# I Homogeneous equation and orthogonality

#### dRic2

Gold Member
Hi, I'm going to cite a book that I'am reading

It is in the general theory of differential equations that the right-hand side of
$$y''''(x) = \rho (x)$$
can be prescribed arbitrary only if the corresponding homogeneous differential equation has no solution except the trivial solution $y=0$. In all the previous cases the boundary conditions were such that the differential equation
$$y''''(x) = 0$$
had no solution under the given boundary conditions. Here, however, two such independent solution exist, namely
$$y = 1$$
$$y = x$$
In such case our boundary value problem is not solvable unless $\rho(x)$ is "orthogonal" to the homogeneous solutions - i.e.
$$\int_0^l \rho(x)dx = 0$$
$$\int_0^l \rho(x)xdx = 0$$
Can anyone provide some simple references where I can find at least an intuition regarding what is stated by the author.

Thanks,
Ric

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#### S.G. Janssens

You would have to make the setting more precise.
1. Which function space? (Presumably $L^2(0,l)$ with standard inner product.)
2. How is the differential operator exactly defined? (In particular, what is its domain?)
Assuming you have checked this, this looks like a direct application of the Closed Range Theorem. (You can search for this term to find references in most FA textbooks. In the simplest setting, it characterizes the closed range of a closed, densely defined operator as the orthoplement of the nullspace of its adjoint.)

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#### dRic2

Gold Member
You would have to make the setting more precise.
1. Which function space? (Presumably $L^2(0,l)$ with standard inner product.)
2. How is the differential operator exactly defined? (In particular, what is its domain?)
Not sure, really. This is from a book on variational mechanics and the problem is introduced very quickly and without any particular definitions. It was something like:

Consider the equilibrium of an elastic bar. For elasticity theory we know that the variational problem takes the following form... the Euler-Lagrange equations thus are....

and very few lines after begins the section I quoted.

Assuming you have checked this, this looks like a direct application of the Closed Range Theorem. (You can search for this term to find references in most FA textbooks.
Thanks. I took an intro course in Functional Analysis (for physics and engineer), but we didn't cover the range of an operator. I will see what I can find, thanks again! :D

#### S.G. Janssens

I took an intro course in Functional Analysis (for physics and engineer), but we didn't cover the range of an operator.
You may also know it as the "image". If $A : D(A) \subseteq X \to X$ is a linear operator on a vector space $X$, then its range is just $R(A) := \{y \in X\,:\, \exists x \in D(A) \text{ such that } Ax = y\}$. (More generally, you can of course do this for any function between sets, but then the term "image" is used more often than "range".)

#### dRic2

Gold Member
Oh, this sounds familiar. Thank you.

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