Use the Laplace transform to solve the heat equationu_t=u_{xx} 0<x<1 text{ and } t>0{u}{left({x},{0}right)}= sin{{left(pi{x}right)}} {u}{left({0},{t}right)}={u}{left({1},{t}right)}={0}

Chesley 2021-02-21 Answered

Use the Laplace transform to solve the heat equation
ut=uxx0<x<1 and t>0
u(x,0)=sin(πx)
u(0,t)=u(1,t)=0

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un4t5o4v
Answered 2021-02-22 Author has 105 answers
Step 1
The given heat equation is ut=uxx0<x<1 and t>0
Also initial and boundary conditions are u(x,0)=sin(πx)  and  u(0,t)=u(1,t)=0
Taking the Laplace transform of heat equation on both sides
L{ut}=L{uxx}
sU(x,s)u(s,0)=d2dx2U(x,s)
d2dx2U(x,s)sU(x,s)=u(x,0)
(d2)(dx2)U(x,s)sU(x,s)=sin(πx)
The Auxiliary equation of the homogeneous part is (D2s)U=0
m2s=0
m=±s
So, the complementary solution of the differential equation is
Uc(x,s)=c1est+c2est
For the particular solution
P.I=1D2ssin(πx)
=sin(πx)π2s       [as  (π2s)!0]
=sin(πx)s+π2
Step 2
So, the complete solution of differential equation is
U(x,s)=Uc(x,s)+Up(x,s)
U(x,s)=c1est+c2est+(sin(πx)s+π2)(1)
We have u(0,t)=u(1,t)=0
Taking Laplace transform we get
U(0,s)=U(1,s)=0(2)
Substituting (2) in (1)
c1+c2=0
c1es+c2es=0
The solution of the above system will give c1=c2=0
Therefore, equation (1) reduces to the form
Step 3

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