How can I solve this rather simple looking integral equation?

Question

I was working on a physics problem and I have reduced it down to a simple integral equation with two boundary conditions:

$$\int_0^{l-t}y(x, t) dx = lh$$

With the conditions:

$$y(0, t) = y(l-t, t) = h$$

I am looking for $y(x, t)$. $l$, $h$ and $t$ are positive real numbers.

Unfortunately, I don't know how to solve it. So, I need a little help there.

Also, I am curious if this same problem could be converted into a differential equation.

Answer 1

Let the antiderivative (wrt $x$) of $y(x,t)$ be $Y(x,t)$. Then

$$Y(l-t,t)-Y(0,t)=lh$$ and deriving on $l$,

$$y(l-t,t)=h.$$

The given conditions are automatically fulfilled, and the last identity establishes an implicit relation between $l$ and $t$, which depends on the particular expression of $y$. But you can't solve for $y$ just with the available information.

Answer 2

Has this something to do with FEM?

I would do:

at $x=0$, $$y(x, t)=h$$

at $x=\frac{1}{2} \cdot(l-t)$, $$y(x, t)=h+\frac{2ht}{l-t}$$

at $x=(l-t)$, $$y(x, t)=h$$

Is this what you are looking for?

Answer 3

There are insufficient conditions to determine a unique solution. You can, however find a particular solution. The most obvious is one that's constant in $x$ $$ y(x,t) = \frac{hl}{l-t} $$

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If you want to be fancy, you can also look for a Fourier-like solution $$ y(x,t) = h + \sum_{n=1}^{\infty} c_n(t) \sin\left( \frac{n\pi}{l-t}x \right) $$

Then $$ \int_0^{l-t} y(x,t)\ dx = h(l-t) + \sum_{k=0}^{\infty}c_n(t)\frac{l-t}{n\pi}(1-(-1)^n) = lh $$

$$ \frac{2(l-t)}{\pi}\sum_{k=0}^\infty \frac{c_{2k+1}(t)}{2k+1} = th $$

which gives a family of solutions $$ y_n(x,t) = h + \frac{n\pi th}{2(l-t)}\sin\left(\frac{n\pi}{l-t}x\right), \quad n = 1,3,5, \dots $$