of asymptotic estimation
A method for determining the asymptotic behaviour as $ 0 < \lambda \rightarrow + \infty $of Laplace integrals
$$ \tag{1 }F ( \lambda ) = \int\limits _ \Omega f ( x) e ^ {\lambda S ( x) } d x ,$$
where $ \Omega = [ a , b ] $is a finite interval, $ S $is a real-valued function and $ f $is a complex-valued function, both sufficiently smooth for $ x \in \Omega $. The asymptotic behaviour of $ F ( \lambda ) $is the sum of the contributions from points at which $ \max _ {x \in \Omega } S ( x) $is attained, if the number of these points is assumed to be finite.
1) If a maximum is attained at $ x = a $and if $ S ^ { \prime } ( a) \neq 0 $, then the contribution $ V _ {a} ( \lambda ) $from the point $ a $in the asymptotic behaviour of the integral (1) is equal to
$$ V _ {a} ( \lambda ) = -\frac{f ( a) + O ( \lambda ^ {-} 1 ) }{\lambda S ^ { \prime } ( a) }e ^ {\lambda S ( a) } .$$
2) If a maximum is attained at an interior point $ x ^ {0} $of the interval $ \Omega $and $ S ^ { \prime\prime } ( x ^ {0} ) \neq 0 $, then its contribution equals
$$ V _ {x ^ {0} } ( \lambda ) = \ \sqrt {- \frac{2 \pi }{\lambda S ^ { \prime\prime } ( x ^ {0} ) } }[ f ( x ^ {0} ) + O ( \lambda ^ {-} 1 ) ] e ^ {\lambda S ( x ^ {0} ) } .$$
This formula was obtained by P.S. Laplace [1]. The case when $ f ( x) $and $ S ^ { \prime } ( x) $have zeros of finite multiplicity at maximum points of $ S $has been completely investigated, and asymptotic expansions have been obtained (see [2]–[8]). The Laplace method can also be extended to the case of a contour $ \Omega $in the complex plane (see Saddle point method).
Let $ \Omega $be a bounded domain in $ \mathbf R _ {x} ^ {n} $and suppose that the maximal $ m $of $ S ( x) $in the closure of $ \Omega $is attained only at an interior point $ x ^ {0} $, where $ x ^ {0} $is a non-degenerate stationary point of $ S $. Then
$$ F ( \lambda ) = \left ( \frac{2 \pi } \lambda \right ) ^ {n/2}| \mathop{\rm det} S _ {xx} ^ { \prime\prime } ( x ^ {0} ) | ^ {-} 1/2[ f ( x ^ {0} ) + O ( \lambda ^ {-} 1 ) ]e ^ {\lambda S ( x ^ {0} ) } .$$
In this case, asymptotic expansions for $ F ( \lambda ) $have also been obtained. All the formulas given above hold for complex $ \lambda $, $ | \lambda | \rightarrow \infty $, $ | \mathop{\rm arg} \lambda | \leq \pi / 2 - \epsilon $. There are also modifications of the Laplace method for the case when the dependence on the parameter is more complicated (see [4], [8]):
$$ F ( \lambda ) = \int\limits _ {\Omega ( \lambda ) }f ( x , \lambda ) e ^ {S ( x , \lambda ) } d x .$$
References
| [1] | P.S. Laplace, "Essai philosophique sur les probabilités" , Oeuvres complètes , 7 , Gauthier-Villars (1886) |
| [2] | A. Erdélyi, "Asymptotic expansions" , Dover, reprint (1956) |
| [3] | N.G. de Bruijn, "Asymptotic methods in analysis" , Dover, reprint (1981) |
| [4] | M.A. Evgrafov, "Asymptotic estimates and entire functions" , Gordon & Breach (1961) (Translated from Russian) |
| [5] | E.T. Copson, "Asymptotic expansions" , Cambridge Univ. Press (1965) |
| [6] | F.W.J. Olver, "Asymptotics and special functions" , Acad. Press (1974) |
| [7] | E. Riekstyn'sh, "Asymptotic expansions of integrals" , 1 , Riga (1974) (In Russian) |
| [8] | M.V. Fedoryuk, "The method of steepest descent" , Moscow (1977) (In Russian) |
Comments
References
| [a1] | N. Bleistein, R.A. Handelsman, "Asymptotic expansions of integrals" , Holt, Rinehart & Winston (1975) pp. Chapt. 5 |
How to Cite This Entry:
Laplace method. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Laplace_method&oldid=47580
This article was adapted from an original article by M.V. Fedoryuk (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
FAQs
Solutions of Laplace's equation are called harmonic functions; they are all analytic within the domain where the equation is satisfied. If any two functions are solutions to Laplace's equation (or any linear hom*ogeneous differential equation), their sum (or any linear combination) is also a solution.
What type of math is Laplace? ›
The Laplace transform is a mathematical technique that changes a function of time into a function in the frequency domain. If we transform both sides of a differential equation, the resulting equation is often something we can solve with algebraic methods.
What is the Laplace formula? ›
Laplace's equation states that the sum of the second-order partial derivatives of R, the unknown function, with respect to the Cartesian coordinates, equals zero: The sum on the left often is represented by the expression ∇2R or ΔR, in which the symbols ∇2and Δ are called the Laplacian or the Laplace operator.
How to learn Laplace transform easily? ›
- Take the Laplace transform of all the terms. You're allowed to do this because an inner product is a linear function of its arguments.
- Replace T(f') with sT(f).
- Solve for T(f) in terms of s.
- Undo the transformation. In other words, try to recognize what function f could be so that T(f) equals the terms of s in step 3.
In general, the Laplace equation can be written as ▽2f=0,where f is any scalar function with multiple variables.
What is simple Laplace equation? ›
As a final example, Laplace's equation appears in two-dimensional fluid flow. For an incompressible flow, ∇·v=0. If the flow is irrotational, then ∇×v=0. We can introduce a velocity potential, v=∇ϕ.
What is the five point formula for Laplace equation? ›
Answer: standard five-point formula is ui,j = 1 4 [ui+1,j + ui-1,j + ui,j+1 + ui,j-1]. the diagonal five-point formula is used to find the values of u2,2,u1,3,u3,3,u1,1, u3,1 and in second step the standard five-point formula is used to find the values of u2,3,u1,2,u3,2, u2,1.
What is the unique solution to the Laplace equation? ›
Finally, we know that Laplace's equation has no maxima or minima except on the boundaries, so that must mean that both the maximum and minimum values of V3 are zero, which means that V3 = 0 everywhere, so V1 = V2. So any solution to the Dirichlet problem with Poisson's (and hence, Laplace's) equation is unique.
Did Laplace believe in God? ›
Views on God
He owned that he was an atheist." Roger Hahn, in his biography of Laplace, mentions a dinner party at which "the geologist Jean-Étienne Guettard was staggered by Laplace's bold denunciation of the existence of God".
What is a Laplace transform for dummies? ›
Used extensively in engineering, the Laplace Transform takes a function of a positive real variable (x or t), often represented as “time,” and transforms it into a function of a complex variable, commonly called “frequency.”
Usually calculus, as well as linear algebra, are assumed to be part of the background and foundation of integral transforms such as Laplace transforms.
What is the solution of the Laplace equation? ›
Since the boundary conditions and Laplace's equation are linear, the solution to the general problem is simply the sum of the solutions to these four problems, u(x,y)=u1(x,y)+u2(x,y)+u3(x,y)+u4(x,y).
What is the math symbol for Laplace? ›
In mathematics, the Laplace operator or Laplacian is a differential operator given by the divergence of the gradient of a function on Euclidean space. It is usually denoted by the symbols ∇·∇, ∇2 or ∆.
What is the formula for the Laplace step function? ›
The Laplace transform of a unit step function is L(s) = 1/s. A shifted unit step function u(t-a) is, 0, when t has values less than a. 1, when t has values greater than a.
How to calculate the Laplace transform? ›
How do you calculate the Laplace transform of a function? The Laplace transform of a function f(t) is given by: L(f(t)) = F(s) = ∫(f(t)e^-st)dt, where F(s) is the Laplace transform of f(t), s is the complex frequency variable, and t is the independent variable.
How is Laplacian calculated? ›
In a Cartesian coordinate system, the Laplacian is given by the sum of second partial derivatives of the function with respect to each independent variable.
How do you solve a system of equations using the Laplace transform? ›
The idea is simple; the Laplace transform of each term in the differential equation is taken. If the unknown function is y(t) then, on taking the transform, an algebraic equation involving Y (s) = L{y(t)} is obtained.
