Solving PDEs using Laplace Transforms, Chapter 15

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Solving PDEs using Laplace Transforms, Chapter 15Given 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.)))}

Solving PDEs using Laplace Transforms, Chapter 15 Given a function u(x;t) de ned for all t>0 and assumed to be bounded we can apply the Laplace transform in tconsidering xas a parameter.

  Using, Chapter, Solving, Transform, Edps, Laplace, The laplace transform, Solving pdes using laplace transforms