# 数学代写|数学建模代写math modelling代考|Integral Equations Arising from a Problem of Elasticity Theory

#### Doug I. Jones

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## 数学代写|数学建模代写math modelling代考|Integral Equations Arising from a Problem of Elasticity Theory

$$f(x)=\int_a^b G(x, \xi) f(\xi) d \xi$$

## 数学代写|数学建模代写math modelling代考|Standard Integral Transform Pairs

$$\begin{gathered} \bar{f}(x)=\int_a^{\infty} e^{-x \xi} f(\xi) d \xi \ f(\xi)=\int_{C-i \infty}^{C+i \infty} e^{-\xi x} \bar{f}(x) d x \end{gathered}$$

$$\begin{gathered} \bar{f}(x)=\int_{\infty}^{\infty} e^{ \pm \xi x} f(\xi) d \xi \ f(\xi)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{ \pm i \xi x} \bar{f}(x) d x \end{gathered}$$

$$\bar{f}(x)=\int_0^{\infty} \sin (\xi x) f(\xi) d \xi$$

$$\begin{gathered} f(\xi)=\frac{2}{\pi} \int_0^{\infty} \sin (\xi x) \bar{f}(x) d x \ f(x)=\int_0^{\infty} \cos (\xi x) f(\xi) d \xi \ f(\xi)=\int_0^{\infty} \cos (\xi x) \bar{f}(x) d x \end{gathered}$$

\begin{aligned} & \bar{f}(x)=\frac{1^}{\pi} \int_{-\infty}^{\infty} \frac{f(\xi)}{x-\xi} d \xi \ & f(\xi)=\frac{1^}{\pi} \int_{-\infty}^{\infty} \frac{\bar{f}(x)}{\xi-x} d x \end{aligned}

$$\int_{-\infty}^{\infty} \frac{f(\xi)}{x-\xi} d \xi=\operatorname{Lt}{\in \rightarrow 0}\left[\int{-\infty}^{x-\in} \frac{f(\xi)}{x-\xi} d \xi+\int_{x+\in}^{\infty} \frac{f(\xi)}{x-\xi} d \xi\right]$$

$$\begin{gathered} \bar{f}(x)=\frac{1^}{\pi} \int_{-1}^1 \frac{f(\xi)}{x-\xi} d \xi \ f(\xi)=\frac{1^}{\pi} \int_{-1}^1 \sqrt{\frac{1-x^2}{1-\xi^2} \frac{f(x)}{\xi-x}} d x \end{gathered}$$

# 数学建模代写

## 数学代写|数学建模代写math modelling代考|Integral Equations Arising from a Problem of Elasticity Theory

Consider an elastic beam $A B$ (Figure 8.1). Let $G(x, \xi)$ denote the displacement at the point $x$ due to unit force applied at $\xi$. This function $G(x, \xi)$ can be found from the equations of elasticity theory. If the force applied is $f(\xi) d \xi$, then in the linear theory of elasticity, the displacement at $x$ would be $G(x, \xi) f(\xi) d \xi$ and if the force is applied all along the beam, the displacement at the point $x$ would be $f(x)$ where
$$f(x)=\int_a^b G(x, \xi) f(\xi) d \xi$$

Knowing $f(\xi)$ at all points of the beam $A B$, Eqn. (55) would enable us to find the displacement $f(x)$ of all points of the beam.

The inverse problem is to determine $f(x)$ when $\bar{f}$ is known, i.e., to find the force distribution which will cause a desired displacement distribution at all points of the beam. The unknown function in this case is $f(\xi)$ and it occurs under the integral sign. Equations like (55) where the unknown function occurs under the integral sign are called integral equations.

The function $G(x, \xi)$ is called the influence function or the kernel function or Green’s function.
Physically, the integral in Eqn. (55) is arising because the effect of the force applied at different points of the beam is being “summed up,” “integrated out,” and there is “accumulation” of all effects.

Thus integral equations are likely to arise in physical, biological, and social problems where there is an “accumulative effect” in operation.

The function $\bar{f}(x)$ is also called the integral transform of $f(\xi)$ through the kernel $G(x, \xi)$. Inverting this integral transform means finding $f(\xi)$ when $f(x)$ is known and this requires the solution of an integral equation.

There are a number of kernels which arise in a large number of applications of mathematics. These give rise to standard integral transforms. Some of these along with the inverse transforms are given in the next section.

## 数学代写|数学建模代写math modelling代考|Standard Integral Transform Pairs

Laplace Transform Pairs
$$\begin{gathered} \bar{f}(x)=\int_a^{\infty} e^{-x \xi} f(\xi) d \xi \ f(\xi)=\int_{C-i \infty}^{C+i \infty} e^{-\xi x} \bar{f}(x) d x \end{gathered}$$
where $C>$ all singularities of $\bar{f}(x)$. Equation (57) is an integral equation in the complex plane.
Fourier Complex Transform Pairs
$$\begin{gathered} \bar{f}(x)=\int_{\infty}^{\infty} e^{ \pm \xi x} f(\xi) d \xi \ f(\xi)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{ \pm i \xi x} \bar{f}(x) d x \end{gathered}$$

Fourier Sine (Cosine) Transform Pairs
$$\bar{f}(x)=\int_0^{\infty} \sin (\xi x) f(\xi) d \xi$$

$$\begin{gathered} f(\xi)=\frac{2}{\pi} \int_0^{\infty} \sin (\xi x) \bar{f}(x) d x \ f(x)=\int_0^{\infty} \cos (\xi x) f(\xi) d \xi \ f(\xi)=\int_0^{\infty} \cos (\xi x) \bar{f}(x) d x \end{gathered}$$
Hilbert Transform Pairs
\begin{aligned} & \bar{f}(x)=\frac{1^}{\pi} \int_{-\infty}^{\infty} \frac{f(\xi)}{x-\xi} d \xi \ & f(\xi)=\frac{1^}{\pi} \int_{-\infty}^{\infty} \frac{\bar{f}(x)}{\xi-x} d x \end{aligned}
where * denotes that we are taking the principal value of the integral concerned
$$\int_{-\infty}^{\infty} \frac{f(\xi)}{x-\xi} d \xi=\operatorname{Lt}{\in \rightarrow 0}\left[\int{-\infty}^{x-\in} \frac{f(\xi)}{x-\xi} d \xi+\int_{x+\in}^{\infty} \frac{f(\xi)}{x-\xi} d \xi\right]$$
If the limits are -1 to 1 , the corresponding transform pair is
$$\begin{gathered} \bar{f}(x)=\frac{1^}{\pi} \int_{-1}^1 \frac{f(\xi)}{x-\xi} d \xi \ f(\xi)=\frac{1^}{\pi} \int_{-1}^1 \sqrt{\frac{1-x^2}{1-\xi^2} \frac{f(x)}{\xi-x}} d x \end{gathered}$$

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