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I need to solve the following equation for $v(x)$: $$\int_0^tv(x)(x+1)dx=f(t)$$ I am given the function $f(t)$. I've done this so far:

If we derive both sides by $t$, we get $v(t)(t+1)=f'(t)$ and $\bar{v}(t)=\frac{f'(t)}{t+1}$. The problem is that I am still off by a constant, i.e., the above only guarantees that : $\int_0^t\bar{v}(x)(x+1)dx+c=f(t)$ which is not enough for me.

doraemonpaul
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Hasanhasan Hasan
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  • Hint: consider taking the Fourier or Laplace transform from both sides, solve algebraic equation and making the inverse transform. – m0nhawk Dec 16 '12 at 19:38
  • @monhawk: how should that help? – Fabian Dec 16 '12 at 19:38
  • If your $f(0)=0$, then $c\equiv0$ for any $\nu(x)$. Isn't it? – 0x2207 Dec 16 '12 at 19:38
  • @0x2207: yes, the constant is in fact $f(0)$. – Fabian Dec 16 '12 at 19:39
  • @Fabian, from definition $\int_{0}^{0} g(x) dx \equiv 0$ – 0x2207 Dec 16 '12 at 19:40
  • @Fabian: The transform would "eat" the integral and we'll have the multiply of two images: $\mathcal{F}[v(x)]\cdot \mathcal{F}[x+1]$. And $\mathcal{F}[f(t)]$. – m0nhawk Dec 16 '12 at 19:41
  • @0x2207: he has $f(t)$ given and searches for $v(x)$. The correct statement is, that the equation does not have a solution unless $f(0)=0$. – Fabian Dec 16 '12 at 19:41
  • @m0nhawk: I do not understand how the transform would eat the integral. Could you expand your comment a bit? – Fabian Dec 16 '12 at 19:42
  • @Fabian: This is the Fredholm equation and the integral represent the convolution with the kernel (kernel and function can be choosed arbitrary in this particular case - no $t$ in under-integral functions). And for every (I don't know a contradiction example) integral transform the convolution of two function transforms into the multiplication of the transforms. See convolution theorem for details. And the constant would naturally appear after transforms. – m0nhawk Dec 16 '12 at 19:50
  • @monhawk: Here is the convolutions theorem for the Laplace transform. I am just not sure how you imagine to choose $f_1$ and $f_2$ such that the integral in the post corresponds to a convolutions for which one can apply this theorem. – Fabian Dec 16 '12 at 19:54
  • @Fabian: I'm also meet so troubles when I start to apply my hint - looks like it was to good to be true. – m0nhawk Dec 16 '12 at 20:18
  • Thanks for this info. The problem does look like Fredholm equation described. My only problem is that $f(0)\neq 0$, does this mean that there is no $v\in L_2[0,1]$ that satisfy the above? – Hasanhasan Hasan Dec 16 '12 at 20:26

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I don't see what is the problem here. Obsviously $f(0)=0$ (for the equation to have a solution) and by deriving as you said, we get $$v(t)=\frac{f^{\prime}(t)}{t+1}$$ If we substitue this back in the orginal equation we have $$f(t)=\int_{0}^{t}f^{\prime}(x)dx=f(t)-f(0)=f(t)$$ which is true

Nameless
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It is easy to see that the constant $c$ in your post is in fact $f(0)$. Furthermore, a little thought shows that your equations cannot be solved unless $f(0)=0$ (just plug in $t=0$ in your equation and you find $0=f(0)$).

Fabian
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