1. ## heat equation

Hi! Can someone please help me with the heat equation, my working out is below. The question is: Solve
$
\frac {du}{dt} = \frac {d^2u}{dx^2}$

with the boundary conditions:
$
0<=x<=L, T>=0, u(x,0) =f(x), u_x(0,t)=u_x(L,t)=0.
$

My working out is:
$
X'' + sX = 0$

So first case is: $s>0$
when $s= 0$ The general Solution of ODE for $x$
$X(x) = A e^{ xs^{\frac {1}{2}}} + Be^{ xs^{\frac {-1}{2}}}$

Then i differentiate it and got
$
X'(x) = A s^{\frac {1}{2}}e^{ xs^{\frac {1}{2}}} - Bs^{\frac {1}{2}}e^{ xs^{\frac {-1}{2}}}$

From the boundary conditions:
$X'(0) = A s^{\frac {1}{2}} - Bs^{\frac {1}{2}}$
$X'(L) = A s^{\frac {1}{2}}e^{ Ls^{\frac {1}{2}}} - Bs^{\frac {1}{2}}e^{ Ls^{\frac {-1}{2}}}$

And this is where i get lost
I tried solving simultaneously but I don't know how to. Can someone please help me? Thank you

2. Originally Posted by gconfused
Hi! Can someone please help me with the heat equation, my working out is below. The question is: Solve
$
\frac {du}{dt} = \frac {d^2u}{dx^2}$

with the boundary conditions:
$
0<=x<=L, T>=0, u(x,0) =f(x), u_x(0,t)=u_x(L,t)=0.
$

My working out is:
$
X'' + sX = 0$

So first case is: $s>0$
when $s= 0$ The general Solution of ODE for $x$
$X(x) = A e^{ xs^{\frac {1}{2}}} + Be^{ xs^{\frac {-1}{2}}}$

Then i differentiate it and got
$
X'(x) = A s^{\frac {1}{2}}e^{ xs^{\frac {1}{2}}} - Bs^{\frac {1}{2}}e^{ xs^{\frac {-1}{2}}}$

From the boundary conditions:
$X'(0) = A s^{\frac {1}{2}} - Bs^{\frac {1}{2}}$
$X'(L) = A s^{\frac {1}{2}}e^{ Ls^{\frac {1}{2}}} - Bs^{\frac {1}{2}}e^{ Ls^{\frac {-1}{2}}}$

And this is where i get lost
I tried solving simultaneously but I don't know how to. Can someone please help me? Thank you
It's actually better to consider $s = - \omega^2, 0 \;\text{and}\; \omega^2$

The first case is $X'' - \omega^2 X = 0$whose solution is $X = A e^{\omega x} + Be^{- \omega x}$ as you said. Sub in the BC gives

$X'(0) = A\omega - B\omega = 0$
$
X'(L) = A \omega e^{\omega L} - B \omega e^{-\omega L}
$

The only solution to this pair is A = B = 0 which leads to the trivial solution $X = 0$ so $u = 0$. So on to the next case s = 0 - you're up.

3. Thanks !!

Are the 3 conditions the same thing that you would do for the wave equation?

4. Originally Posted by gconfused
Thanks !!

Are the 3 conditions the same thing that you would do for the wave equation?
Depends on the boundary conditions but typically yes, there are.