Transcription of Laplace Transform Methods
1 CHAPTER 1 Laplace Transform MethodsLaplace Transform is a method frequently employed by applying the Laplace Transform , one can change an ordinary dif-ferential equation into an algebraic equation, as algebraic equation isgenerally easier to deal with. Another advantage of Laplace transformis in dealing the external force is either impulsive , (the force lasts avery shot time period such as the bat hits a baseball) or the force is onand off for some regular or irregular period of The Laplace TransformIff(t) is defined over interval [0, ), the Laplace Transform off,denoted as f(s),isL(f) = f(s) = 0e stf(t)dtOur first theorem states when Laplace Transform can be performed, (t)is (piecewise) continuous and there are pos-itive numbersM, asuch that|f(t)| Meatfor allt cThen f(s)is defined for alls > cThe next result shows that Laplace Transform is unique in the sensethat different continuous functions will have different Laplace f(s) = g(s)]
2 For alls > c, thenf(t) =g(t)at alltwhere both are iff(t) andg(t) are piecewise continuous (continuous exceptat finite points where left and right limits exists,) their Laplace trans-forms can be same for alls > ceven if they are different at the isolateddiscontinuous point. Since solutions of ordinary equations must becontinuous, so this is of no important Laplace Transform METHODSDue the uniqueness, we can define the inverse Laplace transformL 1asL 1( f)(t) =f(t). both f(s)and g(s)exist for alls > c, thenaf(t)+bg(t)has Laplace Transform for all constantaandband af+bg(s) =a f(s) +b g(s),for alls > cSo to find Laplace Transform of summation, we just need to findLaplace Transform of each next result shows that Laplace Transform changes derivativeinto scalar multiplication, it is this property enable Laplace transformto change ODE into algebraic that the functionf(t)is continuous andpiecewise smooth (f (t)is piecewise continuous) for allt 0and thereare constantsM, asuch that|f(t)| Meatfort T,Then f (s)is defined for alls > aand f (s) =s f(s) f(0).
3 Now if nth derivativefn(t) is piecewise smooth, then f(n)(s) =sn f(s) sn 1f(0) sn 2f (0) f(n 1)(0).For example, Whenn= 2 f(2)(s) =s2 f(s) sf(0) f (0). Whenn= 3 f(3)(s) =s3 f(s) s2f(0) sf (0) f (0). Whenn= 4 f(4)(s) =s4 f(s) s3f(0) s2f (0) sf (0) f(3)(0).The following table list most commonly used functions, andua(t) =u(t a) ={0 ift < a1 ift ais the Heaviside THE Laplace TRANSFORM3f=L 1( f) ff=L 1( f) f11s(s >0)t1s2(s >0)tn(n 0)n!sn+1(s >0)ta(a > 1) (a+1)sa+1(s >0)eat1s a(s >0)sin(kt)ks2+k2(s >0)cos(kt)ss2+k2(s >0)sinh(kt)ks2 k2(s >|k|)cosh(kt)ss2 k2(s >|k|)u(t a)eass(s >0) x(s)ifx +2x 3x=et, x(0) = 1, x (0) = Laplace Transform on both side of the equation,L(x + 2x 3x)(s) =L(et)(s)Using the linear property af+bg(s) =a f(s) +b g(s) we haveL(x + 2x 3x)(s) = x (s) + 2 x (s) 3 x(s).}
4 Together with x (s) =s2 x(s) sx(0) x (0) and x (s) =s x(s) x(0)We have, due tox(0) = 1, x (0) = 1,L(x + 2x 3x)(s) = x (s) + 2 x (s) 3 x(s)= [s2 x(s) sx(0) x (0)] + 2[s x(s) x(0)] 3 x(s)= (s2+ 2s 3) x(s) s 1So we have an algebraic equation for x(s),(s2+ 2s 3) x(s) s 1 = et(s) =1s 1 Solve this equation for x(s),(s2+ 2s 3) x(s) s 1 = et(s) =1s 1(s2+ 2s 3) x(s) =1s 1+s+ 1(s2+ 2s 3) x(s) =1s 1+(s+1)(s 1)s 1(s2+ 2s 3) x(s) =1+(s+1)(s 1)s 1(s2+ 2s 3) x(s) =1+(s2 1)s 1(s2+ 2s 3) x(s) =s2s 141. Laplace Transform METHODSwe get x(s) =s2(s 1)(s2+2s 3)a2. Using MathcadMathcad can help us in find both Laplace Transform and inverseLaplace use Mathcad to find Laplace Transform , we first enter the expres-sion of the function, then press [Shift][Ctrl][.]
5 ], in the place holder typethe key wordlaplacefollowed by comma(,) and the variable example, to find the Laplace off(t) =t2sin(at), you firs enter theexpressiont2sin(at) by typing,t^2*sin(a*t),then press [Shift][Ctrl][.], and entering keywordlaplacefollowed bycomma(,) and t, you will gett2sin(at) Laplace , t 8as2(s2+a2)3 2a(s2+a2)2If you want a simplified result, you type comma(,) after enteringthe variable name and keywordsimplify, so the following inputt^2*sin(a*t)[Shift][Ctrl][.] Laplace ,t,simplifywill produce the resultt2sin(at) Laplace , t, simplify 2a(a2 3s2)(s2+a2)3 Similarly, to use Mathcad to find inverse Laplace Transform , we firstenter the expression, then press [Shift][Ctrl][.], in the place holder typethe key wordinvlaplacefollowed by comma(,) and the variable example, to find the inverse Laplace of8as2(s2+a2)3, you firs enter theexpression8as2(s2+a2)3as,8a*s^2/(s^2+ a^2)^3,then press [Shift][Ctrl][.
6 ], and entering keywordinvlaplacefollowed bycomma(,) and s, and for simplifying the result, by another comma(,)and keywordsimplify, we get8as2(s2+a2)3invlaplace, s, simplify 1a2(sin(at) tacos(at)+t2a2sin(at))2. USING Mathcad5 When using Mathcad together with Laplace Transform to solve anODEanx(n)+an 1x(n 1)+ +a1x +a0x=f(t)we follow these steps, Step One:Apply Laplace to both sides of equation. UsingMathcad to find Laplace Transform off(t). Step Two:Using the linear property and f(n)(s) =sn f(s) sn 1f(0) sn 2f (0) f(n 1)(0).to find an algebraic equation for x(s), Step Three:Solve the equation obtained inStep Twofor x(s) and using Mathcad to find inverse Transform which willbe the solutionx(t).
7 General solution tox + 2x + 3x= we want to find general solution, we setx(0) =aandx (0) =b Step OneApply Laplace Transform to both sides of the equa-tion and find Laplace Transform (x + 2x + 3x)(s) =L(t2e2)(s)Type,t^2*e^t[Shift][Ctrl][.]l aplace,t,simplifywe findL(t2e2)(s) =2(s 1)3 Step Two:Using linear property to find an equation for x(s).L(x + 2x + 3x)(s) = ( x + 2 x + 3 x(s)=s2 x(s) sa b+ 2(s x(s) a) + 3 x(s)= (s2+ 2s+ 3) x(s) sa b 2aThe equation for x(s) is(s2+ 2s+ 3) x(s) sa b 2a=2(s 1)3 Hence, x(s) =sa+b+2a(s2+2s+3)+2(s 1)3(s2+2s+3) Step Three:Using Mathcad to find inverse Laplace trans-form andx(t),we enter,s*a+b+a/(s^2+2s+3)+2/(s-1)^3(s^2+2 s+3)[Shift][Ctrl][.] Laplace ,s61. Laplace Transform METHODSand getx(t) =ae t( 22sin( 2t) + cos( 2t))+b 22e tsin( 2t)+et(13t+29+29cos( 2t))+ 218e tsin( 2t) Solving System of can use Laplace trans-form method to solve system of differential equations.)
8 The procedureis the same as solving a higher order ODE . But, after applying Laplacetransform to each equation, we get a system of linear equations whoseunknowns are the Laplace Transform of the unknown functions. Thefollowing example shows how we can use Laplace method with Mathcadto solve system of differential solution tox = 3x 4y+t2y = x+ 5y+etx(0) = 1, y(0) = 1 Solution Step One:Apply Laplace Transform to both sides of the equa-tions and use Mathcad to find Laplace Transform oft2andet,L(x ) =L(3x 4y+t2)L(y ) =L( x+ 5y+et)andL(t2)(s) =6s3andL(et)(s) =1s 1, Step Two:Apply linear property to get a system of equationsfor x(s) and y(s),due toL(x ) =s x(s) x(0) andL(y ) =s y(s) y(0),s x(s) 1 = 3 x(s) 4 y(s) +6s3s y+ 1 = x(s) + 5 y(s) +1s 1from which we get(s 3) x(s) + 4 y(s) = 1 +6s3 x(s) + (s 5) y= 1 +1s 1 Define x(s) =[ x(s) y(s)],A=[s 3 41s 5],andb(s) =[1 +6s3 1 +1s 1],we have, in matrix form,A x(s) =b(s)3.
9 THE PIECEWISE DEFINED FUNCTIONS7 Step Three:Solve the algebraic equation and apply the in-verse Laplace Transform , using Mathcad we have x(s) =A-1b(s) =1s2 8s+ 11[(s 5)(s3+6)s3 42 ss 1(s 3)(2 s)s 1 s3+6s3]x(t) =[x(t)y(t)]= 1511t2 174121t 10621331+ e4t(37421331cosh( 5t) 23346655sinh( 5t)) et 311t2 48121t 3181331 e4t(6952662cosh( 5t) +81431331sinh( 5t)) 12et a3. The Piecewise defined FunctionsWhen solving an ODEanx(n)+an 1x(n 1)+ +a1x +a0x=f(t)with multiple defined functionf(t), we need to changef(t) into linearcombinations of step functionsua(t).Hereua(t) =u(t a) ={0 ift < a1 ift aandu(t) ={0 ift <01 ift 0is calledHeavisidefunction orunit step have graphed some step functions,Figure Functions81.}}
10 Laplace Transform METHODSIn general, ifb > a, ua(t) ub(t) = 0 if 0< t < a1 ifa < t b0 ift > bTo express a multiple defined function as linear combination of stepfunctions, we need, for each finite interval [a, b] on whichfhas adifferent expression, multiply the expression by the difference of stepfunctionsua(t) ub(t); for the expression defined on an infinite interval[c, ) we just multiply the expression byuc(t). f(t) is the summationof all the products, as illustrated in the following examples, (t) = t2 t+ 3if0< t <2etif2 t <52tsin(t)ift 5as linear combination of step (t) has three different expresses over interval[0, 2), [2, 5) and [5, ), so we have two difference,u(t) u2(t) andu2(t) u5(t),and three products, (t2 t+ 3)(u(t) u2(t)) for (t2 t+ 3) defined on [0,2); et(u2(t) u5(t)) foretdefined on [2, 5); 2tsin(t)u5(t) for 2tdefined on [5, ).]]]]]]]
