The Student Room Group

Method of Characteristics Q

q.JPG

Please see attached.

So after we have inverted x and y in terms of s and t, we have following, x=e^s and y=(e^s)(s+t) so:

s = lnx
t = (y/x) - lnx

Then along the (t)th characteristic curve, du/ds = 1 => u = s + k

This is where the problem arises. We know u(1,y) = y. So do we put that into the above to find k? So i.e. have u = lnx + k. So at x=1, u=y so y=k => u = lnx + y.

OR do we use u(1,y) = y in terms of t and then substitute to find k? If so how do we do it?

Thanks.
Original post by Peter8837
q.JPG

Please see attached.

So after we have inverted x and y in terms of s and t, we have following, x=e^s and y=(e^s)(s+t) so:

s = lnx
t = (y/x) - lnx

Then along the (t)th characteristic curve, du/ds = 1 => u = s + k

This is where the problem arises. We know u(1,y) = y. So do we put that into the above to find k? So i.e. have u = lnx + k. So at x=1, u=y so y=k => u = lnx + y.

OR do we use u(1,y) = y in terms of t and then substitute to find k? If so how do we do it?

Thanks.


The first attempt is wrong because you've forgotten that k is a function of t. Really, you have u(x, y) = log(x) + k(t(x,y)); at x=1, u=y so u(1, y) = log(1) + k(t(1, y)) = k(t(1, y)). You can't deduce that u(x, y) = log(x) + k(t(x,y)) generally.

We have u(s=0, t) = y(s=0, t); y(s=0, t) = t (by the expression in your first line), so u(s=0, t) = t. Hence u(s, t) = s + k(t); u(0, t) = k(t) = t, so k(t) = t.

That is, u(s, t) = s + t.

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