Can You Solve This Complex Nonlinear Differential Equation?

[Luisvzz]

Believe me or not, I've just received a mail from my differential equations teacher asking me for help to solve this DE in wolphram alpha or maple but i think I am doing something wrong because they won't solve it.

It seems to be a

Partial (completely) non linear diferential equation:

u(x,y) = 2a²(u_x)² - 2bu_x-c²u_y

Thank you for your help guys, you are awesome ;)

 

[Vargo]

Well, there are many solutions. In order to get a unique solution you need to first specify what the values of u will be over a "non-characteristic" curve.

Example. Find the solution if u(x,0) = g(x).

We use the method of characteristics and end up with a bunch of equations:

1. x = x_0 + 2bt + 4a^2g'(x_0)(e^{-t}-1), and
2. y = c^2t

You have to solve those equations for t and x_0 in terms of x and y. Then you substitute into
$$ u = 2a^2g'(x_0)^2e^{-2t} - (2bg'(x_0)+c^2h(x_0))e^{-t}.$$
Here, h(x) satisfies the equation g(x)+c^2h(x)+2bg'(x)-2a^2g'(x)^2=0.

Crystal clear, right?

Ok, I'll do a special case so you can at least take home one bona fide solution. Let's suppose the initial condition is u(x,0)= x. So g(x)=x,\, g'(x)=1,\, h(x)=(2a^2-2b-x)/c^2.

From equation 2, we find that t = y/c^2. Substituting that into equation 1, we find x-x_0 = (2b/c^2)y +4a^2(e^{-y/c^2} - 1). Therefore,
$$x_0 = x - (2b/c^2)y -4a^2(e^{-y/c^2} - 1).$$
So we can plug in our expressions for t and x_0 into the formula for u and obtain the solution.

$$ u(x,y) = (x-(2b/c^2)y-4a^2(e^{-y/c^2}-1)-2a^2)e^{-y/c^2} +2a^2e^{-2y/c^2}.$$

[EDIT] I found an error and corrected it, but there might be (probably are) other errors. You'd have to plug that back into the original equation to see.

 

[Vargo]

The reason Wolfram alpha did not work is that it solves Ordinary Differential equations. This is a partial differential equation. As such is it more complex. If you want to refer your teacher to a reference on this, here is a link that explains the theory (but does not offer any concrete examples that I noticed):

http://www.stanford.edu/class/math220a/handouts/firstorder.pdf

 

[Luisvzz]

Well, there are many solutions. In order to get a unique solution you need to first specify what the values of u will be over a "non-characteristic" curve.

Example. Find the solution if u(x,0) = g(x).

We use the method of characteristics and end up with a bunch of equations:

1. x = x_0 + 2bt + 4a^2g'(x_0)(e^{-t}-1), and
2. y = c^2t

You have to solve those equations for t and x_0 in terms of x and y. Then you substitute into
$$ u = 2a^2g'(x_0)^2e^{-2t} - (2bg'(x_0)+c^2h(x_0))e^{-t}.$$
Here, h(x) satisfies the equation g(x)+c^2h(x)+2bg'(x)-2a^2g'(x)^2=0.

Crystal clear, right?

Ok, I'll do a special case so you can at least take home one bona fide solution. Let's suppose the initial condition is u(x,0)= x. So g(x)=x,\, g'(x)=1,\, h(x)=(2a^2-2b-x)/c^2.

From equation 2, we find that t = y/c^2. Substituting that into equation 1, we find x-x_0 = (2b/c^2)y +4a^2(e^{-y/c^2} - 1). Therefore,
$$x_0 = x - (2b/c^2)y -4a^2(e^{-y/c^2} - 1).$$
So we can plug in our expressions for t and x_0 into the formula for u and obtain the solution.

$$ u(x,y) = (x-(2b/c^2)y-4a^2(e^{-y/c^2}-1)-2a^2)e^{-y/c^2} +2a^2e^{-2y/c^2}.$$

[EDIT] I found an error and corrected it, but there might be (probably are) other errors. You'd have to plug that back into the original equation to see.

Well it worked! You are awesome! Thank you a lot.