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# PDE change of variables Watch

1. Hi,

I have a couple of questions remaining on a differential equations example sheet that I can't seem to crack. They have a common theme -- changing variables in a PDE.

Here's the first part of the first one. I'm hoping that with a gentle nudge in the right direction the rest of it should fall into place

Consider the change of variables
such that .

Use the chain rule to express and in terms of and .
What I don't see is how to use the chain rule to get the required expression for the derivatives. I'm also not convinced I understand the meaning of the condition .

Thanks,
2. (Original post by iainfs)
Hi,

I have a couple of questions remaining on a differential equations example sheet that I can't seem to crack. They have a common theme -- changing variables in a PDE.
u(x,y) = v(s,t) means that a function of the variables x and y, after using the given substitution, becomes a (probably) very different function , which now has variables s and t. Do you see why you can't necessarily write u(x,y) = u(s,t)?

For the partial derivative, as s depends on both x and y, you will have both and terms. I will just state the chain rule first:

Someone on here may be able to give you a more rigorous argument for the why the above is true, but the informal way I think of it is: the (du/dx)(dx/ds) sort of 'cancels' to give a (du/ds) term, although we really can't write it this way since u isn't really a function of s. Same with the y part. So we have to split it up into the x and y components, if you will, to get an expression for (dv/ds).
3. (Original post by Dream Eater)
u(x,y) = v(s,t) means that a function of the variables x and y, after using the given substitution, becomes a (probably) very different function , which now has variables s and t.
Thanks. I think I was trying to read too much into that statement.

I think I've done the first two parts; I'd appreciate it if someone could check it over because I've probably ballsed up somewhere!

Then we have

.

(ii) Find, similarly, an expression for
Rewriting the operator , so we find

Further, , so , and similarly .

Then we have

(iii) Hence transform the equation into a partial differential equation for v.
We have

...but I can't see how to get rid of that last pair of terms.

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Updated: December 13, 2010
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