Dirac Delta function/BVP

Question

I am self-studying differential equations using MIT's publicly available materials. One of the recitation questions runs as follows:

The following boundary-value problem models the equation of the central line $y(x)$, $0\leq x \leq 2$, of a uniform weightless beam anchored at one end and carried a concentrated load at its center. \begin{align} y^{(iv)} = 6\delta(x - 1) \qquad y(0) = y'(0) = y''(2) = y'''(2). \end{align} (a) With $y''(0) = 2a$, $y'''(0) = 6b$, find $y$ via $Y(s)$.

(b) Determine $a$ and $b$ from the boundary conditions at $x = 2$.

I'm afraid I don't understand the question. If the beam is anchored at one end, doesn't that imply that $y^{(n)} = 0$ for all $n$, except perhaps for $n = 0$? (In which case, we can just adjust our coordinates accordingly, so that $y(0) = 0$ w.l.o.g.) Then $a = b = y(0) = y'(0) = 0$ trivially, $Y(s) = 6e^{-s}{s^4}$ and $y = h(x-1)(t-1)^3$, where $h(x)$ is the Heaviside step function. However, this seems far too simple; moreover, it ignores the constraints on $y''(2)$ and $y'''(2)$.

I have grown accustomed to the occasional omission/typo in these materials, especially the recitation exercises. Still, it has also been the case that I have misunderstood/misread a problem. I would like to know if this is the case here--does the question make sense as stated, or is there (in the opinion of the community) something missing?

Answer 1

Notice:

• $$\mathcal{L}_x\left[n\delta(x-z)\right]_{(s)}=\frac{n\theta(z)}{e^{sz}}$$
• $$\mathcal{L}_x\left[y^{(n)}(x)\right]_{(s)}=s^n\text{Y}(s)-s^{n-1}y(0)-s^{n-2}y'(0)-\dots-y^{(n-1)}(0)$$

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So, we get:

$$y^{(4)}(x)=n\delta(x-z)\Longleftrightarrow$$ $$\mathcal{L}_x\left[y^{(4)}(x)\right]_{(s)}=\mathcal{L}_x\left[n\delta(x-z)\right]_{(s)}\Longleftrightarrow$$ $$s^4\text{Y}(s)-s\left(s\left(sy(0)+y'(0)\right)+y''(0)\right)-y^{(3)}(0)=\frac{n\theta(z)}{e^{sz}}\Longleftrightarrow$$

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Using the initial conditions (a):

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$$s^4\text{Y}(s)-s\left(s\left(sy(0)+y'(0)\right)+2a\right)-6b=\frac{n\theta(z)}{e^{sz}}\Longleftrightarrow$$ $$\text{Y}(s)=\frac{\frac{n\theta(z)}{e^{sz}}+6b+s\left(s\left(sy(0)+y'(0)\right)+2a\right)}{s^4}$$

Now, with inverse Laplace transform:

$$y(x)=\frac{n(x-z)^3\theta(x-1,z)}{6}+y(0)+x(x(a+bx)+y'(0))$$

Answer 2

@solitaireartist... Yes, the question does make sense. This is a fourth-order ODE describing the bending of a cantilever beam. 'Anchored at one end' you were referring to; in a physical sense means that both the displacement i.e. y(0) and the slope i.e. y'(0) at the clamped end (@x=0) will be equal to zero. Whereas the bending moment which contains the double derivative y''(0) and the shear force which contains the third derivative y'''(0) will both be some finite value and not zero as there is a load applied on the beam. On the other hand at the free end of the beam (@x=2), the bending moment and shear will both be equal to zero. y''(2)=y'''(2)=0

@Jan Eerland has given the solution through the Laplace transform. On solving by the classical method, I ended up getting the solution (centre-line displacement) as y(x)= -x^3 +3x -1 and using the boundary conditions the constants evaluated to a=3 and b=-1.

To get the feel for this problem one needs to have a mechanics background.