I came across this integro-differential equation to solve $$\frac{du(x;t)}{dt}=-\lambda\int_0^xu(\xi;t)\;d\xi\tag{1}$$ under the initial condition $u(x;0)=f(x)$ where $x$ is a parameter, $\lambda$ is a constant, and $0<t<\infty$.

My first thought is that I can just directly integrate the equation to obtain $$u(x;t)=-\lambda\int_0^t\int_0^xu(\xi;\tau)\;d\xi\,d\tau\;.\tag{2}$$ This equation is highly implicit and an explicit expression is desired. So then I thought of using Laplace transforms instead. Let $U(\cdot)$ be the Laplace transform of $u(\cdot)$, and $s$ be the complex co-variable of the real variable $t$, then $$s\,U(x;s)-f(x)=-\lambda\,\int_0^xU(\xi;s)\,d\xi\tag{3}$$ which can be converted to the differential equation $$s\,U'(x;s)+\lambda\,U(x;s)=f'(x)\tag{4}$$ where the derivative is now with respect to $x$. Equation $(4)$ is easily solvable. Though I am uncertain if I did the following Laplace transform correctly, $$\mathscr{L}\left\{\int_0^xu(\xi;t)\,d\xi\right\}=\int_0^xU(\xi;s)\,d\xi\;.\tag{5}$$ I figured since the integration is over the parameter instead of the transforming variable I could bring it into the integral under the heuristic that the Laplace transform of a sum is the sum of the Laplace transforms, but I am unsure of this.

Any insight on any of this or alternative methods to solve Equation $(1)$ is welcome.


2 Answers 2


The solution detailed below is :

With $\quad F(s)=$ Laplace transform of $f(x)$.

$$\Phi(s,t)=e^{-\frac{\lambda \:t}{s}}F(s)$$ $$\boxed{u(x,t)= \text{ Inverse Laplace Transform of } \Phi(s,t)}$$

The result cannot be expressed more explicitly until the function $f(x)$ be explicitly given.

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Incomplete attempt at a solution but thought I would offer it anyways as food for thought.

Differentiate both sides of the integral equation with respect to $x$.

$$ \frac{\partial ^2 u}{\partial x\partial t}=-\lambda u(x,t) $$

$$ u_{\alpha}(x,t)=A(\alpha)\cos(\alpha x+\frac{\lambda}{\alpha} t)+B(\alpha)\sin(\alpha x+\frac{\lambda}{\alpha}t),\;\;\;\alpha>0 $$

Now take advantage of the linearity of the equation, usually we sum but here we have a continuous parameter $\alpha$.

$$ u(x,t)=\int_{0}^{\infty}u_{\alpha}(x,t)\mathbb{d}\alpha $$

$$ f(x)=u(x,0)=\int_{0}^{\infty}u_{\alpha}(x,0)\mathbb{d}\alpha $$

$$ f(x)=\int_{0}^{\infty}A(\alpha)\cos(\alpha x)\mathbb{d}\alpha+\int_{0}^{\infty}B(\alpha)\sin(\alpha x)\mathbb{d}\alpha $$

I would like to now solve for $A(\alpha),B(\alpha)$ in terms of $f(x)$. Perhaps if one converted the above trigonometric expressions to complex exponentials a certain Laplace/Fourier transform structure would emerge that would allow one to use the established formalism to invert (solve for $A,B$ in terms of $f$).

  • $\begingroup$ This is promising. There is an additional boundary condition, however, arising from putting $x=0$ in the original equation: $$\frac{\partial u}{\partial t}(0,t)=0.$$ This yields a type of Goursat problem. $\endgroup$
    – Bob Pego
    Commented Jul 4, 2017 at 9:03

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