<p>I have two functions:</p>
<p><span class="math-container">$$F(x,y) = U_1'(c_1)(a_1-b_1)+U_2'(c_2) = 0$$</span>
<span class="math-container">$$G(x,y) = U_1'(c_1)+U_2'(c_2)(a_2-b_2) = 0$$</span></p>
<p>where <span class="math-container">$c_i=a_i(x-y)+yb_i, i=1,2.$</span> the utility functions <span class="math-container">$U_1,U_2$</span> are twice differentiable, increasing and concave functions, and <span class="math-container">$a-d$</span> are constants.</p>
<p>Computation shows that a solution exists where <span class="math-container">$0&lt;y&lt;1/2&lt;x, x+y&lt;1,F(x,y)=G(x,y)=0$</span>, i.e., the functions intersect at the <span class="math-container">$x,y$</span> state plane:</p>
<p><a href="https://i.sstatic.net/WlnOa.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/WlnOa.png" alt="image of the x,y state plane" /></a></p>
<p>How to prove analytically that the functions intersect in the area <span class="math-container">$0&lt;y&lt;1/2&lt;x, x+y&lt;1?$</span> If we know the following:
<span class="math-container">$$0.5\leq x\leq1; 0\leq y\leq0.5,$$</span></p>
<p><span class="math-container">$$  y\leq x $$</span>
<span class="math-container">$$F(1,0)&gt;0,F(1/2,1/2)&lt;0$$</span>
<span class="math-container">$$G(1,0)&gt;0,G(1/2,0)&gt;0,G(1/2,1/2)&lt;0,$$</span>
<span class="math-container">$$G(1/2,\dot{y})=0,F(1/2,\dot{y})&lt;0$$</span></p>
<p>All help greatly appreciated!</p>

CC BY-SA 4.0
2022-12-15T11:29:34.797Z
What do you denote by $\dot{y}$?
Giskard
CC BY-SA 4.0
2022-12-15T11:55:35.870Z
Hi! its the value of $y$ that satisfies $G(1/2,y)=0$.  By $G(1/2,0)>0,G(1/2,1/2)<0$, we know that a solution exists where $G(1/2,y)=0,  0<y<1/2.$  I denote that value by $\dot{y}$. We also know by other assumptions that at the point $G(1/2, \dot{y})=0$, F is negative: $F(1/2,\dot{y})<0. $ Hope this clarifies!
vonenzo
CC BY-SA 4.0
2022-12-15T11:59:07.717Z
Now I realised my axes legends are wrong in the figure. y-axis is for y and x-axis for x! Sorry for the confusion.
vonenzo
CC BY-SA 4.0
2022-12-15T12:52:45.040Z
Hi vonenzo! I dont' understand what does it mean $U'_i(x,y)$, that is the superscript, I know it could be the sign of a first derivative, but $U_i$ are  functions of two variables, so there should be partial derivatives, so: what are  the $U'_i$?
BakerStreet
CC BY-SA 4.0
2022-12-15T13:04:54.590Z
Hi! It is the total derivative of $U$, so indeed it is the first derivative. Basically, consumption $c_i$ depends on two variables $x,y$. The utility functions $U_i(c_i)$ are here differentiated with respect to $c_i$, and now written explicitly for variables  $x,y$.  I hope this clarifies.
vonenzo
CC BY-SA 4.0
2022-12-15T13:24:18.237Z
So it is $U_i(c_i(x,y))$ and $U'_i$ is the derivative with respect to $c_i$? I think you should write it explicitely, otherwise the two equations make no sense.
BakerStreet
CC BY-SA 4.0
2022-12-15T19:43:22.980Z
Do you mean, instead that $U'_i$ is the total differential of $U_i(x,y)$? This also should be clarified, as  the notation $U'$ for the differential is not standard, and, morover,  a differential is not a first derivative, as you seem to suggest. You should clarify better the formulas, I think, in order to  allow other contributors to answer.
BakerStreet
CC BY-SA 4.0
2022-12-15T19:54:52.753Z
I use *total differential* , *differential*,  *total derivative* as synonimous, *differential* is for me the most usual term.
BakerStreet
CC BY-SA 4.0
2022-12-16T06:12:06.710Z
Thanks for the comments, I clarified the formulas now. Hopefully it makes more sense.
vonenzo
<mathematical-economics><proof>
How to prove that these level curves intersect
vonenzo
# How to prove that these level curves intersect

(Asked by: username_0 on 2022-12-15)

I have two functions:

$$F(x,y) = U\_1'(c\_1)(a\_1-b\_1)+U\_2'(c\_2) = 0$$
$$G(x,y) = U\_1'(c\_1)+U\_2'(c\_2)(a\_2-b\_2) = 0$$

where $c\_i=a\_i(x-y)+yb\_i, i=1,2.$ the utility functions $U\_1,U\_2$ are twice differentiable, increasing and concave functions, and $a-d$ are constants.

Computation shows that a solution exists where $0, i.e., the functions intersect at the $x,y$ state plane:

[![image of the x,y state plane](https://i.sstatic.net/WlnOa.png)](https://i.sstatic.net/WlnOa.png)

How to prove analytically that the functions intersect in the area $0 If we know the following:
$$0.5\leq x\leq1; 0\leq y\leq0.5,$$

$$ y\leq x $$
$$F(1,0)>0,F(1/2,1/2)<0$$
$$G(1,0)>0,G(1/2,0)>0,G(1/2,1/2)<0,$$
$$G(1/2,\dot{y})=0,F(1/2,\dot{y})<0$$

All help greatly appreciated!

**Tags:** mathematical-economics, proof
> **Comments:**
> **username_1**: What do you denote by $\dot{y}$?  
> **username_0**: Hi! its the value of $y$ that satisfies $G(1/2,y)=0$.  By $G(1/2,0)>0,G(1/2,1/2)<0$, we know that a solution exists where $G(1/2,y)=0,  0<y<1/2.$  I denote that value by $\dot{y}$. We also know by other assumptions that at the point $G(1/2, \dot{y})=0$, F is negative: $F(1/2,\dot{y})<0. $ Hope this clarifies!  
> **username_0**: Now I realised my axes legends are wrong in the figure. y-axis is for y and x-axis for x! Sorry for the confusion.  
> **username_2**: Hi vonenzo! I dont' understand what does it mean $U'_i(x,y)$, that is the superscript, I know it could be the sign of a first derivative, but $U_i$ are  functions of two variables, so there should be partial derivatives, so: what are  the $U'_i$?  
> **username_0**: Hi! It is the total derivative of $U$, so indeed it is the first derivative. Basically, consumption $c_i$ depends on two variables $x,y$. The utility functions $U_i(c_i)$ are here differentiated with respect to $c_i$, and now written explicitly for variables  $x,y$.  I hope this clarifies.  
> **username_2**: So it is $U_i(c_i(x,y))$ and $U'_i$ is the derivative with respect to $c_i$? I think you should write it explicitely, otherwise the two equations make no sense.  
> **username_2**: Do you mean, instead that $U'_i$ is the total differential of $U_i(x,y)$? This also should be clarified, as  the notation $U'$ for the differential is not standard, and, morover,  a differential is not a first derivative, as you seem to suggest. You should clarify better the formulas, I think, in order to  allow other contributors to answer.  
> **username_2**: I use *total differential* , *differential*,  *total derivative* as synonimous, *differential* is for me the most usual term.  
> **username_0**: Thanks for the comments, I clarified the formulas now. Hopefully it makes more sense.  

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## 0 answers

