Fundamental solution for unbounded solution

joshmccraney

Gold Member
1,612
42
1. The problem statement, all variables and given/known data
Find the fundamental solution to the unbounded problem $$t u''(t) - u'(t) = \delta_0.$$

2. Relevant equations
Variation of parameters.

3. 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?
 
Last edited:

Ray Vickson

Science Advisor
Homework Helper
10,639
1,667
1. The problem statement, all variables and given/known data
Find the fundamental solution to the unbounded problem $$t u''(t) - u'(t) = \delta_0.$$

2. Relevant equations
Variation of parameters.

3. 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.
 

joshmccraney

Gold Member
1,612
42
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

Science Advisor
Homework Helper
10,639
1,667
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

Gold Member
396
71
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 ?
 

joshmccraney

Gold Member
1,612
42
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

Science Advisor
Homework Helper
10,639
1,667
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

Science Advisor
Homework Helper
10,639
1,667
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
396
71
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

Science Advisor
Homework Helper
10,639
1,667
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.
 

joshmccraney

Gold Member
1,612
42
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

Science Advisor
Homework Helper
10,639
1,667
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.
 
Last edited:

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top