Kernel of integral eq of positive terms only

[sarrah1]

Again it's the Fredholm integral equation of the 2nd type, that is

$y(x)=f(x)+\int_{a}^{b} \,k(x,t) y(t) dt$ , where $a,b>0$

I am taking the norm of the integral operator equal to $||K||= max ([a,b] \int_{a}^{b} \,|k(x,t)| dt$

the kernel is represented by a series of positive terms only or negative terms only. e.g. $e^{xt}$ or $sinhxt$ say.

Let $k(x,t)=k1(x,t)+k2(x,t)$ where $k1$ and $k2$ are any choice of terms from $k(x,t)$ .

my question is does $||K||=||K1||+||K2||$
thanks
sarrah

 

[jvicens]

Hello Sarrah,

Thank you for your question. To answer your question, let's first define the norm of an operator. The norm of an operator is a measure of its size or magnitude, and it is defined as the largest possible value the operator can take when acting on a vector of unit length. In other words, it is the maximum amount by which the operator can change the length of a vector.

Now, let's apply this definition to the integral operator in the Fredholm integral equation of the 2nd type. The norm of this operator, denoted by ||K||, is equal to the maximum value the kernel function can take when integrated over the interval [a,b]. In other words, it is the maximum possible change in the output y(x) when the input y(t) is varied over the interval [a,b].

Now, let's consider the two terms in the kernel function, k1(x,t) and k2(x,t). Each of these terms has its own maximum value when integrated over [a,b], denoted by ||K1|| and ||K2|| respectively. Therefore, when we combine these two terms, the maximum value of the combined kernel function will be the sum of the maximum values of k1(x,t) and k2(x,t). In other words, ||K||=||K1||+||K2||.

To put it simply, the maximum possible change in the output y(x) when the input y(t) is varied over [a,b] will be the sum of the maximum changes caused by k1(x,t) and k2(x,t) individually. Therefore, the norm of the integral operator is indeed equal to the sum of the norms of the individual terms in the kernel function.

I hope this helps to answer your question. If you have any further doubts or queries, please feel free to ask. As a fellow scientist, I am always happy to discuss and clarify any scientific concepts.