Question #24017

Solve the Cauchy problem equation:
u_x + u_y = u^2, u(x,0) = 1.

Expert's answer

u_x + u_y = u^2, u(x,0) = 1.

Method of characteristics:

dx = dy = du/u^2

1) du/u^2 = dx

Integrate:

-1/u + C1=x => C1 = 1/u + x

2) dx = dy

Integrate:

y + C2=x => C2 = x - y

Therefore, the complete integral of the equation:

F ( 1/u + x, x - y) = 0 or in another form:

u = ( f (x- y) - x )^-1 where f - some differentiable function.

The initial condition u(x,0) = 1.

u(x, 0) = ( f (x) - x )^-1 = 1 => f(x) = x+1

u(x,y) = 1/(x-y+1-x) = 1/(-y+1) = 1/(1-y)

Answer: u (x, y) = 1/(1-y)

Method of characteristics:

dx = dy = du/u^2

1) du/u^2 = dx

Integrate:

-1/u + C1=x => C1 = 1/u + x

2) dx = dy

Integrate:

y + C2=x => C2 = x - y

Therefore, the complete integral of the equation:

F ( 1/u + x, x - y) = 0 or in another form:

u = ( f (x- y) - x )^-1 where f - some differentiable function.

The initial condition u(x,0) = 1.

u(x, 0) = ( f (x) - x )^-1 = 1 => f(x) = x+1

u(x,y) = 1/(x-y+1-x) = 1/(-y+1) = 1/(1-y)

Answer: u (x, y) = 1/(1-y)

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