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If $\phi$ is the solution of the integral equation $$\phi(x)=1-2x-4x^2+\int_0^x[3+6(x-t)-4(x-t)^2]\phi(t)dt$$

Then the value of $\phi(\log 2)$ is

(a). 2

(b). 4

(c). 6

(d). 8

I tried this and I get the solution is $23$ but I am not sure about it can anyone please solve this.Thank you

user10354138
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Gopinath
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  • This should be solvable by the Laplace transform, assuming the domain of $\phi$ is $[0,\infty)$. Specifically if $f(x)=1-2x-4x^2$ and $g(x)=3+6x-4x^2$ then $\Phi(s)=F(s) + G(s) \Phi(s)$ so $\Phi(s)=\frac{F(s)}{1-G(s)}$, which will be a rational function since $f,g$ are polynomials, making it easy to handle provided you can find the roots of $1-G(s)$ (which on a first inspection might be a little bit hard, since that reduces to a cubic equation). – Ian Jun 09 '19 at 01:10
  • Actually the cubic equation in question is not hard to solve because the coefficients sum to zero, so $1$ is a root and then the others can be found easily. – Ian Jun 09 '19 at 01:26

2 Answers2

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Differentiating three times gives that $$\phi’’’(x)=3\phi’’(x)+6\phi’(x)-8\phi(x).$$ Solve the characteristic equation and we find it has three distinct roots, so everything will be easy.

The result I get is that $\phi(x)=e^x$ so $\phi(\log(2))=2$.

Thanks to @lan for pointing out my stupid errors.

Feng
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  • It seems to me that you have solved $\phi(x)=1-2x-4x^2+3\int_0^x \phi(t) dt$, which is a different problem. – Ian Jun 09 '19 at 01:18
  • This is by the fundamental theorem of calculus, i.e. $\frac{d}{dx} \int_0^x f(t) dt = f(x)$. So you simply let $t=x$ which cancels to be 0 in your case. – Soby Jun 09 '19 at 01:20
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    @thedilated There is a Leibniz rule term when you differentiate $\int_0^x (x-t) \phi(t) dt$. You do not simply get zero identically, as you can easily see by looking at $\phi(t) \equiv 1$ in which case you have $x^2/2$. To be sure, this problem, being a Volterra type integral equation, can be handled by differentiation, but the Leibniz rule needs to be applied correctly, which this answer did not do. – Ian Jun 09 '19 at 01:21
  • @lan you are right. I will correct my answer. – Feng Jun 09 '19 at 01:24
  • Thanks for fixing it, +1 now. – Ian Jun 09 '19 at 01:40
  • @FengShao how to solve the characteristic eqn – Gopinath Jun 09 '19 at 03:07
  • @Gopinath The characteristic eqn is $x^3-3x^2-6x+8=(x-1)(x-4)(x+2)=0$. – Feng Jun 09 '19 at 03:12
  • @FengShao So how did you get $\phi(x)=e^x$...so can you please explain – Gopinath Jun 09 '19 at 03:14
  • @Gopinath Since the characteristic eqn has three distinct roots, we can write $\phi(x)=c_1e^x+c_2e^{4x}+c_3e^{-2x}$. From $\phi(0)=\phi’(0)=\phi’’(0)=1$ we can get $c_1=1,c_2=c_3=0$. What I used here is a method to solve ODEs. It’s good for to look up some materials to get familiar with it. Regards. – Feng Jun 09 '19 at 03:21
  • @FengShao Thank you so much now I got it... – Gopinath Jun 09 '19 at 03:23
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Just for completeness, you can also do this with the Laplace transform, which uses the property that the Laplace transform of a "causal convolution" $(f*g)(x):=\int_0^x f(x-t) g(t) dt$ is $F(s) G(s)$. Consequently taking the Laplace transform of both sides gives

$$\Phi(s)=F(s) + G(s) \Phi(s)$$

where $F(s)$ is the Laplace transform of $1-2x-4x^2$, i.e. $\frac{1}{s}-\frac{2}{s^2}-\frac{8}{s^3}$, and $G(s)$ is the Laplace transform of $3+6x-4x^2$, i.e. $\frac{3}{s}+\frac{6}{s^2}-\frac{8}{s^3}$. Thus

$$\Phi(s)=\frac{F(s)}{1-G(s)}=\frac{s^2-2s-8}{s^3-3s^2-6s+8}.$$

From here you can carry out a partial fraction decomposition and take an inverse Laplace transform. Actually, you don't even have to carry out a partial fraction decomposition in this particular problem, you can just factor the numerator and denominator and observe the cancellation.

Ian
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