Solving PDEs using Laplace Transforms, Chapter 15

1 Solving PDEs using Laplace Transforms, Chapter 15 Given a functionu(x,t) defined for allt >0 and assumed to be bounded we can apply theLaplace transform intconsideringxas a (u(x,t)) = 0e stu(x,t)dt U(x,s)In applications to PDEs we need the following:L(ut(x,t) = 0e stut(x,t)dt=e stu(x,t) 0+s 0e stu(x,t)dt=sU(x,s) u(x,0)so we haveL(ut(x,t) =sU(x,s) u(x,0)In exactly the same way we obtainL(utt(x,t) =s2U(x,s) su(x,0) ut(x,0).We also need the corresponding transforms of thexderivatives:L(ux(x,t)) = 0e stux(x,t)dt=Ux(x,s)L(uxx(x,t)) = 0e stuxx(x,t)dt=Uxx(x,s)Consider the following 1. u x+ u t=x, x >0, t >0,with boundary and initial conditionu(0,t) = 0t >0,andu(x,0) = 0, x > above we use the notationU(x,s) =L(u(x,t))(s) for the Laplace transform applying the Laplace transform to this equation we havedUdx(x,s) +sU(x,s) u(x,0) =xs dUdx(x,s) +sU(x,s) = is a constant coefficient first order ODE. We solve it by finding the integrating factor =eRsdx=esxThus we haveddx[esxU(x,s)] = integrate both sides to getU(x,s) =e sxs( esrr dr)+Ce can use integration by parts to evaluate the integral: esxxdx= (esxs) xdx=xesxs (esxs)dxxesxs we haveU(x,s) =e sxs(xesxs esxs2)+Ce sx=xs2 1s3+Ce can evaluate the constantCusing the boundary condition0 =U(0,s) = 1s3+C C=1s3so we haveU(x,s) =xs2 1s3+e the inverse Laplace transform we haveu(x,t) =xt t22+H(t x)(t x)22whereHis the unit step function (or Heaviside function)H(x) ={0, x <01, x 2.)))}

2 U x+ u t+u= 0, x >0, t >0,with boundary and initial conditionu(0,t) = 0t >0,andu(x,0) = sin(x), x > above we use the notationU(x,s) =L(u(x,t))(s) for the Laplace transform applying the Laplace transform to this equation we havedUdx(x,s) +sU(x,s) u(x,0) +U(x,s) = 0 dUdx(x,s) + (s+ 1)U(x,s) = sin(x).This is a constant coefficient first order linear ODE. We solve it by finding the integrating factor =eR(s+1)dx=e(s+1)xThus we haveddx[e(s+1)xU(x,s)]=e(s+1)xsin(x).We integrate both sides to getU(x,s) =e (s+1)x( e(s+1)rsin(r)dr)+Ce (s+1) can use integration by parts to evaluate the integral:e (s+1)x( x0e(s+1)rsin(r)dr)=(s+ 1) sin(x) cos(x)s2+ 2s+ we haveU(x,s) =(s+ 1) sin(x) cos(x)s2+ 2s+ 2+Ce (s+1) can evaluate the constantCusing the boundary condition0 =U(0,s) = 1s2+ 2s+ 2+C C=1s2+ 2s+ we haveU(x,s) =(s+ 1) sin(x) cos(x) +e (s+1)x)s2+ 2s+ the inverse Laplace transform we haveu(x,t) =e tcos(t) sin(x) e tsin(t) cos(t) +e tH(t x) sin(t x)This can be written asu(x,t) =e t[sin(x t) +H(t x) sin(t x)].

3 Example 3. u t(x,t) = 2u x2(x,t),0< x <2, t >0,u(0,t) = 0, u(2,t) = 0u(x,0) = 3 sin(2 x).Take the Laplace transform and apply the initial conditiond2 Udx2(x,s) =sU(x,s) u(x,0) =sU(x,s) 3 sin(2 x).We write this equation as a non-homogeneous, second order linear constant coefficient equation forwhich we can apply the methods from Math (x,s) sU(x,s) = 3 sin(2 x).The general solution can be written asU(x,s) =Uh(x,s) +Up(x,s)whereUh(x,s) is the general solution of the homogeneous problemUh(x,s) =c1e sx+c2e sxandUp(x,s) is any particular solution of the non-homogeneous problemUp(x,s) =Acos(2 x) +Bsin(2 x).3We first use the method of undetermined coefficients to findAandB. To this end we haveddxUp(x,s) = 2 Asin(2 x) + 2 Bcos(2 x),d2dx2Up(x,s) = (2 )2 Acos(2 x) + (2 )2 Bsin(2 x).Therefored2dx2Up(x,s) sUp(x,s)= ( (2 )2 s)[Acos(2 x) +Bsin(2 x)]= 3 sin(2 x).From this we conclude that (s+ (2 )2)A= 0,and (s+ (2 )2)B= 3,so thatA= 0, B=3s+ 4 we have the general solutionU(x,s) =c1e sx+c2e sx+3(s+ 4 2)sin(2 x)We note the the Laplace transforms of the boundary conditions giveu(0,t) = 0 U(0,s) = 0,andu(2,t) = 0 U(2,s) = 0So we have0 =U(0,s) =c1+c2,0 =U(2,s) =c1e s2+c2e s2which givesc1=c2= 0 and we haveU(x,s) =3(s+ 4 2)sin(2 x).

4 To find our solution we apply the inverse Laplace transformu(x,t) =L 1(3(s+ 4 2)sin(2 x))= 3e 4 2tsin(2 x).Just as we would have obtained using eigenfunction expansion we consider a similar problem for the 1D wave equation. 2u t2(x,t) =c2 2u x2(x,t) + sin( x),0< x <1, t >0,u(x,0) = 0, ut(x,0) = 0u(0,t) = 0u(1,t) = the Laplace transform and applying the initial conditions we obtaind2 Udx2(x,s) =s2U(x,s) su(x,0) ut(x,0) sin( x)s=s2U(x,s) sin( x) need to solve the constant coefficient non-homogeneous ODEd2 Udx2(x,s) s2U(x,s) = sin( x)sOnce again we know thatU(x,s) =Uh(x,s) +Up(x,s)whereUh(x,s) is the general solution of the homogeneous problemUh(x,s) =c1esx+c2e sxandUp(x,s) is any particular solution of the non-homogeneous problemUp(x,s) =Acos( x) +Bsin( x).We apply the method of undetermined coefficients to findAandB. To this end we haveddxUp(x,s) = Asin( x) + Bcos( x),d2dx2Up(x,s) = 2 Acos( x) + 2 Bsin( x).Therefored2dx2Up(x,s) s2Up(x,s)= ( 2 s2)[Acos( x) +Bsin( x)]= sin( x) this we conclude that (s2+ 2)A= 0,and (s2+ 2)B= 1s,so thatA= 0, B=1s(s2+ 2).

5 So we haveUp(x,s) =sin( x)s(s2+ 2)andU(x,s) =c1esx+c2e sx+sin( x)s(s2+ 2).Next we apply the BCs to =U(0,s) =c1+c2,and 0 =U(1,s) =c1es+c2e swhich impliesc1= 0 andc2= 0. So we arrive atU(x,s) =sin( x)s(s2+ 2).5 Finally we apply the inverse Laplace transform to obtainu(x,t) =L 1(U(x,s)) =L 1(1s(s2+ 2))sin( x)=1 2L 1(1s s(s2+ 2))sin( x)=1 2(1 cos( t)) sin( x).Here we have done partial fractions1s(s2+ 2)=as+bs+c(s2+ 2)=1 2(1s s(s2+ 2)).Example example shows the real use of Laplace transforms in Solving a problem we couldnot have solved with our earlier work. u t(x,t) = 2u x2(x,t), < x < , t >0,u(x,0) =f(x)u(x,t) the assumption thatu(x,t) is bounded we know that the Laplace transform exists and,indeed, we have|u(x,t)| M |U(x,s)| 0e st|u(x,t)|dt M 0e stdt= the Laplace transform we obtaind2 Udx2(x,s) =sU(x,s) u(x,0) =sU(x,s) f(x).We write this equation as a non-homogeneous, second order linear constant coefficient (x,s) sU(x,s) = f(x).The general solution can be written asU(x,s) =Uh(x,s) +Up(x,s)whereUh(x,s) is the general solution of the homogeneous problemUh(x,s) =c1e sx+c2e sxandUp(x,s) is any particular solution of the non-homogeneous problem.

6 We find it using themethod of variation of parameters from Math 3354. For this method we useU1=e sx,U2=e (U1,U2) = U1(x,s)U2(x,s)U 1(x,s)U 2(x,s) = 2 s6Up(x,s) = x0[ U1(x,s)U2( ,s) +U2(x,s)U1( ,s])( f( ))W( ,s)d =12 s x0[ e sxe s +e sxe s ]f( )d = e sx2 s x0e s f( )d +e sx2 s x0e s f( )d So the general solution can be written asU(x,s) =(c1 12 s x0e s f( )d )e sx+(c2+12 s x0e s f( )d )e our assumption thatu(x,t) be bounded for all < x < implies thatU(x,s) is alsobounded for all < x < for any fixeds > in order that the first term in the general solution stays bounded asx we needlimx (c1 12 s x0e s f( )d )= 0which impliesc1=12 s 0e s f( )d .In exactly the same way we must havelimx (c2+12 s x0e s f( )d )= 0which impliesc2= 12 s 0e s f( )d =12 s 0 e s f( )d .ThusU(x,s) =(12 s 0e s f( )d 12 s x0e s f( )d )e sx+(12 s 0 e s f( )d +12 s x0e s f( )d )e sx=(e sx2 s xe s f( )d )+(e sx2 s x e s f( )d )=12 s e s|x |f( )d We want to find the inverse Laplace transformL 1(e s|x |2 s).

7 7 From our table we haveL 1(e a s s)=e a2/(4t) tand if we seta=|x |then we haveL 1(e s|x |2 s)=e |x |2/(4t) 4 t K(|x |,t).So we haveu(x,t) =L 1(U(x,s)) =L 1(12 s e s|x |f( )d )= L 1(e s|x |2 s)f( )d =1 4 t e |x |2/(4t)f( )d = K(|x |,t)f( )d The functionK(x,t) =e x2/(4t) 4 tis called the Fundamental Heat Kernel .8 Table of Laplace Transformsf(t) fort 0 f=L(f) = 0e stf(t)dt11seat1s atnn!sn+1(n= 0,1,..)ta (a+ 1)sa+1(a >0)sinbtbs2+b2cosbtss2+b2sinhbtas2 b2coshbtss2 b2f (t)sL(f) f(0)f (t)s2L(f) sf(0) f (0)tnf(t)( 1)ndnFdsn(s)eatf(t)L(f)(s a)u(t a) = 0t a1t > ae assu(t a)f(t a)e asL(f)(s)u(t a)g(t)e asL(g(t+a))(s) (t a)e as(f g)(t) = t0f(t )g( )d L(f g) =L(f)L(g)9 The error function denoted by erf(x) is given byerf(x) =2 x0e that we can use the properties of integrals to deduce thaterf( x) = erf(x).The complementary error function erfc(x) defined byerfc (x) =2 xe thaterf(x) + erfc(x) =2 ( x0e x2dx+ xe s2ds)= Laplace Transformse a2/(4t) te a s sae a2/(4t)2 t3e a serf (t)es2/4erfc(s/2)serfc(a2 t)e a ss2 t e a2/(4t) a{erfc(a2 t)}e a ss seb2t+ab{erfc(b t+a2 t)}e a s s( s+b) eb2t+ab{erfc(b t+a2 t)}+ erfc(a2 t)be a ss( s+b)10