# Application of Laplace transform to partial differential equations. Am in need of how to use Laplace transforms to solve a Transient convection diffusion equation So any help is appreciated.?

# What is the use of the Laplace transform in industries?

The use of the Laplace transform in industry: The Laplace transform is one of the most important equations in digital signal processing and electronics. The other major technique used is Fourier Analysis. Further electronic designs will most likely require improved methods of these techniques.

# What are Laplace transforms?

Laplace transforms are used in electronics to quickly build a mathematical circuit in the frequency domain (or 's' plane) that can then can be converted quickly into the time domain. The theory of how this works is still a puzzle to me, but the methods used are straightforward. Simply solve the i…ntegral of the function in question multiplied by the exponential function e -st with limits between 0 and infinity. (MORE)

# Is it possible network analysis without use of laplace transform?

In short, yes, it is possible, but much, much more difficult. Laplace transforms turn systems of integro-differential equations into algebraic equations, and give an immediate expression for the frequency response which is very heavily used in design.

# What is the difference between a Fourier transform and a Laplace transform?

Laplace = analogue signal Fourier = digital signal Notes on comparisons between Fourier and Laplace transforms: The Laplace transform of a function is just like the Fourier transform of the same function, except for two things. The term in the exponential of a Laplace transform is a complex numb…er instead of just an imaginary number and the lower limit of integration doesn't need to start at -â. The exponential factor has the effect of forcing the signals to converge. That is why the Laplace transform can be applied to a broader class of signals than the Fourier transform, including exponentially growing signals. In a Fourier transform, both the signal in time domain and its spectrum in frequency domain are a one-dimensional, complex function. However, the Laplace transform of the 1D signal is a complex function defined over a two-dimensional complex plane, called the s-plane, spanned by two variables, one for the horizontal real axis and one for the vertical imaginary axis. If this 2D function is evaluated along the imaginary axis, the Laplace transform simply becomes the Fourier transform. (MORE)

# What is Laplace transform?

A Laplace transform is a mathematical operator that is used to solve differential equations. This operator is also used to transform waveform functions from the time domain to the frequency domain and can simplify the study of such functions. For continuous functions, f(t), the Laplace transform, F(…s), is defined as the Integral from 0 to infinity of f(t)*e -st dt. When this definition is used it can be shown that the Laplace transform, F n (s) of the n th derivative of a function, f n (t), is given by the following generic formula: F n (s)=s n F(s) - s n-1 f 0 (0) - s n-2 f 1 (0) - s n-3 f 2 (0) - s n-4 f 3 (0) - s n-5 f 4 (0). . . . . - s n-n f n-1 (0) Thus, by taking the Laplace transform of an entire differential equation you can eliminate the derivatives of functions with respect to t in the equation replacing them with a Laplace transform operator, and simple initial condition constants, f n (0), times a new variable s raised to some power. In this manner the differential equation is transformed into an algebraic equation with an F(s) term. After solving this new algebraic equation for F(s) you can take the inverse Laplace transform of the entire equation. Since the inverse Laplace transform of F(s) is f(t) you are left with the solution to the original differential equation. (MORE)

# What is the difference between the fourier laplace transform?

They are similar. In many problems, both methods can be used. You can view Fourier transform is the Laplace transform on the circle, that is |z|=1. When you do Fourier transform, you don't need to worry about the convergence region. However, you need to find the convergence region for each Laplace t…ransform. The discrete version of Fourier transform is discrete Fourier transform, and the discrete version of Laplace transform is Z-transform. (MORE)

# Applications of laplace transform in computer science?

We are using integrated circuits inside the CPU. Laplace Transformations helps to find out the current and some criteria for the analysing the circuits... So, in computer field Laplace tranformations plays vital role...

# How do you solve differential equations?

I assume that you mean that you are given a differential equationdy/dx and want to solve it. If that is the case, then you wouldmultiply by dx on both sides and then integrate both the left andright sides of the equation.

# What are the Applications of partial diffrential equations?

in case of finding the center of the ellipse or hyperbola for which axis or non parallel to axis we apply partial differential

# Difference between poisson's equation and Laplace equation?

Carry out mathematical modeling of stress distribution under torsion vase with a cross section of a given shape by solving Poisson's equation (d^2 u)/(dx^2 )+ (d^2 u)/(dx^2 )=-2 When boundary values equal to zero .The shape of the shaft section of an equilateral triangle with sides L 1 = 100 mm

# Advantages of laplace transformation?

Laplace Transformation is modern technique to solve higher order differential equations. It has several great advantages over old classical method, such as:.
In this method we don't have to put the values of constants by our self. .
We can solve higher order differential equations also of more… than second degree equations because using classical mothed we can only solve first or second degree differential equations. (MORE)

# Why do you use laplace transform?

The most generalized reason would be: "To solve initial-valued differential equations of the 2nd (or higher) order." Laplace is a little powerful for 1st order, but it will solve them as well. There is a limitation here: Laplace will only generate an exact answer if initial conditions are provided…. Laplace cannot be used for boundary-valued problems. In terms of electronics engineering, the Laplace transform is used to get your model into the s-domain, so that s-domain analysis may be performed (finding zeroes and poles of your characteristic equation). This is particularly useful if one needs to determine the kind of response an RC, RLC, or LC circuit will provide (i.e. underdamped, overdamped, critically damped). Once in the s-domain, we may begin discussing the components in terms of impedance. Sometimes it is easier to calculate the voltage or current across a capacitor or an inductor in terms of the components' impedances, rather than find it in a t-domain model. The node-voltage and mesh-current methods used to analyze a circuit in the t-domain work in the s-domain as well. (MORE)

# Laplace transformer of sin23t?

laplace of sin(at) = (a ) / (s^2 + a^2).
thus, laplace of sin 23t, just fill in for a=23.
(23) / (s^2 + 23^2).
thats it...

# Using laplace transform find function y given 2nd derivative of y plus y equals 0 with both initial conditions equal to 0?

Solve y+y=0 using Laplace. Umm y=0, 0+0=0, 0.o Oh well here it is. First you take the Laplace of each term, so . . . L(y'')+L(y)=L(0) Using your Laplace table you know the Laplace of all these terms s 2 L(y)-sy(0)-y'(0) + L(y) = 0 Since both initial conditions are 0 this simplifies to. …. . s 2 L(y) + L(y) = 0 You can factor out the L(y) and solve for it. L(y) = 0/(s 2 +1) L(y) = 0 Now take the inverse Laplace of both sides and solve for y. L -1 (L(y)) = L -1 (0) y = 0 (MORE)

# Application of Laplace equation?

The Laplace equation is used commonly in two situations. It is usedto find fluid flow and in calculating electrostatics.

# Why laplace transform?

gonna keep this short and 2 the point .. i had the same question and i found this webpage .. it helped me a lot .. so try reading it hope it helps u out .. www.dartmouth.edu/~sullivan/22files/ Laplace _Transforms.pdf

# Difference between z transform and laplace transform?

z transform is used for the digital signals and laplace is generally used of the contineous signals.

# How do you do inverse laplace transformations?

actually there are specific formulea for inverse LT..it is just like LT..u just have to reverse the process.... eg. L[1]=1/s ILT[1/s]= 1 thats so simple..u just have to remember the formulea....

# Advantages of laplace transform in solving differential equations?

please follow this link, u can see its brief advantage compared to the usual method http://books.google.com/books?id=zMcAXrJpyPkC&pg=PA902&lpg=PA902&dq=advantages+of+laplace+transform+for+solving+differential+equations&source=bl&ots=txzL6fkRMR&sig=rFigMIeYaT3T65q-ydLM8kjioyE&hl=en&ei=KQ3kSrr5C…ILA-Qann4nJCQ&sa=X&oi=book_result&ct=result&resnum=8&ved=0CCMQ6AEwBw#v=onepage&q=&f=false (MORE)

# Uses of partial differential equation in civil engineering?

Civil engineers use partial differential equations in manydifferent situations. These include the following: heating andcooling; motion of a particle in a resisting medium; hangingcables; electric circuits; natural purification in a stream.

# Ten page project on applications of laplace transforms in various engineering fields?

What are the uses of laplace transforms in engineering fields, good luck :) laplace transforms are so boring i dont have a clue what they do.

# What is the fourier transform of the Laplace operator of a function?

Let F(f) be the fourier transform of f and L the laplacian in IR 3 , then F(Lf(x))(xi) = -|xi| 2 F(f)(xi)

# How Laplace Transform is used solve transient functions in circuit analysis?

f(t)dt and when f(t)=1=1/s or f(t)=k=k/s. finaly can be solve:Laplace transform t domain and s domain L.

# Why you use partial differential equation?

PDEs are used in simulation of real life models like heat flow equation is used for the analysis of temperature distribution in a body, the wave equation for the\nmotion of a waveforms, the ï¬ow equation for the ï¬uid flow and Laplaceâs\nequation for an electrostatic potential.

# What is s in Laplace transforms?

The Laplace transform of a function f ( t ), defined for all real numbers t â¥ 0, is the function F ( s ), defined by:.
The parameter s is a complex number:.
with real numbers Ï and Ï. .
A complex number is defined as a number comprising a real number part and an imaginary …number part. An imaginary number is a number in the form bi where b is a real number and i is the square root of minus one ..
(Wiki search) (MORE)

# What are the applications of partial differential equations in computer science?

All the optimization problems in Computer Science have a predecessor analogue in continuous domain and they are generally expressed in the form of either functional differential equation or partial differential equation. A classic example is the Hamiltonian Jacobi Bellman equation which is the precu…rsor of Bellman Ford algorithm in CS. (MORE)

# What is a drawback of trying to solve a partial differential equation explicitly?

As far as I'm aware a PDE can't be solved explicitly because of the nature of the equation. However, you can separate the variables of a PDE essentially turning it into multiple ODEs which then can be solved explicitly. The draw back of this is that most terms in a PDE are a combination of va…riables, such as 3xy+2yz+xyz, rather than being separated neatly already (making separation into ODEs easy), such as x+y+z. In cases where the variables are jumbled up with each other, there are often faster and easier ways of solving the PDE than trying to separate the variables into ODEs and solving explicitly. (MORE)

# Applications of laplace transform in engineering?

Laplace transforms to reduce a differential equation to an algebra problem. Engineers often must solve difficult differential equations and this is one nice way of doing it.

# Why you use a laplace transformation?

It is typically used to convert a function from the time to the frequency domain.

# Difference between laplace transform and z transform?

the difference is the "S" and "Z" parameters. S used for analog computation while Z for digital processing. basically Z is the digital approximation of the analog frequency domain signal. Z=exp(sT) where T is the sampling time.

# The rules of algebra used to transform equations into equivalent equations?

multiply the entire equation by a number divide the entire equation by a number add numbers to both sides of the equation subtract numbers from both sides of the equation use the commutative property to rearrange the equation use the associative property to rearrange the equation factor …a number out of a portion of the equation (MORE)

# Can you list the application of laplace transforms in the field of computer science engineering?

Laplace is used to write algorithms for various programs. More info is available on wiki .

# Can a discontinuous function have a laplace transform?

Sure! The definition of Laplace transform involves the integral of a function, which always makes discontinuous continuous.

# What is relation between laplace transform and fourier transform?

The Laplace transform is related to the Fourier transform, but whereas the Fourier transform expresses a function or signal as a series of modes ofvibration (frequencies), the Laplace transform resolves a function into its moments. Like the Fourier transform, the Laplace transform is used for solvin…g differential and integral equations. (MORE)

# What is the difference between Fourier transform and Laplace transform and z transform?

Fourier transform and Laplace transform are similar. Laplace transforms map a function to a new function on the complex plane, while Fourier maps a function to a new function on the real line. You can view Fourier as the Laplace transform on the circle, that is |z|=1. z transform is the discrete ve…rsion of Laplace transform. (MORE)

# What is laplace equation of continuity?

partial v x w/ respect to x + partial v y w/ respect to y + partial v z w/ respect to z = 0

# How partial differential equations are used in medicine?

Partial differential equations can be used to model physical systems over time and so can for example describe how you walk. In such an application a faulty stride can be found by comparing a patient's walk with a 'normal' walk.

# What is the difference between unilateral and bilateral laplace transform?

unilateral means limit is 0 to infinite and bilateral means -infinite to +infinite in laplace transform

# How are Laplace transforms useful?

Some differential equations can become a simple algebra problem. Take the Laplace transforms, then just rearrange to isolate the transformed function, then look up the reverse transform to find the solution.

# Why laplace transform was used?

Ans: because of essay calucation in s domine rather than time domine and we take inverse laplace transfom

# What is the Laplace transform of unit ramp signal?

normally the unit ramp signal is defined as follows... r(t)= t, t>=0 0,otherwise so the laplace of it is given as R(s)=1/s^2

# How do you apply laplace transform method to solve systems of ordinary DEs?

you apply the Laplace transform on both sides of both equations. You will then get a sytem of algebraic equations which you can solve them simultaneously by purely algebraic methods. Then take the inverse Laplace transform .

# Where you use Partial Differential Equation in your daily life?

It's all around you, starting with equation of diffusion and ending with equation of propagation of sound and EM waves.

# Why is the partial differential equation important?

Partial differential equations are great in calculus for making multi-variable equations simpler to solve. Some problems do not have known derivatives or at least in certain levels in your studies, you don't possess the tools needed to find the derivative. So, using partial differential equations, y…ou can break the problem up, and find the partial derivatives and integrals. (MORE)

# What is the laplace transform of log function?

The Laplace transform is a widely used integral transform inmathematics with many applications in physics and engineering. Itis a linear operator of a function f(t) with a real argument t (t â¥0) that transforms f(t) to a function F(s) with complex argument s,given by the integral F(s) = \int_0^…\infty f(t) e^{-st}\,dt. (MORE)

# Does every continious function has laplace transform?

There are continuous functions, for example f(t) = e^{t^2}, for which the integral defining the Laplace transform does not converge for any value of the Laplace variable s. So you could say that this continuous function does not have a Laplace transform.

# Can a discontinious function have laplace transform?

Yes, but it can be hard to find. Some easier to find examples are: L(Dirac Delta(t-a))=e^(-a*s) L(u(t-a)*f(t))=(e^(-a*s))*L(f(t-a))

# Why Laplace transform not Laplace equation?

Laplace equation: in 3D U_xx+U_yy+U_zz=0 Or in 2D U_xx+U_yy=0 where U is a function of the spatial variables x,y,z in 3D and x,y in 2D.Also, U_xx is the second order partial derivative of u with respect to x, same for y and z. Laplace transform: L(f(t))=integral of (e^(-s*t))*f(t) dt as t go…es from 0 to infinity. Laplace transform is more like an operator rather than an equation. (MORE)

# How do you solve quadratic equations by the new Transforming Method?

Generalties. This new method is may be thesimplest and fastest method to solve quadratic equations instandard form that can be factored. Its solving procees bases on 3features:1 a. The Rule of Signs for real roots of a quadratic equation. b. The Diagonal Sum method (Google or Yahoo Search) to solve…equation type x^2 + bx + c = 0, when a =1. c. The transformation of a quadratic equation type ax^2 + bx + c =0 into the simplified type x^2 + bx + c = 0, with a = 1. The Rule of Signs. a. If a and c have different signs, roots have different signs. b. If a and c have same sign, roots have same sign; - When a and b have diffrent signs, both roots are positive. - When a and b have same sign, both roots are negative. The Diagonal Sum Method to solve equation type x^2 + bx + c= 0, When a = 1, solving results in finding 2 numbers knowing the sum(-b) and the product (c). This method can immediately obtain the 2real roots without factoring by grouping and solving binomials. Itproceeds composing factor pairs of c, following these 3 Tips. TIP 1 . When roots have different signs, composefactor pairs of c with all first numbers being negative. Example 1 . Solve: x^2 - 11x - 102 = 0. Roots havedifferent signs. Compose factor pairs of c = -102 with all firstnumbers being negative. Proceeding: (-1, 102)(-2, 51)(-3, 34)(-6,17). This last sum is -6 + 17 = 11 = -b. Then, the 2 real rootsare: -6 and 17. No factoring and no solving binomials! TIP 2 . When both roots are positive, composefactor pairs of c with all positive numbers. Example 2. Solve: x^2 - 31x + 108 = 0. Both roots are positive.Proceeding:(1, 108)(2, 54)(3, 36)(4, 27). This last sum is 4 + 27 =31 = -b. Then, the 2 real toots are: 4 and 27. Tip 3 . When both roots are negative, composefactor paits with all negative numbers. Example 3 . Solve: x^2 + 62x + 336 = 0. Both roots arenegative. Compose factor pairs of c = 336 with all negativenumbers. Proceeding: (-1, -336)(-2, -168)(-4, -82)(-6, -56). Thislast sum is: -6 - 56 = -62 = -b. Then the 2 real roots are: -6 and-56. The new "Transforming Method" to solvequadratic equations . It proceeds through 3 Steps. STEP 1 . Transform the original equation instandard form ax^2 + bx + c = 0. (1) into a simplified equation,with a = 1 and C = a*c. The transformed equation has the form: x^2+ bx + a*c = 0. (2). STEP 2 . Solve the transformed equation (2) by theDiagonal Sum Method that immediately obtains the 2 real roots.Suppose they are: y1 , and y2 . STEP 3 . Divide both y1, and y2 by the constant (a)to get the 2 real roots x1 and x2 of the original equation (1): x1 = y1/a, and x2 = y2/a. Example 4 . Original equation to solve: 8x^2 - 22x - 13 =0. (1). Transformed equation: x^2 - 22x - 104 = 0.(2). Roots havedifferent signs. Compose factor pairs of a*c = -104.Proceeding:(-1, 104)(-2, 52)(-4, 26). This last sum is 26 - 4 = 22= -b. The 2 real roots of the transformed equation are: y1 = -4 andy2 = 26. Next, find the 2 real roots of the original equation (1):x = y1/8 = -4/8 = -1/2, and x2 = y2/8 = 26/8 = 13/4. Example 5 . Solve: 15x^2 - 53x + 16 = 0. Both roots arepositive. Compose factor pairs of a*c = 240 from the middle of thechain to save time. Proceeding:....(3, 80)(4, 60)(5, 48). This lastsum is 5 + 48 = 53 = -b. Then, y1 = 5 and y2 = 48. Next, find x1 =y1/15 = 5/15 = 1/3, and x2 = y2/15 = 48/15 = 16/5. Conclusion . The strong points of this new methodare: simple, fast, systematic, no guessing, no factoring bygrouping, and no solving binomials. (MORE)

# What is the new Transforming Method to solve quadratic equation?

Solving quadratic equations by the new Transforming Method. It proceeds through 3 Steps. STEP 1. Transform the equation type ax^2 + bx + c = 0 (1) into thesimplified type x^2 + bx + a*c = 0 (2), with a = 1, and with C =a*c. STEP 2. Solve the transformed equation (2) by the Diagonal SumMethod that im…mediately obtains the 2 real roots y1, and y2. STEP 3. Divide both y1, and y2 by the coefficient a to get the 2real roots of the original equation (1): x1 = y1 /a, and x2 = y2/a. Example 1. Solve: 12x^2 + 5x - 72 = 0 (1). Solve the transformedequation: x^2 + 5x - 864 = 0. Roots have different signs (Rule ofSigns). Compose factor pairs of a*c = -864 with all first numbersbeing negative. Start composing from the middle of the factor chainto save time. Proceeding:.....(-18, 48)(-24, 36)(-32, 27). Thislast sum is -32 + 27 = -5 = -b. Then, the 2 real roots of (2) are:y1 = -32, and y2 = 27. Back to the original equation (1), the 2real roots are: x1 = y1/a = -32/12 = -8/3, and x2 = y2/a = 27/12 =9/4. Example 2. Solve 24x^2 + 59x + 36 = 0 (1). Solve the transformedequation x^2 + 59x + 864 = 0 (2). Both roots are negative. Composefactor pairs of a*c = 864 with all negative numbers. To save time,start composing from the middle of the factor chain.Proceeding:....(-18, -48)(-24, -36)(-32, -27). This last sum is -59= -b. Then 2 real roots of equation (2) are: y1 = -32 and y2 = -27.Back to the original equation (1), the 2 real roots are: x1 = y1/24= -32/24 = -4/3, and x2 = y2/24 = -27/24 = -9/8. To know how does this new method work, please read the articletitled: "Solving quadratic equations by the new TransformingMethod" in related links. (MORE)