You are Here: Home >< Maths

# Partial differentiation chain rule Watch

1. 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
2. (Original post by Jaswarbrick)
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
Are you sure it doesn't ask for in terms of ?

and are dependent on .
3. (Original post by notnek)
Are you sure it doesn't ask for in terms of ?

and are dependent on .
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
4. (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

Can you finish it?
5. (Original post by notnek)
You've almost done it:

Substituting ds/dt and ds/dx gives

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
6. (Original post by Jaswarbrick)
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
Actually I might not need partial du/dF after all, I thought I did?
7. (Original post by Jaswarbrick)
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
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...
8. (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 0 <= x < 1, t>= 0]

Thank you very much, or should this be in a new thread?
9. Any help with the above please?

Cheers
10. Anyone?

Cheers
11. (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...
Any help for the above mate?

Cheers
12. (Original post by Jaswarbrick)
Any help for the above mate?

Cheers
I won't be around much to help you with this (plus it's been a while..).

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 17, 2013
Today on TSR

### What is the latest you've left an assignment

And actually passed?

### Simply having a wonderful Christmas time...

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.