Section 15.4

The Riemann Method

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Problem 5841

Integrate by the Riemann method the equation ∂2u∂x2−∂2u∂y2=0\dfrac {\partial^{2} u}{\partial x^{2}} - \dfrac {\partial^{2} u}{\partial y^{2}} = 0 under the conditions

u∣y=0=f(x),∂u∂y∣y=0=F(x). u\Big|_{y=0} = f(x), \quad \frac{\partial u}{\partial y}\Big|_{y=0} = F(x).
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Remark.
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Hint. In this case the Riemann function equals unity.

Problem 5842

Solve by the Riemann method problem 5833 (problem 5833: prove that the equation y2∂2u∂y2−x2∂2u∂x2=0y^{2}\frac{\partial^{2} u}{\partial y^{2}} - x^{2}\frac{\partial^{2} u}{\partial x^{2}} = 0 under the conditions u∣y=1=f(x)u\big|_{y=1} = f(x), ∂u∂y∣y=1=F(x)\frac{\partial u}{\partial y}\big|_{y=1} = F(x) has the solution u(x,y)=12f(xy)+y2f(x/y)+14xy∫xyx/yf(x)x−3/2 dx−12xy∫xyx/yF(x)x−3/2 dxu(x,y) = \frac{1}{2}f(xy)+\frac{y}{2}f(x/y) + \frac{1}{4}\sqrt{xy}\int_{xy}^{x/y} f(x)x^{-3/2}\, dx - \frac{1}{2}\sqrt{xy}\int_{xy}^{x/y} F(x)x^{-3/2}\, dx).

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Remark.
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Hint. As the region of integration one should take the curvilinear triangle formed by the characteristics MPMP, MQMQ and the arc of the equilateral hyperbola xy=1xy = 1 (Fig. 15.13). The Riemann function is determined by the equality:

v(x,y;x0,y0)=x0x. v(x,y;x_{0},y_{0}) = \sqrt{\frac{x_{0}}{x}}.

Fig. 15.13Fig. 15.13

Problem 5843

Show, using the Riemann method, that the integral of the equation ∂2θ∂φ2+θ=1a2∂2θ∂t2\dfrac {\partial^{2} \theta }{\partial \varphi^{2}}+\theta = \dfrac {1}{a^{2}}\dfrac {\partial^{2} \theta }{\partial t^{2}}, satisfying the conditions θ∣t=0=f(φ)\theta \big|_{t=0} = f(\varphi ); ∂θ∂t∣t=0=F(φ)\dfrac {\partial \theta }{\partial t}\Big|_{t=0} = F(\varphi ), has the form

θ(φ,t)=f(φ−at)+f(φ+at)2+12∫φ−atφ+atΦ(φ,t,z) dz, \theta (\varphi ,t) = \frac{f(\varphi -at)+f(\varphi +at)}{2} + \frac{1}{2}\int _{\varphi -at}^{\varphi +at} \Phi (\varphi ,t,z) \, dz,

where

Φ(φ,t,z)=1aF(z)J0((z−φ)2−a2t2)−atf(z)J0′((z−φ)2−a2t2)(z−φ)2−a2t2. \Phi (\varphi ,t,z) = \frac{1}{a}F(z)J_{0}\left(\sqrt{(z-\varphi )^{2}-a^{2}t^{2}}\right) - atf(z)\frac{J_{0}'\left(\sqrt{(z-\varphi )^{2}-a^{2}t^{2}}\right)}{\sqrt{(z-\varphi )^{2}-a^{2}t^{2}}}.
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Remark.
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Hint. The Riemann function is determined by the equality

v(x,y;x0,y0)=J0((x−x0)(y−y0)), v(x,y;x_{0},y_{0}) = J_{0}\left(\sqrt{(x-x_{0})(y-y_{0})}\right),

where J0(x)J_{0}(x) denotes the Bessel function of order zero.

Problem 5844

Applying the Riemann method, prove that, under the conditions u∣t=0=f(x)u\big|_{t=0} = f(x), ∂u∂t∣t=0=φ(x)\dfrac {\partial u}{\partial t}\Big|_{t=0} = \varphi (x), the equation

(l−x)∂2u∂x2−∂u∂x=1a2∂2u∂t2 (l-x)\frac{\partial ^{2} u}{\partial x^{2}} - \frac{\partial u}{\partial x} = \frac{1}{a^{2}}\frac{\partial ^{2} u}{\partial t^{2}}

has the solution

u(x,t)=l−x−at2 f(x+l−x at−a2t24)2l−x4+l−x+at2 f(x−l−x at−a2t24)2l−x4+1l−x4∫l−x−at2l−x+at2Φ(x,t,z) dz; u(x,t) = \frac{\sqrt{\sqrt{l-x}-\frac{at}{2}}\, f\left(x+\sqrt{l-x}\, at-\frac{a^{2}t^{2}}{4}\right)}{2\sqrt[4]{l-x}} + \frac{\sqrt{\sqrt{l-x}+\frac{at}{2}}\, f\left(x-\sqrt{l-x}\, at-\frac{a^{2}t^{2}}{4}\right)}{2\sqrt[4]{l-x}} + \frac{1}{\sqrt[4]{l-x}}\int _{\sqrt{l-x}-\frac{at}{2}}^{\sqrt{l-x}+\frac{at}{2}} \Phi (x,t,z) \, dz; Φ(x,t,z)=zaφ(l−z2)F(12,12,1,a2t24−(z−l−x)24l−x z)+atf(l−z2)8l−xzF′(12,12,1,a2t24−(z−l−x)24l−x z), \Phi (x,t,z) = \frac{\sqrt{z}}{a}\varphi (l-z^{2})F\left(\frac{1}{2},\frac{1}{2},1,\frac{\frac{a^{2}t^{2}}{4}-(z-\sqrt{l-x})^{2}}{4\sqrt{l-x}\, z}\right) + \frac{atf(l-z^{2})}{8\sqrt{l-x}\sqrt{z}}F'\left(\frac{1}{2},\frac{1}{2},1,\frac{\frac{a^{2}t^{2}}{4}-(z-\sqrt{l-x})^{2}}{4\sqrt{l-x}\, z}\right),

where F(α,β,γ,x)=1+α⋅β1⋅γx+α(α+1)β(β+1)1⋅2⋅γ(γ+1)x2+…F(\alpha ,\beta ,\gamma ,x) = 1+\dfrac {\alpha \cdot \beta }{1\cdot \gamma }x+\dfrac {\alpha (\alpha +1)\beta (\beta +1)}{1\cdot 2\cdot \gamma (\gamma +1)}x^{2}+\ldots is the hypergeometric series.

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Remark.
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Hint. The given equation must be reduced to the simplest canonical form (see problem 5820). As the region of integration one should take the triangle formed by the characteristics MPMP, MQMQ and the bisector y=−xy = -x (Fig. 15.14). The Riemann function is determined by the equality

v(x,y;x0,y0)=F(12,12,1,(x−x0)(y−y0)(x0−y0)(x−y)). v(x,y;x_{0},y_{0}) = F\left(\frac{1}{2},\frac{1}{2},1,\frac{(x-x_{0})(y-y_{0})}{(x_{0}-y_{0})(x-y)}\right).

Fig. 15.14Fig. 15.14

Problem 5845

Integrate by the Riemann method the equation

(l2−x2)∂2u∂x2−2x∂u∂x−∂2u∂y2−14u=0,0<x<l, (l^{2}-x^{2})\frac{\partial ^{2} u}{\partial x^{2}} - 2x\frac{\partial u}{\partial x} - \frac{\partial ^{2} u}{\partial y^{2}} - \frac{1}{4}u = 0, \quad 0 < x < l,

subjecting its integral to the conditions

u∣y=0=f(x),∂u∂y∣y=0=φ(x). u\Big|_{y=0} = f(x), \quad \frac{\partial u}{\partial y}\Big|_{y=0} = \varphi (x).
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Remark.
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Hint. The Riemann function has the following form:

v(x,y;x0,y0)=F(12,12,1,sin⁡(x−x0)sin⁡(y−y0)sin⁡(x0−y0)sin⁡(x−y)). v(x,y;x_{0},y_{0}) = F\left(\frac{1}{2},\frac{1}{2},1,\frac{\sin (x-x_{0})\sin (y-y_{0})}{\sin (x_{0}-y_{0})\sin (x-y)}\right).