<aside> 🌲

Kernel: is a function of two variables $K(x,z)$ which we define its operation on $u$ as as

$$ ⁍ $$

</aside>

⚙️ Properties: of the Kernel

• $\mathcal K$ is a linear operator

  $$ \mathcal K (au(x)+bv(x))=a\mathcal K u(x)+b\mathcal K v(x) $$

• The inner product of $\mathcal K$ is

  $$ \braket{v|\mathcal K u } =\int ^b_a v^* (x) \left ( \int ^b_a K(x,z)u(z)\,\text dz \right )\,\text dx $$

• $\mathcal K$ is Hermitian when $K^*(z,x)=K(x,z)$

We can define different types of integral equations

• Fredholm equations of the first and second kind are

  $$ \int ^b_a K(x,z)y(z)\,\text d z=f(x) \qquad y(x)=f(x)+\lambda \int ^b_a K(x,z)y(z) \,\text dz $$

• Replacing the upper limit by $x$ gives Volterra equations of first and second kind

  $$ \int ^x_a K(x,z)y(z)\,\text d z=f(x) \qquad y(x)=f(x)+\lambda \int ^x_a K(x,z)y(z) \,\text dz $$

🗒️ Note: if we set $f=0$ the equations become homogeneous

We can draw parallels between differential equations and integral equations

💃 Example: consider the Volterra equation $y(x)=1+x^2+\int ^x_0 y(z)\,\text d z$

• Since $\lim_{x\to 0} \int ^x_0 y(z)\,\text dz =0$ then $y(0)=1$ and we can differentiate w.r.t. $x$

  $$ y'(x)=2x+y(x) \quad \Rightarrow \quad y'(x)-y(x)=2x $$

💎 Conclusion: all linear differential equations can be turned into integral equations

💃 Example: if we start with $y''(x)+k^2y(x)=f(x)$ with $y(0)=y'(0)=0$

• We can integrate the right hand side twice to give

  $$ \int ^a_0 \left ( \int ^y _0 f(x)\,\text dx \right )\,\text dy =\int ^a_0 \left ( \int ^a_x \,\text dy \right )f(x)\,\text dx = \int ^a_0 (a-x)f(x)\,\text dx $$

• Now taking care of the whole expression we get

  $$ \begin{aligned}&y''(x) + k^2 y(x) = f(x) \\\Rightarrow\quad &y'(x) + \int_0^x \left(k^2 y(z) - f(z)\right) \, \text{d}z + A = 0 \\\Rightarrow\quad &y(x) + \int_0^x \left( \int_0^{x'} \left(k^2 y(z) - f(z)\right) \, \text{d}z \right) \text{d}x' + Ax + B = 0 \\\Rightarrow\quad &y(x) + k^2 \int_0^x (x - z) y(z) \, \text{d}z = \int_0^x (x - z) f(z) \, \text{d}z - Ax - B\end{aligned} $$

  now we can apply the boundary conditions $A$ and $B$ become $0$ and thus we get

  $$ \boxed{y(x)+k^2 \int ^x _0 (x-z)y(z)\,\text dz = g(x)} \quad \text{where} \quad g(x) = \int ^x_0 (x-z) f(z) \,\text dz $$

💼 Case: consider an eigen equation but for integral equations,

$$ y(x)=\lambda \int ^b_a K(x,y) y(z)\,\text dz $$

which we write $y=\lambda \mathcal K y$ where $\lambda \ne 0$ rather than the usual $\mathcal K y=\lambda y$

We recognise this as a homogenous Fredholm equation of the second kind

🗒️ Note: Volterra eigen equations have no solution ie consider

$$ y(x)=\lambda \int ^z_0 K(x,z)y(z)\,\text dz $$

🧽 Assume: differentiable kernel and solution