tearstreakdl

2022-01-15

A question asks us to solve the differential equation
$-u\left(x\right)=\delta \left(x\right)$
with boundary conditions
where $\delta \left(x\right)$ is the Dirac delta function. But inside the same question, teacher gives the solution in two pieces as . I understand when we integrate the delta function twice the result is the ramp function R(x). However elsewhere in his lecture the teacher had given the general solution of that DE as
$u\left(x\right)=-R\left(x\right)+C+Dx$

Dawn Neal

Expert

What you call the Dirac delta function (which is not a function, at least not in the sense of a function from R to R) is a strange object but something about it is clear:
We will not use anything else about the Dirac $\delta$.
If one also asks that ${\int }_{y}^{z}u{}^{″}\left(x\right)dx={u}^{\prime }\left(z\right)-{u}^{\prime }\left(y\right)$ for every $y\le z$, one can integrate once your equation $-u{}^{″}=\delta$, getting that there exists a such that
${u}^{\prime }\left(x\right)=a-\left[x\ge 0\right]$.
Using the facts that ${\int }_{y}^{z}{u}^{\prime }\left(x\right)dx$ should be $u\left(z\right)-u\left(y\right)$ for every $y\le z$, and the value of ${\int }_{y}^{z}\left[x\ge 0\right]dx$, one gets that for every fixed negative number ${x}_{0}$,
$u\left(x\right)=u\left({x}_{0}\right)+a\cdot \left(x-{x}_{0}\right)-x\cdot \left[x\ge 0\right]$.
This means that $b=u\left({x}_{0}\right)-a\cdot {x}_{0}$ does not depend on ${x}_{0}<0$, hence finally, for every x in R,
$u\left(x\right)=a\cdot x+b-x\cdot \left[x\ge 0\right]$.
(And, in the present case, the condition that $u\left(-2\right)=u\left(3\right)=0$ imposes that .)
This is the general solution of the equation $-u{}^{″}=\delta$. Note that every solution u is ${C}^{\mathrm{\infty }}$ on R\0 but only ${C}^{0}$ at 0 hence u' and u'' do not exist in the rigorous sense usually meant in mathematics. Note finally that u is also
$u\left(x\right)=a\cdot x+b-x\cdot \left[x>0\right]$.

lalilulelo2k3eq

Expert

Both describe the same type of function. The ramp function is nothing but
$R\left(x\right)=\left\{\begin{array}{ll}0& x\le 0\\ x& x\ge 0\end{array}$
If you use the general solution and plug in the same boundary conditions,
$u\left(-2\right)=-R\left(-2\right)+C-2D=C-2D=0$
$u\left(3\right)=-R\left(3\right)+C+3D=-3+C+3D=0$
with the solution $C=\frac{6}{5},D=\frac{3}{5}$, and then split it at $x=0$ to get rid of the ramp function and you obtain;
$u\left(x\right)=\left\{\begin{array}{ll}\frac{6}{5}+\frac{3}{5}x& x\le 0\\ -x+\frac{6}{5}+\frac{3}{5}x=\frac{6}{5}-\frac{2}{5}x& x\ge 0\end{array}$
which is exactly the same expression you got by splitting the function earlier (pratically, there is no difference, but its shorter when written with the ramp function).

alenahelenash

Expert

I guess now I understand why and how the teacher came up with the two pieces. Since the form of the solution u(x) is known (it is linear) and because of the presence of the ramp function, one could conclude two pieces could be in the form of because both these functions are zero at the boundaries left and right respectively. About the why: Then he can use simpler integration and derivation:$-\int {u}^{″}\left(x\right)=\int \delta \left(x\right)$$-\left[{u}^{\prime }\left(x\right){\right]}_{L}^{R}=1$${u}_{R}^{{}^{\prime }}\left(x\right)-{u}_{L}^{{}^{\prime }}\left(x\right)=-1$Since we know the pieces, and they are in pretty simple form we can apply the derivatives above and substitute$B-A=-1$The pieces meet at $x=0$, and we know that due to the ramp function, continuing$A\left(0+2\right)=B\left(0-3\right)$$A=-\frac{3}{2}B$Combining$B-\left(-3/2B\right)=-1$$B=-0.4$$A=0.6$