Solve Fredholm equation

Question

Solve the Fredholm integral equation

$$w(x)=x+ k \int_0^1 \left( x^{\frac{1}{2}} t^{\frac{3}{2}}+x^{\frac{3}{2}} t^{\frac{1}{2}}\right) w(t)dt$$

That is what I have been thinking:

Mostly, the solution of Fredholm equations is obtained by the method of successive approximations.

In our case, no information of $y$ is given. I don;t know any other way to find the solution. I would be grateful if you provide me with some ideas.

Answer 1

You can rewrite the equation as $$ w(x) = x + kAx^{1/2} + kBx^{3/2}, $$ where \begin{align} A &= \int_{0}^{1} \mathrm{d} t ~ t^{3/2}w(t) \\ B &= \int_{0}^{1} \mathrm{d} t ~ t^{1/2}w(t). \end{align} Multiplyng both sides of the first equation by $t^{3/2}$ and integrate over $x$, you obtain \begin{align} A &= \int_{0}^{1} \mathrm{d} x ~ x^{5/2} + kA\int_{0}^{1} \mathrm{d} x ~ x^{2} + kB \int_{0}^{1} \mathrm{d} x ~ x^{3} \\ &= \frac{2}{7} + \frac{kA}{3} + \frac{kB}{4}. \end{align} In the same way, \begin{align} A &= \frac{2}{5} + \frac{kA}{2} + \frac{kB}{3}. \end{align} Solving these two equations, you find \begin{align} A &= -\frac{12}{35}\frac{k+60}{k^{2}+48k-72} \\ B &= -\frac{24}{35}\frac{k+42}{k^{2}+48k-72}, \end{align} and you obtain $w(x)$.

Answer 2

To me, this case is known as "degenerate" Fredholm equation. Observe that $$ w(x)=x+ k \int_0^1 \left( x^{\frac{1}{2}} t^{\frac{3}{2}}+x^{\frac{3}{2}} t^{\frac{1}{2}}\right) w(t)dt = x + k x^{\frac{1}{2}} \int_0^1 t^{\frac{3}{2}} w(t) dt + k x^{\frac{3}{2}} \int_0^1 t^{\frac{1}{2}} w(t) dt $$ As $k$, $\int_0^1 t^{\frac{3}{2}} w(t) dt$ and $\int_0^1 t^{\frac{1}{2}} w(t) dt$ are constants, the only possible form of $w(t)$ is $w(x) = x + A x^{\frac{1}{2}} + B x^{\frac{3}{2}}$, where $A, B \in \mathbb{R}$. So, you can substitute it in the equation to detemine $A$ and $B$: $$ x + A x^{\frac{1}{2}} + B x^{\frac{3}{2}} = x + k x^{\frac{1}{2}} \int_0^1 t^{\frac{3}{2}} \left(t + A t^{\frac{1}{2}} + B t^{\frac{3}{2}}\right) dt + k x^{\frac{3}{2}} \int_0^1 t^{\frac{1}{2}} \left(t + A t^{\frac{1}{2}} + B t^{\frac{3}{2}}\right) dt $$ It leads to system of linear equations: $$ \begin{cases} A = k \int_0^1 t^{\frac{3}{2}} \left(t + A t^{\frac{1}{2}} + B t^{\frac{3}{2}}\right) dt \\ B = k \int_0^1 t^{\frac{1}{2}} \left(t + A t^{\frac{1}{2}} + B t^{\frac{3}{2}}\right) dt \end{cases} \Leftrightarrow \\ \Leftrightarrow \begin{cases} A = k \left( \int_0^1 t^\frac{5}{2} dt + A \int_0^1 t^2 dt + B \int_0^1 t^3 dt\right) \\ B = k \left( \int_0^1 t^\frac{3}{2} dt + A \int_0^1 t dt + B \int_0^1 t^2 dt\right) \end{cases} \Leftrightarrow \begin{cases} A = k \left( \frac{2}{7} + \frac{1}{3} A + \frac{1}{4}B\right) \\ B = k \left( \frac{2}{5} + \frac{1}{2} A + \frac{1}{3}B\right) \end{cases} $$ According to Wolfram, it leads to $$ A = -\frac{12}{35} \frac{k^2 + 60k}{k^2 + 48k - 72}, B = -\frac{24}{35} \frac{k^2 + 42k}{k^2 + 48k - 72} $$ So, if $k^2 + 48k - 72 \neq 0$, solution is given by $$ w(x) = x -\frac{12}{35} \frac{k^2 + 60k}{k^2 + 48k - 72} x^\frac{1}{2} - \frac{24}{35} \frac{k^2 + 42k}{k^2 + 48k - 72} x^\frac{3}{2} $$ otherwise (if $k = -24 \pm 18 \sqrt{2}$) there is no solution.