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Laplace Transform:

Laplace Transform Naveen Sihag

The French Newton Pierre-Simon Laplace :

The French Newton Pierre-Simon Laplace Developed mathematics in astronomy, physics, and statistics Began work in calculus which led to the Laplace Transform Focused later on celestial mechanics One of the first scientists to suggest the existence of black holes

History of the Transform :

History of the Transform Euler began looking at integrals as solutions to differential equations in the mid 1700’s: Lagrange took this a step further while working on probability density functions and looked at forms of the following equation: Finally, in 1785, Laplace began using a transformation to solve equations of finite differences which eventually lead to the current transform


Definition The Laplace transform is a linear operator that switched a function f(t) to F(s). Specifically: where: Go from time argument with real input to a complex angular frequency input which is complex.


Restrictions There are two governing factors that determine whether Laplace transforms can be used: f(t) must be at least piecewise continuous for t ≥ 0 |f(t)| ≤ Me γ t where M and γ are constants


Since the general form of the Laplace transform is: it makes sense that f(t) must be at least piecewise continuous for t ≥ 0. If f(t) were very nasty, the integral would not be computable. Continuity


Boundedness This criterion also follows directly from the general definition: If f(t) is not bounded by Me γ t then the integral will not converge.

Laplace Transform Theory :

Laplace Transform Theory General Theory Example Convergence

Laplace Transforms:

Laplace Transforms Some Laplace Transforms Wide variety of function can be transformed Inverse Transform Often requires partial fractions or other manipulation to find a form that is easy to apply the inverse

Laplace Transform for ODEs:

Laplace Transform for ODEs Equation with initial conditions Laplace transform is linear Apply derivative formula Rearrange Take the inverse

PowerPoint Presentation:

Laplace Transform in PDEs Laplace transform in two variables (always taken with respect to time variable, t): Inverse laplace of a 2 dimensional PDE: Can be used for any dimension PDE: ODEs reduce to algebraic equations PDEs reduce to either an ODE (if original equation dimension 2) or another PDE (if original equation dimension >2) The Transform reduces dimension by “1”:

PowerPoint Presentation:

Consider the case where: u x +u t =t with u(x,0)=0 and u(0,t)=t 2 and Taking the Laplace of the initial equation leaves U x + U=1/s 2 (note that the partials with respect to “x” do not disappear) with boundary condition U(0,s)=2/s 3 Solving this as an ODE of variable x, U(x,s)=c(s)e -x + 1/s 2 Plugging in B.C., 2/s 3 =c(s) + 1/s 2 so c(s)=2/s 3 - 1/s 2 U(x,s)=(2/s 3 - 1/s 2 ) e -x + 1/s 2 Now, we can use the inverse Laplace Transform with respect to s to find u(x,t)=t 2 e -x - te -x + t

Example Solutions:

Example Solutions

Diffusion Equation:

Diffusion Equation u t = ku xx in (0,l) Initial Conditions: u(0,t) = u(l,t) = 1, u(x,0) = 1 + sin(πx/l) Using af(t) + bg(t)  aF(s) + bG(s) and df/dt  sF(s) – f(0) and noting that the partials with respect to x commute with the transforms with respect to t, the Laplace transform U(x,s) satisfies sU(x,s) – u(x,0) = kU xx (x,s) With e at  1/(s-a) and a=0, the boundary conditions become U(0,s) = U(l,s) = 1/s. So we have an ODE in the variable x together with some boundary conditions. The solution is then: U(x,s) = 1/s + (1/(s+kπ 2 /l 2 ))sin(πx/l) Therefore, when we invert the transform, using the Laplace table: u(x,t) = 1 + e -kπ 2 t/l 2 sin(πx/l)

Wave Equation:

Wave Equation u tt = c 2 u xx in 0 < x < ∞ Initial Conditions: u(0,t) = f(t), u(x,0) = u t (x,0) = 0 For x  ∞, we assume that u(x,t)  0. Because the initial conditions vanish, the Laplace transform satisfies s 2 U = c 2 U xx U(0,s) = F(s) Solving this ODE, we get U(x,s) = a(s)e -sx/c + b(s)e sx/c Where a(s) and b(s) are to be determined. From the assumed property of u, we expect that U(x,s)  0 as x  ∞. Therefore, b(s) = 0. Hence, U(x,s) = F(s) e -sx/c . Now we use H(t-b)f(t-b)  e -bs F(s) To get u(x,t) = H(t – x/c)f(t – x/c).

PowerPoint Presentation:

Real-Life Applications Semiconductor mobility Call completion in wireless networks Vehicle vibrations on compressed rails Behavior of magnetic and electric fields above the atmosphere

PowerPoint Presentation:

Ex. Semiconductor Mobility Motivation semiconductors are commonly made with superlattices having layers of differing compositions need to determine properties of carriers in each layer concentration of electrons and holes mobility of electrons and holes conductivity tensor can be related to Laplace transform of electron and hole densities

PowerPoint Presentation:

Notation R = ratio of induced electric field to the product of the current density and the applied magnetic field ρ = electrical resistance H = magnetic field J = current density E = applied electric field n = concentration of electrons u = mobility

PowerPoint Presentation:

Equation Manipulation and

PowerPoint Presentation:

Assuming a continuous mobility distribution and that , , it follows:

PowerPoint Presentation:

Applying the Laplace Transform

PowerPoint Presentation:

Johnson, William B. Transform method for semiconductor mobility, Journal of Applied Physics 99 (2006). Source

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