1
$\begingroup$

I'm looking for an explicit example of a BVP for a second order ODE:

$y''+p(x)y'+q(x)y=f(x)$ (where $\,0\leq x\leq L\,$ and $\,y(0)=\alpha\,$ $\,y(L)=\beta$).

If you also have the exact solution, the better. The reason is for test purposes, I've just finished a Mathematica program to solve it (via cubic B-Splines in multiple nodes) and I want to try it! Thank you in advance!

$\endgroup$
1
$\begingroup$

At the very least, your code should reproduce the case with $p(x)=0$, $q(x)=0$ and $f(x)=0$. In that case, the solution is simply $$ y(x)=\frac{\beta-\alpha}{L}x+\alpha $$

Next consider second order ODE with constant coefficients such as the one below $$ y''(x)-4y'+3y=0 $$ whose solution is $y(x)=c_{1}e^{-x}+c_{2}e^{-3x}$.

As a third test, relax the assumption of constant coefficient and try the Euler Cauchy ODE. An example is below: $$ x^2y''-9xy'+25y=0 $$ with solution $y(x)=c_1x^5+c_2\ln|x|x^5$.

As a final test, you can consider the following ODE which can be solved via the variation of parameters: $$ x^2y''-3xy'+4y=x^2\ln(x) $$ with solution $$ y(x)=c_1x^2+c_2x^2\ln(x)+\frac{1}{6}x^2\ln(x)^3 $$

In all these, choose a suitable value of $\alpha$ and $\beta$ to find $c_{1}$ and $c_{2}$.

$\endgroup$
1
$\begingroup$

The function $y(x,t) := \exp\big(tx(x-L)\big)$ solves $y'' - \frac{x(x-L)}{2} y' -\frac{x^2(x-L)^2}{2} y = 0$, $y(0,t) = 1 = y(L,t)$.

$\endgroup$

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.