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Differential equations 2 problem sheet 3

 
  • Jan 14, 2023
  1. Frobenius method. Consider the differential equation$$(x-1)y''(x)-xy'(x)+y(x)=0.$$
    1. Show that $x=1$ is a regular singular point, and determine the appropriate form of the series expansions about $x=1$ for two linearly independent solutions.
    2. By computing the coefficients in the series, find closed form expressions for the two linearly independent solutions.
    $\frac{-x}{x-1}(x-1),\frac1{x-1}(x-1)^2$ are analytic at 1, so $x=1$ is a regular singular point. We seek a solution \(y(x)=\sum_{k=0}^∞a_k(x-1)^{α+k}\)\[\sum_{k=0}^∞a_k(α+k)(α+k-1)(x-1)^{α+k-1}-x\sum_{k=0}^∞a_k(α+k)(x-1)^{α+k-1}+\sum_{k=0}^∞a_k(x-1)^{α+k}=0\]Use $-x=-(x-1)-1$ in the 2nd term. Shift k in the 3rd term\[\sum_{k=0}^∞(α+k-2)a_k(α+k)(x-1)^{α+k-1}-\sum_{k=0}^∞a_k(α+k)(x-1)^{α+k}+\sum_{k=1}^∞a_{k-1}(x-1)^{α+k-1}=0\]Shift k in the 2nd term. Separate k = 0 in the 1st term.\begin{multline*}α(α-2)a_0(x-1)^{α-1}+\sum_{k=1}^∞(α+k-2)a_k(α+k)(x-1)^{α+k-1}-\sum_{k=1}^∞a_{k-1}(α+k-1)(x-1)^{α+k-1}\\+\sum_{k=1}^∞a_{k-1}(x-1)^{α+k-1}=0\end{multline*}Combine these sums\[α(α-2)a_0(x-1)^{α-1}+\sum_{k=1}^∞(α+k-2)[a_k(α+k)-a_{k-1}](x-1)^{α+k-1}=0\tag1\]Indicial equation $α(α-2)=0$ has roots $2,0$.
    Taking $α=0$, solving $a_k(α+k)=a_{k-1}$, we find $a_n={a_0\over\prod_{k=1}^n(2+k)}={2a_0\over(n+2)!}$ Thus a solution is [set a0 = 1/2]\[y_1(x)=\sum_{n=0}^∞{(x-1)^{2+n}\over(n+2)!}=\exp(x-1)-x\]For a second solution, since α1 − α2 = 2 is an integer, we are in Case III, there may or may not be a Frobenius series solution. To find out, we seek a solution \(y(x)=\sum_{k=0}^∞b_k(x-1)^k\), by (1) $b_kk=b_{k-1}⇒b_n=\frac{b_0}{n!}$. The second solution is [set b0 = 1]\[y_2(x)=\sum_{n=0}^∞\frac{(x-1)^n}{n!}=\exp(x-1)\]
  2. The point $x=∞$. Consider the differential equation$$\tag⋆x^3y''(x) + y(x) = 0.$$
    1. Use the transformation of variables x = 1/t to show that (⋆) has a regular singular point at $x=∞$ and determine the indicial exponents.
    2. Obtain the first Frobenius solution in the form of an infinite series in powers of t, solving explicitly for the coefficients.
    3. Find the form of the second Frobenius solution and obtain (but do not attempt to solve) a recurrence relation for the coefficients in the series.
    4. Suppose we seek a particular solution to (⋆) with the leading behaviour y(x) ∼ x as x → ∞. By considering the magnitude of the various terms in the two series solutions found above, determine the next largest term in the expansion of y(x) for large positive x.
    1. $y(x)=u(t)⇒y'(x)=-t^2u'(t),y''(x)=2t^3u'(t)+t^4u''(t)$. Plug in $y''(x)+t^3y(x)=0$, divide by t3\[tu''(t)+2u'(t)+u(t)=0\]has a regular singular point at t = 0, so does (⋆) at x = ∞. Seek a solution \(u(t)=\sum_{k=0}^∞a_kt^{α+k}\)\[t\sum_{k=0}^∞(α+k)(α+k-1)a_kt^{α+k-2}+2\sum_{k=0}^∞(α+k)a_kt^{α+k-1}+\sum_{k=0}^∞a_kt^{α+k}=0\]Sort into powers of t\[α(α+1)a_0t^{α-1}+\sum_{k=1}^∞[(α+k)(α+k+1)a_k+a_{k-1}]t^{α+k-1}=0\]Indicial equation α(α + 1) = 0 has roots α1 = 0, α2 = −1.
    2. Taking $α=0$, solving $(α+k)(α+k+1)a_k+a_{k-1}=0$ [set a0=1] $a_n={(-1)^n\over n!(n+1)!}⇒u_1(x)=\frac{J_1(2\sqrt t)}{\sqrt t}$
    3. Taking $α=-1$, for $k=1$, the equation $(k-1)k=0$, we're in Case IIIa We seek a solution of the form \(u_2(t)=u_1(t)\log t+\sum_{k=0}^∞b_kt^{k-1}=u_1(t)\log t+b_0t^{-1}+\sum_{k=1}^∞b_kt^{k-1}\) \begin{multline*}t\left(u_1''(t)\log t+2u_1'(t)t^{-1}-u_1(t)t^{-2}+2b_0t^{-3}+\sum_{k=3}^∞b_k(k-1)(k-2)t^{k-3}\right)\\+2\left(u_1'(t)\log t+u_1(t)t^{-1}-b_0t^{-2}+\sum_{k=2}^∞b_k(k-1)t^{k-2}\right)\\+u_1(t)\log t+b_0t^{-1}+\sum_{k=1}^∞b_kt^{k-1}=0\end{multline*}Using $tu_1''(t)+2u_1'(t)+u_1(t)=0$,\[2u_1'(t)+(b_0+u_1(t))t^{-1}+\sum_{k=2}^∞[k(k-1)b_k+b_{k-1}]t^{k-2}=0\]At $O(t^{-1})$ we obtain $b_0+u_1(0)=0⇒b_0=-a_0=-1$. Plug in $u_1=\sum_{k=1}^∞{(-t)^{k-1}\over k!(k-1)!}$\[2\sum_{k=2}^∞{(-1)^{k-1}t^{k-2}\over k!(k-2)!}+\sum_{k=2}^∞{(-1)^{k-1}t^{k-2}\over k!(k-1)!}+\sum_{k=2}^∞[k(k-1)b_k+b_{k-1}]t^{k-2}=0\]Merge 1st and 2nd terms\[\sum_{k=2}^∞\frac{(-1)^k (2k-1)}{k!(k-1)!}t^{k-2}=\sum_{k=2}^∞[k(k-1)b_k+b_{k-1}]t^{k-2}\]We obtain a recurrence relation $k(k-1)b_k+b_{k-1}=\frac{(-1)^k (2k-1)}{k!(k-1)!}$.
    4. y(x) ∼ x as x → ∞ is equivalent to u(t) ∼ t−1 as t → 0 Since u1(t) ∼ 1, u2(t) ∼ −t−1, we have u(t) = C u1(t) −u2(t) = −u2(t) + O(1) −u2(t) = −u1(t)log(t) − t−1 + O(1) = −log(t) − t−1 + O(1). So the next largest term is log(x).
  3. Bessel functions. Consider Bessel’s equation (of order n):\[x^2y''(x)+xy'(x)+(x^2-n^2)y(x)=0,\tag⋆\]for integer n ≥ 0.
    1. Find the indicial exponents α1, α2 (Re α1 > Re α2) for the local series expansion of (⋆) about x = 0.
    2. Determine the series $y(x)=∑^∞_{k=0}a_kx^{k+α_1}$ that solves (⋆), giving the coefficients ak in closed form. Find a0 such that the series is the expansion of the Bessel functions of first kind,$$\tag#J_n(x)=\left(x\over2\right)^n\sum_{k=0}^∞\frac{\left(-x^2/4\right)^k}{k!(k+n)!}$$
    3. Using (#), show that the following recursion relation is true for all integers n ≥ 0:$$J_{n+1}(x)=\frac{2n}xJ_n(x)-J_{n-1}(x)$$
    4. For any integer n ≥ 0, show that$$\int_0^1x\left[J_n(αx)\right]^2\,dx=\frac12\left[J_n'(α)\right]^2$$where α is any zero of Jn.
    1. Plug in \(y(x)=\sum_{k=0}^∞a_kx^{α+k}\)\[x^2\sum_{k=0}^∞(α+k)(α+k-1)a_kx^{α+k-2}+x\sum_{k=0}^∞(α+k)a_kx^{α+k-1}+(x^2-n^2)\sum_{k=0}^∞a_kx^{α+k}=0\]Combine coefficients of $x^{α+k}$\[\sum_{k=0}^∞[(α+k)^2-n^2]a_kx^{α+k}+\sum_{k=0}^∞a_kx^{α+k+2}=0\]Shift index in the 2nd term\[\sum_{k=0}^∞[(α+k)^2-n^2]a_kx^{α+k}+\sum_{k=2}^∞a_{k-2}x^{α+k}=0\]The lowest power is xα. The indicial equation α2 − n2 = 0 has solution α1 = n, α2 = −n.
    2. The coefficient of xn+1 is a1 ⇒ a1 = 0 The coefficient of xn+k (k > 1) is $((n+k)^2-n^2)a_k+a_{k-2}=0⇒a_k=-\frac{a_{k-2}}{k(k+2n)}$ a1 = 0 ⇒ ak = 0 for odd k. $a_{2k}=\frac{(-1)^ka_0}{\prod_{i=1}^k(2i)(2i+2n)}=\frac{(-1)^k2^nn!a_0}{2^{2k+n}k!(k+n)!}⇒$For $a_0=\frac1{2^nn!}$ we get (#).
    3. For n = 0, we have $J_{-1}(x)=-J_1(x)$ so equality holds. For n ≥ 1,\begin{align*}\frac{2n}xJ_n(x)-J_{n-1}(x)&=n\left(x\over2\right)^{n-1}\sum_{k=0}^∞\frac{\left(-x^2/4\right)^k}{k!(k+n)!}-\left(x\over2\right)^{n-1}\sum_{k=0}^∞(k+n)\frac{\left(-x^2/4\right)^k}{k!(k+n)!}\\&=-\left(x\over2\right)^{n-1}\sum_{k=0}^∞k\frac{\left(-x^2/4\right)^k}{k!(k+n)!}\\&=-\left(x\over2\right)^{n-1}\sum_{k=1}^∞\frac{\left(-x^2/4\right)^k}{(k-1)!(k+n)!} \text{since }k=0\text{ term is 0}\\&=-\left(x\over2\right)^{n-1}\left(-x^2/4\right)\sum_{k=0}^∞\frac{\left(-x^2/4\right)^k}{k!(k+n+1)!} \text{shift }k\\&=\left(x\over2\right)^{n+1}\sum_{k=0}^∞\frac{\left(-x^2/4\right)^k}{k!(k+n+1)!}=J_{n+1}(x)\end{align*}
    4. If α = 0 then n = 0, $J_n'(0)=0$, equality holds. If α ≠ 0, substitute z = αx, integrate by parts\begin{align*} I&=\int_0^1x[J_n(αx)]^2\,dx \\ &=\frac{1}{α^2}\int_0^αz[J_n(z)]^2\,dz\\ &=\frac{1}{2α^2}\left[z^2J_n(z)^2\right]_0^α-\frac{1}{α^2}\int_0^αz^2J_n(z)J'_n(z)\,dz\\ &=-\frac{1}{α^2}\int_0^αz^2J_n(z)J'_n(z)\,dz \text{since }J_n(α)=0 \end{align*}Since Jn satisfies Bessel’s equation, $z^2J_n=n^2J_n(z)-zJ'_n(z)-z^2J''_n(z)$\begin{align*} I&=-\frac{1}{α^2}\int_0^αJ'_n(z)\left[n^2J_n(z)-zJ'_n(z)-z^2J''_n(z)\right]\,dz\\ &=-\frac{1}{α^2}\left[\frac{n^2}{2}J_n^2(z)|_0^α-\int_0^α\left( zJ'_n(z)+z^2J''_n(z) \right)J_n'(z)\,dz\right]\\ &=\frac{n^2}{2α^2}J_n^2(0)+\frac{1}{α^2}\int_0^α\left( zJ'_n(z)+z^2J''_n(z) \right)J'_n(z)\,dz \text{since $J_n(α)=0$}\\ &=\frac{n^2}{2α^2}J_n^2(0)+\frac{1}{2α^2}\int_0^α\frac{d}{dz}\left( z^2 [J'_n(z)]^2\right)\,dz\\ &=\frac{1}{2}[J_n'(α)]^2 \text{since $J_n(0)=0$ if $n≠0$} \end{align*}
  4. Bessel functions in a Sturm-Liouville problem.
    1. Determine the bounded eigenfunctions yj and eigenvalues λj of the following singular Sturm–Liouville problem on 0 ≤ x ≤ 1:\[\left(-xy'(x)\right)'=λxy(x), y(1) = 0.\]
    2. Use (a) to obtain the eigenfunction expansion for the bounded solution of the following inhomogeneous problem on 0 ≤ x ≤ 1:\[\left(xy'(x)\right)'=x, y(1) = 0.\]Leave the coefficients ck in your final answer in terms of integrals containing Bessel functions.
    1. Substitute $r=\sqrtλx,y(x)=u(r)$ By chain rule $y'(x)=\sqrtλu'(r)$, $\frac{d}{dx}\left(-xy'(x)\right)=\sqrtλ\frac{d}{dr}\left(-\frac{r}{\sqrtλ}\sqrtλu'(r)\right)=\sqrtλ\frac{d}{dr}\left(-ru'(r)\right)$\[\left(-ru'(r)\right)'=ru(r), u(\sqrtλ) = 0.\]This Bessel equation has bounded solution $u(r)=J_0(r)⇒λ_k=j_{0,k}^2,y_k=J_0(j_{0,k}x)$
    2. Eigenfunction expansion $y(x)=\sum_{k∈ℕ}c_ky_k(x)$$$c_k=\frac{⟨f,y_k⟩}{λ_k⟨y_k,ry_k⟩}=\frac{-\int_0^1xJ_0(j_{0,k}x)\,dx}{j_{0,k}^2\int_0^1xJ_0(j_{0,k}x)^2\,dx}$$Using 3(d)$\int_0^1xJ_0(j_{0,k}x)^2\,dx=\frac12[J_0'(j_{0,k})]^2=\frac12[J_1(j_{0,k})]^2$ By 6.10 recurrence formulas$$\frac{2n}{x}J_n(x)=J_{n-1}(x)+J_{n+1}(x)\tag1$$and$$2J'_n(x)=J_{n-1}(x)-J_{n+1}(x)\tag2$$we have\begin{align*} [uJ_1(u)]'&=J_1(u)+uJ_1'(u)\\&\overset{(2)}=J_1(u)+\frac u2J_0(u)-\frac u2J_2(u)\\&\overset{(1)}=J_1(u)+\frac u2J_0(u)-\frac u2\left[\frac2uJ_1(u)-J_0(u)\right]\\&=uJ_0(u)\\⇒\int_0^xuJ_0(u)\,du&=xJ_1(x)\\⇒\int_0^1uJ_0(j_{0,k}u)\,du&=j_{0,k}^{-1}J_1(j_{0,k})\\⇒c_k&=\frac{-j_{0,k}^{-1}J_1(j_{0,k})}{j_{0,k}^2\frac12[J_1(j_{0,k})]^2}\\&=-2j_{0,k}^{-3}[J_1(j_{0,k})]^{-1}\end{align*}The equation has bounded solution $y(x)=\frac14\left(x^2-1\right)$ and we compute $c_k$\[c_k=\frac{\int_0^1\frac14(x^2-1)J_0(j_{0,k}x)x\,dx}{\int_0^1[J_0(j_{0,k}x)]^2x\,dx}\]Two formulas for $c_k$ should match.
      Table[NIntegrate[1/4 (x^2-1)x BesselJ[0,x BesselJZero[0,k]],{x,0,1}]/NIntegrate[x BesselJ[0,x BesselJZero[0,k]]^2,{x,0,1}],{k,1,5}]
Table[N[-2BesselJZero[0,k]^(-3)/BesselJ[1,BesselJZero[0,k]]],{k,1,5}]
  5. Legendre functions and associated Legendre functions. Consider Legendre’s equation$$\tag{⋆}\left(1-x^2\right) y''(x)-2 x y'(x)+\left(ℓ(ℓ+1)-\frac{m^2}{1-x^2}\right)y(x)=0$$and let $P_ℓ^m(x)$ denote the solution for integers 0 ≤ m ≤ ℓ. Show that$$\int_{-1}^1P_k^m(x) P_ℓ^m(x) \,dx=\begin{cases}0&\text{ if }ℓ≠k \\\frac2{(2k+1)}\frac{(k+m)!}{(k-m)!}&\text{ if }ℓ=k\end{cases}$$
    Write (⋆) in Sturm-Liouville form\[ℒy =-((1-x^2) y')' + \frac{m^2}{1-x^2}y = λ y\]Eigenvalues \(λ_ℓ=ℓ(ℓ+1)\), and eigenfunctions \(P_ℓ^m (x)\) satisfy orthogonality conditions$$\int_{-1}^1P_k^m(x) P_ℓ^m(x) \,dx=0 \text{ if }ℓ≠k$$For $ℓ=k$, let$$𝒜_k^m=\int_{-1}^1 P_k^m(x)^2\,dx $$Substituting 6.3.31a\[P_k^m (x)=(-1)^m(1-x^2)^{\frac m2}\frac{d^mP_k(x)}{dx^m}\]we have$$𝒜_k^m=\int_{-1}^1(1-x^2)^m\left(\frac{d^mP_k(x)}{dx^m}\right)^2\,dx$$Integrating by parts $m$ times, each time $u$ contains the factor $\left(1-x^2\right)$, so boundary term vanishes\begin{align*}𝒜_k^m&=\int_{-1}^1(1-x^2)^m\frac{d^mP_k(x)}{dx^m}\frac{d^mP_k(x)}{dx^m}dx\\&=\left[(1-x^2)^m\frac{d^mP_k(x)}{dx^m}\frac{d^{m-1}P_k(x)}{dx^{m-1}}\right]_{-1}^1-\int_{-1}^1\frac{d}{dm}\left((1-x^2)^m\frac{d^mP_k(x)}{dx^m}\right)\frac{d^{m-1}P_k(x)}{dx^{m-1}}dx\\&=-\int_{-1}^1\frac{d}{dx}\left((1-x^2)^m\frac{d^mP_k(x)}{dx^m}\right)\frac{d^{m-1}P_k(x)}{dx^{m-1}}dx\\&=⋯\\&=(-1)^m\int_{-1}^1\frac{d^m}{dx^m}\left((1-x^2)^m\frac{d^mP_k(x)}{dx^m}\right)P_k(x)\,dx\end{align*}Substituting 6.3.29$$P_k(x)=\frac{(-1)^k}{2^kk!}\frac{d^k}{dx^k}[(1-x^2)^k]$$we get$$𝒜_k^m=\frac{(-1)^m}{2^{2k}(k!)^2}\int_{-1}^1\frac{d^m}{dx^m}\left((1-x^2)^m\frac{d^{k+m}P_k(x)}{dx^{k+m}}\right)\frac{d^k}{dx^k}[(1-x^2)^k]\,dx$$Integrating by parts $k$ times, each time $v$ contains the factor $\left(1-x^2\right)$, so boundary term vanishes$$𝒜_k^m=\frac{(-1)^{k+m}}{2^{2k}(k!)^2}\int_{-1}^1 \frac{d^{k+m}}{dx^{k+m}}\left(\left(1-x^2\right)^m \frac{d^{k+m}}{dx^{k+m}}\left(1-x^2\right)^k \right)\;\left(1-x^2\right)^kdx$$Using Leibniz rule$$\frac{d^{k+m}}{dx^{k+m}}\left(\left(1-x^2\right)^m \frac{d^{k+m}}{dx^{k+m}}\left(1-x^2\right)^k\right)=\sum_{r=0}^{k+m}\binom{k+m}r\frac{d^r}{dx^r}\left(1-x^2\right)^m\frac{d^{2k+2 m-r}}{dx^{2k+2 m-r}}\left(1-x^2\right)^k$$ By definition $m≤k$, so $2m∈[0,k+m]$ $\frac{d^r}{dx^r}\left(1-x^2\right)^m≠0$ when $r≤2 m$ $\frac{d^{2k+2 m-r}}{dx^{2k+2 m-r}}\left(1-x^2\right)^k≠0$ when $2k+2m-r≤k⇔r≥2m$. So the only non-zero term in the sum occurs when $r=2m$.$$𝒜_k^m=\frac{(-1)^{k+m}}{2^{2k}(k!)^2}\binom{k+m}{2m}\int_{-1}^1\left(1-x^2\right)^k \frac{d^{2m}}{dx^{2m}}\left(1-x^2\right)^m\frac{d^{2k}}{dx^{2k}}\left(1-x^2\right)^kdx$$Using $\frac{d^{2m}}{dx^{2m}}\left(1-x^2\right)^m=(-1)^m(2m)!$ and $\frac{d^{2k}}{dx^{2k}}\left(x^2-1\right)^k=(-1)^k(2k)!$ we have$$𝒜_k^m=\frac1{2^{2k}(k!)^2}\frac {\left(2k\right)!(k+m)!}{(k-m)!}\int_{-1}^1\left(1-x^2\right)^k\,dx$$By definition of Beta function $B(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1}dt=\frac{Γ(a)Γ(b)}{Γ(a+b)}$. Substituting $t=\frac{1+x}2$$$\int_{-1}^1\left(1-x^2\right)^k\,dx=2^{2k+1}B(k+1,k+1)=\frac{2^{2k+1}(k!)^2}{\left(2k+1\right)!}$$Therefore we have$$𝒜_k^m=\frac2{(2k+1)}\frac{(k+m)!}{(k-m)!}$$
    Associated Legendre Polynomials Orthogonality Proof