3 Method of separation of variables - specific solutions

We shall now study some specific problems which can be fully solved by the separation of variables method.

Example

3 Solve the heat conduction equation

$\frac{{\partial }^{2}u}{\partial x^2}=\frac{1}{2}\frac{\partial u}{\partial t}$

over     $0<x<3,t>0$    for the boundary conditions

$u\left(0,t\right)=u\left(3,t\right)=0$

and the initial condition

$u\left(x,0\right)=5sin4\pi x.$

Solution

Assuming $u\left(x,t\right)=X\left(x\right)T\left(t\right)$ gives rise to the differential equations (4) and (5) with the parameter $k=2$ :

$\frac{dT}{dt}=2KT\frac{d^{2}X}{d{x}^2}=KX$ The $T$ equation has general solution

$T=A{e}^{2Kt}$

which will increase exponentially with increasing $t$ if $K$ is positive and decrease with $t$ if $K$ is negative. In any physical problem the latter is the meaningful situation. To emphasise that $K$ is being taken as negative we put

$K=-{\lambda }^2$

so

$T=A{e}^{-2{\lambda }^{2}t}.$

The $X$ equation then becomes

$\frac{d^{2}X}{d{x}^2}=-{\lambda }^{2}X$

which has solution

$X\left(x\right)=Bcos\lambda x+Csin\lambda x.$

Hence

$u\left(x,t\right)=X\left(x\right)T\left(t\right)=\left(Dcos\lambda x+Esin\lambda x\right)e^{-2{\lambda }^{2}t}$ (11)

where $D=AB$ and $E=AC$ .

(You should always try to keep the number of arbitrary constants down to an absolute minimum by multiplying them together in this way.)

We now insert the initial and boundary conditions to obtain the constant $D$ and $E$ and also the separation constant $\lambda$ .

The initial condition $u\left(0,t\right)=0$ gives

$\left(Dcos0+Esin0\right)e^{-2{\lambda }^{2}t}=0$ for all $t$ .

Since $sin0=0$ and $cos0=1$ this must imply that $D=0$ .

The other initial condition $u\left(3,t\right)=0$ then gives

$Esin\left(3\lambda \right)e^{-2{\lambda }^{2}t}=0\text{for all}t.$

We cannot deduce that the constant $E$ has to be zero because then the solution (11) would be the trivial solution $u\equiv 0$ . The only sensible deduction is that

$sin3\lambda =0$ i.e. $3\lambda =n\pi$ (where $n$ is some integer).

Hence solutions of the form (11) satisfying the 2 boundary conditions have the form

$u\left(x,t\right)=E_{n}sin\left(\frac{n\pi x}{3}\right)e^{-\frac{2n^2{\pi }^{2}t}{9}}$

where we have written $E_n$ for $E$ to allow for the possibility of a different value for the constant for each different value of $n$ .

We obtain the value of $n$ by using the initial condition $u\left(x,0\right)=5sin4\pi x$ and forcing this solution to agree with it. That is,

$u\left(x,0\right)=E_{n}sin\left(\frac{n\pi x}{3}\right)=5sin4\pi x$

so we must choose $n=12$ with $E_{12}=5$ .

Hence, finally,

$u\left(x,t\right)=5sin\left(\frac{12\pi x}{3}\right)e^{-\frac{2}{9}{\left(12\right)}^2{\pi }^{2}t}=5sin\left(4\pi x\right)e^{-32{\pi }^{2}t}.$

Task!

Solve the $1$ -dimensional wave equation $\frac{{\partial }^{2}u}{\partial x^2}=\frac{1}{16}\frac{{\partial }^{2}u}{\partial t^2}$   for  $0<x<2,t>0$

The boundary conditions are

$u\left(0,t\right)=u\left(2,t\right)=0$

The initial conditions are

(i)   $u\left(x,0\right)=6sin\pi x-3sin4\pi x$ (ii)   $\frac{\partial u}{\partial t}\left(x,0\right)=0$

Firstly, either using (7) and (8) or by working from first principles assuming the product solution

$u\left(x,t\right)=X\left(x\right)T\left(t\right),$

write down the ODEs satisfied by $X\left(x\right)$ and $T\left(t\right)$ :

Answer

Now decide on the appropriate sign for $K$ and then write down the solution to these equations:

Answer

Now obtain the general solution $u\left(x,t\right)$ by multiplying $X\left(x\right)$ by $T\left(t\right)$ and insert the two boundary conditions to obtain information about two of the constants:

Answer

$\frac{\partial u}{\partial t}\left(x,0\right)=0\text{for all}x0<x<2.$

and deduce the value of $F$ :

Answer

Answer

The above solution perhaps seems rather involved but there is a definite sequence of logical steps which can be readily applied to other similar problems.