# another differentiation question ( PDE )

• Apr 9th 2008, 04:03 PM
Richyie
another differentiation question ( PDE )
hi sorry for the insurgent of questions ..
anyways we have :

du/dx + x*du/dy = 0

with boundary conditions
i) u = sinh(y) on x=0
ii) u = cosh(x) on y=0

question: solve pde above and show solution is single valued throughout (x,y) plane.

heres what i have so far:
I get 2 integrals I = u and J = y-(x^2)/2 so general solution to pde is
u = f(y-(x^2)/2)

but not sure where to go from here : /
• Apr 9th 2008, 04:16 PM
mr fantastic
Quote:

Originally Posted by Richyie
hi sorry for the insurgent of questions ..
anyways we have :

du/dx + x*du/dy = 0

with boundary conditions
i) u = sinh(y) on x=0
ii) u = cosh(x) on y=0

question: solve pde above and show solution is single valued throughout (x,y) plane.

heres what i have so far:
I get 2 integrals I = u and J = y-(x^2)/2 so general solution to pde is
u = f(y-(x^2)/2)

but not sure where to go from here : /

Look for a seperable solution: u = f(x) g(y).

Substitute into the PDE. After a modest amount of algebra you get:

$\displaystyle \frac{1}{x f} \frac{df}{dx} = - \frac{1}{g} \frac{dg}{dy}$ ......
• Apr 9th 2008, 04:42 PM
Richyie
according to the solutions i have for
the condition u(0,y) = sinh(y)

u = sinh(y-x^2/2) is a solution

i dont think i get the logic of how he got that tho,

i guess u = sinh(y-x^2/2) would satisfy the general solution u(x,y) = f(y-x^2/2) and the condition u(0,y) = sinh(y), but then could i say
u = sinh(y-x^2/2)+ b(x) is a solution aswell ? ? ( b(x) being an arbitary function of x)
one of the things i am confused about here is what the f means in u(x,y) = f(y-x^2/2)

for example if u(x,y) = f(x+y) .. since f is an arbitary function. .. . cant this imply: u(x,y) = f(x+y) => u(x,y) = (x+y)^2 => u(x,y) = f(x^2+y^2+2xy) . ..