Slightly different results to an ODE system - hand calculation vs Mathematica

Question

This has been driving me mad for the last few days. I have a a pair of ODEs:

$$\frac{d^2 M_N}{d x^2}=\lambda_{N}^2 M_N$$

$$\frac{d^2 M_{N-1}}{d x^2}=\lambda_{N-1}^2 M_{N-1}-\frac{f}{d_{N-1}}M_N$$

With BCs $$M_N(1)=M_{N-1}(1)=0,\ \frac{d M_{N}}{d x}(0)=h,\ \frac{d M_{N-1}}{d x}(0)=0$$

The solution to $M_N$ is: $$M_N(x)=\frac{h}{\lambda_N}\text{sech}(\lambda_N)\sinh(\lambda_N(x-1))$$ No surprises there. The trouble comes when I solve $M_{N-1}$. I get: $$M_{N-1}=\frac{fh}{d_{N-1}\lambda_{N}^2}\left(\frac{1}{\lambda_N}\text{sech}(\lambda_N)\sinh(\lambda_N(x-1))-\frac{1}{\lambda_{N-1}}\text{sech}(\lambda_{N-1})\sinh(\lambda_{N-1}(x-1))\right)$$ However, Mathematica gets $$M_{N-1}=\frac{fh}{d_{N-1}(\lambda_{N-1}^2-\lambda_{N}^2)}\left(\frac{1}{\lambda_N}\text{sech}(\lambda_N)\sinh(\lambda_N(x-1))-\frac{1}{\lambda_{N-1}}\text{sech}(\lambda_{N-1})\sinh(\lambda_{N-1}(x-1))\right)$$

Now, I can say for certain that $|\lambda_N| \ne |\lambda_{N-1}|$. I realise Mathematica doesn't know this, and I suspect that could have something to do with this. But I simply can't find a way to come up with the same result by hand.

I know this is a selfish question, but can someone step through the process and show me how to come up with Mathematica's answer please!

Answer 1

The general solution to the ODE

$$ u''(x) = a^2\,u(x) $$ for $a \in \mathbb{R}$ is $$ u(x) = A\sinh(ax) + B\cosh(ax) $$

The BC

$$u(1) = 0$$

then requires $$ B = -A\tanh(a) $$ so $$ u(x) = A \left[\, \sinh(ax) - \tanh(a)\cosh(ax) \,\right] = A\operatorname{sech}(a) \sinh[\,a\,(x-1)\,] $$ and $$ u'(x) = Aa\operatorname{sech}(a) \cosh[\,a\,(x-1)\,] $$

The BC

$$u'(0) = h$$

then implies $$ h = Aa\operatorname{sech}(a) \cosh(a) = Aa $$ so $A = h/a$ and the solution is:

$$ u(x) = \frac{h}{a}\,\operatorname{sech}(a) \sinh[\,a\,(x-1)\,] $$

The solution to the non-homogeneous ODE $$ v''(x) = b^2\,v(x) + c\,u(x) $$ where $b,c \in \mathbb{R}$, subjected to the BCs $v(1) = v'(0) = 0$, is the sum of the general solution of its homogeneous counterpart and a particular solution. The general solution of the homogeneous ODE $v''(x) = b^2\,v(x)$ is, as before, $$ v(x) = C\sinh(bx) + D\cosh(bx) $$

As for a particular solution of $v''(x) = b^2\,v(x) + c\,u(x)$, let's try $$ v(x) = E\,u(x) $$ where $E$ is a constant. Then, $$ v''(x) = b^2\,v(x) + c\,u(x) \implies E\,u''(x) = b^2\,E\,u(x) + c\,u(x) $$ But $u''(x) = a^2\,u(x)$ and we find $$ E = \frac{c}{a^2 - b^2} $$

Thus, the general solution of $v''(x) = b^2\,v(x) + c\,u(x)$ is $$ v(x) = C\sinh(bx) + D\cosh(bx) + \frac{c}{a^2 - b^2}\,u(x) $$ assuming $a \ne b$. Applying the BC $v(1) = 0$, and using $u(1) = 0$, it follows $$ D = -C\tanh(b) $$ so $$ v(x) = C\operatorname{sech}(b) \sinh[\,b\,(x-1)\,] + \frac{c}{a^2 - b^2}\,u(x) $$ and $$ v'(x) = Cb\operatorname{sech}(b) \cosh[\,b\,(x-1)\,] + \frac{c}{a^2 - b^2}\,u'(x) $$

The BCs $v'(0) = 0$ and $u'(0) = h$ then imply $$ 0 = Cb + \frac{ch}{a^2 - b^2} $$ from which $$ C = - \frac{ch}{b}\frac{1}{a^2 - b^2} = \frac{ch}{b}\frac{1}{b^2 - a^2} $$

Therefore, $$ v(x) = \frac{ch}{b}\frac{1}{b^2 - a^2}\,\operatorname{sech}(b) \sinh[\,b\,(x-1)\,] + \frac{c}{a^2 - b^2}\,u(x) $$ which can be rearranged as

$$ v(x) = -\frac{ch}{b^2 - a^2}\, \left\{\, \frac{1}{a}\,\operatorname{sech}(a) \sinh[\,a\,(x-1)\,] - \frac{1}{b}\,\operatorname{sech}(b) \sinh[\,b\,(x-1)\,] \,\right\} $$ where the first term comes from $u(x)$.

The identifications $$ u \quad\leftrightarrow\quad M_N $$ $$ a \quad\leftrightarrow\quad \lambda_N $$ $$ v \quad\leftrightarrow\quad M_{N-1} $$ $$ b \quad\leftrightarrow\quad \lambda_{N-1} $$ $$ c \quad\leftrightarrow\quad -\frac{f}{d_{N-1}} $$ then lead to the result obtained by Mathematica.