Section 16.9

Miscellaneous Problems

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Problem 6128

The functions y(x)y(x) and z(x)z(x) satisfy the equation

y2+y′2=z′. y^{2}+y'^{2} = z'.

Moreover, it is given that

y(0)=a+b,y(n)=aen+be−n,z(0)=a2−b2. y(0) = a+b, \quad y(n) = ae^{n}+be^{-n}, \quad z(0) = a^{2}-b^{2}.

Find yy and zz from the condition that z(n)z(n) is a maximum.

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Problem 6129

Consider the problem of the shortest distance between two points M1M_{1} and M2M_{2} in space as a Mayer problem.

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Problem 6130

Among all curves joining the given points M1(x1,y1)M_{1}(x_{1},y_{1}) and M2(x2,y2)M_{2}(x_{2},y_{2}), lying in the vertical plane XOYXOY (OYOY --- downward), find the one along which, moving, a heavy point will arrive from M1M_{1} to M2M_{2} in the shortest time. The initial speed is v1v_{1}; besides gravity, the point is acted on by a resistance R=kvnR=kv^{n}.

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Problem 6131

Among the curves joining the given points M1(x1,y1)M_{1}(x_{1},y_{1}) and M2(x2,y2)M_{2}(x_{2},y_{2}), lying in the plane XOYXOY (OYOY --- downward), find the one along which, moving, a point will arrive at M2M_{2} with the greatest speed. Here the speed of the point at M1M_{1} equals v1v_{1}; the point is acted on by gravity and by a resistance force R=kvnR=kv^{n}. Compare the normal reaction of the curve with the centrifugal force.

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Problem 6132

Join the points M1(0,0,0)M_{1}(0,0,0) and M2(a,0,0)M_{2}(a,0,0) by a curve having given curvature 1ρ\frac{1}{\rho } and the greatest length.

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Problem 6133

An airplane departs from the point A(x1,y1)A(x_{1},y_{1}) at the moment t=0t=0, intending to reach the point B(x2,y2)B(x_{2},y_{2}) in the shortest time. Find how the airplane must be directed for this, given that the distribution of wind in space and time is known, i.e., the components of the wind velocity ww along the axes OXOX and OYOY are known: wx=u(x,y,t)w_{x}=u(x,y,t), wy=v(x,y,t)w_{y}=v(x,y,t). The airplane's own speed is VV, and w2=u2+v2<V2w^{2}=u^{2}+v^{2} < V^{2}.

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Problem 6134

An airplane wing of length ll is located along the axis OXOX and moves parallel to the axis OYOY with speed v0v_{0} (the plane XOYXOY is horizontal). Denote by f(x) dxf(x) \, dx the lift of an element of the wing of length dxdx. In aerodynamics it is shown, under certain assumptions, that under these conditions a vortex of intensity dΓ=1ρv0f′(x) dxd\Gamma = \frac{1}{\rho v_{0}}f'(x) \, dx is shed from the element of the wing. Here ρ\rho is the density of air, taken to be constant. (This formula is a consequence of the basic formula of N.E.Zhukovsky.) These vortices induce at every point ξ\xi of the wing a downward air velocity

w(ξ)=14πρv0∫−l/2l/2f′(x)ξ−x dx. w(\xi ) = \frac{1}{4\pi \rho v_{0}} \int _{-l/2}^{l/2} \frac{f'(x)}{\xi -x} \, dx.

Here the axis OZOZ is directed downward, and the integral is understood everywhere, here and below, in the principal-value sense. The induced drag of the wing is determined by the formula

B=∫−l/2l/2ω(ξ)v0f(ξ) dξ=14πρv02∫−l/2l/2∫−l/2l/2f′(x)f(ξ)ξ−x dx dξ. B = \int _{-l/2}^{l/2} \frac{\omega (\xi )}{v_{0}}f(\xi ) \, d\xi = \frac{1}{4\pi \rho v_{0}^{2}} \int _{-l/2}^{l/2}\int _{-l/2}^{l/2} \frac{f'(x)f(\xi )}{\xi -x} \, dx \, d\xi .

It is required to determine, for a given total lift of the wing

A=∫−l/2l/2f(x) dx A = \int _{-l/2}^{l/2} f(x) \, dx

the distribution of lift, i.e., the function f(x)f(x), for which the induced drag is smallest. Here it should be taken into account that, because of the absence of vortices outside the interval −l2≤x≤l2-\frac{l}{2} \leq x \leq \frac{l}{2}, the condition f(±l2)=0f\left(\pm \frac{l}{2}\right) = 0 must hold.

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Problem 6135

Two equations are given

u′=f(x,y,y′,z,u),z′=g(x,y,y′,z,u). u' = f(x,y,y',z,u), \quad z' = g(x,y,y',z,u).

Find yy as a function of xx under the condition that the values of xx, yy, zz, and uu are given at x=x0x=x_{0}, and yy is also given at x=x1x=x_{1}, while u(x1)u(x_{1}) must be an extremum. Example:

u(x1)=∫x0x1f(x,y,y′,z) dx;z=∫x0x1+y′2 dx. u(x_{1}) = \int _{x_{0}}^{x_{1}} f(x,y,y',z) \, dx; \quad z = \int _{x_{0}}^{x} \sqrt{1+y'^{2}} \, dx.

Form Euler's differential equations.

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Problem 6136

Find the extremum of the function W=f(u,v)W = f(u,v), where

u=∫t1t2F(x,y,x′,y′) dt,v=∫t1t2G(x,y,x′,y′) dt u = \int _{t_{1}}^{t_{2}} F(x,y,x',y') \, dt, \quad v = \int _{t_{1}}^{t_{2}} G(x,y,x',y') \, dt

for given values of xx and yy at the ends of the interval (t1,t2)(t_{1},t_{2}).

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Problem 6137

Find the form of the extremals for the problem of the minimum of the integral

∫ABy dx⋅∫AB1+y′2 dx, \int _{A}^{B} y \, dx \cdot \int _{A}^{B} \sqrt{1+y'^{2}} \, dx,

where A(x1,y1)A(x_{1},y_{1}) and B(x2,y2)B(x_{2},y_{2}) are given points.

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