Heat Equation trouble

**CooperCripps** · September 14th 2010, 11:50 PM

I'm going through the following heat equation problem, and I understand how the solution is constructed except for one algebraic re-arrangement that I can't make sense of (and its probably really obvious) which is right at the start of the solution. I was hoping someone could enlighten me.

Solve $u_t = ku_{xx}$ on $0 \leq x \leq L$ with IC's $u(x,0) = f(x)$ and BC's $u_x(0,t) = u_x(L,t) = 0$ .

The example says:

Let $u = XT$ .

Confused from here:
Sub in $XT' = kX''$

$\frac{X''}{X} = \frac{T'}{kT} = - \lambda$

Then $X'' + \lambda X = 0$ .
To the line above

From here I have no problems understanding and solving this problem.

This is probably embarrassingly simple, but I keep getting

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

and why is this equal to $-\lambda$

**TheEmptySet** · September 15th 2010, 12:31 AM

Originally Posted by CooperCripps

I'm going through the following heat equation problem, and I understand how the solution is constructed except for one algebraic re-arrangement that I can't make sense of (and its probably really obvious) which is right at the start of the solution. I was hoping someone could enlighten me.

Solve $u_t = ku_{xx}$ on $0 \leq x \leq L$ with IC's $u(x,0) = f(x)$ and BC's $u_x(0,t) = u_x(L,t) = 0$ .

The example says:

Let $u = XT$ .

Confused from here:
Sub in $XT' = kX''$

$\frac{X''}{X} = \frac{T'}{kT} = - \lambda$

Then $X'' + \lambda X = 0$ .
To the line above

From here I have no problems understanding and solving this problem.

This is probably embarrassingly simple, but I keep getting

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

and why is this equal to $-\lambda$

You are assuming the production from of the solutions

$\displaystyle u(x,t)=X(x)T(t)$ this gives that

$\displaystyle u_t(x,t)=X(x)T'(t)$ and

$\displaystyle u_{xx}(x,t)=X''(x)T(t)$

So the PDE gives us

$\displaystyle u_{t}=ku_{xx} \iff X(x)T'(t)=kX''(x)T(t)$

Now isolating the T's and X's gives

$\displaystyle \frac{T'(t)}{kT(t)}=\frac{X''(x)}{X(x)}$

Now since this equality must hold for all values of x and t each side of the equation must be constant this gives

$\displaystyle \frac{T'(t)}{kT(t)}=\frac{X''(x)}{X(x)}=c$ or the 2 ODE's

$\displaystyle \frac{X''(x)}{X(x)}=c \text{ and } \frac{T'(t)}{kT(t)}=c$

Now to answer you second question:

First note the the boundary conditions

$u_x(0,t)=u_x(L,t)=0 \implies \text{ that } X'(0)=X'(L)=0$

Using the X equation above gives

$X''(x)-cX(x)=0$ Now if we multiply this by $X(x)$ and integrate we get

$\displaystyle \int_{0}^{L}X(x)X''(x)dx -c\int_{0}^{L}[X(x)]^2dx=0$

If we integrate the first integral by parts with

$u=X(x) \implies du=X'(x) \text{ and } dv=X''(x) \implies v=X'(x)$ we get

$\displaystyle X(x)X'(x)\bigg|_{0}^{L}-\int_{0}^{L}[X'(x)]^2dx -c\int_{0}^{L}[X(x)]^2dx=0$

Now using the boundary conditions we get

$\displaystyle X(L)X'(L)-X(0)X'(0)-\int_{0}^{L}[X'(x)]^2dx -c\int_{0}^{L}[X(x)]^2dx=0$

$\displaystyle -\int_{0}^{L}[X'(x)]^2dx -c\int_{0}^{L}[X(x)]^2dx=0$

Notice that both of the integrands are nonnegative, So the only way this equation can be true is if $c < 0$ . This is why the separation constant is $-\lambda$

**CooperCripps** · September 15th 2010, 01:03 AM

Originally Posted by TheEmptySet

Now to answer you second question:

First note the the boundary conditions

$u_x(0,t)=u_x(L,t)=0 \implies \text{ that } X'(0)=X'(L)=0$

Using the X equation above gives

$X''(x)-cX(x)=0$ Now if we multiply this by $X(x)$ and integrate we get

$\displaystyle \int_{0}^{L}X(x)X''(x)dx -c\int_{0}^{L}[X(x)]^2dx=0$

If we integrate the first integral by parts with

$u=X(x) \implies du=X'(x) \text{ and } dv=X''(x) \implies v=X'(x)$ we get

$\displaystyle X(x)X'(x)\bigg|_{0}^{L}-\int_{0}^{L}[X'(x)]^2dx -c\int_{0}^{L}[X(x)]^2dx=0$

Now using the boundary conditions we get

$\displaystyle X(L)X'(L)-X(0)X'(0)-\int_{0}^{L}[X'(x)]^2dx -c\int_{0}^{L}[X(x)]^2dx=0$

$\displaystyle -\int_{0}^{L}[X'(x)]^2dx -c\int_{0}^{L}[X(x)]^2dx=0$

Notice that both of the integrands are nonnegative, So the only way this equation can be true is if $c < 0$ . This is why the separation constant is $-\lambda$

Thanks EmptySet. This may be a dumb question, but apart from using integration as you have a done above, is their an faster method of determining the separation constant? For example, if in an exam I don't think I would like to use this method because it may take too long.

Would you know the separation constant to be $- \lambda$ by intuition?

**Jester** · September 15th 2010, 03:59 AM

You can look at cases

$c = \omega^2, c = 0, c = -\omega^2$ (i.e. $c > 0, = 0, < 0$ ),

and solve $X'' - c X = 0$ subject to $X'(0) =0, X'(L)=0$ .

The only case that gives a nonzero answer is $c = -\omega^2$

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