Fundamental solution for unbounded solution

[member 428835]

Homework Statement

Find the fundamental solution to the unbounded problem $$t u''(t) - u'(t) = \delta_0.$$

Homework Equations

Variation of parameters.

The Attempt at a Solution

I'm not sure how to use variation of parameters on this since it's an unbounded problem, so I'm not even trying to use it. The homogenous solution is $u_h = c_1 t^2 + c_2$. So I'm thinking the fundamental solution should look something like
$$u =
\left\{
\begin{array}{ll}
c_1 t^2 + c_2 & t\leq 0 \\
c_3 t^2 + c_2 & t>0
\end{array}
\right.
$$
where I make both constant terms $c_2$ to ensure continuity at $t=0$. But now I'm really not sure, is this even right?

 

[Ray Vickson]

Homework Statement

Find the fundamental solution to the unbounded problem $$t u''(t) - u'(t) = \delta_0.$$

Homework Equations

Variation of parameters.

The Attempt at a Solution

I'm not sure how to use variation of parameters on this since it's an unbounded problem, so I'm not even trying to use it. The homogenous solution is $u_h = c_1 t + c_2$. So I'm thinking the fundamental solution should look something like
$$u =
\left\{
\begin{array}{ll}
c_1 t + c_2 & t\leq 0 \\
c_3 t + c_2 & t>0
\end{array}
\right.
$$
where I make both constant terms $c_2$ to ensure continuity at $t=0$. But now I'm really not sure, is this even right?

Is $\delta_0$ a constant, of do you really mean $\delta(t)?$

Your fundamental solutions are missing some important terms.

 

[member 428835]

Is $\delta_0$ a constant, of do you really mean $\delta(t)?$

Your fundamental solutions are missing some important terms.

Sorry, it's the dirac delta function, so $\delta(t)$. Yea what should I add?

 

[Ray Vickson]

Sorry, it's the dirac delta function, so $\delta(t)$. Yea what should I add?

When I try to substitute your $u_h(t)$ into the (homogeneous) DE, it does not work.

PF rules forbid me from telling you the solution, but I can tell you how I would look for it if I were doing the problem.

I would set $v(t) = u'(t)$ and then solve the de $t v'(t) - v(t) = \delta(t).$ From there, $u(t) = \int v(t) \, dt.$

BTW: there seem to be serious problems if your right-hand-side is $\delta(t)$ exactly; I think it would be better to use $\delta(t-t_0)$, then see if there is a sensible limit of the solution when you try to take $t_0 \to 0.$ Alternatively, you could take the de to be $(t-t_0) u''(t) - u'(t) = \delta(t),$ then look at what happens to the solution as $t_0 \to 0.$

 

[dRic2]

I've never studied differential equation with a Dirac delta function, but I have encountered them several times in my studies. In the books I read authors usually change the delta into a "source condition", that is they change the equation into something like this:
$$t u'' - u' = 0$$
with $\lim_{t \rightarrow 0} u(t) = 1$. Is this wrong ? I am asking because I'm interested and I do not know if I remember it correctly.

PS: BTW shouldn't it be easier to use Fourier or Laplace transform ?

 

[member 428835]

When I try to substitute your $u_h(t)$ into the (homogeneous) DE, it does not work.

PF rules forbid me from telling you the solution, but I can tell you how I would look for it if I were doing the problem.

I would set $v(t) = u'(t)$ and then solve the de $t v'(t) - v(t) = \delta(t).$ From there, $u(t) = \int v(t) \, dt.$

BTW: there seem to be serious problems if your right-hand-side is $\delta(t)$ exactly; I think it would be better to use $\delta(t-t_0)$, then see if there is a sensible limit of the solution when you try to take $t_0 \to 0.$ Alternatively, you could take the de to be $(t-t_0) u''(t) - u'(t) = \delta(t),$ then look at what happens to the solution as $t_0 \to 0.$

Sorry, I forgot to square it! I had it written on paper but forgot to type it. I edited my first post to reflect the change.

You say there are serious problems: can you elaborate? Are you using variation of parameters to solve, or something else?

 

[Ray Vickson]

Sorry, I forgot to square it! I had it written on paper but forgot to type it. I edited my first post to reflect the change.

You say there are serious problems: can you elaborate? Are you using variation of parameters to solve, or something else?

I have already suggested how you should proceed: solve the problem with the right-hand-side changed to $\delta(t-t_0), \: t_0 \neq 0$ then try to see what happens when you attempt to take $t_0 \to 0.$

 

[Ray Vickson]

I've never studied differential equation with a Dirac delta function, but I have encountered them several times in my studies. In the books I read authors usually change the delta into a "source condition", that is they change the equation into something like this:
$$t u'' - u' = 0$$
with $\lim_{t \rightarrow 0} u(t) = 1$. Is this wrong ? I am asking because I'm interested and I do not know if I remember it correctly.

PS: BTW shouldn't it be easier to use Fourier or Laplace transform ?

If the DE has constant coefficients then the Laplace transform method would work well. However, this problem has a term $t u''(t)$ in it, and gettimg the Laplace transform of that would not be easy. I think you would end up needing to solve and integral equation to determine the Laplace transform (because the transform of $t u''(t)$ would be the convolution of the transforms of $t$ and $u''(t)$).

 

[dRic2]

If the DE has constant coefficients then the Laplace transform method would work well. However, this problem has a term $t u''(t)$ in it, and gettimg the Laplace transform of that would not be easy. I think you would end up needing to solve and integral equation to determine the Laplace transform (because the transform of $t u''(t)$ would be the convolution of the transforms of $t$ and $u''(t)$).

What about the first part of my post: is it correct? Or am I completely wrong ?

 

[Ray Vickson]

What about the first part of my post: is it correct? Or am I completely wrong ?

Perhaps wrong, but the question was not specific enough for me to really be sure. Try actually doing what you suggest and then see how it works out.

 

[member 428835]

I have already suggested how you should proceed: solve the problem with the right-hand-side changed to $\delta(t-t_0), \: t_0 \neq 0$ then try to see what happens when you attempt to take $t_0 \to 0.$

I don't know how to solve because the problem is unbounded. For me, I'm at the same spot if the RHS was $\delta(t)$ or $\delta(t-t_0)$. Can you give me an idea for direction?

Typically I would use BCs at left and right endpoints and construct the Green's function via variation of parameters. But I can't do this since I don't know how to handle the unbounded nature of the problem.

 

[Ray Vickson]

I don't know how to solve because the problem is unbounded. For me, I'm at the same spot if the RHS was $\delta(t)$ or $\delta(t-t_0)$. Can you give me an idea for direction?

Typically I would use BCs at left and right endpoints and construct the Green's function via variation of parameters. But I can't do this since I don't know how to handle the unbounded nature of the problem.

Here is how I was taught this stuff back in the Stone Age.

Let us take the DE to be $t u''(t) - u'(t) = \delta(t-a), \; a \neq 0.$

We want a particular solution $u_p(t).$ Furthermore, let's require that $u_p(t)$ be continuous at $t=a$ and that $u'_p(t)$ remain finite in a neighborhood of $t=a.$ Then, we must have
$$\begin{array}{rcr}
u_p(a+) - u_p(a-) & =0& \hspace{3ex}(1) \\
a[u'_p(a+) - u'_p(a-)]& = 1& \hspace{3ex}(2)
\end{array}$$ The first condition is just continuity of $u_p(t)$ at $t=a$, while the second one is a jump condition on $u_p'(t)$ at $t=a$:
$$\lim_{\epsilon \to 0} \int_{a-\epsilon}^{a+\epsilon} t u_p''(t) \, dt = \lim_{\epsilon \to 0} \int_{a-\epsilon}^{a+\epsilon} \delta(t-a) \, dt = 1$$ where we use the fact that the other terms $\int_{a-\epsilon}^{a+\epsilon} u_p'(t)(t) \, dt \to 0$ as $\epsilon \to 0$ because of finiteness of $u'_p(t).$

If we write
$$u_p(t) = \begin{cases} b_1 + c_1 t^2 , & t < a\\
b_2 + c_2 t^2 , & t \geq a
\end{cases}$$
then conditions (1) and (2) give
$$\begin{array}{ccr}
b_1+c_1 a^2 - b_2 - c_2 a^2 &=0& \hspace{3ex}(1a) \\
2 a^2 c_2 - 2 a^2 c_1 &=1 & \hspace{3ex}(2a)
\end{array}$$
We can place two additional restrictions on the $b_i, c_i$ in order to get a unique solution. For example, we could take $b_1 = 0, c_1 = 0$.

We can get a general $u(t)$ by adding the one-piece homogeneous solution $b+c t^2$ onto this specific $u_p(t).$

Now we can see what would go wrong if we try taking $a=0$: the jump condition would be $0 [u_p'(0+) - u_p'(0-)] = 1,$ which is not possible if both $u_p'(0\, \pm)$ are finite.