Variation of Parameters and Duhamel’s Principle Date: Feb 26, 2002 Last Revision: Feb 26, 2002 Maple 6Bent E. [email protected]@math.orst.eduCourse: Mth 256Term: Winter 2002File name: 256w2002_varparm_duhamel.mws> restart;In this worksheet we illustrate Duhamel’s principle and variation of parameters. We use Maple to ease the calculations though, of course, if one is going to use Maple there is generally little reason to do variation of parameters. Our reason here is pedagogical.We consider only second order linear differential equations. Corresponding results hold for higher order.Consider a linear inhomogeneous ordinary differential equation of the form> ode00:=diff(y(t),t,t)+p(t)*diff(y(t),t)+q(t)*y(t)=g(t); := ode00 = + + ∂∂2t2()y t ()p t∂∂t()y t ()q t ()y t ()g tThe method of variation of parameters gives us a particular solution of ode00 in the form> y(t)=Int(K(t,s)*g(s),s=a..t); = ()y t d⌠⌡at()K,ts ()g sswhere> K(t,s)=(y1(s)*y2(t)-y1(t)*y2(s))/(y1(s)*D(y2)(s)-D(y1)(s)*y2(s)); = ()K,ts − ()y1 s ()y2 t ()y1 t ()y2 s − ()y1 s ()()D y2 s ()()D y1 s ()y2 sand y1, y2 is a fundamental solution set for the associated homogeneous equation.For each s we see K(t,s) is a linear combination of y1(t) and y2(t) and so is a solution of the associated linear homogeneous ordinary differential equation> ode00h:=lhs(ode00)=0; := ode00h = + + ∂∂2t2()y t ()p t∂∂t()y t ()q t ()y t 0Moreover, it is clearly the solution with initial values> y(s)=0,D(y)(s)=1;, = ()y s 0 = ()()D ys 1This characterization of K(t,s) as the solution (for each s) of an initial value problem for a homogeneous linear ordinary differential equation, and the corresponding result for higher order equations, is known as Duhamel’s principle. Note while K(t,s) is defined in terms of a fundamental solution set, Duhamel’s principle shows us that it doesn’t actually depend on the choice of fundamental solution set.Duhamel’s principle has a physical interpretation. It also has theoretical applications.Here then is a procedure to calculate K(t,s):> duhamel:=proc(ode,y,t,s)> local odeh,inith,solnh,K;> odeh:=lhs(ode)=0;> inith:=y(s)=0,D(y)(s)=1;> K:=rhs(dsolve({odeh,inith},y(t)));> end:Note this procedure is not very robust. We make no error checking and we make strong assumptions about the way our differential equation is written.Let’s look at some examples:Example 1> ode01:=(x-1)*diff(y(x),x,x)-x*diff(y(x),x)+y(x)=(x-1)^2/x; := ode01 = − + () − x 1∂∂2x2()y xx∂∂x()y x ()y x() − x 12x> K01:=duhamel(ode01,y,x,s); := K01 − + x − s 1sexes() − s 1> g01:=subs(x=s,rhs(ode01)/(x-1)); := g01 − s 1s> int(K01*g01,s): y01:=simplify(subs(s=x,%)); := y01 − − x ()ln x 1Let’s compare our result with Maple’s solution of ode01.> dsolve(ode01,y(x)); = ()y x − − + + x ()ln x 1 _C1 x _C2 exSetting the parameters _C1 and _C2 to 0 we a particular solution and it is the same as the one we found by Duhamel’s principle.Example 02> ode02:=diff(y(x),x,x)-y(x)=exp(2*x)/(1+exp(x)); := ode02 = − ∂∂2x2()y x ()y xe()2 x + 1 ex> K02:=duhamel(ode02,y,x,s); := K02 − 12exes12ese()−x> g02:=subs(x=s,rhs(ode02)); := g02e()2 s + 1 es> int(K02*g02,s): y02:=simplify(subs(s=x,%)); := y02 − + − 12ex()ln + 1 ex14ex1212e()−x()ln + 1 exAgain let’s compare our result with Maple’s solution of ode02.> dsolve(ode02,y(x)); = ()y x − + − + + 12ex()ln + 1 ex14ex1212e()−x()ln + 1 ex_C1 ex_C2 e()−xExample 03> ode03:=diff(y(t),t,t)+y(t)=tan(t); := ode03 = + ∂∂2t2()y t ()y t ()tan t> K03:=duhamel(ode03,y,t,s); := K03 − ()cos s ()sin t ()sin s ()cos t> g03:=subs(t=s,rhs(ode03)); := g03 ()tan s> int(K03*g03,s): y03:=simplify(subs(s=t,%)); := y03 − ()cos tln + 1()sin t()cos tAgain let’s compare our result with Maple’s solution of ode03.> dsolve(ode03,y(t)); = ()y t − + + ()cos t ()ln + ()sec t ()tan t_C1()sin t_C2()cos tAgain we have agreement.Example 04> ode04:=x^2*diff(y(x),x,x)+x*diff(y(x),x)+(x^2-1/4)*y(x)=x^(5/2)*cos(x); := ode04 = + + x2∂∂2x2()y xx∂∂x()y x − x214()y xx() / 52()cos x> K04:=duhamel(ode04,y,x,s); := K04 − s ()cos s ()sin xxs ()sin s ()cos xx> g04:=subs(x=s,rhs(ode04)/x^2); := g04 s ()cos s> int(K04*g04,s): y04:=simplify(subs(s=x,%)); := y0414x () + ()cos x ()sin xxAgain let’s compare our result with Maple’s solution of ode04.> dsolve(ode04,y(x));= ()y x + + + _C1 ()cos xx_C2 ()sin xx14x ()cos x14x() / 32()sin xAgreement