# Homework Help: Seperation of variables

1. Apr 25, 2006

This is the the first time I've encountered seperation with partial differential equations. There are no worked examples, so I need some help to work through this problem. The question seems to be somewhat hand holding, since it seems to be THE introduction.

Q: Apply seperation of variables $u_t = u_x$ by substituting $u=A(x)B(t)$ and then dividing by AB. If one side depends only on $t$ and the other only on $x$, they must equal a constant $k$; what are $A$ and $B$?

$$\frac{\partial u}{\partial t}-\frac{\partial u}{\partial x} = 0$$

$$u = A(x)B(t)$$

$$\frac{\partial}{\partial t} \left[ A(x)B(t) \right] - \frac{\partial}{\partial x} \left[ A(x)B(t) \right] = 0$$

$$A(x)B'(t)-A'(x)B(t)=0$$

$$\frac{A(x)B'(t)-A'(x)B(t)}{A(x)B(t)}$$

$$\frac{B'(t)}{B(t)}-\frac{A'(x)}{A(x)}=0$$

Now I was reading on various websites, that I can set each independent term equal to seperation constants to make two coupled (is this the proper word to use?) differential equations. I don't understand where this step comes from.

but...

$$\frac{B'(t)}{B(t)}=k$$

$$\frac{A'(x)}{A(x)}=k$$

Now solving for $A(x)$ and $B(t)$. I'm a little rusty here, so I don't know if this part is correct.

Rewriting the two equations above in Leibniz notation

$$\frac{dB(t)}{dt} \cdot \frac{1}{B(t)} = k$$

Seperating:

$$\frac{dB(t)}{B(t)} = k dt$$

$$\int \frac{dB(t)}{B(t)} = \int k\,\,dt$$

$$\ln B(t) = kt +c$$

$$B(t) = e^{kt+c}$$

And subsequently:

$$A(x) = e^{kx+c}$$

Does this make sense? :)

Last edited: Apr 25, 2006
2. Apr 25, 2006

The part where I said I wasn't unsure. Is this line of reasoning correct?

$$\frac{B'(t)}{B(t)}-\frac{A'(x)}{A(x)}=0$$

Since the "A term" is equal to the "B term" for some value of x,t we can seperate them as a system of equations. Thus, allowing the ODE to be solved with seperation of variables.

Last edited: Apr 25, 2006
3. Apr 26, 2006

### HallsofIvy

More to the point, "A term" is equal to the "B term" for all values of x,t, so we can separate them.

Write this as
$$\frac{B'(t)}{B(t)}=\frac{A'(x)}{A(x)}$$
Fix x and change t. The left side of the equation does not change because x does not change. But the equation is valid for all x and t- therefore the right side must not have changed even though t changes:
we must have
$$\frac{B'(t)}{B(t)}= k$$
for some constant k. Now let x change while t is fixed. We get
$$\frac{A'(x)}{A(x)}= k$$.
Of course, since they are equal, they must be equal to the same constant.

You have the equations
$$\frac{dB}{B}= kdt$$
and
$$\frac{dA}{A}= kdt$$ so
as, you had before,
$$B(t)= e^{kt}+ c$$
which you can rewrite as
$$B(t)= Ce^{kt}$$
and
$$A(x)= De^{kx}$$
(be careful to use different symbols for the "undetermined constant" in each- they are not necessarily the same)

Finally, since you started by assuming that u(x,y)= A(x)B(t),
a solution to the differential equation is u(x,y)= Cekxekt= Cek(x+t). (The C here is the product of the constants C and D.) Of course, nothing has been said about what k might be. That would be determined by the additional requirements.

4. Apr 27, 2006