Solving a PDE System with Arbitrary g(s): Help for Senior Thesis

[Lucradan]

Hello Everyone!

I'm a newb to these forums, and I'm having some difficulties with my senior thesis and was hoping someone here might be able to help or at least point me in the right direction.

I have the following PDE system:

dx/dp = - g(s) * dy/ds
dy/dp = g(s) * dx/ds

with initial conditions x(s,0) = x0(s) and y(s,0) = y0(s) and arbitrary g(s).

I can solve it in the case of g(s) = 1, but am not sure how to solve it under any other value of g(s).

Thank You.,
Eric

 

[Clausius2]

Hello Everyone!

I'm a newb to these forums, and I'm having some difficulties with my senior thesis and was hoping someone here might be able to help or at least point me in the right direction.

I have the following PDE system:

dx/dp = - g(s) * dy/ds
dy/dp = g(s) * dx/ds

with initial conditions x(s,0) = x0(s) and y(s,0) = y0(s) and arbitrary g(s).

I can solve it in the case of g(s) = 1, but am not sure how to solve it under any other value of g(s).

Thank You.,

Derive the first equation respect to p, substitute the second in the first and you'll have an homogeneous second order equation for x that you can integrate directly two times.

 

[Lucradan]

I thank you for your help. I'm still a little stuck on what to do with this PDE now:

d2x/dp2 + g(s)g'(s)*dx/ds+g(s)^2*d2x/ds2 = 0

with initial condition x(0,s) = x0(s).

I bought a book on PDEs but I've only gotten as far as characteristic curves. (PDE is a graduate course, so I've not taken it yet).

I tried solving the equation in Maple 10, and it gave me the solution was:

x(s,p) = F(s)*H(p)

I can solve for F(s) and H(p), but it doesn't fit with the initial condition of x(0,s) = x0(s).

Again, I'm sorry to bother, I'm having to do the PDEs in my thesis with little experience in the area, so details, even minute, are very helpful.

Plus, if anyone knows of a great PDE guide on line, please post or PM me.

 

[J77]

What does g'(s) represent?

 

[Lucradan]

g(s) is an arbitrary function just like x0(s) and y0(s).

 

[J77]

And the dash - differentitation with respect to s?

 

[Lucradan]

The PDE equation system is

with intial conditions

x(s,0) = x0(s)
y(s,0) = y0(s)

and g(s) is positive function of s, by the way and has domain s>=0

g'(s) is shorthand for dg/ds

 

[BoTemp]

I thank you for your help. I'm still a little stuck on what to do with this PDE now:

d2x/dp2 + g(s)g'(s)*dx/ds+g(s)^2*d2x/ds2 = 0

with initial condition x(0,s) = x0(s).

I bought a book on PDEs but I've only gotten as far as characteristic curves. (PDE is a graduate course, so I've not taken it yet).

I tried solving the equation in Maple 10, and it gave me the solution was:

x(s,p) = F(s)*H(p)

I can solve for F(s) and H(p), but it doesn't fit with the initial condition of x(0,s) = x0(s).

Pretty standard, separation of variables. Have you been given x0(s)? From what I can find, H(p) = A*p + B (A&B constants). H(0) = B, a constant. Should be able to fit anything, just a matter of making F(s) fit.

edit: I'm not sure how far it's possible to go as long as g(s) is arbitrary. One may be able to set up x and y as derivatives and/or (more likely) integrals of g(s), but I don't know if that's guaranteed to be possible in general.

edit 2: If one guesses d2x/dp2 = a*x (a is some scalar, pos, neg, or complex) then the above equation can be placed in Sturm-Liouville form. Remember that despite appearances this equation is linear in x and y, so any superposition of solutions will also be a solution. Very helpful for the boundaries.