�Laplace Transform

• BRANCH-E&TC AND ELECTRICAL ENGINEERING
• SEM -3rd
• SUBJECT-MATHEMATICS
• TOPIC-LAPLACE TRANSFORM
• FACULTY-tapas ranjan si

The French Newton�Pierre-Simon Laplace�

• Developed mathematics in astronomy, physics, and statistics

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• Began work in calculus which led to the Laplace Transform

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• Focused later on celestial mechanics

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• One of the first scientists to suggest the existence of black holes

History of the Transform�

• Euler began looking at integrals as solutions to differential equations in the mid 1700’s:

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• Lagrange took this a step further while working on probability density functions and looked at forms of the following equation:

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• 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^γt where M and γ are constants

Continuity

• Since the general form of the Laplace transform is:

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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.

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Boundedness

• This criterion also follows directly from the general definition:

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• If f(t) is not bounded by Me^γt then the integral will not converge.

Laplace Transform Theory

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• General Theory

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• Example

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• Convergence

Laplace Transforms

• Some Laplace Transforms
• Wide variety of function can be transformed

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• Inverse Transform

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• Often requires partial fractions or other manipulation to find a form that is easy to apply the inverse

Laplace Transform for ODEs

• Equation with initial conditions

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• Laplace transform is linear

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• Apply derivative formula

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• Rearrange

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• Take the inverse

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”:

Consider the case where:

u_x+u_t=t with u(x,0)=0 and u(0,t)=t² and

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Taking the Laplace of the initial equation leaves U_x+ U=1/s² (note that the partials with respect to “x” do not disappear) with boundary condition U(0,s)=2/s³

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Solving this as an ODE of variable x, U(x,s)=c(s)e^-x + 1/s²

Plugging in B.C., 2/s³=c(s) + 1/s² so c(s)=2/s³ - 1/s²

U(x,s)=(2/s³ - 1/s²) e^-x + 1/s²

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Now, we can use the inverse Laplace Transform with respect to s to find

u(x,t)=t²e^-x - te^-x + t

Example Solutions

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)

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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)

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With e^at 🡪 1/(s-a) and a=0,

the boundary conditions become U(0,s) = U(l,s) = 1/s.

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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+kπ²/l²))sin(πx/l)

Therefore, when we invert the transform, using the Laplace table:

u(x,t) = 1 + e^-kπ2t/l2sin(πx/l)

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Wave Equation

u_tt = c²u_xx in 0 < x < ∞

Initial Conditions:

u(0,t) = f(t), u(x,0) = u_t(x,0) = 0

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For x 🡪 ∞, we assume that u(x,t) 🡪 0. Because the initial conditions vanish, the Laplace transform satisfies

s²U = c²U_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 🡪 ∞.

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Therefore, b(s) = 0. Hence, U(x,s) = F(s) e^-sx/c. Now we use

H(t-b)f(t-b) 🡪 e^-bsF(s)

To get

u(x,t) = H(t – x/c)f(t – x/c).

Real-Life Applications

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

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

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

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Equation Manipulation

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

Applying the Laplace Transform

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

Source