c****w
发帖数: 565 | 1 how to solve:
a*dS/dx+b*dS/dy=beta(x,y)
know: a,b,beta
solve: S
3Q~! | B********e
发帖数: 10014 | 2 any textbook covering monge cector or integral curve of Monge vector works f
or you
pretty standard method, but not easy to write it down here
【在 c****w 的大作中提到】
: how to solve:
: a*dS/dx+b*dS/dy=beta(x,y)
: know: a,b,beta
: solve: S
: 3Q~!
| c****w
发帖数: 565 | 3 thanks for reply. I know the standard method using characteristics. I just
got confused with some steps. Take easy approach:
a*dU/dx+b*dU/dy=beta(x,y)
dx/ds=a
dy/ds=b
dU/ds=beta(x,y)
x=a*s+c1
y=b*s+c2
U=\int{beta(x,y)}ds+c3
I know c3=F(bx-ay), but I get confused when working out \int{beta(x,y)}ds.
1. Is it a double integral of both x and y?
2. or just put x=as+c1, y=bs+c2 into the integral, then, replace s in the
integral results using either x or y relation to s?
===============================
【在 B********e 的大作中提到】
: any textbook covering monge cector or integral curve of Monge vector works f
: or you
: pretty standard method, but not easy to write it down here
| B********e
发帖数: 10014 | 4
it will be more clear if you write it as
U=int_0^s{beta(at+c1,bt+c2)}dt+c3
why not F(bx-ay)+Const.? if you don't give the initial data
it's clear now
aoh, you actually got it already
you can do it now. just make sure you need initial data to cancel the
constants c1,c2,c3, then come up with a equation system about x,y,u,s
solve for s,plug into u...
since your a,b are constants, there is a much easier way to do
do independant variable transform
\xi=bx-ay
\eta=bx+ay
change the pde to a ode, it's p
【在 c****w 的大作中提到】
: thanks for reply. I know the standard method using characteristics. I just
: got confused with some steps. Take easy approach:
: a*dU/dx+b*dU/dy=beta(x,y)
: dx/ds=a
: dy/ds=b
: dU/ds=beta(x,y)
: x=a*s+c1
: y=b*s+c2
: U=\int{beta(x,y)}ds+c3
: I know c3=F(bx-ay), but I get confused when working out \int{beta(x,y)}ds.
| c****w
发帖数: 565 | 5 Thanks a lot for the reply! However, I still have one point unclear.
\int_0^s{beta(at+c1,bt+c2)}dt
Here, how to express s in the final solution? use x=a*s+c1 or y=b*s+c2?
Either way will only give one variable (x or y) in the result of the
integration on beta. How can U produce dU/dx+dU/dy=beta, where beta is
function of both x and y?
where am I wrong?
【在 B********e 的大作中提到】
:
: it will be more clear if you write it as
: U=int_0^s{beta(at+c1,bt+c2)}dt+c3
: why not F(bx-ay)+Const.? if you don't give the initial data
: it's clear now
: aoh, you actually got it already
: you can do it now. just make sure you need initial data to cancel the
: constants c1,c2,c3, then come up with a equation system about x,y,u,s
: solve for s,plug into u...
: since your a,b are constants, there is a much easier way to do
| B********e
发帖数: 10014 | 6 hehe,
keep in mind one thing:in general the initial data will be a curve,
(for example: x=1,y=u when s=0)
which means when you try to get c2,you'll not get a number,you
will instead get c2(u)---a relation connects y and u,
in the meantime, from x=as+c1,you get s as funtion of x,plug into
3rd equation,you get everything in the final equation.
the key is:c1,c2 might not be exact constants/numbers.
find a book and practise a couple easier exercise ,you will find the
subtle point;)
【在 c****w 的大作中提到】
: Thanks a lot for the reply! However, I still have one point unclear.
: \int_0^s{beta(at+c1,bt+c2)}dt
: Here, how to express s in the final solution? use x=a*s+c1 or y=b*s+c2?
: Either way will only give one variable (x or y) in the result of the
: integration on beta. How can U produce dU/dx+dU/dy=beta, where beta is
: function of both x and y?
: where am I wrong?
| c****w
发帖数: 565 | 7 Thanks a lot for the reply, that's exactly the point I missed. Thank you
very much!
【在 B********e 的大作中提到】
: hehe,
: keep in mind one thing:in general the initial data will be a curve,
: (for example: x=1,y=u when s=0)
: which means when you try to get c2,you'll not get a number,you
: will instead get c2(u)---a relation connects y and u,
: in the meantime, from x=as+c1,you get s as funtion of x,plug into
: 3rd equation,you get everything in the final equation.
: the key is:c1,c2 might not be exact constants/numbers.
: find a book and practise a couple easier exercise ,you will find the
: subtle point;)
| r*******y
发帖数: 1081 | 8 characteristic line ?
【在 c****w 的大作中提到】
: how to solve:
: a*dS/dx+b*dS/dy=beta(x,y)
: know: a,b,beta
: solve: S
: 3Q~!
|
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