# Laplace Transform to solve ODE

I'm trying to solve the BVP $$f''(x)=\delta(x-a)$$ where $$0 and $$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.

• 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 $$L(u''(t))(s)=s\cdot(s\cdot L(u(t))(s)-u'(0))-u(0)$$ Jan 28, 2023 at 15:09
• That's not quite right. I think you have your u and u' mixed up
– jet
Jan 28, 2023 at 15:15
• 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 $$\hat u(s) = \frac{s^2 u_0 + su_0' + e^{-as}}{s^3}.$$ Jan 28, 2023 at 17:22

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 ?

• I got a different RHS after applying LT. Isn't the Laplace transform of Heaviside $H(x-a)$ equal to $e^{-ap}$?
– jet
Jan 28, 2023 at 15:36
• You are right. I was distracted ! Jan 28, 2023 at 15:37
• Fixed now...... Jan 28, 2023 at 15:39
• 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. Jan 28, 2023 at 15:43
• Actually, even I'm distracted since Laplace Transform of $H(x-a)$ is not $e^{-ap}$ but actually $\frac{e^{-ap}}{p}$
– jet
Jan 28, 2023 at 15:46