 # 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

$u\left(x,0\right)=\mathrm{sin}\left(\pi x\right)$
$u\left(0,t\right)=u\left(1,t\right)=0$

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Step 1
The given heat equation is
Also initial and boundary conditions are
Taking the Laplace transform of heat equation on both sides
$L\left\{ut\right\}=L\left\{uxx\right\}$
$sU\left(x,s\right)-u\left(s,0\right)=\frac{{d}^{2}}{{dx}^{2}}U\left(x,s\right)$
$\frac{{d}^{2}}{{dx}^{2}}U\left(x,s\right)-sU\left(x,s\right)=-u\left(x,0\right)$
$\left({d}^{2}\right)\left({dx}^{2}\right)U\left(x,s\right)-sU\left(x,s\right)=-\mathrm{sin}\left(\pi x\right)$
The Auxiliary equation of the homogeneous part is $\left({D}^{2}-s\right)U=0$
${m}^{2}-s=0$
$m=±\sqrt{s}$
So, the complementary solution of the differential equation is
${U}_{c}\left(x,s\right)={c}_{1}{e}^{\sqrt{s}t}+{c}_{2}{e}^{-\sqrt{s}t}$
For the particular solution
$P.I=-\frac{1}{{D}^{2}-s}\mathrm{sin}\left(\pi x\right)$

$=\frac{\mathrm{sin}\left(\pi x\right)}{s+{\pi }^{2}}$
Step 2
So, the complete solution of differential equation is
$U\left(x,s\right)={U}_{c}\left(x,s\right)+{U}_{p}\left(x,s\right)$
$U\left(x,s\right)={c}_{1}{e}^{\sqrt{s}t}+{c}_{2}{e}^{-\sqrt{s}t}+\left(\frac{\mathrm{sin}\left(\pi x\right)}{s+{\pi }^{2}}\right)\dots \left(1\right)$
We have $u\left(0,t\right)=u\left(1,t\right)=0$
Taking Laplace transform we get
$U\left(0,s\right)=U\left(1,s\right)=0\dots \left(2\right)$
Substituting (2) in (1)
${c}_{1}+{c}_{2}=0$
${c}_{1}{e}^{\sqrt{s}}+{c}_{2}{e}^{-\sqrt{s}}=0$
The solution of the above system will give ${c}_{1}={c}_{2}=0$
Therefore, equation (1) reduces to the form
Step 3