Question

state and prove Hpf -Lax formula​

Answer 1

Answer:

Step-by-step explanation:

I think this is the answer. .

Let M be a differentiable manifold, of dimension n, and v a vector field on M. Suppose that x is an isolated zero of v, and fix some local coordinates near x. Pick a closed ball D centered at x, so that x is the only zero of v in D. Then we define the index of v at x, indexx(v), to be the degree of the map u:∂D→Sn-1 from the boundary of D to the (n-1)-sphere given by u(z)=v(z)/| v(z) |.

Theorem. Let M be a compact differentiable manifold. Let v be a vector field on M with isolated zeroes. If M has boundary, then we insist that v be pointing in the outward normal direction along the boundary. Then we have the

where the sum of the indices is over all the isolated zeroes of v and } is the Euler characteristic of M. A particularly useful corollary is when there is a non-vanishing vector field implying Euler characteristic 0.

The theorem was proven for two dimensions by Henri Poincaré and later generalized to higher dimensions by Heinz Hopf.

Answer 2

Prove Hpf - Lax formula: ​

The Hpf-Lax formula is a formula in mathematical analysis that relates a solution to a hyperbolic partial differential equation (PDE) to a solution of a first-order ordinary differential equation (ODE).

Specifically, the formula relates the solution of the hyperbolic PDE to the solution of a first-order ODE known as the characteristic equation.

The Hpf-Lax formula states that if u(x,t) is a solution of the hyperbolic PDE:

∂²u/∂t² = c² ∂²u/∂x²

Hence

=> u(x,t) = F(x - ct) + G(x + ct)

where F and G are arbitrary functions.

To prove the formula,

Assume that u(x,t) has a solution of the form:

u(x,t) = f(x-ct) + g(x+ct)

Where f and g are arbitrary functions.

Take the partial derivatives of u with respect to x and t:

∂u/∂x = f'(x-ct) - g'(x+ct)

∂u/∂t = -cf'(x-ct) + cg'(x+ct)

Compute the second partial derivatives of u with respect to x and t:

∂²u/∂x² = f''(x-ct) + g''(x+ct)

∂²u/∂t² = c²f''(x-ct) + c²g''(x+ct)

Substitute  these expressions into the hyperbolic PDE

c²f''(x-ct) + c²g''(x+ct) = ∂²u/∂t² = ∂²u/∂x² = f''(x-ct) + g''(x+ct)

Which simplifies to:

f''(x-ct) - c²f''(x-ct) = g''(x+ct) - c²g''(x+ct)

Divide both sides by -c²

f''(x-ct)/c² - f'(x-ct)/c = g''(x+ct)/c² - g'(x+ct)/c

Now define two new functions, F and G, as:

F(y) = (1/c)f(y)

G(z) = (1/c)g(z)

Where y = x- ct and z = x + ct.

Substitute these expressions into the previous equation gives:

=> F''(y) = G''(z)

Which is a first-order ordinary differential equation known as the characteristic equation.

Therefore, the solution to the hyperbolic PDE  is :

u(x,t) = f(x-ct) + g(x+ct) = c[F(y) + G(z)] = F(x-ct) + G(x+ct)

Where F and G are solutions to the characteristic equation and can be arbitrary functions.

Hence, It is proved the Hpf-Lax formula.

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