Partial differentiation chain rule

Hi there,

I am given that u = F(x - ct), where F() is ANY function.

I have to calculate partial du/dt and partial du/dx [for the adjective eqn.)

So i started off with letting s = x - ct

So then u = F(s),

and partial ds/dt = -c, and partial ds/dx = 1

I think I have to work out partial du/dF, and partial du/ds,

so then partial du/dt = [partial ds/dt] . [partial du/ds]

and partial du/dx = [partial ds/dx] . [partial du/ds]

but im unsure how to go about doing these, as the F being ANY function is really confusing me.

Thank you very much

Reply 1

Notnek

Original post by Jaswarbrick

Are you sure it doesn't ask for

\displaystyle \frac{\partial u}{\partial t}

in terms of

\displaystyle \frac{\partial u}{\partial x}

\displaystyle \frac{\partial u}{\partial t}

and

\displaystyle \frac{\partial u}{\partial x}

are dependent on

F

Reply 2

Jaswarbrick

Original post by notnek

Are you sure it doesn't ask for

\displaystyle \frac{\partial u}{\partial t}

in terms of

\displaystyle \frac{\partial u}{\partial x}

\displaystyle \frac{\partial u}{\partial t}

and

\displaystyle \frac{\partial u}{\partial x}

are dependent on

F

The question is:

"Verify (not prove) that u = F(x - ct) is the general solution of the adjective equation 'partial du/dt + c. partial du/dx = 0 where F() is ANY function."

Thank you

Reply 3

Notnek

Original post by Jaswarbrick

The question is:

"Verify (not prove) that u = F(x - ct) is the general solution of the adjective equation 'partial du/dt + c. partial du/dx = 0 where F() is ANY function."

Thank you

You've almost done it:

...and partial ds/dt = -c, and partial ds/dx = 1...
...so then partial du/dt = [partial ds/dt] . [partial du/ds]

and partial du/dx = [partial ds/dx] . [partial du/ds]...

Substituting ds/dt and ds/dx gives

\displaystyle \frac{\partial u}{\partial t}=-c\frac{\partial u}{\partial s}

\displaystyle \frac{\partial u}{\partial x}=\frac{\partial u}{\partial s}

Can you finish it?

Reply 4

Jaswarbrick

Original post by notnek

You've almost done it:

Substituting ds/dt and ds/dx gives

\displaystyle \frac{\partial u}{\partial t}=-c\frac{\partial u}{\partial s}

\displaystyle \frac{\partial u}{\partial x}=\frac{\partial u}{\partial s}

Can you finish it?

Well i'm getting:

partial du/ds = s. partial dF/ds + F.1 (using product rule, but I don't know if this is right), and

partial du/dF = s. partial dF/dF (which doesn't make sense) + F. partial dF/ds

cheers

Reply 5

Jaswarbrick

Original post by Jaswarbrick

Actually I might not need partial du/dF after all, I thought I did?

Reply 6

Notnek

Original post by Jaswarbrick

It's easier than you're making it - you don't need to work out du/ds.

Look again at the equations I gave in my last post and then look at what you're trying to prove...

Reply 7

Jaswarbrick

Original post by notnek

It's easier than you're making it - you don't need to work out du/ds.

Look again at the equations I gave in my last post and then look at what you're trying to prove...

Ah I see, so because:

partial du/ds = partial du/dx,

I can substitute that into the first of your equations to give

partial du/dt = -c. [partial du/ds]

=> partial du/dt = -c. [partial du/dx]

so it's just a substitution? Thank you I think I have it now,

also part 2 of the question is:

2. Assume periodic boundary conditions u(0,t) = u(1,t) and following initial conditions:

I1: u(x,0) = exp[-100(x-1/4)^2] (Gaussian)

I2: u(x,0) = 1 when 1/4 <= x <= 1/2 (0 otherwise, Rectangular)

Describe how an initial shape evolves and write down the exact solution for the initial conditions I1 and I2.
[domain]

Thank you very much, or should this be in a new thread?

Reply 8

Jaswarbrick

Any help with the above please?

Cheers

Anyone?

Cheers

Original post by notnek

It's easier than you're making it - you don't need to work out du/ds.

Look again at the equations I gave in my last post and then look at what you're trying to prove...