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I'm trying to solve the BVP $f''(x)=\delta(x-a)$ where $0<a<1$ and $f(0)=f(1)=1$ but I'm really not sure where to start. I tried taking the Laplace Transform of the equation, to get (after applying $f(0)=0$):

$p^2\bar{f}(p)-f'(0)=e^{-ap}$

But I'm not sure how to proceed. I need to somehow eliminate the $f'(0)$ term, and I'm sure I have to do this via the other condition $f(1)=0$, but not sure how to apply it. Then I'd invert $\bar{f}(p)$.

Some ideas: Maybe a change of variables $x\to 1-x$? and appeal to symmetry of delta function. Any help would be great. Thanks.

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    $\begingroup$ You have to be careful: The Laplace transform of $u''(t)$ is $$L(u''(t))(s) = s\cdot L(u'(t))(s)-u'(0),$$ so in terms of just $u$ you get $$L(u''(t))(s)=s\cdot(s\cdot L(u(t))(s)-u'(0))-u(0)$$ $\endgroup$ Jan 28, 2023 at 15:09
  • $\begingroup$ That's not quite right. I think you have your u and u' mixed up $\endgroup$
    – jet
    Jan 28, 2023 at 15:15
  • $\begingroup$ Because I can't edit my comment (you are right, I mixed up $u(0)$ and $u'(0)$ there): One has $$L(u''(t))(s) = s( sL(u(t))(s)-u(0))-u'(0),$$ so when writing $\hat u := L(u(t))$, $u_0 := u(0)$, $u_0':= u'(0)$, your ODE becomes $$s^2 \hat u(s) - su_0-u'_0 = \frac{e^{-a s}}{s},$$ which is of course equivalent to $$\hat u(s) = \frac{s^2 u_0 + su_0' + e^{-as}}{s^3}.$$ $\endgroup$ Jan 28, 2023 at 17:22

1 Answer 1

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Do a first of integration "by hand":

$$f'(t)=Y(t-a)+k$$

where $Y$ is Heaviside function, and $k$ a constant to be obtained later (by Initial Conditions).

Then only, apply Laplace Transform :

$$sF(s)-\underbrace{f(0)}_1=\frac{e^{-sa}}{s}+ks$$

$$F(s)=\dfrac{e^{-sa}}{s^2}+\dfrac{1}{s}+k$$

Can you proceed from here ?

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  • $\begingroup$ I got a different RHS after applying LT. Isn't the Laplace transform of Heaviside $H(x-a)$ equal to $e^{-ap}$? $\endgroup$
    – jet
    Jan 28, 2023 at 15:36
  • $\begingroup$ You are right. I was distracted ! $\endgroup$
    – Jean Marie
    Jan 28, 2023 at 15:37
  • $\begingroup$ Fixed now...... $\endgroup$
    – Jean Marie
    Jan 28, 2023 at 15:39
  • $\begingroup$ Applying directly Laplace Transform to the second order equation, you have a $f'(0)$ to which you cannot give at once a value, but only at the very end. $\endgroup$
    – Jean Marie
    Jan 28, 2023 at 15:43
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    $\begingroup$ Actually, even I'm distracted since Laplace Transform of $H(x-a)$ is not $e^{-ap}$ but actually $\frac{e^{-ap}}{p}$ $\endgroup$
    – jet
    Jan 28, 2023 at 15:46

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