I have stumbled upon a situation where I need to find out the relationship between two different functions $b(x)$ and $D(x)$ which follow the equation,
$$ \frac{\partial^2 }{\partial x^2}\bigg[b(x)^2p(x,t)\bigg] = \int_{-\infty}^\infty D(x-x')\frac{\partial^2 p(x',t)}{\partial x'^2}dx' $$
My question, is there a definitive and unique map between $b(x)$ and $D(x)$?
This is my approach :
Since $p(x,t)$ is only depending on $x$ in this problem, let's suppose $p(x,t)=u(x), \forall t$. Let also $V(x)=b^2(x)p(x,t)=b^2(x)u(x)$. Which means that $$ \frac{\partial^2 }{\partial x^2}\bigg[b(x)^2p(x,t)\bigg] =V''(x).$$ You can observe that the right hand side is a convolution product. We have $$\int_{-\infty}^\infty D(x-x')\frac{\partial^2 p(x',t)}{\partial x'^2}dx'=\int_{-\infty}^\infty D(x-x')u''(x')dx'=(D\ast u'')(x) .$$ We get then the equation $$V''(x)=(D\ast u'')(x)$$ By applying the Laplace transform we get the auxiliary equation $$\mathcal L\{V''\}(z)=\mathcal L\{D\ast u''\}(z)$$ By using Laplace transform properties $$\mathcal L\{V''\}(z)=\mathcal L\{D\ast u''\}(z)$$ $$\implies z^2L\{V\}(z)-zV(0)-V'(0)=L\{D\}(z)L\{u''\}(z)$$ $$\implies z^2L\{V\}(z)-zV(0)-V'(0)=L\{D\}(z)\left[z^2L\{u\}(z)-zu(0)-u'(0)\right]$$ I guess here everything depends on your initial conditions of $V$ and $u$, and since $V$ itself depends on $u$ then it's all about the initial conditions of $u=p(x,t)$. You will have to use laplace inverse transform to have a simplified equation from which I'm sure you'll get a formula.