Section 14.10

Differential Equations in a Complex Variable

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Problem 5674

Solve, by Cauchy's method, the equation y′′−2y′+y=f(x)y''-2y'+y = f(x); y0=ay_{0} = a, y0′=by_{0}' = b, given these initial conditions.

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Remark.
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Cauchy's method for finding the integrals of a linear system of differential equations with constant coefficients

dxmdt+∑ν=1nαmνxν=φm(t),m=1,2,…,n(*) \frac{dx_{m}}{dt}+\sum _{\nu =1}^{n} \alpha _{m\nu }x_{\nu } = \varphi _{m}(t), \quad m = 1,2,\ldots ,n \tag {*}

consists in the following:

  1. Introducing the symbol δμν=0\delta_{\mu \nu } = 0 for μ≠ν\mu \neq \nu and δνν=1\delta_{\nu \nu } = 1, we solve the equations

    ∑ν=1n(αμν+sδμν)Aν=ξμ;μ=1,2,…,n;(**) \sum _{\nu =1}^{n} (\alpha _{\mu \nu }+s\delta _{\mu \nu })A_{\nu } = \xi _{\mu }; \quad \mu = 1,2,\ldots ,n; \tag {**}

    which gives:

    Aν=∑μ=1nBνμ(s)ξμΔ(s), A_{\nu } = \frac{\sum _{\mu =1}^{n} B_{\nu \mu }(s)\xi _{\mu }}{\Delta (s)},

    where Δ(s)\Delta (s) is the determinant of the system (∗∗)(**).

  2. We form the functions

    ωm(s)=∑μ=1nBmμ(s)(ξμ+∫0tφμ(τ)e−sτ dτ)esτ. \omega _{m}(s) = \sum _{\mu =1}^{n} B_{m\mu }(s)\left(\xi _{\mu }+\int _{0}^{t} \varphi _{\mu }(\tau )e^{-s\tau }\, d\tau \right)e^{s\tau }.

Then the equations (∗)(*) have the system of particular integrals:

xm=12πi∫Cωm(s)Δ(s) ds,m=1,2,…,n. x_{m} = \frac{1}{2\pi i}\int _{C} \frac{\omega _{m}(s)}{\Delta (s)}\, ds, \quad m = 1,2,\ldots ,n.

Here the integration is taken in the positive direction around the circle ∣s∣=R\left|s\right| = R, large enough that all the roots of the equation Δ(s)=0\Delta (s) = 0 lie inside the circle. At t=0t = 0 the quantities xmx_{m} reduce to ξm\xi_{m}.

For a single equation

y(n)+a1y(n−1)+…+any=f(x) y^{(n)}+a_{1}y^{(n-1)}+\ldots +a_{n}y = f(x)

Cauchy's rule becomes the following:

  1. We form the function

    φ(s)=∑ν=0n−1y0(ν)ψν(s), \varphi (s) = \sum _{\nu =0}^{n-1} y_{0}^{(\nu )}\psi _{\nu }(s),

    where ψν(s)=sn−ν−1+a1sn−ν−2+…+an−ν−1\psi_{\nu }(s) = s^{n-\nu -1}+a_{1}s^{n-\nu -2}+\ldots +a_{n-\nu -1}.

  2. Then

    y=12πi∫∑ν=0n−1y0(ν)ψν(s)⋅esx+∫0xe(x−t)sf(t) dtsn+a1sn−1+…+an ds. y = \frac{1}{2\pi i}\int \frac{\sum _{\nu =0}^{n-1} y_{0}^{(\nu )}\psi _{\nu }(s)\cdot e^{sx}+\int _{0}^{x} e^{(x-t)s}f(t)\, dt}{s^{n}+a_{1}s^{n-1}+\ldots +a_{n}}\, ds.

Here y0,y0′,…,y0(n−1)y_{0}, y_{0}', \ldots , y_{0}^{(n-1)} are the values at x=0x = 0, and the integration proceeds around the circle ∣s∣=R\left|s\right| = R, inside which lie all the roots of the denominator.

Problem 5675

Solve, by Cauchy's method (problem 5674), the system of equations x′−x+y=0x'-x+y = 0, y′+x+z=0y'+x+z = 0, z′−x+y=0z'-x+y = 0; x0=ax_{0} = a, y0=by_{0} = b, z0=cz_{0} = c, given these initial conditions.

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Problem 5676

Solve, by Cauchy's method (problem 5674), the system of equations x′+x+y=etx'+x+y = e^{t}, y′−x+3y=e−ty'-x+3y = e^{-t}; x0=−y0=1x_{0} = -y_{0} = 1, given these initial conditions.

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Problem 5677

Solve, by Cauchy's method (problem 5674), the system of equations x′=4x+2y+2zx' = 4x+2y+2z, y′+x+z=0y'+x+z = 0, z′+2x+y=0z'+2x+y = 0; −x0=y0=z0=1-x_{0} = y_{0} = z_{0} = 1, given these initial conditions.

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Problem 5678

Solve, by Cauchy's method (problem 5674), the system of equations x′′=3x+4yx'' = 3x+4y, y′′+x+y=0y''+x+y = 0; x0=y0′=1x_{0} = y_{0}' = 1, y0=x0′=0y_{0} = x_{0}' = 0, given these initial conditions.

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Problem 5679

Find the general integral by means of power series for the equation y′′−xy=0y''-xy = 0.

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Problem 5680

Find the general integral by means of power series for the equation y′′−x2y=0y''-x^{2}y = 0.

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Problem 5681

Find the general integral by means of power series for the equation y′′′+xy=0y'''+xy = 0.

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Problem 5682

Find the general integral by means of power series for the equation y′′′+x3y=0y'''+x^{3}y = 0.

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Problem 5683

Find the general integral by means of power series for the equation (1−x2)y′′−4xy′−2y=0(1-x^{2})y''-4xy'-2y = 0.

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Problem 5684

Find the general integral by means of power series for the equation xy′′+y′+y=0xy''+y'+y = 0.

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Problem 5685

Find the general integral by means of power series for the equation xy′′+2y′+xy=0xy''+2y'+xy = 0.

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Problem 5686

Prove that the equation (1−x2)y′′−5xy′−4y=0(1-x^{2})y''-5xy'-4y = 0 is satisfied by the series

∑n=0∞(n!)2(2n)!(2x)2nand∑n=0∞(2n+1)!(n!)2⋅(x2)2n+1, \sum _{n=0}^{\infty } \frac{(n!)^{2}}{(2n)!}(2x)^{2n} \quad \text{and} \quad \sum _{n=0}^{\infty } \frac{(2n+1)!}{(n!)^{2}}\cdot \left(\frac{x}{2}\right)^{2n+1},

the first of which equals 12((arcsin⁡x)2)′′\dfrac {1}{2}\left((\arcsin x)^{2}\right)''.

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Problem 5687

Prove that the general integral of the equation

(1−x2)y′′−(α+β+1)xy′−αβy=0 (1-x^{2})y''-(\alpha +\beta +1)xy'-\alpha \beta y = 0

equals C1y1+C2y2C_{1}y_{1}+C_{2}y_{2}, where

y1=1+∑n=1∞α(α+2)⋯(α+2n−2)β(β+2)⋯(β+2n−2)(2n)!x2n, y_{1} = 1+\sum _{n=1}^{\infty } \frac{\alpha (\alpha +2)\cdots (\alpha +2n-2)\beta (\beta +2)\cdots (\beta +2n-2)}{(2n)!}x^{2n}, y2=∑n=0∞(1(2n−1)!(α+1)(α+3)⋯(α+2n−1)(β+1)(β+3)⋯(β+2n−1)x2n+1). y_{2} = \sum _{n=0}^{\infty } \left(\frac{1}{(2n-1)!}(\alpha +1)(\alpha +3)\cdots (\alpha +2n-1)(\beta +1)(\beta +3)\cdots (\beta +2n-1)x^{2n+1}\right).
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Problem 5688

Prove that the equation x2y′+(x−1)y+1=0x^{2}y'+(x-1)y+1 = 0, whose general integral is

y=Cxe−1x−1xe−1x∫xe1x dx, y = \frac{C}{x}e^{-\frac{1}{x}}-\frac{1}{x}e^{-\frac{1}{x}}\int xe^{\frac{1}{x}}\, dx,

is formally satisfied by the series ∑n=0∞n!xn\sum_{n=0}^{\infty } n!x^{n}, which diverges for every x≠0x \neq 0.

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Problem 5689

Find the integral of the equation

(1−x2)y′′−2xy′+n(n+1)y=0, (1-x^{2})y''-2xy'+n(n+1)y = 0,

where 2n2n is a non-integer, as a series in positive powers of xx.

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Problem 5690

Find the integral of the equation (1−x2)y′′−2xy′+n(n+1)y=0(1-x^{2})y''-2xy'+n(n+1)y = 0, where 2n2n is a non-integer (problem 5689), as a series in negative powers of xx.

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Problem 5691

Find the general integral of the equation

4x(x−1)y′′+4((a+2)x−a−1)y′+(2a+1)y=0, 4x(x-1)y''+4\left((a+2)x-a-1\right)y'+(2a+1)y = 0,

representing it by the identity y=C1φ(x,a)+C2x−aφ(−a,x)y = C_{1}\varphi (x,a)+C_{2}x^{-a}\varphi (-a,x), where φ(a,x)\varphi (a,x) is a series in powers of xx, equal to 11 at x=0x = 0.

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Problem 5692

Find the integrals of the previous equation (problem 5691) for a=0a = 0 as series converging in a neighborhood of the point x=1x = 1.

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Problem 5693

Find the integral of Bessel's equation

x2y′′+xy′+(x2−n2)y=0, x^{2}y''+xy'+(x^{2}-n^{2})y = 0,

setting y=C1y1+C2y2y = C_{1}y_{1}+C_{2}y_{2}, where y1y_{1} and y2y_{2} are generalized power series of the form xσ(a0+a1x+a2x2+⋯ )x^{\sigma }(a_{0}+a_{1}x+a_{2}x^{2}+\cdots ), where σ\sigma is some constant, not necessarily an integer. The number nn is assumed to be a non-integer.

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Problem 5694

Show that the general integral of Gauss's equation

x(x−1)y′′+((α+β+1)x−γ)y′+αβy=0 x(x-1)y''+\left((\alpha +\beta +1)x-\gamma \right)y'+\alpha \beta y = 0

can, for ∣x∣<1\left|x\right| < 1, be represented by the formula

y=C1F(α,β,γ;x)+C2x1−γF(α−γ+1,β−γ+1,2−γ;x), y = C_{1}F(\alpha ,\beta ,\gamma ;x)+C_{2}x^{1-\gamma }F(\alpha -\gamma +1,\beta -\gamma +1,2-\gamma ;x),

where F(α,β,γ;x)F(\alpha ,\beta ,\gamma ;x) denotes the hypergeometric series:

F(α,β,γ;x)=1+α1⋅βγx+α(α+1)1⋅2⋅β(β+1)γ(γ+1)x2+…. F(\alpha ,\beta ,\gamma ;x) = 1+\frac{\alpha }{1}\cdot \frac{\beta }{\gamma }x+\frac{\alpha (\alpha +1)}{1\cdot 2}\cdot \frac{\beta (\beta +1)}{\gamma (\gamma +1)}x^{2}+\ldots .
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Problem 5695

By means of the Laplace transform, integrate the equation xy′′+(2n+1)y′+xy=0xy''+(2n+1)y'+xy = 0.

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Problem 5696

By means of the Laplace transform (problem 5695), integrate the equation xy′′−(α+β)(1+x)y′+αβxy=0xy''-(\alpha +\beta )(1+x)y'+\alpha \beta xy = 0.

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Problem 5697

By means of the Laplace transform (problem 5695), integrate the equation xy′′+2ay′−q2xy=0xy''+2ay'-q^{2}xy = 0; a>0a > 0.

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Problem 5698

Prove that for 0<a<20 < a < 2 the general integral of the equation xy′′+ay′−q2xy=0xy''+ay'-q^{2}xy = 0 can be represented by the formula

y=∫0πexqcos⁡φ(Asin⁡a−1φ+Bx1−asin⁡1−aφ) dφ, y = \int _{0}^{\pi } e^{xq\cos \varphi }(A\sin ^{a-1}\varphi +Bx^{1-a}\sin ^{1-a}\varphi )\, d\varphi ,

and for a=1a = 1 by the formula:

y=∫0πexqcos⁡φ(A+Bln⁡(xsin⁡2φ)) dφ. y = \int _{0}^{\pi } e^{xq\cos \varphi }\left(A+B\ln (x\sin ^{2}\varphi )\right)\, d\varphi .
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Problem 5699

Find two particular integrals of the equation xy′′′−y=0xy'''-y = 0, setting:

y1=∫sin⁡xt⋅u(t) dt;y2=∫cos⁡xt⋅v(t) dt. y_{1} = \int \sin \frac{x}{t}\cdot u(t)\, dt; \quad y_{2} = \int \cos \frac{x}{t}\cdot v(t)\, dt.
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Problem 5700

Find two pairs of particular integrals of the system

2xy′+2z′+y=0;2y′=2xz′+z;Im(x)<1, 2xy'+2z'+y = 0; \quad 2y' = 2xz'+z; \quad \mathrm{Im}\left(x\right) < 1,

seeking one of the unknowns in the form of the integral ∫u(t)sin⁡xt dt\int u(t)\sin xt\, dt, and the other in the form of the integral ∫v(t)cos⁡xt dt\int v(t)\cos xt\, dt.

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Problem 5701

Prove that the integral of Legendre's equation

(1−x2)y′′−2xy′+n(n+1)y=0 (1-x^{2})y''-2xy'+n(n+1)y = 0

can be represented in one of the forms:

y=∫01t−n+12(1−t)−n+22(x2−t)n2 dt;n>0, y = \int _{0}^{1} t^{-\frac{n+1}{2}}(1-t)^{-\frac{n+2}{2}}(x^{2}-t)^{\frac{n}{2}}\, dt; \quad n > 0, y=∫01tn2(1−t)n−12(x2−t)−n+12 dt;n>−1. y = \int _{0}^{1} t^{\frac{n}{2}}(1-t)^{\frac{n-1}{2}}(x^{2}-t)^{-\frac{n+1}{2}}\, dt; \quad n > -1.
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Problem 5702

Let φ(x,t)=tβ−1(1−t)γ−β−1(1−tx)−α\varphi (x,t) = t^{\beta -1}(1-t)^{\gamma -\beta -1}(1-tx)^{-\alpha }. Prove that each of the integrals

∫1∞φ(x,t) dt;α+1>γ>β; \int _{1}^{\infty } \varphi (x,t)\, dt; \qquad \alpha +1 > \gamma > \beta ; ∫01/xφ(x,t) dt;β>0, α<1; \int _{0}^{1/x} \varphi (x,t)\, dt; \qquad \beta > 0, \ \alpha < 1; ∫01φ(x,t) dt;γ>β>0, α<1; \int _{0}^{1} \varphi (x,t)\, dt; \qquad \gamma > \beta > 0, \ \alpha < 1; ∫0−∞φ(x,t) dt;β>0, α+1>γ, α<1; \int _{0}^{-\infty } \varphi (x,t)\, dt; \qquad \beta > 0, \ \alpha +1 > \gamma , \ \alpha < 1; ∫11/xφ(x,t) dt;γ>β, α<1; \int _{1}^{1/x} \varphi (x,t)\, dt; \qquad \gamma > \beta , \ \alpha < 1; ∫1/x∞φ(x,t) dt;α+1>γ, α<1 \int _{1/x}^{\infty } \varphi (x,t)\, dt; \qquad \alpha +1 > \gamma , \ \alpha < 1

is a particular integral of the hypergeometric equation

x(x−1)y′′+((α+β+1)x−γ)y′+αβy=0. x(x-1)y''+\left((\alpha +\beta +1)x-\gamma \right)y'+\alpha \beta y = 0.
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Problem 5703

Show that the equation

(Ax2+Bx+C)y′′+(Dx+E)y′+Fy=0 (Ax^{2}+Bx+C)y''+(Dx+E)y'+Fy = 0

is reduced, by a simple change of variables, to the hypergeometric equation.

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Problem 5704

Prove the identity: (1+x)n=F(−n,β,β,−x)(1+x)^{n} = F(-n,\beta ,\beta ,-x), where FF is the hypergeometric series (problem 5694).

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Problem 5705

Prove the identity: ex=lim⁡n→∞F(1,n,1,xn)e^{x} = \lim_{n\to \infty } F\left(1,n,1,\dfrac {x}{n}\right), where FF is the hypergeometric series (problem 5694).

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Problem 5706

Prove the identity: ln⁡(1+x)=xF(1,1,2,−x)\ln (1+x) = xF(1,1,2,-x), where FF is the hypergeometric series (problem 5694).

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Problem 5707

Prove the identity: 12xln⁡1+x1−x=F(12,1,32,x2)\dfrac {1}{2x}\ln \dfrac {1+x}{1-x} = F\left(\dfrac {1}{2},1,\dfrac {3}{2},x^{2}\right), where FF is the hypergeometric series (problem 5694).

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Problem 5708

Prove the identity: (1+x)n+(1−x)n=2F(−n2,1−n2,12,x2)(1+x)^{n}+(1-x)^{n} = 2F\left(-\dfrac {n}{2},\dfrac {1-n}{2},\dfrac {1}{2},x^{2}\right), where FF is the hypergeometric series (problem 5694).

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Problem 5709

Prove the identity: ch⁡x=lim⁡α,β→∞F(α,β,12,x24αβ)\ch x = \lim_{\alpha ,\beta \to \infty } F\left(\alpha ,\beta ,\dfrac {1}{2},\dfrac {x^{2}}{4\alpha \beta }\right), where FF is the hypergeometric series (problem 5694).

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Problem 5710

Prove the identity: cos⁡narccos⁡x=F(n,−n,12,1−x2)\cos n\arccos x = F\left(n,-n,\dfrac {1}{2},\dfrac {1-x}{2}\right), where FF is the hypergeometric series (problem 5694).

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Problem 5711

Prove the identity:

Jn(x)=lim⁡α,β→∞(x2)n1Γ(n+1)F(α,β,n+1,−(x2αβ)2), J_{n}(x) = \lim _{\alpha ,\beta \to \infty } \left(\frac{x}{2}\right)^{n}\frac{1}{\Gamma (n+1)}F\left(\alpha ,\beta ,n+1,-\left(\frac{x}{2\sqrt{\alpha \beta }}\right)^{2}\right),

where FF is the hypergeometric series (problem 5694), and Jn(x)J_{n}(x) is the Bessel function, equal to

∑ν=0∞(−1)ν(x2)n+2νν!Γ(n+ν+1). \sum _{\nu =0}^{\infty } \frac{(-1)^{\nu }\left(\dfrac {x}{2}\right)^{n+2\nu }}{\nu !\Gamma (n+\nu +1)}.
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Problem 5712

Denoting F(α+n,β+n,γ+n,1−x2)F\left(\alpha +n,\beta +n,\gamma +n,\dfrac {1-x}{2}\right) (problem 5694) by FnF_{n}, prove that

((1−x)γ(1+x)α+β−γ+1F1)′=−2γ(1−x)γ−1(1+x)α+β−γF0. \left((1-x)^{\gamma }(1+x)^{\alpha +\beta -\gamma +1}F_{1}\right)' = -2\gamma (1-x)^{\gamma -1}(1+x)^{\alpha +\beta -\gamma }F_{0}.
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Problem 5713

Using the preceding (problem 5712), show that

((1−x)γ+n−1(1+x)α+β−γ+nFn)(n)=(−1)n2nγ(γ+1)⋯(γ+n−1)(1−x)γ−1(1+x)α+β−γF0. \left((1-x)^{\gamma +n-1}(1+x)^{\alpha +\beta -\gamma +n}F_{n}\right)^{(n)} = (-1)^{n}2^{n}\gamma (\gamma +1)\cdots (\gamma +n-1)(1-x)^{\gamma -1}(1+x)^{\alpha +\beta -\gamma }F_{0}.
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Problem 5714

Setting γ=β+1\gamma = \beta +1, β=−n\beta = -n, α=p+q+n+1\alpha = p+q+n+1 for the Jacobi polynomial (problem 5581)

Q=n!(p+1)⋯(p+n)Pn(p,q)(x)=F(p+q+n+1,−n,p+1,1−x2), Q = \frac{n!}{(p+1)\cdots (p+n)}P_{n}^{(p,q)}(x) = F\left(p+q+n+1,-n,p+1,\frac{1-x}{2}\right),

prove the identity:

(1−x)p(1+x)qQ=(−1)n2n(p+1)⋯(p+n)dndxn((1−x)p+n(1+x)q+n). (1-x)^{p}(1+x)^{q}Q = \frac{(-1)^{n}}{2^{n}(p+1)\cdots (p+n)}\frac{d^{n}}{dx^{n}}\left((1-x)^{p+n}(1+x)^{q+n}\right).
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Problem 5715

In the particular case p=q=0p = q = 0 (problem 5714), obtain the Rodrigues formula for the Legendre polynomials (problem 5582):

Pn(x)=(−1)n2n⋅n!dn(1−x2)ndxn. P_{n}(x) = \frac{(-1)^{n}}{2^{n}\cdot n!}\frac{d^{n}(1-x^{2})^{n}}{dx^{n}}.
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Problem 5716

Prove the identity for the Chebyshev polynomials (problem 5583):

Tn(x)=cos⁡narccos⁡x2n−1=12n−1F(n,−n,12,1−x2)=(−1)n1⋅3⋯(2n−1)⋅1−x22n−1⋅dn(1−x)n−12dxn. T_{n}(x) = \frac{\cos n\arccos x}{2^{n-1}} = \frac{1}{2^{n-1}}F\left(n,-n,\frac{1}{2},\frac{1-x}{2}\right) = \frac{(-1)^{n}}{1\cdot 3\cdots (2n-1)}\cdot \frac{\sqrt{1-x^{2}}}{2^{n-1}}\cdot \frac{d^{n}(1-x)^{n-\frac{1}{2}}}{dx^{n}}.
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Problem 5717

Prove the identity: ∫0xup−1(1−u)q−1 du=xppF(1−q,p,p+1,x)\displaystyle \int_{0}^{x} u^{p-1}(1-u)^{q-1}\, du = \frac{x^{p}}{p}F(1-q,p,p+1,x), where FF is the hypergeometric series (problem 5694).

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Problem 5718

Prove the identity: ∫0xup−1e−u du=xpplim⁡n→∞F(1−n,p,p+1,xn)\displaystyle \int_{0}^{x} u^{p-1}e^{-u}\, du = \frac{x^{p}}{p}\lim_{n\to \infty } F\left(1-n,p,p+1,\frac{x}{n}\right), where FF is the hypergeometric series (problem 5694).

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Problem 5719

Prove the identity: ∫01du(1−u2)(1−k2u2)=π2F(12,12,1,k2)\displaystyle \int_{0}^{1} \frac{du}{\sqrt{(1-u^{2})(1-k^{2}u^{2})}} = \frac{\pi }{2}F\left(\frac{1}{2},\frac{1}{2},1,k^{2}\right), where FF is the hypergeometric series (problem 5694).

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