Thread: Heat Equation trouble

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

The example says:

Let $\displaystyle u = XT$.

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

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

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

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

This is probably embarrassingly simple, but I keep getting

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

and why is this equal to $\displaystyle -\lambda$

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

The example says:

Let $\displaystyle u = XT$.

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

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

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

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

This is probably embarrassingly simple, but I keep getting

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

and why is this equal to $\displaystyle -\lambda$

You are assuming the production from of the solutions

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

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

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

So the PDE gives us

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

Now isolating the T's and X's gives

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

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


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

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

Now to answer you second question:

First note the the boundary conditions

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

Using the X equation above gives

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

$\displaystyle \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

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

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

Now using the boundary conditions we get

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

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

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

Originally Posted by TheEmptySet

Now to answer you second question:

First note the the boundary conditions

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

Using the X equation above gives

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

$\displaystyle \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

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

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

Now using the boundary conditions we get

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

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

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

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 $\displaystyle - \lambda $ by intuition?

You can look at cases

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

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

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