Chesley

2021-02-21

Use the Laplace transform to solve the heat equation

un4t5o4v

Skilled2021-02-22Added 105 answers

Step 1

The given heat equation is${u}_{t}={u}_{xx}0<x<1\text{and}t0$

Also initial and boundary conditions are$u(x,0)=\mathrm{sin}\left(\pi x\right)\text{}\text{and}\text{}u(0,t)=u(1,t)=0$

Taking the Laplace transform of heat equation on both sides

$L\left\{ut\right\}=L\left\{uxx\right\}$

$sU(x,s)-u(s,0)=\frac{{d}^{2}}{{dx}^{2}}U(x,s)$

$\frac{{d}^{2}}{{dx}^{2}}U(x,s)-sU(x,s)=-u(x,0)$

$\left({d}^{2}\right)\left({dx}^{2}\right)U(x,s)-sU(x,s)=-\mathrm{sin}\left(\pi x\right)$

The Auxiliary equation of the homogeneous part is$({D}^{2}-s)U=0$

${m}^{2}-s=0$

$m=\pm \sqrt{s}$

So, the complementary solution of the differential equation is

${U}_{c}(x,s)={c}_{1}{e}^{\sqrt{s}t}+{c}_{2}{e}^{-\sqrt{s}t}$

For the particular solution

$P.I=-\frac{1}{{D}^{2}-s}\mathrm{sin}\left(\pi x\right)$

$=\frac{-\mathrm{sin}\left(\pi x\right)}{-{\pi}^{2}-s}\text{}\text{}\text{}\text{}\text{}\text{}\text{}[\text{as}\text{}(-{\pi}^{2}-s)!\ne 0]$

$=\frac{\mathrm{sin}\left(\pi x\right)}{s+{\pi}^{2}}$

Step 2

So, the complete solution of differential equation is

$U(x,s)={U}_{c}(x,s)+{U}_{p}(x,s)$

$U(x,s)={c}_{1}{e}^{\sqrt{s}t}+{c}_{2}{e}^{-\sqrt{s}t}+\left(\frac{\mathrm{sin}\left(\pi x\right)}{s+{\pi}^{2}}\right)\dots \left(1\right)$

We have$u(0,t)=u(1,t)=0$

Taking Laplace transform we get

$U(0,s)=U(1,s)=0\dots (2)$

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

The given heat equation is

Also initial and boundary conditions are

Taking the Laplace transform of heat equation on both sides

The Auxiliary equation of the homogeneous part is

So, the complementary solution of the differential equation is

For the particular solution

Step 2

So, the complete solution of differential equation is

We have

Taking Laplace transform we get

Substituting (2) in (1)

The solution of the above system will give

Therefore, equation (1) reduces to the form

Step 3

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