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Show that $$F(f)(t) = t^2 + \frac{t}{3}f(t) + \frac{1}{5}\int_0^t e^uf(u) du$$ is a contraction on $(C[0, 1), d_u)$.

Deduce that the differential equation $$(15 − 5t)\frac{df}{dt} = (5 + 3e^{t})f + 30t$$ has a unique solution in $C[0, 1]$.

Answer I have completed the first part and shown that $F$ is a contraction on $C[0, 1]$. Then by the contraction mapping theorem there exists a fixed point $f \in C[0, 1]$ such that $F(f)(t) = f(t)$

So I have $f(t) = t^2 + \frac{t}{3}f(t) + \frac{1}{5}\int_0^t e^uf(u) du$

and if I rearrange and differentiate I get $$(15 − 5t)\frac{df}{dt} = (5 + 3e^{t})f + 30t$$

So I know $f$ is a solution to the differential equation. But I have to now show it is the only solution. Normally that is done by taking another function, $g$, from $C[0, 1]$ that satisfies the equation and showing that it must equal $f$. However I am unable to make that work here.

And in fact, I don't understand how there can be a unique solution anyway when we are given no initial condition such as $x(0) = 1$...and then if we had $f(t) = t^2 + \frac{t}{3}f(t) + \frac{1}{5}\int_0^t e^uf(u)du$ + K, where $K$ is any constant, this will also satisfy the equation. So is this a trick question and the differential equation doesn't ac

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The contraction has a unique fixed point. –  Pragabhava Nov 19 '12 at 1:42
    
Yes and this is used to show that a solution does exist for the differential equation. But I need to show that this solution is unique. –  sonicboom Nov 19 '12 at 7:54
    
The fixed point is the solution ;) –  Pragabhava Nov 19 '12 at 17:16
    
Yes, it is the solution. But it has to be shown that it is the only solution. For instance, if I had $f(t) = t^2 + \frac{t}{3}f(t) + \frac{1}{5}\int_0^t e^uf(u) du + 9999999999$ I could rearrange and differentiate (which would get rid of the constant), and it would satisfy the differential equation. So why is it that $f(t) = t^2 + \frac{t}{3}f(t) + \frac{1}{5}\int_0^t e^uf(u) du$, i.e. with no constant, is the unique solution to the equation? –  sonicboom Nov 19 '12 at 18:10
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No is not, as it doesn't satisfy the initial condition. The solution is unique when initial conditions are imposed. You should really study the Piccard's Theorem. The uniqueness is consequence of the contraction. As the operator is a contraction, the iterations converge to the fixed point, which is unique, i.e., they cannot converge to any other point. Also, the addition of the constant works because this ode is linear; Piccard's is more general and it works whit nonlinear odes, where you cannot add constants blindly. –  Pragabhava Nov 19 '12 at 18:24

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