D’Alembert’s formula; x </math>

Tuesday, July 1st, 2008

In mathematics, and specifically partial differential equations, d´Alembert’s formula is the general solution to the one-dimensional wave equation: <math>u_{tt}-c^2u_{xx}=0, u(x,0)=g(x), u_t(x,0)=h(x)\,</math>. It is named after the mathematician Jean le Rond d’Alembert.

The characteristics of the PDE are <math>x\pm ct=\mathrm{const}\,</math>, so use the change of variables <math>\mu=x+ct, \eta=x-ct\,</math> to transform the PDE to <math>u_{\mu\eta}=0\,</math>. The general solution of this PDE is <math>u(\mu,\eta) = F(\mu) + G(\eta)\,</math> where <math>F\,</math> and <math>G\,</math> are <math>C^1\,</math> functions. Back in <math>x,t\,</math> coordinates,

<math>u(x,t)=F(x+ct)+G(x-ct)\,</math>
<math>u\,</math> is <math>C^2\,</math> if <math>F\,</math> and <math>G\,</math> are <math>C^2\,</math>.

This solution <math>u\,</math> can be interpreted as two waves with constant velocity <math>c\,</math> moving in opposite directions along the x-axis.

Now consider this solution with the Cauchy data <math>u(x,0)=g(x), u_t(x,0)=h(x)\,</math>.

Using <math>u(x,0)=g(x)\,</math> we get <math>F(x)+G(x)=g(x)\,</math>.

Using <math>u_t(x,0)=h(x)\,</math> we get <math>cF’(x)-cG’(x)=h(x)\,</math>.

Integrate the last equation to get

<math>cF(x)-cG(x)=\int_{-\infty}^x h(\xi) d\xi + c_1\,</math>

Now solve this system of equations to get

<math>F(x) = \frac{-1}{2c}\left(-cg(x)-\left(\int_{-\infty}^x h(\xi) d\xi +c_1 \right)\right)\,</math>
<math>G(x) = \frac{-1}{2c}\left(-cg(x)+\left(\int_{-\infty}^x h(\xi) d\xi +c_1 \right)\right)\,</math>

Now, using

<math>u(x,t) = F(x+ct)+G(x-ct)\,</math>

d´Alembert’s formula becomes:

<math>u(x,t) = \frac{1}{2}\left[g(x-ct) + g(x+ct)\right] + \frac{1}{2c} \int_{x-ct}^{x+ct} h(\xi) d\xi</math>


External links

  • An example of solving a nonhomogeneous wave equation from www.exampleproblems.com

Feynman-Kac formula; The notion of ‘expectation

Wednesday, February 20th, 2008

The Feynman-Kac formula, named after Richard Feynman and Mark Kac, establishes a link between partial differential equations (PDEs) and stochastic processes. It offers a method of solving certain PDEs by simulating random paths of a stochastic process.

Suppose we are given the PDE

<math>\frac{\partial f}{\partial t} + \mu(x,t) \frac{\partial f}{\partial x} + \frac{1}{2} \sigma^2(x,t) \frac{\partial^2 f}{\partial x^2} = 0 </math>

subject to the terminal condition

<math>\ f(x,T)=\psi(x) </math>

where <math>\mu,\ \sigma,\ \psi</math> are known functions, <math>\ T</math> is a parameter and <math>\ f</math> is the unknown. This is known as the (one-dimensional) Kolmogorov backward equation. Then the Feynman-Kac formula tells us that the solution can be written as an expectation:

<math>\ f(x,t) = E[ \psi(X_T) | X_t=x ] </math>

where <math>\ X</math> is an Itō process driven by the equation

<math>dX = \mu(X,t)\,dt + \sigma(X,t)\,dW,</math>

where <math>\ W(t)</math> is a Wiener process (also called Brownian motion) and the initial condition for <math>\ X(t)</math> is <math>\ X(t) = x</math>. This expectation can then be approximated using Monte Carlo or quasi-Monte Carlo methods.

Contents


Proof

Applying Itō’s lemma to the unknown function <math>\ f</math> one gets

<math>df=\left(\mu(x,t)\frac{\partial f}{\partial x}+\frac{\partial f}{\partial t}+\frac{1}{2}\sigma^2(x,t)\frac{\partial^2 f}{\partial x^2}\right)\,dt+\sigma(x,t)\frac{\partial f}{\partial x}\,dW.</math>

The first term in parentheses is the above PDE and is zero by hypothesis. Integrating both sides one gets

<math>\int_t^T df=f(X_T,T)-f(x,t)=\int_t^T\sigma(x,t)\frac{\partial f}{\partial x}\,dW.</math>

Reorganising and taking the expectation of both sides:

<math>f(x,t)=\textrm{E}\left[f(X_T,T)\right]-\textrm{E}\left[\int_t^T\sigma(x,t)\frac{\partial f}{\partial x}\,dW\right]</math>

Since the expectation of an Itō integral with respect to a Wiener process <math>\ W</math> is zero, one gets the desired result:

<math>f(x,t)=\textrm{E}\left[f(X_T,T)\right]=\textrm{E}\left[\psi(x)\right]=\textrm{E}\left[\psi(X_T)|X_t=x\right]</math>


Remarks

When originally published by Kac in 1949, the Feynman-Kac formula was presented as a formula for determining the distribution of certain Wiener functionals. Suppose we wish to find the expected value of the function

<math> e^{-\int_0^t V(x(\tau))\, d\tau} </math>

in the case where <math>\ x(\tau)</math> is some realization of a diffusion process starting at <math>\ x(0) = 0</math>. The Feynman-Kac formula says that this expectation is equivalent to the integral of a solution to a
diffusion equation. Specifically, under the conditions that <math>\ u V(x) \geq 0</math>,

<math> E\left( e^{- u \int_0^t V(x(\tau))\, d\tau} \right) = \int_{-\infty}^{\infty} w(x,t)\, dx </math>

where <math>\ w(x,0) = V(x)</math> and

<math>

\frac{\partial w}{\partial t} = \frac{1}{2} \frac{\partial^2 w}{\partial x^2} - u V(x) w.
</math>

The Feynman-Kac formula can also be interpreted as a method for evaluating functional integrals of a certain form. If

<math> I = \int f(x(0)) e^{-u\int_0^t V(x(t))\, dt} g(x(t))\, Dx </math>

where the integral is taken over all random walks, then

<math> I = \int w(x,t) g(x)\, dx </math>

where <math>\ w(x,t)</math> is a solution to the parabolic partial differential equation

<math> \frac{\partial w}{\partial t} = \frac{1}{2} \frac{\partial^2 w}{\partial x^2} - u V(x) w </math>

with initial condition <math>\ w(x,0) = f(x)</math>.


See also

  • Richard Feynman
  • Mark Kac
  • Itō’s lemma
  • Kunita-Watanabe theorem
  • Girsanov theorem
  • Kolmogorov forward equation (also known as Fokker-Planck equation)


References

D’Alembert’s formula; w </math>

Friday, December 14th, 2007

In mathematics, and specifically partial differential equations, d´Alembert’s formula is the general solution to the one-dimensional wave equation: <math>u_{tt}-c^2u_{xx}=0, u(x,0)=g(x), u_t(x,0)=h(x)\,</math>. It is named after the mathematician Jean le Rond d’Alembert.

The characteristics of the PDE are <math>x\pm ct=\mathrm{const}\,</math>, so use the change of variables <math>\mu=x+ct, \eta=x-ct\,</math> to transform the PDE to <math>u_{\mu\eta}=0\,</math>. The general solution of this PDE is <math>u(\mu,\eta) = F(\mu) + G(\eta)\,</math> where <math>F\,</math> and <math>G\,</math> are <math>C^1\,</math> functions. Back in <math>x,t\,</math> coordinates,

<math>u(x,t)=F(x+ct)+G(x-ct)\,</math>
<math>u\,</math> is <math>C^2\,</math> if <math>F\,</math> and <math>G\,</math> are <math>C^2\,</math>.

This solution <math>u\,</math> can be interpreted as two waves with constant velocity <math>c\,</math> moving in opposite directions along the x-axis.

Now consider this solution with the Cauchy data <math>u(x,0)=g(x), u_t(x,0)=h(x)\,</math>.

Using <math>u(x,0)=g(x)\,</math> we get <math>F(x)+G(x)=g(x)\,</math>.

Using <math>u_t(x,0)=h(x)\,</math> we get <math>cF’(x)-cG’(x)=h(x)\,</math>.

Integrate the last equation to get

<math>cF(x)-cG(x)=\int_{-\infty}^x h(\xi) d\xi + c_1\,</math>

Now solve this system of equations to get

<math>F(x) = \frac{-1}{2c}\left(-cg(x)-\left(\int_{-\infty}^x h(\xi) d\xi +c_1 \right)\right)\,</math>
<math>G(x) = \frac{-1}{2c}\left(-cg(x)+\left(\int_{-\infty}^x h(\xi) d\xi +c_1 \right)\right)\,</math>

Now, using

<math>u(x,t) = F(x+ct)+G(x-ct)\,</math>

d´Alembert’s formula becomes:

<math>u(x,t) = \frac{1}{2}\left[g(x-ct) + g(x+ct)\right] + \frac{1}{2c} \int_{x-ct}^{x+ct} h(\xi) d\xi</math>


External links

  • An example of solving a nonhomogeneous wave equation from www.exampleproblems.com