Conversion of second order ode into integral equation

Question

The second order differential equation is $$-\phi''(x,\psi)+g(x)\phi(x,\psi)=\lambda^2\phi(x,\psi) $$ where $\lambda\in\mathbb{C}$ and $x>0$. with conditions $$\phi(0,\psi)=0,\;\;\phi'(0,\psi)=1$$ can be converted into following Volterra equation $$\phi(x,\psi)=\lambda^{-1}\sin(\lambda x)+\int_0^x\lambda^{-1}\sin(\lambda(x-y)g(y)\phi(y,\psi))dy$$ I don't know how this ode is converted in the above equation: any help will be highly appreciated. In particular I don't know how trigonometry involved and how to tackle $g(x)$, while converting. I don't know much about integral equations. Thanks in advance.

Answer 1

The basic method applied in such cases consists in of selecting a linear part of the given differential equation and then calculating a fundamental solution and a particular solution of the chosen part which satisfies the Dirichlet/Neumann/Cauchy or other conditions imposed on the original equation: then, applying the convolution with the remaining part and adding it the particular solution gives you the sought for integral equation. The method is particularly useful when one tries to solve a differential equation which is linear except for a few terms: in this case, the whole equation is linear, but it has a variable $x$-dependent coefficient which make it difficult to solve in "closed form", so transforming it in an integral equation can lead to a more tractable problem. By rearranging the terms of the present equation to be solved, it can be written in the form $$ \phi''(x,\psi)+\lambda^2\phi(x,\psi)=g(x)\phi(x,\psi)\tag{1}\label{1} $$ which resembles the non-homogeneous standard second order linear ODE with constant coefficients, i.e. $$ u''(x)+\lambda^2u(x)=f(x)\tag{2}\label{2} $$ with the Cauchy conditions $$ \begin{cases} u(0)=0\\ u'(0)=1 \end{cases}.\tag{3}\label{3} $$ The fundamental solution of \eqref{2} i.e. its solution when $f(x)=\delta(x)$ is the Dirac delta distribution (Vladimirov [1], §1.1 pp. 5-6) is $$ \mathscr{E}(x)=\lambda^{-1}H(x)\sin\lambda x , $$ (Vladimirov [1], §4.96 pp. 73-74 and §15.4.4 p. 200) where $H(x)$ is the Heaviside function (Vladimirov [1], §0.1 p. 2).

The solution of its homogeneous version ( $f(x)\equiv 0$) which satisfies the Cauchy conditions \eqref{3} is $$ u_{ho}(x)=\lambda^{-1}\sin\lambda x $$ Now, for an arbitrary given function (distribution) $f$, the general solution of \eqref{2} satisfying \eqref{3} has the form $$ u(x)=u_{ho}(x)+\mathscr{E}\ast f(x)=\lambda^{-1}\sin\lambda x +\lambda^{-1}\int\limits_o^x\sin\lambda (x-y)f(y)\mathrm{d}y\tag{4}\label{4} $$ The term $$ \mathscr{E}\ast f(x)=\lambda^{-1}\int\limits_o^x\sin\lambda (x-y)f(y)\mathrm{d}y $$ is sometimes called the Volterra convolution: and the Volterra type integral equation above, equivalent to \eqref{1}, is obtained from by applying the following substitutions $$ u(x)=\phi(x,\psi)\qquad f(x)=g(x)\phi(x,\psi) $$

[1] Vladimirov, V. S. (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR 2012831, Zbl 1078.46029.

Answer 2

I will consider the simplified case for the IVP $$u''(t)+\lambda^2u(t)=f(t,u(t)),\qquad u(0)=0,\quad u'(0)=1, \qquad\lambda>0.$$ You can apply these methods to your equation, I hope it helps.

1st way: Apply the Laplace transform to this equation, $$s^2\mathcal{L}\{u(t)\}(s)-su(0)-u'(0)+\lambda^2\mathcal{L}\{u(t)\}(s)=\mathcal{L}\{f(t,u(t)\}(s)$$ then rearrange the equality, $$\mathcal{L}\{u(t)\}(s)=\frac{1}{\lambda}\frac{\lambda}{s^2+\lambda^2}+\frac{1}{\lambda}\frac{\lambda}{s^2+\lambda^2}\mathcal{L}\{f(t,u(t)\}(s),$$ finally apply inverse Laplace transform and use the convolution operation.

2nd way: Multiply the equation with $\sin\lambda(t-s)$ and integrate, use integration by parts and do some calculation (as mentioned in this question).

3rd way: Use variations of parameters formula see [F. Brauer, J. A. Nohel, Qualitaive Theory of Differential Equations: An Introduction, Dover Publ., pp:110].