# Thread: help solving the Heat equation

1. ## help solving the Heat equation

$\displaystyle u = X(x) T(t)$

using seperation of variables

$\displaystyle \frac{X''}{x} = \frac{T'}{KT} = -\lambda$

$\displaystyle X'' + \lambdax = 0$

$\displaystyle T' + \lambdaKT = 0$

when $\displaystyle \lambda = 0$

$\displaystyle X'' = 0$

Am not sure how to proceed from here. if anyone can show or help. thank you.

2. ## Re: help solving the Heat equation

Once you get to this stage,

$\displaystyle \frac{X''}{X} = \frac{T'}{T}$

it is expert advice to set both these equations equal to $\displaystyle -\lambda^2,\lambda>0$. This is because all other options for the constant lead to trivial solutions, something you can easily check for yourself, while the square simplifies calculations a bit.

Now, we get the equations

$\displaystyle X'' + \lambda^2 x = 0 \ (1)$

$\displaystyle T' + \lambda^2 T = 0 \ (2)$

Equation (1) is coupled with the boundary conditions $\displaystyle X(0)=X(1)=0$, which are derived from the conditions $\displaystyle u(0,t)=X(0)T(t)=0,u(1,t)=X(1)T(t)=0.$
This is a Sturm-Liouville problem, with a solution of the form $\displaystyle X_n(t)= c_n \sin(n\pi x), \ c_n$ constant, for every $\displaystyle n \in \mathbb{N}.$

Subsequently, equation (2) has a solution of the form $\displaystyle T_n(x)=d_n\exp(-n^2 \pi^2 t),$ for every $\displaystyle n \in \mathbb{N}.$

So, the general solution of the initial equation is $\displaystyle u(x,t)=\sum_n X_n(x)T_n(x)=\sum_n A_n \exp(-n^2 \pi^2 t) \sin(n\pi x), \ A_n$ constants to be
determined in order that $\displaystyle u(0,x)=\sin(\pi x).$

3. ## Re: help solving the Heat equation

Originally Posted by Tweety
$\displaystyle u = X(x) T(t)$

using seperation of variables

$\displaystyle \frac{X''}{x} = \frac{T'}{KT} = -\lambda$

$\displaystyle X'' + \lambda x = 0$

$\displaystyle T' + \lambda KT = 0$
(You didn't have a space between "\lambda" and x and between "lambda" and KT

when $\displaystyle \lambda = 0$
?? You cannot assume that $\displaystyle \lambda= 0$! part of the problem is to determine possible values for $\displaystyle \lambda$.

$\displaystyle X'' = 0$

Am not sure how to proceed from here. if anyone can show or help. thank you.
For any $\displaystyle \lambda$, $\displaystyle X''+ \lambda X= 0$ is a "linear ordinary equation with constant coefficients".
You are also told that X(0, t)= X(1, t)= 0. Using your separation that says X(0)T(t)= 0 so that either X(0)= 0 or T(t)= 0 for all t. If T(t)= 0 for all t, we have the "trivial solution", u(x, t)= 0 for all x and t which cannot satisfy the initial condition, $\displaystyle u(x,0)= sin(\pi x)$. Similarly we must have $\displaystyle X(1)= 0$.

So we have the "linear ordinary equation with constant coefficients", $\displaystyle X''+ \lambda X= 0$ with the conditions X(0)= X(1)= 0. What is the general solution to that linear equation? What must $\displaystyle \lambda$ be in order that there be a non-trivial solution satisfying those boundary conditions?

4. ## Re: help solving the Heat equation

Originally Posted by Tweety
$\displaystyle u = X(x) T(t)$

using seperation of variables

$\displaystyle \frac{X''}{x} = \frac{T'}{KT} = -\lambda$

$\displaystyle X'' + \lambda x = 0$

$\displaystyle T' + \lambda KT = 0$
You didn't have a space after "\lambda".

when $\displaystyle \lambda = 0$
Why would you set $\displaystyle \lambda$ equal to 0? Finding values of $\displaystyle \lambda$ so that this problem has a solution is part of solving it.

$\displaystyle X'' = 0$

Am not sure how to proceed from here. if anyone can show or help. thank you.
As Rebesques said, you have to use the boundary conditions as well. Since we are given that u(0, t)= 0 we must have X(0)T(t)= 0 for all t. In order that the product of two numbers be 0, one or the other must be 0. So we must have either X(0)= 0 or T(t)= 0 for all t. But "T(t)= 0" for all t would give u(x, t)= 0 for all t and so could not satisfy the initial condition "$\displaystyle u(x, 0)= sin(\pi x)$". So we must have X(0)= 0 and, similarly, X(1)= 0.

So we have the "linear ordinary differential equation with constant coefficients" problem $\displaystyle X''+ \lambda X= 0$ with boundary conditions X(0)= 0, X(1)= 0. Rebesques says "it is expert advice to set both these equations equal to $\displaystyle -\lamba$, $\displaystyle \lambda> 0$". You can work that out for yourself by considering the possible cases.

1) Suppose $\displaystyle \lambda= 0$ (as you assume). Then the differential equation is X''= 0. Integrating, X'= C, a constant. Integrating again, X= Cx+ D where D is another constant. Now we want to find C and D so that X(0)= C(0)+ D= D= 0. Obviously, D must be 0. But then X(x)= Cx and when x= 1, X(1)= C= 0 so C= 0 also. But then X(x)= 0x+ 0= 0 for all x. That is the "trivial solution". We can't have the trivial solution because, as I said above, we could not then satisfy the initial condition.

2) Suppose $\displaystyle \lamba$ is positive- to make that clear, write $\displaystyle \lambda= \alpha^2$ where $\displaystyle alpha> 0$. The differential equation is now $\displaystyle X''- \alpha^2 X= 0$. That differential equation has "characteristic equation" $\displaystyle r^2- \alpha^2= 0$ which has roots $\displaystyle \alpha$ and $\displaystyle -\alpha$. So the general solution to the differential equation is $\displaystyle X(x)= Ce^{\alpha x}+ De^{-\alpha x}$. Now look at the boundary conditions: $\displaystyle X(0)= C+ D= 0$ and $\displaystyle X(1)= Ce^{\alpha}+ De^{-\alpha}= 0$. From the first equation, $\displaystyle D= -C$. That means we can write the second equation as $\displaystyle X(1)= Ce^{\alpha}- Ce^{-\alpha}= C(e^{\alpha}- e^{-\alpha})= 0$. But e to a positive power is larger than 1 while e to negative power is less than 1. Since $\displaystyle \alpha$ is not 0, those two exponential are NOT equal, their difference is NOT 0, and so we must have C= 0. But then D= -C= 0 also so we have $\displaystyle X(x)= 0e^{\alpha x}+ 0e^{-\alpha x}= 0$ for all x. That is again the trivial solution.

3) $\displaystyle \lambda< 0$. In that case, we can write $\displaystyle \lambda= -\alpha^2$ again with $\displaystyle \alpha> 0$. Now the differential equation is $\displaystyle X''+ \alpha^2X= 0$. That has characteristic equation $\displaystyle r^2+ \alpha^2= 0$ which has imaginary roots, $\displaystyle \alpha i$ and $\displaystyle -\alpha i$. So the general solution to the differential equation in this case is $\displaystyle X(x)= C cos(\alpha x)+ D sin(\alpha x)$ and the boundary conditions are $\displaystyle X(0)= C cos(0)+ Dsin(0)= C= 0$ and $\displaystyle X(1)= D sin(\alpha)= 0$ so that either D= 0, giving the trivial solution again, or $\displaystyle sin(\alpha)= 0$ which will be true as long as $\displaystyle \alpha= n\pi$ for n any imteger.

So we can say that $\displaystyle X''+ \lambda X= 0$ with boundary conditions X(0)= X(1)= 0 has a non-trivial solution if and only if $\displaystyle \lambda= -\alpha^2= -n^2\pi^2$ and, i that case, $\displaystyle X(x)= C sin(n\pi x)$ for any integer n and any constant C.