Partial Differential Equation Urgent Help

[utorstudent]

I have a PDE problem set due on monday and as I look at the teacher's method for solving this problem it doesn't make any sense to me. I'll show what I have so far.


Find the solution of
U^2U_x + U_y = 0
that satisfies u(x,0)=x


So I started out with the characteristic equations the way my professor did it in his notes

x_t = u^2
y_t = 1
z_t = 1

then integrating these I get

t = \int 1/u^2 dx
t = y + y_0
t = z + z_o


I don't know how to integrate the u in the first equation (as its a function of x - and I don't think that integrating w.r.t t will help.) and using the intial conditions

x_o = s, y_o = 0, z_0 = s

so that reduces them to


t = \int 1/u^2 dx
t = y
t = z + s


and this is where I got stuck. I know I have to solve for x and y as functions of s and t and substitute them into the third equation to get the solution but I can't figure out how to do this while eliminating the s and t (I mean I could solve for s if I could do the integral in one but I can't.)


Anyways, all help would be greatly appreciated.

 

[utorstudent]

could I please get a hand with this. I still don't know where to go from the above.

 

[utorstudent]

Does anyone know of a good website which explains the method to solving PDE's of this form. I still can't see a way to solve it (I tried to replace u with x and that doesn't work) and I'm really worried about finishing. Thanks to anyone who replies.

 

[Pythagorean]

I'm not that far in diff eq yet, but it looks a lot like parametric equations from calculus with diff eq included.

Have you tried using parametric methods?

Also, you have three equations for t, so you can set them all equal to each other if that helps:

since t=a and t=b, b=a

You also may need a laplace transform.

Sorry I can't be more of a help

 

[saltydog]

Day late and a dollar short no doubt. Here goes anyway:

First put the equation into the form:

F(x,y,u,u_x,u_y)=F(x,y,u,p,q)=0

with:

u(x,0)=x

Then to solve parametrically, first set the initial condition as:

u(f(s),g(s))=G(s)

and then solve the system:

\frac{dx(s,t)}{dt}=F_p

\frac{dy(s,t)}{dt}=F_q

\frac{du(s,t)}{dt}=pF_p+qF_q

Doing this I get:

\frac{dx(s,t)}{dt}=u^2

\frac{dy(s,t)}{dt}=1

\frac{du(s,t)}{dt}=pu^2+q=0

So:

y(s,t)=t+y(s,0)=t+g(s)=t

u(s,t)=u(s,0)=s

x(s,t)=s^2t+x(s,0)=s^2t+f(s)=s^2t+s

I haven't checked this yet. What did you get?

 

[saltydog]

Day late and a dollar short no doubt. Here goes anyway:

First put the equation into the form:

F(x,y,u,u_x,u_y)=F(x,y,u,p,q)=0

with:

u(x,0)=x

Then to solve parametrically, first set the initial condition as:

u(f(s),g(s))=G(s)

and then solve the system:

\frac{dx(s,t)}{dt}=F_p

\frac{dy(s,t)}{dt}=F_q

\frac{du(s,t)}{dt}=pF_p+qF_q

Doing this I get:

\frac{dx(s,t)}{dt}=u^2

\frac{dy(s,t)}{dt}=1

\frac{du(s,t)}{dt}=pu^2+q=0

So:

y(s,t)=t+y(s,0)=t+g(s)=t

u(s,t)=u(s,0)=s

x(s,t)=s^2t+x(s,0)=s^2t+f(s)=s^2t+s

I haven't checked this yet. What did you get?

Just for the record, I mis-interpreted the initial conditions when solving the ODEs. The solutions I get are:

y(s,t)=t+y(s,0)=t+g(s)=t

u(s,t)=u(s,0)=s

x(s,t)=s^2t+x(s,0)=s^2t+f(s)=s^2t

Solving for u:

u(x,y)=\pm\sqrt{\frac{x}{y}}

Which when substituted into the PDE, satisfies the relation. Utorstudent, if you use any of this, then kindly explain how the initial conditions for the ODEs are arrived at.

 

[saltydog]

Just for the record, I mis-interpreted the initial conditions when solving the ODEs. The solutions I get are:

y(s,t)=t+y(s,0)=t+g(s)=t

u(s,t)=u(s,0)=s

x(s,t)=s^2t+x(s,0)=s^2t+f(s)=s^2t

Solving for u:

u(x,y)=\pm\sqrt{\frac{x}{y}}

Which when substituted into the PDE, satisfies the relation. Utorstudent, if you use any of this, then kindly explain how the initial conditions for the ODEs are arrived at.

Alright so I made another mistake. The formula for calculating x(s,t) is:

\frac{dx(s,t)}{dt}=u^2=s^2;\quad x(s,0)=s

so that leads to:

x(s,t)=s^2t+s

Then solving for u(x,y) yields:

u(x,y)=\frac{-1\pm\sqrt{1+4xy}}{2y}

Note:

\mathop\lim\limits_{y\to 0}\left(\frac{-1+\sqrt{1+4xy}}{2y}<br /> \right)=x

but the negative one goes to minus infinity. So the solution is:

u(x,y)=\frac{-1+\sqrt{1+4xy}}{2y}

Thanks Qbert.