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Math Help - Existence and Uniqueness Question

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    Existence and Uniqueness Question

    Can the fundamental theorem of existence and uniqueness be applied to the initial value problem: x' = t^2/x^2, x(0) = 0? In other words, if f(t,x) = t^2/x^2, is f(t,x) and its partial derivative with respect to x continuous on some interval containing the initial conditions? My initial reaction is to say that it is not continuous since there is a zero in the denominator in both equations, however, when (0, 0) is plugged into both equations the result is 0/0. Looking at the limit as (t, x) approaches (0, 0) tells me that it is not continuous. I guess my question is, is it sufficient to just say that there is a zero in the denominator (since x(0) = 0) and the theorem cannot be applied here, or would I need to go through showing the nonexistence of the limit as (t, x) approaches (0, 0)? Thanks.
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    Re: Existence and Uniqueness Question

    Is x a function of t? If so, there should be a solution because the DE is separable...
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    Re: Existence and Uniqueness Question

    Yes, x is a function of t. This is what I'm thinking: the fundamental existence and uniqueness theorem applies only if the function and its partial derivative with respect to x are continuous in t at t(initial) and in x at x(initial) for some interval. In this case, the function is not continuous at x(initial) since t^2/0 is undefined, therefore the theorem does not apply and nothing can be deduced from the theorem. However, upon solving through separation of variables we discover that a solution does indeed exist.

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    Re: Existence and Uniqueness Question

    Yes, that is a separable equation and it can be written as x^2dx= t^2dt and then integrated as \frac{x^3}{3}= \frac{t^3}{3}+ C which is equivalent to x^3= t^3+ c where c= 3C. With the condition that x= 0 when x= 0 we have x^3= t^3. If we are only interested in real number solutions we can write x= t.

    However, the question was whether the "existence and uniqueness theorem" applies here. The answer is "No, it does not" for exactly the reason you give. The right side of the equation is NOT continuous in x and t at t= 0. However, the theorem only says that if the conditions apply, then we can be certain there exist a unique solution. It does NOT say anything about what happens if the conditions are not satisfied. There might exist a unique solution, there might exist more than one solution, or there might be no solution at all.
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