I have the following equation and want to find $f(x)$

$f(x)=x $ + $\int_{0}^{1} (xy^2 + yx^2)$ $f(y)dy$

When i tried to get a solution from wolfram alpha, it gave me an answer but says it is solving a Fredholm Integral Equation. I am a high schooler and have no idea what that means. Is there any simpler method to solve the above problem for $f(x)$?
Also if it is convenient can one please explain what a fredholm integral equation is and how to solve it?

Edit: Can someone help me?

  • $\begingroup$ If you look at the RHS, you can see that it has the form $ax^2+bx$. Try plugging in $f\left(x\right)=ax^2+bx$ and see what happens. $\endgroup$ – Joshua Tilley Jul 24 '18 at 12:54

This is not a topic you'd understand unless you know some graduate level analysis, so don't feel bad!

This is a special type of equation called a "Fredholm" Integral Equation. These arise in many areas of engineering and physics, and are a very interesting mathematical topic unto themselves.

Generally speaking, a Fredholm integral is written in the following form:

$u(x)=f(x)+\lambda \int_{a}^{b}K(x,y)u(y)dy$. The problem is you are given a function $f(x)$, and another function $K(x,y)$, and your goal is to solve for $u(x)$.

Now let's first discuss $K(x,y)$. This is something called the "Kernel", and the solution of the Integral Equation depends heavily on this Kernel. Now the Kernel notation $K(x,y)$ may lead you to think this a function of two variables (not encountered until calculus III). It's not necessarily wrong to think of it this way, but essentially, all you are doing is fixing an $x$ (treating it like a constant), which comes from your interval $[a,b]$ and you integrate with respect to $y$.

Remember integrating with respect to $y$ means you have a function of the variable $y$ and you want to find the anti-derivative/area under the curve. Example: $\int y^2 dy=y^3/3+C$.

Your Kernel is $K(x,y)=xy^2+x^2y$, and has a special property. It is a "separable" Kernel, which means it can be written as a sum and product of functions $g_n(x), h_n(y)$. When a Kernel is separable, it can be solved using linear algebra methods (also a slightly more advanced course than calculus).

I will edit my answer with the full solution included if you wish.

One final point: sometimes, Integral equations have no solutions, it depends on $\lambda$, and also on the "energy" of the Kernel. It turns out that if $\lambda$ obeys a certain inequality, then the solution to the Integral Equation not only exists, it is the only one, sometimes there are none.

There a lot of other possibilities, for example, if your Kernel is not separable. The Kernel function $K(x,y)$ doesn't even need to be continuous. The solution depends a lot on what kind of function you are working with. You also can have your function $u(y)=tan(e^y)$ which is a non-linear function. When you throw in non-linear terms, the equations become harder to solve in general, and we need to use computational methods to get an approximate solution (which is yet another topic unto itself).


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