# PDE question

Greger
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 can't 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?

voko
So what are the characteristic curves and how did you get them?

Greger
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

voko
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.

Greger
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 I am 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 separate u from everything else without having to define F

is it possible?

voko
I am afraid in this equation it is possible only if u(x, 0) = p(x) is nice.

Greger
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

voko
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.

Greger
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 I am not sure about is,

what is the significance of saying that p(x) is smooth?

voko
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 I am 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.