Answer to Question #252342 in Differential Equations for pde

Question

Use the laplace transform to solve the one dimensional boundary value problem

∂u/∂t=∂²u/∂x², 0<x<1, t>0

subject to u(x,0)=sin πx, u(0,t)=0, u(1,t)=0

Answer 1

Solution;

Take the Laplace transform of both the left hand side as;

$L[\frac{\partial u}{\partial t}]=sU(x,s)-u(x,0)$

And of the right hand side as;

$L[\frac{\partial^2u}{\partial x^2}]=\frac{\partial^2U}{{\partial x^2}}(x,s)$

We have;

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

From the problem;

$u(x,o)=sin(πx)$

Substitute;

$\frac{\partial^2U}{\partial x^2}(x,s)-sU(x,s)=-sin(πx)$

This is a ODE whose general solution can be written as;

$U(x,s)=U_h(x,s)+U_p(x,s)$

where $U_h(x,s)$ is the general solution of the homogeneous problem obtained as;

$m^2-s=0$

$m=\displaystyle_-^+\sqrt s$

Hence;

$U_h(x,s)=c_1e^{\sqrt s x}+c_2e^{-\sqrt s x}$

and $U_p(x,s)$ is any particular solution of the non-homogeneous problem,which is;

$U_p(x,s)=Acos(πx)+Bsin(πx)$

We first use the method of undetermined coefficients to find A and B. To this end we have;

$\frac{d}{dx}U_p(x,s)=-πAsin(πx)+πBcos(πx)$

The second derivative;

$\frac{d^2}{dx^2}U_p(x,s)=-π^2Acos(πx)-π^2Bcos(πx)$

Therefore;

$\frac{d^2}{dx^2}U_p(x,s)-U_p(x,s)=-sin(πx)$

By substitution;

$(-π^2-s)(Acos(πx)+Bsin(πx))=-sin(πx)$

From which we conclude that;

$-(π^2+s)A=0$ ; A=0

Also;

$-(π^2+s)B=-1$ ;$B=\frac{1}{s+π^2}$

Now we have the general solution as;

$U(x,s)=c_1e^{\sqrt sx}+c_2e^{-\sqrt sx}+\frac{1}{s+π^2}sin(πx)$

We note the the Laplace transforms of the boundary conditions give;

$u(0,t)=0 \implies U(0,s)=0$

$u(1,t)=0\implies U(1,s)=0$

So we have;

$U(0,s)=0=c_1+c_2$

$U(1,s)=0=c_1e^{\sqrt s}+c_2e^{-\sqrt s}$

Which gives $c_1=c_2=0$ and we have;

$U(x,s)=\frac{1}{s+π^2}sin(πx)$

To find our solution we apply the inverse Laplace transform;

$u(x,t)=sin(πx)L^{-1}[\frac{1}{s+π^2}]$

$u(x,t)=e^{-π^2t}sin(πx)$