image equation


In solving an initial value problemMathworldPlanetmathPlanetmath leading to an ordinary differential equationMathworldPlanetmath, the Laplace transformMathworldPlanetmath offers often a way to simplify the equation: both sides are Laplace transformed.  The transformed equation, the so-called image equation, is in many cases simplier than the original differential equation, since it does not contain the derivatives of the unknown function y(t).  From the image equation one may solve the Laplace transform Y(s) of y(t) and then inverse transform Y(s) getting y(t).

Let’s consider e.g. the ordinary n’th order linear differential equation

a0dnydt+a1dn-1ydxn-1++an-1dydt+any(t)=f(t) (1)

subject to the initial conditions

y(0)=y0,y(0)=y0,,yn-1(0)=y0(n-1). (2)

Due to the linearity of the Laplace transform the image equation of (1) is

a0{dnydxn}+a1{dn-1ydxn-1}++an{y(t)}={f(t)}. (3)

Denote  {y(t)}=:Y(s)  and  {f(t)}=:F(s).  We put into (3) the expressions of the Laplace transforms of the derivatives on the left hand side (see “Laplace transforms of derivatives (http://planetmath.org/LaplaceTransformsOfDerivatives)”) getting

a0[snY(s)-(sn-1y0+sn-2y0++y0(n-1))]
+a1[sn-1Y(s)-(sn-2y0+sn-3y0++y0(n-2))]
+      
+an-1[sY(s)-(y0)]
=F(s).

This equation is simplified to

(a0sn+a1sn-1++an-1s+an)Y(s)=
a0[y0sn-1+y0sn-2++y0(n-1)]+
+a1[y0sn-2+y0sn-3++y0(n-2)]+
+      +
+an-2[y0s+y0]+an-1[y0]+F(s).

For brevity, denote in the last equation the polynomial multiplier of Y(s) by φ(s) and the sum preceding F(s) by ψ(s).  Then the equation can be written as

φ(s)Y(s)=ψ(s)+F(s),

i.e.

Y(s)=ψ(s)φ(s)+F(s)φ(s). (4)

The function Y(s) defined by (4) is the Laplace transform of the solution y(t) of the differential equation (1) which satisfies the initial conditions (2).  If we now find a function y*(t) the Laplace transform of which is the function Y(s) defined by (4), then y*(t) will do for y(t) due to the uniqueness property of Laplace transform expressed in the entry “Mellin’s inverse formula (http://planetmath.org/MellinsInverseFormula)”.
If we seek the solution of (1) satisfying the zero initial conditions

x0=x0=x0′′==x0(n-1)=0,

then  ψ(s)0  and

Y(s)=F(s)φ(s),

i.e.

Y(s)=F(s)a0sn+a1sn-1++an.

Example.  The 4’th order differential equation

y′′′′(t)+y(t)= 0 (5)

should be solved with the initial conditions

y(0)=y′′′(0)=1,y(0)=y′′(0)=0.

The image equation of (5) is

s4Y(s)-s3y(0)-s2y(0)-sy′′(0)-y′′′(0)+Y(s)= 0,

i.e.

(s4+1)Y(s)=s3+1.

Thus one needs to determine the inverse Laplace transform of

Y(s)=144s3s4+1+1s4+1. (6)

The zeroes of the numerator s4+1 are the eighth roots of unity eiπ4, ei3π4, ei5π4, ei7π4, in other words the complex numbersPlanetmathPlanetmath ±1±i2.  By the special case (3) of the Heaviside formula, the first addend of (6) corresponds the original function

14±e±1±i2t=et2+e-t22eit2+e-it22=cosht2cost2.

Utilizing also the general Heaviside formula (http://planetmath.org/HeavisideFormula) (1), one can get from (6) the result

y(t):=cosht2cost2+12(cosht2sint2-sinht2cost2).

References

  • 1 N. Piskunov: Diferentsiaal- ja integraalarvutus kõrgematele tehnilistele õppeasutustele. Teine köide. Viies trükk.  Kirjastus Valgus, Tallinn (1966).

Title image equation
Canonical name ImageEquation
Date of creation 2014-03-20 20:16:56
Last modified on 2014-03-20 20:16:56
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 13
Author pahio (2872)
Entry type Topic
Classification msc 34A05
Classification msc 44A10