# Fundamental solution for unbounded solution

Gold Member

## 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?

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Science Advisor
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## 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.

Gold Member
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?

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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.##

Gold Member
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 ?

Gold Member
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?

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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.##

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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
Gold Member
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 ?

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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.

Gold Member
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.

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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.

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