In the plane $(x,y)$ find two solutions of initial value problem:

$\frac14u^2_x+uu_y=u $


I am trying to use the method of characteristics, but i am not getting right way

please could you help me.. Thanks in Advance..


Assume a solution u of the form f(x) g(y) to get

(f')^2 g^2 / 4 + f^2 g g' = f g.

Divide by f^2 g to get

1/4 (f'/f)^2 g + g' = 1/f, so

g' = 1/f - 1/4 (f'/f)^2 g.

Think of y as the independent variable and x being held constant. This last equation is a simple differential equation in g which is easily solved. In fact the solution to

g' = b - a g is g = b/a + c exp(- a y), where c is any constant.

Apply this to the formula above to get a general solution to the original pde. Then substitute for the specific condition.

Justin: When I look back at this I see that there is a problem with it, namely g is assumed to be a function only of y, not of x, but it is clearly a function of x. This solution is flawed. Sorry. If I see how to fix it I will write more.


  • $\begingroup$ could you please do it explicitly , because i m preparing for qualifying and it is important for exam too, sorry for bothering you.. $\endgroup$ – Toeplitz Apr 28 '14 at 14:57

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