integration of $(r\,du/dx)^2$

Question

I have been trying to find a way to integrate $$\int\rho\left(\frac{\partial u}{\partial x}\right)^2 \, dx$$ but have not managed to find a way. $\rho$ is density and $u$ is displacement. They both vary in space $(x)$. I tried in Mathematica and it does not give me an answer? Is there anyway to evaluate it? Thanks.

Answer 1

Assuming $\rho$ and $u$ are arbitrary smooth functions then $$\int \rho(x)\,\left(\frac{\partial u}{\partial x}(x)\right)^2 dx = \sum_{n=0}^\infty\frac{x^{n+1}}{n+1}\sum_{k=0}^n\frac{\rho^{(k)}(0)}{k!}\sum_{j=0}^{n-k}\frac{u^{(j+1)}(0)\,\,u^{(n-j-k+1)}(0)}{j!(n-j-k)!}$$ Where $f^{(n)}(x)$ is the $n$-th derivative of $f$ at $x$. You can approximate this by truncating the first sum at a finite value of $n$.

To obtain this we expand both $u^{(1)}(x)$ and $\rho(x)$ using their Taylor expansions then multiply these three series together (two for $u^{(1)}$ and one for $\rho$) and gather together terms of the same order in $x$ giving $$\rho(x)\,\left(\frac{\partial u}{\partial x}(x)\right)^2 = \sum_{n=0}^\infty x^{n}\sum_{k=0}^n\frac{\rho^{(k)}(0)}{k!}\sum_{j=0}^{n-k}\frac{u^{(j+1)}(0)\,\,u^{(n-j-k+1)}(0)}{j!(n-j-k)!}$$ Then we can integrate the series easily by bringing the integral inside the first sum $$\int \rho(x)\,\left(\frac{\partial u}{\partial x}(x)\right)^2 dx = \sum_{n=0}^\infty\int x^n dx \sum_{k=0}^n\frac{\rho^{(k)}(0)}{k!}\sum_{j=0}^{n-k}\frac{u^{(j+1)}(0)\,\,u^{(n-j-k+1)}(0)}{j!(n-j-k)!}$$ which gives us the answer above. I do not know if this is useful but it will depend on what you know about the two functions.