# Why can't a second order differential equation have 2 Neuman conditions? [closed]

I read that a problem such as: $$\begin{equation} \begin{cases} -\frac{\partial^2 y}{\partial x^2} + y = f(x)\\ \mathrm{condition} \ n°1\\ \mathrm{condition} \ n°2 \end{cases} \end{equation}$$ Could have a unique solution if $$\mathrm{condition} \ 1$$ and $$\mathrm{condition} \ 2$$ were:

• Initial conditions (Cauchy problem)
• 2 Dirichlet conditions ($$y(a) = y_a, \ y(b) = y_b \ | \ b > a, \ x \in [a, b]$$)
• 1 Dirichlet condition + 1 Neuman condition ($$y'(a) = y_a, \ y(b) = y_b \ | \ b > a, \ x \in [a, b]$$ or the contrary)

Why can't we have 2 Neuman conditions ?

• Your title mentions homogeneous equations while the one you wrote isn't homogeneous May 17 at 16:33
• Right, I'll change that
– user1057995
May 17 at 16:35
• The actual general theory here is called the Fredholm alternative, which relates existence of solution to $Lu=f$ to existence of nontrivial solution to $L^*u=0$, where $L^*$ is the adjoint of the linear differential operator $L$. The quirky thing about this theory in the setting of differential equations is that what $L^*$ actually looks like depends on both the operator $L$ and the boundary conditions you're looking at.
– Ian
May 17 at 17:19

In elementary terms, whether the solution to the BVP with boundary conditions given in terms of operators $$L_a[f]$$ and $$L_b[f]$$ is guaranteed to exist uniquely comes down to whether the matrix

$$\begin{bmatrix} L_a[f_1] & L_a[f_2] \\ L_b[f_1] & L_b[f_2] \end{bmatrix}$$

is invertible or not, where $$f_1,f_2$$ are two linearly independent solutions to the homogeneous equation. If it's invertible, then if a solution exists then it is unique, and it turns out that for a "typical" operator the solution exists as well (although this part is more complicated to show in general). If it's singular, then the solution either doesn't exist or isn't unique, and which case you are in depends on what $$f$$ is.

In the special case of $$Lu=u''$$, you can choose $$f_1=1,f_2=x$$. In the pure Dirichlet case there is no problem as you have

$$\begin{bmatrix} 1 & a \\ 1 & b \end{bmatrix}$$ and the determinant is $$b-a \neq 0$$. In the half-Neumann case you also have no problem; supposing the Neumann condition is on the left, you're looking at

$$\begin{bmatrix} 0 & 1 \\ 1 & b \end{bmatrix}$$

which is again invertible. But in the pure Neumann case you're looking at

$$\begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}$$

which is singular. This corresponds to the fact that $$f(x)=c$$ is a solution to the fully homogeneous problem (homogeneous ODE+ homogeneous BCs) for any $$c$$.

The situation changes with $$y''=\lambda y$$ with nonzero $$\lambda$$. In this case the pure Neumann matrix is $$\begin{bmatrix} \sqrt{\lambda} e^{\sqrt{\lambda} a} & -\sqrt{\lambda} e^{-\sqrt{\lambda} a} \\ \sqrt{\lambda} e^{\sqrt{\lambda} b} & -\sqrt{\lambda} e^{-\sqrt{\lambda} b} \end{bmatrix}$$ and the determinant is $$\lambda(e^{\sqrt{\lambda}(b-a)}-e^{-\sqrt{\lambda}(b-a)})$$. So the requirement now is is that $$2\sqrt{\lambda}(b-a)$$ isn't an integer multiple of $$2\pi i$$. If instead it is, then you have nontrivial homogeneous solutions given by cosines.

The general theory underlying this kind of question is called the Fredholm alternative, but it comes with some functional analysis baggage to understand, especially because we usually care about non-compact operators.

Because you can’t have all your conditions with derivates. You need at least 1 normal condition because of the integration constant.

• I think this is not exactly correct. It is correct for $y''=f$ with Neumann BCs; there is a nontrivial solution to $y''=0$ with homogeneous Neumann BCs namely any constant, so the solution if it exists is not unique. But with $y''+\lambda y=0$ I think it depends on what $\lambda$ is. In particular $\begin{bmatrix} \sqrt{\lambda} e^{\sqrt{\lambda} a} & -\sqrt{\lambda} e^{-\sqrt{\lambda} a} \\ \sqrt{\lambda} e^{\sqrt{\lambda} b} & -\sqrt{\lambda} e^{-\sqrt{\lambda} b} \end{bmatrix}$ is sometimes invertible.