# Thread: First Order Semi-Linear PDE

1. ## First Order Semi-Linear PDE

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

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

2. 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.

3. 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 $u(x,y)$ ?

4. 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 $u(x,0) = x^4$

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

5. 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.

6. 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 . . .

7. 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 $v(r,s,t)$:


\begin{aligned}
2yw_x-xw_y-4xyz^{1/2}w_z&=2y\left(v_r r_x+v_s s_x+v_t t_x\right)\\
&-x\left(v_r r_y+v_s s_y+v_t t_y\right)\\
&-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:

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

8. 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 $u(x,0) = x^4$

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

9. 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.