A Narrow, Technical Problem in Partial Differential Equations

While I was in school, one of my professors set this problem to me and my classmates and challenged us to solve it over the next few days. I found the challenge intriguing and it fascinated me, so I thought it was worth sharing. The problem was this:

Show that

$\displaystyle v(x,t) = \int_{-\infty}^{\infty} f(x-y,t)g(y)dy, (1.1)$

where $\displaystyle g(y)$ has finite support and also satisfies the PDE

$\displaystyle \frac{\partial v}{\partial t} = -\kappa \frac{\partial^{2}v}{\partial x^{2}}. (1.2)$

First off, what does finite support mean? Mathematically speaking, a function has support which is characterized by a subset of its domain whose members do not map to zero, and yet are finite. (Just as a quick note: much of the proper definitions require an understanding in mathematical analysis and measure theory, something which I have not studied in detail, so take that explanation with a grain of salt.)

As for the solution, we can rewrite the given PDE as

$\displaystyle \frac{\partial v}{\partial t} - \kappa \frac{\partial^{2}v}{\partial x^{2}} = 0. (2)$

The PDE requires a first-order time derivative and a second-order spatial derivative.

$\displaystyle \therefore \frac{\partial v}{\partial t} = \frac{\partial}{\partial t}\int_{-\infty}^{\infty} f(x-y,t)g(y)dy, (3.1)$

and

$\displaystyle \frac{\partial^{2} v}{\partial x^{2}} = \frac{\partial^{2}}{\partial x^{2}}\int_{-\infty}^{\infty} f(x-y,t)g(y)dy. (3.2)$

Next, we substitute Eqs. (3.1) and (3.2) into Eq.(2), yielding

$\displaystyle \frac{\partial}{\partial t}\int_{-\infty}^{\infty} f(x-y,t)g(y)dy -\kappa \frac{\partial^{2}}{\partial x^{2}}\int_{-\infty}^{\infty} f(x-y,t)g(y)dy = 0. (4)$

Note that taking the derivative of a function and then integrating that function is equivalent to integrating the function and differentiating the same function, in conjunction with the fact that the sum or difference of the integrals is the integral of the sum or difference (proofs of these facts are typically covered in a course in real analysis). Taking advantage of these gives

$\displaystyle \int_{-\infty}^{\infty} \bigg\{\frac{\partial}{\partial t}f(x-y,t)-\kappa\frac{\partial^{2}}{\partial x^{2}}f(x-y,t)\bigg\}g(y)dy = 0. (5)$

Notice that the terms contained in the brackets equate to $\displaystyle 0$. This means that

$\displaystyle \int_{-\infty}^{\infty} 0 \cdot g(y)dy = 0. (6)$

This implies that the function $\displaystyle v(x,t)$ does satisfy the given PDE (Eq.(2)).

References:

Definition of Support in Mathematics: https://en.wikipedia.org/wiki/Support_(mathematics)

Method of Successive Approximations-Elementary Methods to Solving Integral Equations

SOURCES: Tricomi, F.G., Integral Equations. 1985. Dover. 1.

Wazwaz, A.M., A First Course in Integral Equations. 2015. 3.6

In regards to solving integral equations, there are two main types of equations: Volterra and Fredholm equations. The first of which is the topic of this post. First, a brief introduction. Just as there are equations which deal with unknown functions and their derivatives, there also exists another type of equation which involves the same function but include the integral of this function. There are also a class of equations called integro-differential equations (see my series on Monte Carlo and Radiative transfer), in which the equation deals with the function, its derivative, and its integral.

Here, I will be dealing with Volterra integral equations. More specifically, I will be considering the Volterra equation of the second kind (VESK) of the form:

$\displaystyle \alpha(x)=h(x)+\lambda\int_{0}^{x}K(x,y)\alpha(y)dy, (1)$

where $y\in [0,x]$. The first term of the integrand $K(x,y)$ denotes the kernel. The kernel of the integral arises from its conversion from an initial value problem. Indeed, solving the integral equation is equivalent to solving the initial value problem of a differential equation. The integral equation includes the initial conditions instead of being added in near the end of the solution of an IVP.

The fact that we are dealing with Volterra equations means that the kernel is subject to the condition:

$\displaystyle K(x.y); y > x.$

The typical way to solve this involves the method of successive approximations (some call this method Picard’s method of successive approximation). I first came across this method whilst taking my differential equations course. The context in which this arose was that of the existence and uniqueness of solutions to first order differential equations.

The method is as follows:

Suppose we have the equation given in (1). We then define an initial function $\alpha_{0}(x)=h(x).$ Then the next iteration of $\alpha$ is

$\displaystyle \alpha_{1}(x)=h(x)+\lambda\int_{0}^{x}K(x,y)h(y)dy. (2)$

Naturally, the next term $\alpha_{2}(x)$ is

$\displaystyle \alpha_{2}(x)=h(x)+\lambda\int_{0}^{x}K(x,y)f(y)dy+\lambda^{2}\int_{0}^{x}K(x,z)dz \int_{0}^{z}K(z,y)f(y)dy, (3)$

or more simply

$\displaystyle \alpha_{2}(x)=h(x)+\lambda\int_{0}^{x}K(x,y)\alpha_{1}(y)dy. (4)$

The reason I chose to include Eq.(3) is because while reading about this method I found that the traditional expression (Eq.(4)) left much to be desired. For me, it didn’t demonstrate the “successive” part of successive approximations. In general, we may become convinced that

$\displaystyle \alpha_{n}(x)=h(x)+\lambda\int_{0}^{x}K(x,y)\alpha_{n-1}(y)dy. (5)$

Once the general expression for $\alpha_{n}(x)$ is determined, we may determine the exact solution $\alpha(x)$ via

$\displaystyle \alpha(x)=\lim_{n\rightarrow \infty}\alpha_{n}(x). (6)$

This is probably one of the simpler methods of solving integral equations, since we do not require any real analysis, but we can obtain solutions for simple integral equations. I will discuss a few other methods of solution in future posts in this series.