0

This question is from a mathematics competition question paper.

We are given the integral equation $$ y(x)= f(x) + \int_{0}^x \sin(x-t)y(t) dt $$

Then $y$ is given by:

  1. $y(x) = f(x) + \int_{0}^x (x-t)f(t) dt$
  2. $y(x) = f(x) - \int_{0}^x (x-t)f(t) dt$
  3. $y(x) = f(x) - \int_{0}^x \cos(x-t)f(t) dt$
  4. $y(x) = f(x) - \int_{0}^x \sin(x-t)f(t) dt$

From what I know, the solution is $$y(x) = f(x) + \int_{0}^x R(x,t,\lambda)f(t) dt,$$ where $R(x,t,\lambda)$ is the resolvent kernel.

Which is given by $$R(x,t,\lambda)= \sum_{n=0}^\infty \lambda^n K_{n+1}(x,t),$$ and $K_{n+1}(x,t)= \int_{t}^x K(x,t)K_n(x,t) $ with $K_1(x,t) = K(x,t)$.

But in this case, the resolvent kernel is not easy to find so I suppose there is something else used here. Please guide me.

sunspots
  • 762
Kashif
  • 710
  • 1
    Wouldn't it make the most sense to take Laplace transforms on both sides and use the Convolution Theorem? – Stephen Donovan Oct 31 '21 at 10:52
  • I am not very familiar with the Laplace transforms, can you please share something where I can educate myself, also most probably we don't need to apply that. Since most of the things taught on these topics didn't used Laplace transforms. – Kashif Oct 31 '21 at 10:55
  • You can learn about Laplace transforms here: https://tutorial.math.lamar.edu/Classes/DE/LaplaceIntro.aspx – Stephen Donovan Oct 31 '21 at 10:57
  • Thanks @StephenDonovan – Kashif Oct 31 '21 at 10:59

2 Answers2

2

Use the Laplace transform. Let $Y(p) = \mathcal{L}\left\{y(x)\right\}$ (Laplace transform of $y(x)$), $F(p) = \mathcal{L}\left\{f(x)\right\}$ and $S(p)=\mathcal{L}\left\{\sin(x)\right\} = \frac{1}{p^2+1}$. As $$ y(x)= f(x) + \int_{0}^x \sin(x-t)y(t) dt \Leftrightarrow y(x) = f(x) + (\sin * y)(x) $$ where $*$ denotes convolution. From the properties of the Laplace transform, $\mathcal{L}\left\{(\sin * y)(x)\right\} =\mathcal{L}\left\{\sin(x)\right\} \cdot \mathcal{L}\left\{y(x)\right\}$, so the Laplace transform of given equation is $$ Y(p) = F(p) + S(p) Y(p) \Leftrightarrow Y(p) = \frac{F(p)}{1-S(p)} $$ As $1 - S(p) = 1 - \frac{1}{p^2+1} = \frac{p^2}{p^2+1}$, we get $Y(p) = \left(1 + \frac{1}{p^2}\right)F(p) = F(p) + \frac{1}{p^2}F(p)$. Inverse Laplace transform of $\frac{1}{p^2}$ equal $g(x)=x$, so using convolution property again, one can get $$ y(x) = f(x) + \int_0^x (x-t)f(t)dt $$

Yalikesifulei
  • 2,550
  • It might sound ignorant, but I am not much familiar with these. Can you please tell me are the operations we are using one-one? – Kashif Oct 31 '21 at 11:07
2

The integral on the right-hand side of the equation is the convolution $\sin(x) * y(x),$ where

$$f(x) * g(x) = \int_0^x f(x - t)g(t) dt$$

A very useful property of the convolution is how it behaves with the Laplace transform:

$$\mathcal L\{f(x) * g(x)\} = \mathcal L\{f(x)\} \cdot \mathcal L\{g(x)\}$$

So, let's try taking Laplace transforms of both sides. (Laplace transform table for convenience)

Let $F(s) = \mathcal L\{f(x)\}$ and $Y(s) = \mathcal L\{y(x)\}$:

$$\mathcal L\{y(x)\} = \mathcal L\{f(x) + sin(x) * y(x)\} = \mathcal L\{f(x)\} + \mathcal L\{sin(x)\} \cdot \mathcal L\{y(x)\} \\ Y(s) = F(s) + \frac{1}{s^2 + 1} Y(s) \\ \frac{s^2}{s^2 + 1}Y(s) = F(s) \\ Y(s) = \frac{s^2 + 1}{s^2} F(s) = F(s) + \frac{1}{s^2}F(s)$$

Now we can take inverse Laplace transforms on both sides and reapply the Convolution Theorem to obtain:

$$y(x) = f(x) + x * f(x) = f(x) + \int_0^x (x - t)f(t) dt$$

which lines up with answer choice 1.

Hope this helps!

Stephen Donovan
  • 5,163