Integro-differential equation in one dimensional linear thermo-elasticity

Question

I have this system of coupled pde's:

\begin{equation} \frac{\partial^2\theta}{\partial x^2}=\frac{\partial \theta}{\partial t}+\sqrt{a}\frac{\partial^2 u}{\partial x\, \partial t} \end{equation} \begin{equation} \frac{\partial^2 u}{\partial x^2}=\sqrt{a}\frac{\partial \theta}{\partial x} \end{equation} with $a$ a positive real constant, $x\in[0,1]$, $t\in[0,\infty)$ and $\theta$ and $u$ are real valued functions of $x$ and $t$. This equation arises in one dimensional linear thermo-elasticity. It has been stated in a book that $\theta$ satisfies the following integro-differential equation: \begin{equation} \frac{\partial^2\theta}{\partial x^2}=(1+a)\frac{\partial\theta}{\partial t}-a\,\frac{d}{dt}\int^{1}_{0}\theta(y,t)\,dy. \end{equation} Can somebody explain how do they get the second term? The only condition specified is that $u(0,t)=u(1,t)=0$.

Answer 1

Integrating along $x$ direction for fixed $t$ once, we get: $$\frac{\partial^2 u}{\partial x^2}=\sqrt{a}\frac{\partial \theta}{\partial x} \implies \frac{\partial u}{\partial x}(x,t) = \sqrt{a}\,\theta(x,t) + K(t) $$ where $K(t)$ is some function independent of $x$. Integrating along $x$ again, we have:

$$0 = u(1,t) - u(0,t) = \int_{0}^{1} \frac{\partial u}{\partial x}(x,t) dx = \sqrt{a}\int_{0}^{1}\theta(x,t) dx + K(t)$$

Combine these two, we obtain: $$\frac{\partial u}{\partial x}(x,t) = \sqrt{a}\left(\theta(x,t) - \int_{0}^{1}\theta(y,t) dy\right)$$

Substitute this back into the first equation gives the integro-differential equation.

Answer 2

Solving an integro-differential equation should be more difficult than solving a pde.

So you are better to modify to follow this approach:

$\begin{cases}\dfrac{\partial^2\theta}{\partial x^2}=\dfrac{\partial\theta}{\partial t}+\sqrt{a}\dfrac{\partial^2u}{\partial x\partial t}\\\dfrac{\partial^2u}{\partial x^2}=\sqrt{a}\dfrac{\partial\theta}{\partial x}\end{cases}$

$\therefore\dfrac{\partial^3\theta}{\partial x^3}=\dfrac{\partial^2\theta}{\partial x\partial t}+\sqrt{a}\dfrac{\partial^3u}{\partial x^2\partial t}$

$\dfrac{1}{\sqrt{a}}\dfrac{\partial^4u}{\partial x^4}=\dfrac{1}{\sqrt{a}}\dfrac{\partial^3u}{\partial x^2\partial t}+\sqrt{a}\dfrac{\partial^3u}{\partial x^2\partial t}$

$\dfrac{\partial^4u}{\partial x^4}=(a+1)\dfrac{\partial^3u}{\partial x^2\partial t}$

Note that this PDE is separable.

Let $u(x,t)=X(x)T(t)$ ,

Then $X''''(x)T(t)=(a+1)X''(x)T'(t)$

$\dfrac{(a+1)T'(t)}{T(t)}=\dfrac{X''''(x)}{X''(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-\dfrac{s^2}{a+1}\\X''''(x)+s^2X''(x)=0\end{cases}$

$\begin{cases}T(t)=c_5(s)e^{-\frac{ts^2}{a+1}}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs+c_3(s)x+c_4(s)&\text{when}~s\neq0\\c_1x^3+c_2x^2+c_3x+c_4&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-\frac{ts^2}{a+1}}\sin xs~ds+\int_0^\infty C_2(s)e^{-\frac{ts^2}{a+1}}\cos xs~ds+\int_0^\infty C_3(s)xe^{-\frac{ts^2}{a+1}}~ds+\int_0^\infty C_4(s)e^{-\frac{ts^2}{a+1}}~ds$

$u(0,t)=0$ :

$\int_0^\infty C_2(s)e^{-\frac{ts^2}{a+1}}~ds+\int_0^\infty C_4(s)e^{-\frac{ts^2}{a+1}}~ds=0$

$\int_0^\infty C_4(s)e^{-\frac{ts^2}{a+1}}~ds=-\int_0^\infty C_2(s)e^{-\frac{ts^2}{a+1}}~ds$

$C_4(s)=-C_2(s)$

$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-\frac{ts^2}{a+1}}\sin xs~ds+\int_0^\infty C_2(s)e^{-\frac{ts^2}{a+1}}\cos xs~ds+\int_0^\infty C_3(s)xe^{-\frac{ts^2}{a+1}}~ds-\int_0^\infty C_2(s)e^{-\frac{ts^2}{a+1}}~ds=\int_0^\infty C_1(s)e^{-\frac{ts^2}{a+1}}\sin xs~ds+\int_0^\infty C_2(s)e^{-\frac{ts^2}{a+1}}(\cos xs-1)~ds+x\int_0^\infty C_3(s)e^{-\frac{ts^2}{a+1}}~ds$

$u(1,t)=0$ :

$\int_0^\infty C_1(s)e^{-\frac{ts^2}{a+1}}\sin s~ds+\int_0^\infty C_2(s)e^{-\frac{ts^2}{a+1}}(\cos s-1)~ds+\int_0^\infty C_3(s)e^{-\frac{ts^2}{a+1}}~ds=0$

$\int_0^\infty C_3(s)e^{-\frac{ts^2}{a+1}}~ds=-\int_0^\infty C_1(s)e^{-\frac{ts^2}{a+1}}\sin s~ds-\int_0^\infty C_2(s)e^{-\frac{ts^2}{a+1}}(\cos s-1)~ds$

$\int_0^\infty C_3(s)e^{-\frac{ts^2}{a+1}}~ds=\int_0^\infty(C_2(s)(1-\cos s)-C_1(s)\sin s)e^{-\frac{ts^2}{a+1}}~ds$

$C_3(s)=C_2(s)(1-\cos s)-C_1(s)\sin s$

$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-\frac{ts^2}{a+1}}\sin xs~ds+\int_0^\infty C_2(s)e^{-\frac{ts^2}{a+1}}(\cos xs-1)~ds+x\int_0^\infty(C_2(s)(1-\cos s)-C_1(s)\sin s)e^{-\frac{ts^2}{a+1}}~ds=\int_0^\infty C_1(s)e^{-\frac{ts^2}{a+1}}(\sin xs-x\sin s)~ds+\int_0^\infty C_2(s)e^{-\frac{ts^2}{a+1}}(\cos xs-x\cos s+x-1)~ds$

$\dfrac{\partial^2u}{\partial x^2}=-\int_0^\infty C_1(s)s^2e^{-\frac{ts^2}{a+1}}\sin xs~ds-\int_0^\infty C_2(s)s^2e^{-\frac{ts^2}{a+1}}\cos xs~ds$

$\therefore\dfrac{\partial\theta}{\partial x}=-\dfrac{1}{\sqrt{a}}\int_0^\infty C_1(s)s^2e^{-\frac{ts^2}{a+1}}\sin xs~ds-\dfrac{1}{\sqrt{a}}\int_0^\infty C_2(s)s^2e^{-\frac{ts^2}{a+1}}\cos xs~ds$

$\theta(x,t)=\dfrac{1}{\sqrt{a}}\int_0^\infty C_1(s)se^{-\frac{ts^2}{a+1}}\cos xs~ds-\dfrac{1}{\sqrt{a}}\int_0^\infty C_2(s)se^{-\frac{ts^2}{a+1}}\sin xs~ds+C(t)$