Applications of Laplace transforms

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Applications of Laplace transforms

1. Solving initial value problems: solutions of linear differential equations with constant coefficients

step 1: Perform Laplace transform on both sides of the differential equation.

step 2: The initial value problem is converted to an algebraic equation.

step3: Algebraic equation is solved.

step4: Consider the inverse Laplace transform of this solution step 5: This is the solution of the initial value problem.

Example 1:

Solve the initial value problem which is a first order differential equation

y^'(t)+5y(t)=t ; y(t=0)=2;

Considering Laplace transform on both the sides and applying the formula for Laplace transform of a derivative , we obtain

sY(s)−2+5Y(s)=1

s² ... step 1 ( ^s⁺⁵^¿^Y⁽^s⁾⁼_s¹²⁺²

^Y⁽^s⁾⁼₍_s+5¹_)s²⁺_s₊₅² ... step 2

We solve for ^Y⁽^s) using partial fractions. ₍_s+5¹_)s²⁼^A_s ⁺_s^B²⁺_s+^C₅

A s(s+5)+B(s+5)+C s²=1 ; putting ^s=0 , ^B⁼¹₅ ; putting ^s=−5,^C=₂₅¹ ; putting ^s=1, ⁶^A⁺⁶₅⁺₂₅¹ ⁼¹

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or ^A⁼⁻¹₂₅ ; Substituting the values of ^{A , B , C}

Y(s)=−1 25s+ 1

5s²+ 1

25(s+1) ... step3

y(t)=L⁻¹[Y(s)]

y(t)=−1 25 +1

5t+ 1

25e⁻⁵^t ...step4 Example 2:

We now solve } left (t right ) +2 {y} ^ {'} left (t right ) +6= {t} ^ {2

y^¿

y(t=0)=2; y^'(t=0)=0

} left (t right ) + {2y} ^ {'} left (t right ) +6 right rbrace = L\{ {t} ^ {2} \ y^¿

L¿ s²Y(s)−sy(t=0)−y^'(t=0)+2sY(s)−4+6

s=2 s³ Y(s)(s²+2s)=2s+4−6

s+2 s³ Y(s)=2s+4

s²+2s− 6

s(s²+2s)+ 2 s³(s²+2s) y(t)=2−(3t+3e⁻²^t

2 −3

2) +1

8 e⁻²^t+t 4−t²

4+t³ 6−1

8

The significance of Laplace transform for solving differential equations is that it can be used

a) For discontinuous forcing functions while the conventional method cannot be used for solving such differential equations.

b) Initial conditions are used as a part of the transformation from differential to algebraic equation unlike the conventional method which uses these initial conditions for evaluating the constants for complete solution of the differential equations. For example consider the conventional simple harmonic motion with damping. The differential equation representing it is

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md²x dt² +bdx

dt+kx=f(t)

With ^{m ,b , k} respectively as the mass ,damping constant and the force constant. If the forcing term ^f^(t⁾ is not a continuous function, it is still possible to solve this differential equation using Laplace transforms.

Example 3:

Solve the following initial value problem using Laplace transform.

d²x

dt² +4x=sin 3t , x(0) ¿x^'(0)=0 ; ^x^'⁽⁰⁾⁼^dx_dt

L

{

^ddt²^x² +4x

}

⁼^L^{{sin 3}^t^}

s²X(s)+4X(s)−sx(0)−¿ x^'(0)= 3

s²+9 ;

s

(s²+4)(¿ ¿2+9) X(s)=3

¿

;

Thus

s

(s²+4)(¿ ¿2+9) 2

¿ x(t)=L⁻¹¿

; Using partial fraction decomposition we obtain

x(t)= 3

10sin 2t−1 5sin 3t

Physical significance: The above differential equation represents the motion of a simple harmonic motion with an external driving force. It could represent a spring mass system which is subjected to an external force ^{sin 3}^t . The resultant displacement is represented by ^x^(t⁾ which represents a motion with natural frequency ( ^ω=2^¿ and the forcing term ( ^ω=3^¿^.

Example 4:

We next a differential equation for a spring mass system with damping.

x^''−3x^'+2x=0 ; ^x⁽⁰⁾⁼⁰^{; x}^'⁽⁰⁾⁼⁰^;

L{x^{' '}−3x^'+2x}=s²X(s)−1−3sX(s)+2X(s)=0

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X(s)= 1

(s²−3s+2)= 1

(s−2)− 1

s−1 ; ^x⁽^t^)=e²^t^−e^t Example 5:

This example is that of a series LCR circuit with an external source

V(t) and ^{L ,} ^R , ^C as the inductance , resistance and capacitance respectively. The instantaneous charge and the current are ^q^(t⁾ and ^i(t⁾ respectively. At ^t^=0, charge on the capacitor and the current in the circuit is ^0.

Ldi

dt+R i+q

C=V

since ^dq_dt⁼ⁱ , we obtain ^L^d_dt²^q2 +Rdq dt+ q

C=V

Considering the Laplace transform on both the sides we obtain ,

L

[

^s²^Q⁽^s⁾^−sq⁽⁰⁾⁻^dq^dt ^t=0

]

⁺^¿ ^R^[^sQ⁽^s)−q(⁰⁾^]⁺^Q^C^(s⁾⁼^V^s

Assigning values to the constants, ^R=10^{ohms , L=1}^{H ,C}^=.01^F ,

V=220V , we obtain

s²Q(s)+10s Q(s)+100Q(s)=220/s ; Solving for ^Q⁽^s⁾ ,

Q(s)= 220

s(s²+10s+100) ;

q(t) ¿11/5−(11∗(cos(5∗3^(1/2⁾∗t)+(3^(1/2)∗sin(5∗3^(1/2)∗t))/3))/(5∗exp(5∗t))

2. Application of Laplace transform for solving simultaneous differential equations.

This finds applications in coupled circuits like those of two inductive coils with mutual inductance between them. Determination of currents in such circuits involves solving simultaneous differential equations.The second application is coupled oscillators like coupled spring mass systems.

x^'=y+3x (1)

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y^'=4 y−x (2)

x(t=0)=1 ; ;

y(t=0) =0

Taking Laplace transform of (1) , we obtain

L{x^'(t)}⁼L{ y+3x} or ^sX⁽^s^)−1=Y⁽^s⁾⁺³^X^(s) or

X(s)[s−3]−Y(s)=1 (3)

L{y^'(t)}⁼L{4y−x} or ^sY⁽^s⁾⁻⁰⁼⁴^Y⁽^s)−^X⁽^s)

X(s)+Y(s){s−4}=0 (4) substituting for ^Y⁽^s⁾ from eq. (3) in eq. (4) ,we obtain

X(s)+{s−4}{ X(s)[s−3] -1} ^¿⁰

X(s){s²−7s+13}=s−4 or

X(s)= s−4

s²−7s+13 ;

x(t)=¿ exp((7t)/2)(cos((3^(1/2)t)/2) - (3^(1/2)sin((3^(1/2)t)/2))/3) 2. We apply Laplace transform for solving the family of differential equations for successive disintegration in radioactivity. Let ^N^A⁽^t⁾ be the parent nucleus which disintegrates into the daughter nucleus

N_B(t) which further disintegrates into the stable nucleus ^N^c^(t⁾ .

N_A→ N_B→ N_C . Determine the number of nuclides ^N^A^{, N}^B and ^N^C at any instant of time ^{t ,} given the initial number of nuclides are

N_A=N₀; N_B=0; N_C=0

d N_A

dt =¿ ⁻^λ^A^N^A (5)

d N_B

dt =−λ_BN_B+λ_AN_A (6) ^{d N}_dt^C⁼^λBN_B

Taking Laplace transform of eq(5) , we obtain

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s L{^NA}⁻^N0=−L{^λAN_A} ; ^(s⁺^λ)^L^{^N^A^} ^¿^N⁰^; ^L^{^N^A^}⁼_(s^N₊⁰_λ)

N_A=N₀e^−λ^A^t

Considering the Laplace transform of eq.(6) we obtain

s L{^NB}⁻⁰⁼^λAL{N_A}−λ_BL{N_B} ; ^L^{^N^B^}^=¿ ₍_S+_λ^λ^A^N⁰

B)⁽S+λA)

L{^NB}⁼^λAN₀ 1

λ_A−λ_B

[

^S⁺¹^λ^A⁻^S+¹^λ^B

]

^; ^N^B⁼^λ^λA^A−^Nλ⁰_B[e⁻^λ^B^t−e^−λ^A^t] We repeat the process for eq(3) and obtain ^N^C .

s L{N_C}−0=λ_AN₀ λ_B

λ_A−λ_B[e^−λ^B^t−e^−λ^A^t] ; Let ^k⁼^λ^A^N⁰ _λ ^λ^B

A−λ_B ;

L{N_C}=k

s [e^−λ^B^t−e^−λ^A^t]; N_C=k

[

^∫⁰^t ê^−λ^B^v^dv+^∫⁰^t ê^−λÂ^v^dv

]

Using ^L⁻¹

{

^F⁽s^s)

}

⁼^∫₀^t ^f^(v^)dv ^;

Thus ^N^C^=k

[

^λ¹^B⁻ê^−λ^λ^B^B^t⁺ê^−λ^λÂÂ^t⁻^λ¹Â

]

3.

Application of Laplace transform for solving partial differential equations

a) Solution of heat flow along semi infinite bar The heat diffusion equation is

¿

∇²u¿ r ^{, t}^¿=^cρ_k ^∂u_{∂ t}^{(r ,t}⁾ with _u^¿_¿ r ^{, t}^¿^{, c ,} ^{ρ , k} respectively as the temperature distribution in the material, specific heat capacity, density of the material and its thermal conductivity. This equation assumes that there are no external heat sources.

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For a semi infinite bar , ^x>⁰ , we consider the 1-d form of the heat diffusion equation. Let ^u⁽^{x , t}⁾ represent the temperature at a distance

x from one end of the bar at time ^t . The boundary condition is that ^u⁽⁰^{, t}^)=u⁰ and the initial condition is ^u⁽^{x ,}⁰⁾⁼⁰

As it is a semi infinite bar, we solve this problem using Laplace transform ( ^x>⁰^¿ . For an infinite bar , we employ Fourier transform.

Let _cρ^k ^=α , the thermal diffusivity , We consider the Laplace transform on both the sides of the heat diffusion equation

L

{

^α ^∂²^u(∂ x^{x ,t}² ⁾

}

⁼^L^{^{∂ u(}^∂t^{x ,t}⁾^} ^or

α∫

0

∞ ∂²u(x , t)

∂ x² e^−stdt=sU(x , s)−u(x ,0) with ^U⁽^{x , s}⁾ as the Laplace transform of ^u^{(x ,t}⁾

The LHS integration is with respect to ^t while the double differentiation is with respect to ^x .

α∫

0

∞ ∂²u(x , t)

∂ x² e^−stdt=α ∂²

∂ x²∫

0

∞

u(x , t)e^−stdt=α∂²U(x , s)

∂ x²

Using the initial condition ^u⁽^{x ,}⁰⁾⁼⁰ , we obtain

α ∂²U(x , s)

∂ x² =sU(x , s) , let _α^s⁼^p²

U(x , s)=A e^px+B e⁻^px with ^A and ^B as constants to be evaluated.

u(x , t)=0 as ^{x → ∞} for the boundedness of the temperature. Thus

U(x , s) = 0 as ^{x → ∞} which implies that ^A⁼⁰

Since û⁽⁰^{, t}^)=u⁰ therefore Û⁽⁰^{, s)=}^L^{û0}⁼û_s⁰ with û⁰ as the constant temperature at ^x=0 .

u₀ s =B

and the complete solution is Û⁽^{x , s}⁾⁼û_s⁰ê^−px = û_s⁰ê^−x^√^s^/∝

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The solution ^u⁽^{x , t}⁾ is the Laplace inverse of ^U⁽^{x , s}⁾^.

x

√⁴^αt

1−erf ⁡¿ L⁻¹{u0

s e^−px}=u₀¿

We first discuss the Laplace inverse of the function ^e⁻^√^s/∝^/^s Using the property of the Laplace transform

If L f (v)=F(s), L

{

^∫⁰^t ^f^(v⁾^dv

}

⁼^F⁽^s^s) ^or

L⁻¹

{

^F^s^(s⁾

}

^=¿ ^∫0 t

f(v)dv , ^F⁽_s^s)^=e⁻^√^s^/∝^/^s , we obtain from eqn(4)

∫0 t v

−3 2 e

−1 4v

2√^π ^dv , substituting ^v^=1/⁴^y² we get

∫

1/2√t

∞ 2

√^π ^e

−y²dy = erfc ( ^1/2√^t^¿ We next consider ^L⁻¹^{^e^−x_s^√^s^}

Using the property that ^L

{

^f

⁽

^a^t

⁾ }

⁼âF(âs) and putting â=^x²

L⁻¹

{

^e^{s x}⁻^√^x²²^s

}

^=¿ ^x¹²^erfc

(

²

_√

¹^x^t²

)

L⁻¹

{

^e^−x^s^√^s

}

^=erfc

(

2^x√t

)

Example 2:

Consider the 1-d wave equation,

∂²u(x , t)

∂ x² =1 v²

∂²u(x , t)

∂ t²

u(x , t) is the displacement of the string form the equilibrium position at position ^x at time ^t , ^v represents the velocity of the wave.

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We consider the Laplace transform of the wave equation to obtain

∫0

∞ ∂²u(x , t)

∂ x² e^−stdt ¿ ∂²

∂ x²∫

0

∞

u(x , t) e^−stdt as the integration is only over

t so _{∂ x}^∂²2 can be outside the integral. Secondly after integration, the dependence on ^t vanishes and _{∂ x}^∂²2 can be converted into an ordinary differential.

L

{

^∂²^u^{∂ x}^{(x , t}² ⁾

}

⁼^d²^U^{d x}⁽^{x , s)}²

Example 3:

Determine the ^L

{

^∂u∂ t⁽^{x , t}⁾

}

By definition , ^L

{

^∂u∂ t⁽^{x , t}⁾

}

⁼^∫^∞₀ ^∂u∂ t⁽^{x , t}⁾e^−st dt

integrating by parts [^u(^{x , t}⁾^e^−st]0

∞+s∫

0

∞

u(x , t)e^−stdt Assuming that ^u^{(x ,t}⁾ is bounded at ^{t → ∞} ,

L

{

^∂u∂ t⁽^{x , t)}

}

⁼^sU⁽^{x , s}^)−u⁽^{x ,}⁰⁾ ( 7)

L

{

^∂∂ t²^u²

}

can be determined by considering ^∂u_{∂ t}⁼^f^(t⁾ and using eq.( ) ^L

{

^∂∂ t²^u²

}

⁼^{s F}⁽^s)−f⁽^{x ,0}^)=s^[^sU⁽^{x , s}^)−u(^{x ,}⁰⁾^]⁻^∂u^{∂ t}t=0

L

{

^∂∂ t²^u²

}

⁼^s²^U⁽^{x , s}⁾⁻^su⁽^{x ,0}⁾⁻^∂u^{∂ t}t=0

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Exercises

1. Solve the wave equation for a string fixed at both ends

x=0∧x=1 . ^u⁽⁰^{, t)=}⁰^,u⁽¹^{, t}⁾⁼⁰^;u⁽^{x ,}^{0)=sin 4}^{πx , u}^t⁽^{x ,}⁰⁾⁼⁰ with ^u^t⁼^∂u_{∂ t}

2. Consider 2 masses ^m¹ and ^m² which are coupled by a massless spring, of spring constant ^k . Solve the simultaneous differential equations and hence determine the displacement of the masses at any instant of time ^t . Initial conditions are

x₁=x₂=0;d x₁

dt =u;d x₂ dt =0

3. Solve the following simultaneous differential equation.

d x

dt =4x−y ;dy

dt = ^x+⁴^y

4. In a series L C R circuit , at ^t⁼⁰ a voltage ^V=20 sin 400t volts is suddenly switched on. Assume the values of L ^¿¹^{H ,} R= 1000ohms and C ^¿^6.25micro Farads . Determine the value of current ⁱ⁽^t⁾ and charge ^q^(t⁾ at any instant of time.

5. In an open circuit with L ^¿¹⁰^{mH ;} R ^¿²⁵⁰^Ω , C ^¿¹^μF ,at ^t⁼⁰ , the capacitor has a charge of ¹⁰⁻⁵^coulombs . At this instant the circuit is closed. Determine the charge ^Q (t) .

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6. Solve the second order differential equation

d²y

d t²+9y=9U(t−3) , ^y⁽⁰⁾⁼^dy_dt

t=0

=0

7. Using Laplace transform , solve

∂u(x , t)

∂ t =∂²u(x , t)

∂ x² , ^u⁽^{x ,}^{0)=3 sin2}^πx , ^u⁽⁰^{, t)=0,u}⁽¹^{, t)=0} (D.U.)

0<x<1;t>0

8. Solve the simultaneous differential equations using Laplace transforms.

dx

dt + ^y=0^; ^dy_dt⁻^x=0^; ^x⁽⁰⁾⁼¹^{, y}⁽⁰⁾⁼⁰ (D.U.) 9. Solve using Laplace transform

dx

dt=−x−y ; ^dy_dt⁼⁻⁴^x−^y ; ^x⁽^t⁼⁰⁾⁼¹^{; y}⁽^t=0⁾⁼⁰