# Solving 2nd order with boundary conditions

$$f''+\lambda f = 0, \quad f=f(y), \quad a \leq y \leq b, \quad f(a)=f(b)=0$$

In this case, I tried

$$f(y)=sin(k_ny), \quad \lambda=k_n^2$$ $$f(a)=0 \Longrightarrow k_na = n\pi, \quad n=0,1,2,...$$ $$k_n=\frac{n\pi}{a}, \quad n=0,1,2,...$$

The problem is that I don't know how to apply the second boundary condition here, so I know something in my approach is wrong. The answer is stated below. I can see why this solution works, but I just don't know how to derive it myself. Also I'm wondering if there is a solution that works from $$n=0$$.

$$f(y)=sin(k_n(y-a)), \quad k_n=\frac{n\pi}{b-a}, \quad n=1,2,...$$

EDIT: I tried setting $$f(y) = Acos(k_ny) + Bsin(k_ny)$$ $$f(a) = Acos(k_na) + Bsin(k_na) = 0$$ $$f(b) = Acos(k_nb) + Bsin(k_nb) = 0$$

But I'm not sure where to go from here. $$f(a)$$ can be zero if $$A=-B, k_na = \pi/4 + n\pi$$, but it can also be zero if $$A=0, k_na=n\pi$$ or $$B=0, k_na=\pi/2+n\pi$$. Do I have to check the second condition for all these 3 solutions, or is there any way to know which one I should go with?

• The function you should try is $f(y) = C_1 \cos (k_n y) + C_2 \sin (k_n y)$ – JoseSquare Feb 25 at 20:12
• $f_0 (y)=0$ obviously satisfies both the equation and boundary conditions. – user Feb 25 at 20:54
• @user but in the answer, $n=0$ isn't defined. so my second question was if there's a way to rewrite the answer so it's defined for $n=0$ aswell. this is not as important as understanding how they reach the answer though, it's 2nd prio – armara Feb 26 at 7:47
• @JoseSquare hey thx for the tip, I updated the answer with an EDIT – armara Feb 26 at 7:47
• Contrary to what you think, there are no distinct $k_n$ other than $\sqrt\lambda$. – Yves Daoust Feb 26 at 8:10

Assuming $$\lambda>0$$, the general solution is

$$f(y)=c_c\cos(\sqrt\lambda y)+c_s\sin(\sqrt\lambda y).$$

Plugging the boundary conditions,

$$0=c_c\cos(\sqrt\lambda a)+c_s\sin(\sqrt\lambda a), \\0=c_c\cos(\sqrt\lambda b)+c_s\sin(\sqrt\lambda b).$$

The determinant of this homogeneous linear system is

$$\sin(\sqrt\lambda(b-a)).$$

So

• if $$\sqrt\lambda(b-a)$$ is not a multiple of $$\pi$$, only the trivial solution $$f=0$$ is possible;

• if $$\sqrt\lambda(b-a)$$ is a multiple of $$\pi$$, the solutions are

$$c\sin(\sqrt\lambda(x-a))$$ where $$c$$ is free.

• this was an approach i hadn't thought about, need to read a bit more about how determinants can be used to solve system of equations. thanks for bringing this to my attention, i gave you +1! – armara Feb 26 at 8:16
• @armara: you'd better accept this answer. It does give the complete solution. [I never ask such things, but here it's too blatant.] You seem to be stuck on the idea that there are several modes, which is false. – Yves Daoust Feb 26 at 8:19
• sure I can do that, you obviously know this better than me. however, i'm personally going to use the other answer for now since I don't quite understand this one. i'll be studying this answer tho, seems like it's good :) if you have any pointers on where i should read to understand these concepts better, i would gladly go there – armara Feb 26 at 8:26
• @armara: I can't believe that you don't know the resolution of a $2\times2$ system. You are just disconcerted by the trigonometric functions. – Yves Daoust Feb 26 at 8:29

Hint: Solve the related problem

$$f''(x) + \lambda f(x) = 0, \quad 0 \le x \le 1, \quad f(0)=f(1)=0$$

Then make the coordinate transform $$y = a + (b-a)x$$

Edit: You can think about the transform organically. The general solution is

$$f(y) = A\cos(\sqrt{\lambda}y) + B\sin(\sqrt{\lambda}y) \tag{1}$$

This would be very convenient if one of the B.C.s was at $$x=0$$ since you have $$f(0) = A$$ and $$f'(0)=\sqrt{\lambda}B$$. Since this is not the case, you can try to see what would happen if the solution was shifted to $$x=a$$ instead

$$f(y) = \widetilde{A}\cos(\sqrt{\lambda}(y-a)) + \widetilde{B}\sin(\sqrt{\lambda}(y-a)) \tag{2}$$

This is also a solution of the ODE since the operation $$f''$$ is shift-invariant, and the shifted sinusoids are still a linear combination of the original $$\sin$$ and $$\cos$$ functions.

Using $$(2)$$ as the solution form gives $$\widetilde{A} = 0$$ and $$\sqrt{\lambda} = \dfrac{n\pi}{b-a}$$.

Edit 2: I will expand on Yves' answer further to show how you can arrive at the same answer

Using $$(1)$$ as the solution form, we need to solve

\begin{align} A\cos(\sqrt{\lambda}a) + B\sin(\sqrt{\lambda}a) &= 0 \\ A\cos(\sqrt{\lambda}b) + B\sin(\sqrt{\lambda}b) &= 0 \end{align}

We'll get a non-trivial solution if the determinant of this system is $$0$$. Another way to understand this without using the determinant is to solve the system normally. Find $$B$$ in terms of $$A$$ in the second equation and plug it into the first, you get

$$A\cos(\sqrt{\lambda}a) - A\frac{\cos(\sqrt{\lambda}b)}{\sin(\sqrt{\lambda}b)}\sin(\sqrt{\lambda}a) = 0$$

$$\implies A\big[\cos(\sqrt{\lambda}a)\sin(\sqrt{\lambda}b) - \cos(\sqrt{\lambda}b)\sin(\sqrt{\lambda}a)\big] = A\sin(\sqrt{\lambda}(b-a)) = 0$$

We can't have $$A=0$$, so we require

$$\sin\sqrt{\lambda}(b-a) = 0 \implies \sqrt{\lambda}(b-a) = n\pi$$

where $$n=1,2,3,\dots$$, $$n$$ can't be $$0$$ here since it would also give a trivial solution (you should've excluded the cases $$\lambda=0$$ and $$\lambda < 0$$ earlier on)

Now, $$A$$ and $$B$$ form a family of solutions satisfying

$$\frac{A}{B} = -\frac{\sin(\sqrt{\lambda}a)}{\cos(\sqrt{\lambda}a)} = -\frac{\sin(\sqrt{\lambda}b)}{\cos(\sqrt{\lambda}b)}$$

we can arbitrarily set $$A = -c\sin(\sqrt{\lambda}a)$$, $$B=c\cos(\sqrt{\lambda}a)$$ (which is fine since we know the final solution still has one free constant). Thus

$$f(y) = -c\sin(\sqrt{\lambda}a)\cos(\sqrt{\lambda}y) + c\cos(\sqrt{\lambda}a)\sin(\sqrt{\lambda}y) = c\sin(\sqrt{\lambda}(y-a))$$

• hey and thanks for the tip, i'll definitely remember this approach. however, i'd like to understand the regular way aswell (setting $f(x)=Acos..+Bsin..$) since I won't always think of a good transformation :) – armara Feb 26 at 7:49
• You seem to be very focused on the normalization of the variable. This is neither necessary, nor really helpful. It doesn't change the nature of the resolution. – Yves Daoust Feb 26 at 8:15
• (after your edit): this makes so much sense... you know that you want either $A=0$ or $B=0$, so you shift both your sine and your cosine to make that happen. and then it's trivial. big thanks, this helped a lot! – armara Feb 26 at 8:15
• @YvesDaoust You're right. It was mostly the shifting I wanted to focus on. The scaling just makes it look "nice" – Dylan Feb 26 at 8:20