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PDE question

  • Thread starter Greger
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  • #1
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Hey,

I'm trying to solve the following pde,

u(x,y) u_x + u_y =0 with u(x,0) = p(x) for some known p(x)

where u_x defines the partial derivative of u(x,y) wrt x

after finding the characteristic curves and the first integrals i get the general solution is

F(x^2 - zy^2, z) = 0

(note z=u(x,y))

At this point i'm not sure what to do next,

usually you can rewrite this as

f(x^2-zy^2) + z =0, however as z is contained in the argument you cant solve this for z without knowing what f is.

One thing i was thinking was just to make up some F and continue on, but that does not seem correct,

can anyone push me in the right direction?
 

Answers and Replies

  • #2
6,054
390
So what are the characteristic curves and how did you get them?
 
  • #3
46
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dx/dt = u = z => x=zt + a

t=(x-a)/c

dy/dt = 1 => y=t+b

t=y-b

dz/dt = 0 => z=c


then dx/dt - z dy/dt = 0 => 1/2 d/dt(x^2 + zy^2) = 0

so x^2 + zy^2 = k


unless i made a dumb error, i thought it was pretty straight forward so left it out
 
  • #4
6,054
390
I get dx/dy = u and du/dy = 0, which together give u = const and x - uy = const, so u(x, y) = F(x - uy).

I don't understand how you end up with the squares.
 
  • #5
46
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oh ok,

the way they taught our class was what i posted above

dx/dt = u = z

dy/dt = 1

dz/dt = 0 => z=c


then dx/dt - z dy/dt = 0 => 1/2 d/dt(x^2 + zy^2) = 0

since dx/dt - z dy/dt= 1/2 d/dt(x^2 + zy^2)

so d/dt(x^2 + zy^2) = 0

integrating both sides gives x^2 + zy^2 = k

but yea there's lots of possibly first integrals,

the problem im having is the next step,

so using what you got

the general solution would be

F(x-uy, u) = 0

how can i solve this for u?

this is like

h(x-uy) + g(u) = 0 or something right?

im having trouble working out how to seperate u from everything else without having to define F

is it possible?
 
  • #6
6,054
390
I am afraid in this equation it is possible only if u(x, 0) = p(x) is nice.
 
  • #7
46
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Oh,

Sorry I didn't mention it states in the question that p(x) is a smooth function,

Does this mean that you apply the initial condition to F(....)=0?

So for x=x y=0 you get


F(x,p(x)) = 0

f(x) + p(x) = 0

f(x) = -p(x)

f(x-uy) = -p(x-uy)

or something like that? i still see myself getting stuck alone the line
 
  • #8
6,054
390
If u(x, 0) = p(x), then what happens with u(x, y) = F(x - uy) when y = 0? You get p(x) = F(x), so u(x, y) = p(x - uy).

For some particular p(x), you might be able to express u explicitly as a function of (x, y), but I do not see what can be done further in the general case. I may be missing something, though.
 
  • #9
46
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how come you write it as F(x-uy) and not F(x-uy,u) like i did below?

F(x-uy, u)=0

f(x-uy) = -u

at y=0

f(x)=-p(x)

i guess you can absorb the negative into p(x)

f(x-uy) = p(x-uy)

i guess one important question that im not sure about is,

what is the significance of saying that p(x) is smooth?
 
  • #10
6,054
390
how come you write it as F(x-uy) and not F(x-uy,u) like i did below?

F(x-uy, u)=0

f(x-uy) = -u

at y=0

f(x)=-p(x)

i guess you can absorb the negative into p(x)

f(x-uy) = p(x-uy)
That's just differences in notation. We have two integrals x - uy = const and u = const, so we can immediately conclude that u = F(x - uy). Put another way, for any quasilinear homogeneous equation we know that u = const is an integral, so we know in advance that u is a function of the other integral, so we can skip the more general form.

i guess one important question that im not sure about is,

what is the significance of saying that p(x) is smooth?
Because u = p(x - uy), p has to have a continuous derivative. This is guaranteed if it is smooth.
 

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