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what can you say about the existance and uniqness of solution for the below initial value problem without solving it,

first derivative of y(x)=Absolute value of y(x) and y(0)=0.

I didn't know the answer, so I forwarded your question to a math group and got these responses:

1:

The unique solution is y = 0.

I can't do this problem without solving it: it's plain that the unique solution family is y = Ce^x where C >= 0. But y(0) = 0 forces C = 0.

2:

If f(x, y) and ∂f/∂y are both continuous in a neighbourhood of (x₀, y₀), then dy/dx = f(x, y), y(x₀) = y₀ has a unique solution in a neighbohood of (x₀, y₀). This is the fundamental existence theorem for differential equations. Now f(x, y) = |y| is continuous in some open inteval about 0, but ∂f/∂y isn't. So we cannot conclude from the existence theorem that a solution exists, much less that there is a unique solution. The trouble here, however, is that the existence theorem is not "if and only if". It has an "if" part but no "only if". Intuitively I don't think your equation has a solution in any open interval about 0.

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Comment | Dear Janet, Thank you for your kind response. But as you said, the problem has only one solution, it is y(x)=0. It can be seen by solving it. But my problem is how can we prove it without solving it. There is another theorem about it except second item you mentioned. It is Picard- Lindolorf theorem and related with Lipschitz continuity, but I do not know how can I apply it. But I really thank for your great help and effort. |

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