First Order Semi-Linear PDE

**Jason Bourne** · November 14th 2008, 01:56 AM

$2yu_x - xu_y = -4xyu^{\frac{1}{2}}$

Where $u_x$ and $u_y$ denotes partial differentiation with respect to x and y respectively of $u(x,y)$

For the above PDE how do I find the characteristic curves in the xy-plane?

How do I show that along these characteristics $u(x,y) = (y^2 +C)^2$ ? (C is a constant)

I think I found that $\frac{dx}{dt} = 2y$ , and $\frac{dy}{dt} = -x$ and $\frac{du}{dt} = -4xyu^{\frac{1}{2}}$

and that on the characterisitcs $y^2 + \frac{x^2}{2} = C_1$ but I don't know where to go from here

Thanks

**shawsend** · November 14th 2008, 05:29 AM

This is how I'd approach it:

I'd first write it in standard form for ``quasi-linear PDEs'':

$2yu_x-xu_y-(-4xyu^{1/2})=0$

We now first solve the equation:

$2yw_x-xw_y-4xyz^{1/2}w_z=0;\quad w(x,y,z)$

The solution of the first is then found to be:

$w(x,y,u)=0$

Solving for the characteristic curves:

$\frac{dy}{dx}=-\frac{x}{2y};\quad\quad \frac{dz}{dx}=-\frac{4xyz^{1/2}}{2y}=-2xz^{1/2}$

I get:

$y^2=-\frac{x^2}{2}+a\quad\quad z=\left(b/2-x^2/2\right)^2$

Solving for a and b, I get the characteristic curves are then the intersections of the surfaces:

$a=1/2(x^2+2y^2),\quad b=3x^2/4$

Been a while for me though. Need to double check this.

**Jason Bourne** · November 14th 2008, 06:35 AM

We now first solve the equation:

$2yw_x-xw_y-4xyz^{1/2}w_z=0;\quad w(x,y,z)$

I don't understand how are getting this equation. Please explain.

The solution of the first is then found to be:

$w(x,y,u)=0$

What do you mean here?

I get:

$y^2=-\frac{x^2}{2}+a\quad\quad z=\left(b/2-x^2/2\right)^2$

Solving for a and b, I get the characteristic curves are then the intersections of the surfaces:

$a=1/2(x^2+2y^2),\quad b=3x^2/4$

How did you solve for b here? I don't see how you arrived at that answer for b.

So how do I get back to $u(x,y)$ ?

Thanks for your help

**Jason Bourne** · November 14th 2008, 07:16 AM

I think I understand how to do this now.

$\frac{dx}{dt} = 2y$ , and $\frac{dy}{dt} = -x$ and $\frac{du}{dt} = -4xyu^{\frac{1}{2}}$

so $\frac{du}{dx} = -2xu^{\frac{1}{2}}$

thus $u = (\frac{-x^2}{2} + c_1)^2$

and $y^2 + \frac{x^2}{2} = c_2$

from $\frac{dy}{dx}$

so $u(x,y) = (y^2 + C)^2$

with the initial condition $u(x,0) = x^4$

$u(x,y) = (y^2 + x^2)^2$

**shawsend** · November 14th 2008, 07:20 AM

Hi. The solution of:

$a(x,y,u)u_x+b(x,y,u)u_y-c(x,y,u)=0;\quad u=u(x,y)$

can be expressed implicitly as $w(x,y,u)=0$ where $w(x,y,z)$ is a solution of the following:

$a(x,y,z)w_x+b(x,y,z)w_y+c(x,y,z)w_z=0$

However I'm having problems implementing it. Also, I used Mathematica to solve for a and b: Personally I prefer to get the answer any sleazy way I can get it, then go back and figure out how to do it for the purist in the class.

See "Basic Partial Differential Equations" by Bleecker and Csordas. This is covered and is what I'm using for this problem.

**shawsend** · November 14th 2008, 07:48 AM

Alright, alright, I made a mistake: I'm relying too much on Mathematica. We need to solve the equations:

$y^2=-x^2/2+a$

$z^{1/2}=-x^2/2+b/2$

for a and b. This by inspection is obviously:

$a=1/2(x^2+2y^2)$

$b=2(z^{1/2}+x^2/2)$

I don'w know what Mathematica is doing. Anyway, when I use those substitutions, the requisite terms cancel as expected, leaving the desired $w_z$ term. I think we got this one . . .

**shawsend** · November 15th 2008, 04:56 AM

I'd like to finish this as best as I can (it's not rigorous, some holes remain but it's a start):

We now make the transformation:

$r=y^2+\frac{x^2}{2}$

$s=2z^{1/2}+x^2$

$t=z$

and let $v(r,s,t)\equiv w(x,y,z)$

and express the PDE in terms of $v(r,s,t)$ :

$<br /> \begin{aligned}<br /> 2yw_x-xw_y-4xyz^{1/2}w_z&=2y\left(v_r r_x+v_s s_x+v_t t_x\right)\\<br /> &-x\left(v_r r_y+v_s s_y+v_t t_y\right)\\<br /> &-4xyz^{1/2}\left(v_r r_z+v_s s_z+v_t t_z\right)=0\end{aligned}$

$=2xyv_r+4xyv_s-2xyv_r-4xyv_s-4xyz^{1/2}v_t=0$

$-4xyz^{1/2}v_t=0$

Note I left x, y, z untransformed into the r,s, and t variables since this is not necessary at this point.

I'm left then with the PDE: $v_t=0$ which is easily integrated to a constant function in the variables r and s:

$v(r,s,t)=C(r,s)$

A solution for the PDE in terms of w is then:

$w(x,y,x)=C(y^2+x^2/2,2z^{1/2}+x^2/2)$ where C is any $C^1$ function. For example, the function:

$w(x,y,z)=y^2+x^2/2$ is a solution. Another one is:

$w(x,y,z)=C(\alpha,\beta)=\alpha+\beta=y^2+x^2/2+2z^{1/2}+x^2/2$

The solution of the original PDE is then $w(x,y,u)=0$ In the last case, that would be:

$y^2+3/2 x^2+2u^{1/2}=0$ is an implicit solution to the PDE. We can check this easily:

$u^{1/2}=-\frac{y^2+3/2x^2}{2}$
(slight problem here with the negative sign but let's continue anyway). Let a solution to the original PDE be:

$u=\frac{(y^2+3/2x^2)^2}{4}$

back-substituting I get:

$\frac{2y}{4}\left[2(y^2+3/2x^2)3x\right]-\frac{x}{4}\left[2(y^2+3/2x^2)2y\right]-\frac{4xy}{2}(y^2+3/2x^2)=0$

**Jason Bourne** · November 15th 2008, 08:10 AM

Originally Posted by Jason Bourne

I think I understand how to do this now.

$\frac{dx}{dt} = 2y$ , and $\frac{dy}{dt} = -x$ and $\frac{du}{dt} = -4xyu^{\frac{1}{2}}$

so $\frac{du}{dx} = -2xu^{\frac{1}{2}}$

thus $u = (\frac{-x^2}{2} + c_1)^2$

and $y^2 + \frac{x^2}{2} = c_2$

from $\frac{dy}{dx}$

so $u(x,y) = (y^2 + C)^2$

with the initial condition $u(x,0) = x^4$

$u(x,y) = (y^2 + x^2)^2$

Thanks for your help.

**shawsend** · November 15th 2008, 10:26 AM

Originally Posted by Jason Bourne

I think I understand how to do this now.

$\frac{dx}{dt} = 2y$ , and $\frac{dy}{dt} = -x$ and $\frac{du}{dt} = -4xyu^{\frac{1}{2}}$

so $\frac{du}{dx} = -2xu^{\frac{1}{2}}$

thus $u = (\frac{-x^2}{2} + c_1)^2$

and $y^2 + \frac{x^2}{2} = c_2$

from $\frac{dy}{dx}$

so $u(x,y) = (y^2 + C)^2$

with the initial condition $u(x,0) = x^4$

$u(x,y) = (y^2 + x^2)^2$

Ok, I see what you mean. Sorry about that.

Thread: First Order Semi-Linear PDE

LinkBack

Thread Tools

Display

First Order Semi-Linear PDE

Similar Math Help Forum Discussions

[SOLVED] 1st Order Semi Linear PDE xu_x+yu_y=u+3

[SOLVED] Form of Semi Linear Linear PDE's

[SOLVED] Semi Linear PDE

How can first-order logic be at the same time complete and only semi-decidable?

Third Order Semi Implicit Runge Kutta Method

Search Tags