# Solution of a simple integral equation

Problem Statement
(a) find all continuous functions $f$ satisfying $\int_0^x f(x) \, dx = (f(x))^2+C$ for some constant $C≠0$ assuming that $f$ has at most one 0.

(b) Also find a solution that is 0 on an interval $(-\infty,b]$ with $0 \lt b$, but nonzero for $x \gt b$
Relevant Equations
$\frac d {dx} \int_a^x {f(t)} \, dt = f(x)$
I did the first part, it is part (b) that I'm having trouble understanding. For any $x \lt b$, $f(x)=0$ and $\int_0^x {f(t)} \, dt = 0$ (since $f$ is 0 everywhere from 0 to $b$), which turns the equation $\int_0^x f(t) \, dt = (f(x))^2+C$ into $0=0+C$, which implies $C=0$. But that is the one value that $C$ cannot have, so I don't see how such a function can be a solution to the equation for $x \lt b$...

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#### fresh_42

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Combine the two solutions of a.) at $x=b$.

I'm sorry, I don't really understand what you mean by combining them or how that would help. The solutions of a) are $f(x)=\frac 1 2 x±\sqrt {-c}$. No matter what I define $f$ as for $x=b$ or $x>b$, I just don't see how the function can be a solution of the equation when it is 0 for $x<b$.

The way I see it, if such a solution $f$ to the equation $\int_0^x f(x) \, dx = (f(x))^2+C$ exists such that $f(x)=0$ for $x<b$, then inserting some $x'<b$ into the equation yields $0=C$, which is a contradiction. What am I missing here?...

#### fresh_42

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You are missing an entire solution. I would have told you where you were mistaken if you had written what you did.

You are missing an entire solution. I would have told you where you were mistaken if you had written what you did.
My solution for (a) is as follows:
Differentiating the equation yields $2f(x)f'(x)=f(x)$, so in points where $f(x)≠0$, $f'(x)=\frac 1 2$. So in any interval where $f$ is nonzero, $f(x)=\frac 1 2 x+a$, and since $f$ has only 1 zero this gives 2 possible values of $a$, one for the interval before the 0 of $f$ and one for the interval after it. However, since $f$ is continuous, these 2 values must be the same, therefore $f(x)=\frac 1 2 x+a$ for all $x$. Inserting this into the original equation yields $\int_0^x {\frac 1 2 x+a} \, dx = (\frac 1 2 x+a)^2+C$ or $\frac 1 4 x^2+ax=\frac 1 4 x^2+ax+a^2+C$ or $a=±C$.

These are the only solutions I reached. The solution book does not show any additional solutions, just this.

#### fresh_42

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You made an assumption in the first line which is not necessarily justified! What if there is no point where $f(x)\neq 0\,?$

You made an assumption in the first line which is not necessarily justified! What if there is no point where $f(x)\neq 0\,?$
Part (a) states that $f$ has at most one zero. Even if $f$ is the zero function, the problem states that $C≠0$, so I don't see how such a function can be a solution.

#### fresh_42

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2018 Award
Sorry, I missed that. I thought $f=0$ would be a solution and could be combined with a straight for a suitable constant. Anyway, part b) requires a solution with definitely more than one zero. Both together is impossible, so you need to drop this requirement.

I just realised I copied the equation wrong... it needs to be:

$\int_0^x f(t) \, dt = (f(x))^2+C$

But this still brings me to my original problem, I don't see how you can have any solution at all for part b). It seems I am missing something very basic here, which is not uncommon for me...

The solution $f$ in part b) has to answer the requirement that $f(x)=0$ for all $x≤b$, and the equation has to answer the requirement that $C\neq0$, but that seems to cause a contradiction whenever inserting any $x<b$ into the equation. Say we insert $x=\frac b 2$, then by the equation:

$\int_0^{\frac b 2} f(t) \, dt = (f({\frac b 2}))^2+C$

But since $f(x)=0$ on $[0,\frac b 2]$, $\int_0^{\frac b 2} f(t) \, dt=0$ and $(f({\frac b 2}))^2$, which yields $0=C$, which contradicts the condition...

#### fresh_42

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I get two solutions $F(x) = \left( \dfrac{x}{2} \pm \gamma \right)^2 - \gamma^2$ with $\gamma = +\sqrt{-C} > 0$ for a negative value $C$. These are two parabolas, opened at the top. Both satisfy a.).

For part b.) we need an intersection of these parabolas with $G(x)\equiv 0$ for $x \leq b<0$. This can only be done with one of the parabolas at $x=-4\gamma$ which yields $C= -\dfrac{b^2}{16}$. The combined function of $G(x)$ for $x\leq b$ and $F(x)$ for $x>b$ is the only possibility I can imagine.

#### pasmith

Homework Helper
I get two solutions $F(x) = \left( \dfrac{x}{2} \pm \gamma \right)^2 - \gamma^2$ with $\gamma = +\sqrt{-C} > 0$ for a negative value $C$. These are two parabolas, opened at the top. Both satisfy a.).

For part b.) we need an intersection of these parabolas with $G(x)\equiv 0$ for $x \leq b<0$. This can only be done with one of the parabolas at $x=-4\gamma$ which yields $C= -\dfrac{b^2}{16}$. The combined function of $G(x)$ for $x\leq b$ and $F(x)$ for $x>b$ is the only possibility I can imagine.
If $f$ is quadratic then its integral is cubic but its square is quartic, so it can't satisfy the integral equation. This suggests the existence of affine solutions, so the easiest way to solve this is to assume $f(x) = Ax + B$, substitute that into the integral equation, and compare coefficients of powers of $x$.

Don't forget that if $x < 0$ then the left hand side is $-\int_{x}^0 f(x)\,dx$.

The second part is asking for an $f$ which is zero on $[0,b)$ and for $x \geq b$ must satisfy $$\int_b^x f(t)\,dt = (f(x))^2 + C$$ which can be solved in the same manner as part (a).

However for $x < b$ if $f$ is identically zero then we must have $C = 0$. There is no avoiding that.

#### fresh_42

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2018 Award
If fff is quadratic then its integral is cubic but its square is quartic, so it can't satisfy the integral equation.
$F(x) = \int_0^x f(t)\,dt$ can, which is why I switched to capital letters.

It seems my misunderstanding of part b) just boils down to poor articulation of the question. I thought part b) requested a function that is 0 on $(-\infty, b]$ and nonzero on $(b,\infty)$ and is a solution to the equation no matter what $x$ is inserted into it. But it seems the function only needs to be a solution for $x>b$, which is rather weird since I assume a condition like this needs to be explicitly stated, but I see no other way for this question to make sense...

However, the solution for part b) in the answer book goes as follows:
"On the other hand, for $b=\sqrt {-C}$ we can also choose $f$ to be 0 on $(-\infty,b]$, and $f(x)=\frac x 2 - \frac b 2$ for $x\geq b$"

Which also seems to not make sense, since even for $x\geq b$, $\int_0^x f(x) \, dx = (f(x))^2$, which yet again means C=0! And the question can't want me to drop the C=0 condition since it asks to drop it only on part c). It can't be that the author forgot to mention it because then he wouldn't have bothered to say $b=\sqrt {-C}$ in the solution if $C=0$. So I don't know anymore...

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Well i think I pretty much got it. I just assumed C=0 on part b) and ignored the answer book, everything seemed to have worked out when I did that.

#### fresh_42

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My proposal would have been
$$f(x) = \begin{cases} 0& \text{ if } x<b\\ \frac{1}{2}x + \frac{1}{4}b&\text{ if } x \geq b \end{cases}$$
with $F(x)=\int_0^x f(t)\,dt = \frac{1}{4}\left( x^2+bx \right)$ and $C=-\frac{1}{16}b^2\,.$

#### MathematicalPhysicist

Gold Member
My solution for (a) is as follows:
Differentiating the equation yields $2f(x)f'(x)=f(x)$, so in points where $f(x)≠0$, $f'(x)=\frac 1 2$. So in any interval where $f$ is nonzero, $f(x)=\frac 1 2 x+a$, and since $f$ has only 1 zero this gives 2 possible values of $a$, one for the interval before the 0 of $f$ and one for the interval after it. However, since $f$ is continuous, these 2 values must be the same, therefore $f(x)=\frac 1 2 x+a$ for all $x$. Inserting this into the original equation yields $\int_0^x {\frac 1 2 x+a} \, dx = (\frac 1 2 x+a)^2+C$ or $\frac 1 4 x^2+ax=\frac 1 4 x^2+ax+a^2+C$ or $a=±C$.

These are the only solutions I reached. The solution book does not show any additional solutions, just this.
They asked you in (a) to find all the continuous functions that satisfy this equation, which means there might be functions that aren't differentiable, so you found some functions that are $C^1$.

But for checking for all $f\in C^0$ you need perhaps to invoke iteration:

$$f(x) = \sqrt{\int_0^x f(t) -C} = \sqrt{\int_0^x \sqrt{\int_0^t f(s)ds-C}dt-C}$$
Now you could square everything and get: $$f^2(x)+C = \int_0^x \sqrt{\int_0^t f(s)ds-C}dt = \int_0^x f(t)dt$$.

But I don't see how to find from this method all the continuous functions.

My proposal would have been
$$f(x) = \begin{cases} 0& \text{ if } x<b\\ \frac{1}{2}x + \frac{1}{4}b&\text{ if } x \geq b \end{cases}$$
with $F(x)=\int_0^x f(t)\,dt = \frac{1}{4}\left( x^2+bx \right)$ and $C=-\frac{1}{16}b^2\,.$
That solution seems only to work for $x>b$, else the equation reads $-\frac 1 {16} b^2=0$. Also the function is not continuous at $b$, so I don't think it is viable, considering the conditions for the solution. I just assumed that $C=0$ and everything worked out with $f(x)=\frac 1 2 x - \frac 1 2 b$ for $x>b$, $f(x)=0$ for $x\leq b$, it doesn't seem there is any other viable way of getting a solution that answers the conditions.

They asked you in (a) to find all the continuous functions that satisfy this equation, which means there might be functions that aren't differentiable, so you found some functions that are $C^1$.

But for checking for all $f\in C^0$ you need perhaps to invoke iteration:

$$f(x) = \sqrt{\int_0^x f(t) -C} = \sqrt{\int_0^x \sqrt{\int_0^t f(s)ds-C}dt-C}$$
Now you could square everything and get: $$f^2(x)+C = \int_0^x \sqrt{\int_0^t f(s)ds-C}dt = \int_0^x f(t)dt$$.

But I don't see how to find from this method all the continuous functions.
As far as I'm aware, the solution for the equation is necessarily differentiable at all points but one. Since $\int_0^x f(x) \, dx$ is differentiable by the first FTOC, and differs only by a constant from $f(x)^2$, the latter must also be differentiable, which only occurs if $f'(x)$ exists unless $f(x)=0$, which only occurs at a single point in part a).

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#### fresh_42

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2018 Award
That solution seems only to work for $x>b$, ...
No problem. $b < 0$ and the integral from $0$ to $x$. So as long as we don't think of a reverse integration path, there will be no conflict. And as signs are so important in this problem, a reversed integration would be quite an arbitrary objection.

No problem. $b < 0$ and the integral from $0$ to $x$. So as long as we don't think of a reverse integration path, there will be no conflict. And as signs are so important in this problem, a reversed integration would be quite an arbitrary objection.
Well that addresses one issue, unfortunately the question explicitly says $b$ needs to be positive, so we are brought into another one, and the discontinuity at $b$ still exists. Honestly, this whole problem seems to be a very simple exercise becoming overly complicated probably because of an author's mistake, which would not be the first (Spivak).

However I'll take the opportunity to educate myself a bit more. For any $x<b$ in your function, $\int_0^x f(t)\,dt = \int_0^b f(t)\,dt =\frac 1 2 b^2$, so the equation reads $\frac 1 2 b^2=\int_0^x f(t)\,dt =((f(x))^2+C=C=-\frac 1 {16} b^2$. So how does that work out for those x's?

#### MathematicalPhysicist

Gold Member
@Adgorn , apparently it's a popular question:
I believe there's a solution manual in the web for the third edition, if you want to check your work.

@Adgorn , apparently it's a popular question:
I believe there's a solution manual in the web for the third edition, if you want to check your work.
I never manage to find copies of my questions, perhaps I require more training in the art of forum digging. I actually have the physical copies of both the 4th addition and the answer book, I only resort to online help if the answer book does not provide a sufficient explanation, as was the case here. At this point I'm fairly certain there is some mistake in the question. The author probably forgot to mention that C can be equal to 0 in part b). That would explain all my troubles, or at least so it seems.

"Solution of a simple integral equation"

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