By Goursat E.

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Every year younger mathematicians congregate in Saint Flour, France, and hear prolonged lecture classes on new subject matters in chance concept. The aim of those notes, representing a direction given by means of Terry Lyons in 2004, is to supply an easy and self assisting yet minimalist account of the main effects forming the root of the idea of tough paths.

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**Extra resources for A course in mathematical analysis. - part.2 Differential equations**

**Example text**

The equation is x¨ = −1 so that dy/dx = − 1/y and the phase paths are given by the parabolas y 2 = −2x + C1 √ • 2|x|1/2 sgn (x) + y < 0. In this case x¨ = 1 so that the phase paths are given by y 2 = 2x + C2 . When the parabolic paths reach the switching curve their only exit is along the switching curve into the equilibrium point at the origin. 28 The relativistic equation for an oscillator is m0 x˙ + kx = 0, [1 − (x/c) ˙ 2] d dt √ |x| ˙

2 ∂x ∂t ∂x where α, β and γ are positive constants. Show that there exist travelling wave solutions of the form u(x, t) = U (x − ct) for any c, where U (ζ ) satisﬁes d2 U dU + βU 3 = 0. 21, show that when c = α/γ , all such waves are periodic. 22. The wave function u(x, t) satisﬁes the partial differential equation ∂u ∂u ∂ 2u + βu3 + γ = 0. +α ∂x ∂t ∂x 2 Let u(x, t) = U (x − ct) and ζ = x − ct. Then ∂u dU = , ∂x dζ ∂ 2 u d2 U = , ∂x 2 dζ 2 dU ∂u = −c , ∂t dζ so that the partial differential equation becomes the ordinary differential equation d2 U dU + βU 3 = 0.

35. The signiﬁcant feature of the equation x¨ + g(x)x˙ 2 + h(x) = 0 (i) is the x˙ 2 term. Let z = f (x), where f (x) is twice differentiable and it is assumed that z = f (x) can be uniquely inverted into x = f −1 (z). Differentiating ˙ z˙ = f (x)x, Therefore x˙ = z˙ , f (x) x¨ = z¨ = f (x)x¨ + f (x)x˙ 2 . z¨ f (x)x˙ 2 z¨ f (x)˙z2 − = − . f (x) f (x) f (x) f (x)3 Substitution of these derivatives into (i) results in z¨ − f (x) 2 g(x) 2 z˙ + f (x)h(x) = 0. z˙ + f (x) f (x)2 50 Nonlinear ordinary differential equations: problems and solutions The z˙ 2 can be eliminated by choosing f (x) so that f (x) = g(x).

### A course in mathematical analysis. - part.2 Differential equations by Goursat E.

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