Fredholm Integral Equation Numerically

[member 428835]

Homework Statement

A specific problem of the Fredholm integral equation is given as

$$\phi(x) = \pi x^2+\int_0^\pi3(0.5\sin(3x)-tx^2)\phi(t)\,dt$$

and the exact solution is $\phi(x) = \sin 3x$.

Homework Equations

Nothing comes to mind.

The Attempt at a Solution

I'm unsure how to approach solving this numerically. I can immediately tell that $\phi(0) = 0$. I'm trying to think of a simple procedural way of numerically solving this equation. Something like an Euler scheme. Any ideas or where to start?

 

[eys_physics]

You can, e.g., use iteration. That is,
$$\phi_n(x)=\pi x^2+\int_0^{\pi} 3(0.5 \sin (3x)-tx^2)\phi_{n-1}(t) dt$$
with $n=1, 2, ...$ and $\phi_0=0$. In this way you can compute the solution for the required values of $x$. One alternative is to discretize the integral which leads to a linear system of equations.

 

[Ray Vickson]

Homework Statement

A specific problem of the Fredholm integral equation is given as

$$\phi(x) = \pi x^2+\int_0^\pi3(0.5\sin(3x)-tx^2)\phi(t)\,dt$$

and the exact solution is $\phi(x) = \sin 3x$.

Homework Equations

Nothing comes to mind.

The Attempt at a Solution

I'm unsure how to approach solving this numerically. I can immediately tell that $\phi(0) = 0$. I'm trying to think of a simple procedural way of numerically solving this equation. Something like an Euler scheme. Any ideas or where to start?

An approach I saw once in a book long ago was a finite-difference scheme, where you approximate the integral by a sum, for example, using the tranpezoidal rule:
$$h(t) + \int_0^t f(x) \,dx \approx h(n \Delta x) +\frac{1}{2} f(0) \Delta x + \sum_{i=1}^{n-1} f(i \Delta x) \Delta x + \frac{1}{2} f(n \Delta x) \Delta x,$$
when $t = n \Delta x$. Using that you can get a set of coupled linear equations for $f(\Delta x), f(2 \Delta x) , \ldots, f(n \Delta x)$. The equations have, essentially, a triangular form, so we get $f(\Delta x)$ in terms of $f(0)$ and $h(\Delta x)$, get $f(2 \Delta x)$ in terms of $h(2 \Delta x), f(0), f(\Delta x)$, etc.

 

[member 428835]

You can, e.g., use iteration. That is,
$$\phi_n(x)=\pi x^2+\int_0^{\pi} 3(0.5 \sin (3x)-tx^2)\phi_{n-1}(t) dt$$
with $n=1, 2, ...$ and $\phi_0=0$. In this way you can compute the solution for the required values of $x$.

I like the second suggestion that you suggest and also that Ray Vickerson suggested. However, are you saying in your first suggestion that $\phi(x) = \lim_{n\to\infty}\phi_n(x)$? If so can you explain how?

 

[member 428835]

An approach I saw once in a book long ago was a finite-difference scheme, where you approximate the integral by a sum, for example, using the tranpezoidal rule:
$$h(t) + \int_0^t f(x) \,dx \approx h(n \Delta x) +\frac{1}{2} f(0) \Delta x + \sum_{i=1}^{n-1} f(i \Delta x) \Delta x + \frac{1}{2} f(n \Delta x) \Delta x,$$
when $t = n \Delta x$. Using that you can get a set of coupled linear equations for $f(\Delta x), f(2 \Delta x) , \ldots, f(n \Delta x)$. The equations have, essentially, a triangular form, so we get $f(\Delta x)$ in terms of $f(0)$ and $h(\Delta x)$, get $f(2 \Delta x)$ in terms of $h(2 \Delta x), f(0), f(\Delta x)$, etc.

Hey Ray VIckerson, so in my situation wouldn't I would have something like $$
\phi(x) = \pi x^2 +\frac{1}{2} 3(0.5\sin(3x)-\pi x^2))\phi(\pi)+\sum_{i=1}^{n-1}3\left(0.5\sin(3x)-\frac{i\pi}{n} x^2\right)\phi\left(\frac{i\pi}{n}\right)
$$
How does this give us a system of equations?

 

[Ray Vickson]

Homework Statement

A specific problem of the Fredholm integral equation is given as

$$\phi(x) = \pi x^2+\int_0^\pi3(0.5\sin(3x)-tx^2)\phi(t)\,dt$$

and the exact solution is $\phi(x) = \sin 3x$.

Homework Equations

Nothing comes to mind.

The Attempt at a Solution

I'm unsure how to approach solving this numerically. I can immediately tell that $\phi(0) = 0$. I'm trying to think of a simple procedural way of numerically solving this equation. Something like an Euler scheme. Any ideas or where to start?

Hey Ray VIckerson, so in my situation wouldn't I would have something like $$
\phi(x) = \pi x^2 +\frac{1}{2} 3(0.5\sin(3x)-\pi x^2))\phi(\pi)+\sum_{i=1}^{n-1}3\left(0.5\sin(3x)-\frac{i\pi}{n} x^2\right)\phi\left(\frac{i\pi}{n}\right)
$$
How does this give us a system of equations?

If we write the equation as
$$\phi(x) = f(x) + \int_0^{\pi} K(x,t) \phi(t) \, dt, $$
then when we discretize $x$ and $t$ to $0,h, 2h, \cdots, Nh = \pi$ and use an integration scheme of the form
$$\int_0^{\pi} F(t) \, dt \approx \sum_{i=0}^N c_i h F(ih), \hspace{4em} (1)$$
then the integral equation is approximated as
$$p_i = f_i + \sum_{j=0}^N h C_{ij} p_j,$$
where $p_i = \phi(ih), f_i = f(ih)$ and $C_{ij} = c_j K(ih,jh)$. The variables to be solved for are the $p_0, p_1, \ldots, p_N.$

Note that different integration schemes just correspond to different constants $c_i$ in (1). The trapezoidal rule uses $c_0=c_N = 1/2$ and $c_i = 1, \,1 \leq i \leq N-1$. Simpson's rule would correspond to some other set of constants $c_i$.

However, that is not the way I would finally decide to solve your specific problem. Because of the special structure of your right-hand-side, your equation is easily solvable by elementary methods. Unfortunately, I cannot say more without giving away the whole solution. All I can suggest is that you look very carefully at the actual form of your equation. If you still have not got it a week or two after turning in your assignment for marking, I will be happy to outline the best method then.

 

[eys_physics]

I like the second suggestion that you suggest and also that Ray Vickerson suggested. However, are you saying in your first suggestion that ϕ(x)=limn→∞ϕn(x)ϕ(x)=limn→∞ϕn(x)\phi(x) = \lim_{n\to\infty}\phi_n(x)? If so can you explain how?

Hey, Josh. I'm sorry for the delay of my reply, but I was busy with other things yesterday. You are correct in that $\phi(x)=\lim_{n \rightarrow \infty}\phi_n(x)$.
I cannot give a formal proof of that but I can give you some ideas. For this purpose, I will consider the slightly more general equation
$$f(x)=g(x)+\int_a^b h(x,t) f(t) dt$$.
We can do one iteration by inserting again the definition (given by the equation) of $f(t)$ on the right-hand side, i.e.
$$f(x)=g(x)+\int_a^b h(x,t)[g(t)+\int_a^b h(t,t_1)f(t_1)dt_1]$$.This same process can now be continued arbitrarily number of times leading to more integrals on the right-hand side.

Consider now the method based on iteration I proposed. Then,
$$f_0=0,$$
$$f_1=g(x)+\int_a^b h(x,t) f_0(y) dy$$,
$$f_2=g(x)+\int_a^b h(x,t) f_1(y)dy=g(x)+\int_a^b h(x,t)[g(t)+\int_a^b h(t,t_1)f_0(y_1)dy_1]$$, etc

One can therefore see, that the iteration leads to an equation of the type as above. It can also be shown that $\phi(x)=\lim_{n \rightarrow \infty}\phi_n(x)$.

The second method I proposed is quite similar to the one proposed by Ray but a bit simpler. Consider again the equation
$$f(x)=g(x)+\int_a^b h(x,t) f(t) dt$$.
By Gauss-Legendre integration or other quadrature method, one can write
$$f(x)=g(x)+\sum_{k=1}^n w_k h(x,t_k) f(t_k)$$
where $t_k$ are the nodes and $w_k$ the corresponding weights. By considering a mesh of discrete points $x_i, i=1, N$, you obtain a system of linear equations which can be solved by standard methods.

 

[member 428835]

If we write the equation as
$$\phi(x) = f(x) + \int_0^{\pi} K(x,t) \phi(t) \, dt, $$
then when we discretize $x$ and $t$ to $0,h, 2h, \cdots, Nh = \pi$ and use an integration scheme of the form
$$\int_0^{\pi} F(t) \, dt \approx \sum_{i=0}^N c_i h F(ih), \hspace{4em} (1)$$
then the integral equation is approximated as
$$p_i = f_i + \sum_{j=0}^N h C_{ij} p_j,$$
where $p_i = \phi(ih), f_i = f(ih)$ and $C_{ij} = c_j K(ih,jh)$. The variables to be solved for are the $p_0, p_1, \ldots, p_N.$

Note that different integration schemes just correspond to different constants $c_i$ in (1). The trapezoidal rule uses $c_0=c_N = 1/2$ and $c_i = 1, \,1 \leq i \leq N-1$. Simpson's rule would correspond to some other set of constants $c_i$.

However, that is not the way I would finally decide to solve your specific problem. Because of the special structure of your right-hand-side, your equation is easily solvable by elementary methods. Unfortunately, I cannot say more without giving away the whole solution. All I can suggest is that you look very carefully at the actual form of your equation. If you still have not got it a week or two after turning in your assignment for marking, I will be happy to outline the best method then.

Hey, Josh. I'm sorry for the delay of my reply, but I was busy with other things yesterday. You are correct in that $\phi(x)=\lim_{n \rightarrow \infty}\phi_n(x)$.
I cannot give a formal proof of that but I can give you some ideas. For this purpose, I will consider the slightly more general equation
$$f(x)=g(x)+\int_a^b h(x,t) f(t) dt$$.
We can do one iteration by inserting again the definition (given by the equation) of $f(t)$ on the right-hand side, i.e.
$$f(x)=g(x)+\int_a^b h(x,t)[g(t)+\int_a^b h(t,t_1)f(t_1)dt_1]$$.This same process can now be continued arbitrarily number of times leading to more integrals on the right-hand side.

Consider now the method based on iteration I proposed. Then,
$$f_0=0,$$
$$f_1=g(x)+\int_a^b h(x,t) f_0(y) dy$$,
$$f_2=g(x)+\int_a^b h(x,t) f_1(y)dy=g(x)+\int_a^b h(x,t)[g(t)+\int_a^b h(t,t_1)f_0(y_1)dy_1]$$, etc

One can therefore see, that the iteration leads to an equation of the type as above. It can also be shown that $\phi(x)=\lim_{n \rightarrow \infty}\phi_n(x)$.

The second method I proposed is quite similar to the one proposed by Ray but a bit simpler. Consider again the equation
$$f(x)=g(x)+\int_a^b h(x,t) f(t) dt$$.
By Gauss-Legendre integration or other quadrature method, one can write
$$f(x)=g(x)+\sum_{k=1}^n w_k h(x,t_k) f(t_k)$$
where $t_k$ are the nodes and $w_k$ the corresponding weights. By considering a mesh of discrete points $x_i, i=1, N$, you obtain a system of linear equations which can be solved by standard methods.

The iteration technique does not work. Regardless of $\phi_0$ we find $\lim_{n\to\infty}\phi_n$ does not converge.

 

[eys_physics]

The iteration technique does not work. Regardless of $\phi_0$ we find $\lim_{n\to\infty}\phi_n$ does not converge.

I don't know why your iteration did not converge. Can you give some more details about your implementation?
From the formal point of view the iteration should converge. But, sometimes it is difficult in practice. Usually, the second method I proposed works well.
But, I'm surprised that the iteration doesn't work for such a simple equation.

 

[member 428835]

I don't know why your iteration did not converge. Can you give some more details about your implementation?
From the formal point of view the iteration should converge. But, sometimes it is difficult in practice. Usually, the second method I proposed works well.
But, I'm surprised that the iteration doesn't work for such a simple equation.

Yea I did it in Mathematica, but no luck. The coefficients for $\sin(3x)$ grow very big very fast. I decided to use Ray Vickerson's approach, and it worked great.

 

[eys_physics]

Okay, good that it worked out good for you.

 

[Ray Vickson]

Yea I did it in Mathematica, but no luck. The coefficients for $\sin(3x)$ grow very big very fast. I decided to use Ray Vickerson's approach, and it worked great.

Yes, indeed, the iterative method diverges. Looking at the right-hand-side of the integral equation, we have
$$\phi(x) = \pi x^2 + \frac{3}{2} \sin(3x) \underbrace{\int_0^{\pi} \phi(t) \, dt}_{\text{constant}} -
3 x^2 \underbrace{\int_0^{\pi} t \phi(t) \, dt}_{\text{constant}}, $$
so we can write the iterative results as
$$\phi_n(x) = \pi A_n x^2 + B_n \sin(3x),$$
with
$$A_{n+1} = 1-B_n - \frac{3}{4} \pi^4 A_n, \: B_{n+1} = B_n + \frac{1}{2} \pi^4 A_n \hspace{4em}(1)$$
and $A_0 = 1, B_0 = 0$.

I solved the recursion (1) with the given initial conditions using Maple, and obtained
$$\begin{array}{rcr} A_n &=& 0.0137696\, (0.3363889)^n - 0.0137696\, (-72.393206)^n \\
B_n &=& 1 + 1.0091243\, (0.3363889)^n + 0.0091243\, (-72.393206)^n
\end{array}
$$
This obviously explodes exponentially.

Nevertheless, a non-iterative solution is easy to get: writing $\phi(x) = \pi A x^2 + B \sin(3x)$, the integral equation yields the equations
$$A = 1 - \frac{3}{4} \pi^4 A - B, \; B = B + \frac{1}{2} \pi^4 A,$$
so we have $A=0, B=1$. This gives $\phi(x) = \sin(3x)$ exactly, as claimed.

 

[member 428835]

Yes, indeed, the iterative method diverges. Looking at the right-hand-side of the integral equation, we have
$$\phi(x) = \pi x^2 + \frac{3}{2} \sin(3x) \underbrace{\int_0^{\pi} \phi(t) \, dt}_{\text{constant}} -
3 x^2 \underbrace{\int_0^{\pi} t \phi(t) \, dt}_{\text{constant}}, $$
so we can write the iterative results as
$$\phi_n(x) = \pi A_n x^2 + B_n \sin(3x),$$
with
$$A_{n+1} = 1-B_n - \frac{3}{4} \pi^4 A_n, \: B_{n+1} = B_n + \frac{1}{2} \pi^4 A_n \hspace{4em}(1)$$
and $A_0 = 1, B_0 = 0$.

I solved the recursion (1) with the given initial conditions using Maple, and obtained
$$\begin{array}{rcr} A_n &=& 0.0137696 (0.3363889)^n - 0.0137696 (-72.393206)^n \\
B_n &=& 1 + 1.0091243 (0.3363889)^n + 0.0091243 (-72.393206)^n
\end{array}
$$
This obviously explodes exponentially.

Nevertheless, a non-iterative solution is easy to get: writing $\phi(x) = \pi A x^2 + B \sin(3x)$, the integral equation yields the equations
$$A = 1 - \frac{3}{4} \pi^4 A - B, \; B = B + \frac{1}{2} \pi^4 A,$$
so we have $A=0, B=1$. This gives $\phi(x) = \sin(3x)$ exactly, as claimed.

Was this the elementary solution you proposed? To be honest I like the linear system much more

 

[Ray Vickson]

Was this the elementary solution you proposed? To be honest I like the linear system much more

Basically, yes. It works for any equation of the form
$$f(x) = g(x) + \int_a^b K(x,y) g(y) \, dy$$
with
$$K(x,y) = u_1(x) v_1(y) + u_2(x) v_2(y) + \cdots + u_r(x) v_r(y)$$
The reason is that the quantities $C_i = \int_a^b v_i(y) g(y) \, dy$ are constants (although unknown), so the solution must be of the form
$$f(x) = g(x) + C_1 u_1(x) + C_2 u_2(x) + \cdots + C_r u_r(x).$$
Plugging that form into the integral equation gives you a set of linear equations for finding the coefficients $C_i$.