Can Separation of Variables Solve u_t = u_x for u=A(x)B(t)?

[FrogPad]

This is the the first time I've encountered separation 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 separation 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 separation 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? :)
Thanks in advance.

 

[FrogPad]

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 separate them as a system of equations. Thus, allowing the ODE to be solved with separation of variables.

 

[HallsofIvy]

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 separate them as a system of equations. Thus, allowing the ODE to be solved with separation of variables.

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)= Ce^kxe^kt= Ce^k(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.

 

[FrogPad]

Could you be more awesome?
no... no you can't. Thank you!

It's always fun when something makes sense.