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Show that there is no solution to:

$$v(x)=1+3 \int_0^1 xy v(y) dy$$ When the solution $v(x)$ must be of the form $1+cx$ for some constant $c$.

I tried solving this I am getting $x=0$. What am I doing wrong?

kiwifruit
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3 Answers3

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I assume $y$ is a dummy variable for the integration and independent of $x$. Pull the $x$ out of the integral, and write $v(y)=1+cy$; then this says

$$1+cx=1+3x\int_0^1 y+cy^2dy$$

Integrating gives $\frac {y^2}2+c\frac{y^3}3$, so after evaluating this becomes $$1+cx = 1+3x\left(\frac 13 c+\frac 12\right)=1+cx+\frac 32x$$ So what we've said here is that $v(x)=v(x)+\frac 32x$ as functions (in other words, this equality is true for all values of $x$.) But that's just silly: it says, for instance, that $v(2)=v(2)+3$, so $3=0$.

  • But why don't we get 3/2 x=0 rather than just 3/2=0? What happened to the x? – kiwifruit Mar 22 '14 at 22:08
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    @kiwifruit You can substitute in $x=1$ to get that. $x$ is a variable, so that equation should be true for all values of $x$. –  Mar 22 '14 at 22:08
  • Oh, so the goal is to find a c such that this equation is true for all values of x, rather than taking a specific x? – kiwifruit Mar 22 '14 at 22:10
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    @kiwifruit Right - that $v(x)$ is a function is the thing to remember. We just showed that if $v$ was of your form, then $v(x)=v(x)+\frac 32x$, which is nonsense when we're talking about functions. (In particular, $v(1)=v(1)+\frac 32$ is just silly.) –  Mar 22 '14 at 22:13
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Let $$\frac c3=\int_0^1 yv(y)dy$$ then $$v(x)=1+cx$$ hence we find $$\frac c3=\int_0^1y(1+cy)dy=\frac12+\frac c3$$ which impossible.

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See that

LHS: $1+cx$

RHS: $$1+3\int_{0}^{1}xy(1+cy)dy=1+3x\int_{0}^{1}y+cy^2dy=1+x[\frac{1}{2}y^2+\frac{c}{3}y^3]^{1}_{0}=1+3x(\frac{1}{2}+\frac{c}{3})$$So if it should be equal we have that $$ c=3(\frac{1}{2}+\frac{c}{3})\Rightarrow 0=\frac{3}{2} $$ which is clearly impossible

Kaladin
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