1
$\begingroup$

This should be easy, but for some reason I don't succeed in determining the constants of the solution of a differential equation. This general solution is $$\theta(x) = C_1e^{mx}+C_2e^{-mx}$$ or $$\theta(x) = A\cosh(mx)+B\sinh(mx)$$ And the following conditions apply: $$\theta(0) = \theta_0\\ \left .-k\frac{d\theta}{dx}\right |_{x=L}=h\theta_L = h\theta(L)$$ Where $m^2 = \frac{hP}{kA}$ and the solution should be: $$\theta = \theta_0\frac{\cosh(m[L-x])+(h/mk)\sinh[m(L-x)]}{\cosh(mL)+(h/mk)\sinh(mL)}$$ But I don't succeed in obtaining this solution. The question comes from a physical situation, but I thought that, although all the symbols have a physical meaning, it was better to ask it here than in the physics part of the site. It comes from this paper, at page 236-237: $\theta$ stands for $T-T_\infty$. The solution is equation (17-40). But I don't think that these physical details are needed to solve the question.

In my attempts I solved it already until I get: $$\theta = \theta_0 \left [\cosh(mx)-\frac{h\theta_L\sinh(mx)}{\theta_0km\cosh(mL)}-\frac{\sinh(mL)}{\cosh(mL)}\right ]$$

but now it feels like it is impossible to proceed, because I see no way to get rid of the $\theta_L$. I don't see any mistakes though in the calculations and I hope that someone can clarify this for me.

$\endgroup$
1
$\begingroup$

You need $\theta'(L)=-h\theta(L)/k$, so calculate $$\theta'(x)=Am\sinh (mx) + Bm\cosh (mx)$$ and $\theta'(L)=Am\sinh(mL)+Bm\cosh(mL)$, whereas $$\theta(L)=A\cosh(mL)+B\sinh(mL)$$ So you need $$A(m\sinh(mL)+\frac{h}{k}\cosh(mL))=-B(\frac{h}{k}\sinh(mL)+m\cosh(mL))$$ On the other side, $$\theta(0)=A$$ so $A=\theta_0$ and $$B=-\theta_0\frac{m\sinh(mL)+\frac{h}{k}\cosh(mL)}{\frac{h}{k}\sinh(mL)+m\cosh(mL)}$$ So $$\theta=\frac{\theta_0}{\frac{h}{k}\sinh(mL)+m\cosh(mL)}(\frac{h}{k}\cosh(mx)\sinh(mL)+m\cosh(mx)\cosh(mL) -m\sinh(mL)\sinh(mx)-\frac{h}{k}\cosh(mL)\sinh(mx))=$$ $$=\frac{\theta_0}{\frac{h}{k}\sinh(mL)+m\cosh(mL)}(\frac{h}{k}\sinh(m(L-x))+m\cosh(m(L-x)))$$ which is yours up to dividing denominator and numerator by $m$. I used just the standard formulae of addiction and subtraction for hyperbolic functions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.