^{-2ξ}π\cos(πξ)\right)\\&-A\left(-2e^{-2ξ}(\sin(πξ)+π\cos(πξ))+e^{-2ξ}π(\cos(πξ)-π\sin(πξ))\right)\end{align*}Solving the linear equationsFullSimplify[Solve[{a*(Sin[π*ξ] + π*Cos[π*ξ]) == c*Sin[π*ξ], 4 == c*D[Exp[-2*ξ]*Sin[π*ξ], ξ] - a*D[Exp[-2*ξ]*(Sin[π*ξ] + π*Cos[π*ξ]), ξ]}, {a, c}]]\begin{cases}A=\frac4{π^2}e^{2ξ}\sin(πξ)\\C=\frac4{π^2}e^{2ξ}(\sin(πξ)+π\cos(πξ))\end{cases}So\[g(x,ξ)=\begin{cases}\frac4{π^2}e^{2(ξ-x)}\sin(πξ)(\sin(πx)+π\cos(πx))&0<x<ξ<1\\\frac4{π^2}e^{2(ξ-x)}\sin(πx)\left(\sin(πξ)+π\cos(πξ)\right)&0<ξ<x<1\end{cases}\] - $My=-\fracλxy⇔x^2y''+xy'+(4+λ)y=0$ Cauchy-Euler equation
characteristic equation $m(m-1)+m+4+λ=0⇒m=±i\sqrt{4+λ}$ for $4+λ≥0$ [If $4+λ<0$ only trivial solution]$$y=A\cos(\sqrt{4+λ}\ln x)+B\sin(\sqrt{4+λ}\ln x)$$$y(1)=0⇒A=0$
$y(e^π)=0⇒\sin(π\sqrt{4+λ})=0$ for non-trivial solution $B≠0$
$⇒\sqrt{4+λ}=k∈ℤ⇒λ_k=k^2-4$ is eigenvalue, $y_k=B\sin(k\ln x)$ is eigenfunction.
Let $y(1)=w(1)=0,y(e^π)=w(e^π)=0$ \begin{split} ⟨My, w⟩&=∫_1^{e^π} (ty''+y'+\frac4ty) w \mathrm{d} t\\ &=∫_1^{e^π} (ty')'w+\frac4tyw \mathrm{d} t\\ &=[ty'w]_1^{e^π}+∫_1^{e^π}(-ty'w'+\frac4tyw) \mathrm{d} t\\ &=[t(y'w-yw')]_1^{e^π}+∫_1^{e^π}(tw')'y+\frac4tyw \mathrm{d} t=⟨y,Mw⟩ \end{split}So the problem is self-adjoint.
- characteristic equation $\frac14m^2+m+1+{π^2\over4}=0$ has solution $m=-2+iπ$
- At 0, $P(x)=2-\frac1x,Q(x)=\frac1{x^2}$ not analytic but $xP(x),x^2Q(x)$ are, so 0 is regular singular.
Substitute $x=1/t,y(x)=u(t)$\begin{array}l y'=\dot u{dt\over dx}=-t^2\dot u\\y''=-2t{dt\over dx}\dot u-t^2{dt\over dx}\ddot u=2t^3\dot u+t^4\ddot u\end{array}the ODE becomes\[2t^2\dot u+t^3\ddot u-t(2-t)\dot u+tu=0⇔P(t)=(3t^{-1}-2t^{-2}),Q(t)=t^{-2}\]$tP(t)$ not analytic, so $∞$ is irregular singular.
For $x=0$, indicial equation $α(α-1)-α+1=0⇒$double root $α=1$ Case II See §5.2.6
$y_1(x)=x\sum_{k=0}^∞a_nx^k=\sum_{k=0}^∞a_kx^{k+1}$
$y_2(x)=y_1(x)\ln(x)+\sum_{k=0}^∞b_kx^{k+1}$
Because (5) is homogeneous, let $a_0=1$
$\lim_{x→0}y_1'(x)=1$ bounded
$\lim_{x→0}y_2'(x)=\lim_{x→0}\ln(x)+x\frac1x+b_0=∞$
$y'(0)=1⇒y=y_1(x)$.
For $k≥1:a_k=-\frac2ka_{k-1}⇒a_k=\frac{(-2)^k}{k!}⇒y_1(x)=\sum_{k=0}^∞\frac{(-2)^k}{k!}x^{k+1}=xe^{-2x}$ - Substitute $x=\cos z,z∈(0,π),y(x)=u(z),\frac{dy}{dx}=u_z\frac{dz}{dx}=-\frac{u_z}{\sin z},$\[\frac{u_{zz}}{\sin z}+n^2\frac{u}{\sin z}=0⇒u_{zz}+n^2u=0\]$⇒u=A\cos(nz)+B\sin(nz)⇒y(x)=A\cos(n\arccos x)+B\sin(n\arccos x)$
$$
I(n,m)≔∫_{-1}^1 T_n(x) T_m(x)\left(1-x^2\right)^{-1/2} \mathrm{~d} x=∫_0^π \cos(nz)\cos(mz)\mathrm{~d}z
$$
Sturm-Liouville problem, eigenfunctions $T_n$, weighting function $r(x)=(1-x^2)^{-1/2}$
By §4.7.2 Orthogonality $I(n,m)=0$ for $n≠m$.
$T_0(x)=1⇒I(0,0)=π$
For $n>0$, $I(n,n)=\int_0^π\cos^2(nz)\mathrm{~d}z=\fracπ2$
To find eigenfunction expansion of $g(x)=\left(1-x^2\right)^{1/2}$, let $g(x)=\sum_{n=0}^∞c_nT_n(x)$\begin{align*} c_0&=\frac1{I(0,0)}∫_{-1}^1T_0(x)g(x)\left(1-x^2\right)^{-1/2}\mathrm{~d}x=\frac1π∫_{-1}^1\mathrm{~d}x=\frac2π\\ c_n&=\frac1{I(n,n)}∫_{-1}^1T_n(x)g(x)\left(1-x^2\right)^{-1/2}\mathrm{~d}x=\frac2π∫_0^π\cos(nz)\sin(z)\mathrm{~d}z=\frac{(-1)^{n+1}-1}π\frac2{n^2-1} \end{align*}Therefore\begin{align*}g(x)&=\frac2π+\sum_{n=1}^∞\frac{(-1)^{n+1}-1}π\frac2{n^2-1}\cos(n\arccos x)\\&=\frac2π-\frac2π\sum_{n=1}^∞\frac2{4n^2-1}\cos(2n\arccos x)\end{align*}FourierCosCoefficient[(1 - Cos[x]^2)^(1/2), x, n]
- At 0, $P(x)=2-\frac1x,Q(x)=\frac1{x^2}$ not analytic but $xP(x),x^2Q(x)$ are, so 0 is regular singular.
- Substitute $x∼x_0+ϵx_1+…$
$O(1):x_0^5+x_0^2=0⇒x_0=0,0,-1,e^{±iπ/3}$
$O(ϵ):5x_0^4x_1+2x_0x_1-1=0⇒x_0(5x_0^3x_1+2x_1)=1$ [At $x_0=0$ this is 0=1, need rescale]
For $x_0^3=-1$ this is $x_0x_1(-3)=1⇒x_1=-\frac1{3x_0}$ we get three solutions\begin{array}l x∼-1+\fracϵ3\\x∼e^{iπ\over3}+{e^{i{2π\over3}}\over3}\\x∼e^{-iπ\over3}+{e^{-i{2π\over3}}\over3} \end{array}For $x_0=0$ need rescale $x∼ϵ^{1/2}$, let $x=Xϵ^{1/2}$, equation becomes $$ϵ^{5/2}X^5+ϵX^2-ϵ=0⇒ϵ^{3/2}X^5+X^2-1=0$$ Let $X=X_0+X_1ϵ^{1/2}+X_2ϵ+X_3ϵ^{3/2}$ \begin{array}l O(1):&X_0^2-1=0⇒X_0=±1\\ O(ϵ^{1/2}):&2X_0X_1=0⇒X_1=0\\ O(ϵ):&X_1^2+2X_0X_2=0⇒X_2=0\\ O(ϵ^{3/2}):&X_0^5+2X_1X_2+2X_0X_3=0⇒X_3=-\frac12 \end{array} $⇒X=±1-\frac12ϵ^{3/2}+O(ϵ^2)⇒x=±ϵ^{1/2}-\frac{ϵ^2}2+O(ϵ^{5/2})$
- $ϵy''+(x+1)y'+y=0(ϵ→0)$
Taking $ϵ→0$ for leading order expansion: $(x+1)y_0'+y_0=0⇒y_0=\frac A{x+1}$
$P_1(x)>0∀x⇒$boundary layer at left ($x=0$)
Outer expansion: At $x=1$, $y_0(1)=1⇒A=2⇒y_0(x)=\frac2{x+1}$
Inner expansion: Let $x=δX,y(x)=Y(X)$ for $δ≪1$, substitute in\[\fracϵ{δ^2}Y''(X)+\frac{δX+1}δY'(X)+Y(X)=0\]Dominant balance: for $δ=ϵ$,\[Y_0''(X)+Y_0'(X)=0,Y_0(0)=0⇒Y_0(X)=β(1-e^{-X})\]From matching $\lim_{X→∞}Y_0(X)=\lim_{x→0^+}y_0(x)⇒β=2⇒Y_0(X)=2(1-e^{-X})$ \begin{align*} y_\text{comp}&=y_0(x)+Y_0(X)-\text{common limit}\\ &=\frac2{x+1}+2(1-e^{-X})-2\\&=\frac2{x+1}-2e^{-x/ϵ} \end{align*}
- Substitute $x∼x_0+ϵx_1+…$