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Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$, $g:\mathbb{R}^2 \rightarrow \mathbb{R}$ and $h:\mathbb{R}^2 \rightarrow \mathbb{R}$ be twice continuously differentiable functions such that, \begin{align*} f(x_1,x_2)+g(a_1t+x_1,a_2t+x_2)+h(b_1t+x_1,b_2t+x_2) = t^2 \end{align*} and $f(0,0)=g(0,0)=h(0,0)=0$ and $\bf b \neq 0$ and $\bf a \neq b$.

$\textbf{Edit:}$ Assume $a_1b_2=b_1a_2$.

$\textbf{Question:}$ I would be happy if someone is able to (1) find solutions for $f$, $g$, and / or $h$ or (2) find solutions for $g(a_1t,a_2t)$ or $h(b_1t,b_2t)$ or (3) provide some ideas for potential ways to find solutions.

$\textbf{What I've tried:}$ Take first-order partial derivatives with respect to $t$, $x_1$, and $x_2$, \begin{align*} a_1 g_1(a_1t+x_1,a_2t+x_2)+ a_2g_2(a_1t+x_1,a_2t+x_2) +b_1h_1(b_1t+x_1,b_2t+x_2) & \\ +b_2h_2(b_1t+x_1,b_2t+x_2) &= 2t && (1)\\ f_1(x_1,x_2)+g_1(a_1t+x_1,a_2t+x_2)+h_1(b_1t+x_1,b_2t+x_2) &= 0 && (2)\\ f_2(x_1,x_2)+g_2(a_1t+x_1,a_2t+x_2)+h_2(b_1t+x_1,b_2t+x_2) &= 0 && (3) \end{align*} where $g_j(a_1t+x_1,a_2t+x_2)=\left.\frac{\partial g(y_1,y_2)}{\partial y_j}\right|_{(y_1,y_2)=(a_1t+x_1,a_2t+x_2)}$. Substitute (2) and (3) into (1), \begin{align*} (a_1-b_1) g_1(a_1t+x_1,a_2t+x_2)+ (a_2-b_2)g_2(a_1t+x_1,a_2t+x_2) \\ -b_1f_1(x_1,x_2) -b_2f_2(x_1,x_2)&= 2t \end{align*} Let $x_1=x_2=0$. Then, \begin{align*} (a_1-b_1) g_1(a_1t,a_2t)+ (a_2-b_2)g_2(a_1t,a_2t) &=b_1f_1(0,0) +b_2f_2(0,0)+ 2t \\ \frac{dg(a_1t,a_2t)}{dt} -b_1 g_1(a_1t,a_2t)-b_2 g_2(a_1t,a_2t) &=b_1f_1(0,0) +b_2f_2(0,0)+ 2t && (4) \end{align*} Without the term $ -b_1 g_1(a_1t,a_2t)-b_2 g_2(a_1t,a_2t)$ then this would imply that $g(a_1t,a_2t)=c_1t^2+c_2t$. This is why I suspect that the only solutions may be that all the functions are quadratic (but I'm not sure about this). I'm not sure how to approach the problem except for taking derivatives (I also tried taking all second-order partial derivatives, which led to something similar to Equation (4)).

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I apologize in advance for any error(s) I may have made.

For notation, let, $x = (x_1, x_2)$, $a = (a_1,a_2)$, and $b = (b_1,b_2)$. I will use the subscript notation that you use to denote partials. I will also use italicized numbers in parenthesis to reference certain lines.

Starting with, $$f(x) + g(at+x) + h(bt+x) = t^2 \qquad \textit{(0)}$$ evaluate the identity at $t := 0$. Then, $$f(x) + g(x) + h(x) = 0 \qquad \ \\ f(x) = -g(x)-h(x) \qquad \textit{(1)}$$ substituting this into $\textit{(0)}$ gives, $$ g(at+x)-g(x) + h(bt+x)-h(x) = t^2 $$ Which allows us to obtain your results, independent of f, $$ a_1 g_1(at+x) + b_1 h_1(bt+x) + a_2 g_2(at+x) + b_2 h_2(bt+x) = 2t \\ g_1(at+x) - g_1(x) + h_1(bt+x) - h_1(x) = 0\\ g_2(at+x) - g_2(x) + h_2(bt+x) - h_2(x) = 0\\ a_1 g_1(x) + b_1 h_1(x) + a_2 g_2(x) + b_2 h_2(x) = 0 $$ Where the fourth of these is obtained from the first by replacing $t$ with $0$. Subtract the fourth equation from the first, rearrange the second and third equations and disregard the fourth to get $$ a_1 (g_1(at+x)-g_1(x)) + b_1 (h_1(bt+x)-h_1(x)) + a_2 (g_2(at+x)-g_2(x)) + b_2 (h_2(bt+x)-h_2(x)) = 2t \qquad \ \\ g_1(at+x) - g_1(x) = -( h_1(bt+x) - h_1(x)) \qquad \textit{(2)} \\ g_2(at+x) - g_2(x) = -( h_2(bt+x) - h_2(x)) \qquad \textit{(3)} $$ The second and third equations tell us that we can substituted the $g$ differences with negative $h$ differences into the first equation to obtain: $$(b_1 - a_1) (h_1(bt+x)-h_1(x)) + (b_2 - a_2) (h_2(bt+x)-h_2(x)) = 2t $$ Now this is a bit similar to what you obtained, but we can do even better. Divide both sides of this equation by $t$ and take the limit as $t \to 0$ to find that $$ (b_1 - a_1)( \nabla h_1(x) \cdot b ) + (b_2 - a_2)( \nabla h_2(x) \cdot b ) = 2 $$ Where $\nabla$ denotes the gradient. Your initial conditions tell you that at least one of $b_1 - a_1$,$b_2-a_2$ is non-zero. Furthermore since h is twice continuously differentiable its second order mixed partials are equal. so our final equation can be written as a linear combination of $h_{1,1},h_{1,2},h_{2,2}$: $$ b_1 (b_1 - a_1) h_{1,1} + (b_2 (b_1 - a_1) + b_1 (b_2 - a_2) ) h_{1,2} + b_2 (b_2 - a_2) h_{2,2} = 2 \\ \iff b_1 (b_1 - a_1) h_{1,1} + (2 b_2 b_1 - (b_2 a_1 + b_1 a_2)) h_{1,2} + b_2 (b_2 - a_2) h_{2,2} = 2 $$ To make this a little cleaner, let $c_1 := b_1 (b_1 - a_1)$, $ c_2 := 2 b_2 b_1 - (b_2 a_1 + b_1 a_2)$, and $c_3 := b_2 (b_2 - a_2)$. Then $$ c_1 h_{1,1} + c_2 h_{1,2} + c_3 h_{2,2} = 2 \qquad \textit{(4)} $$

At this point, what's left to do is solve $\textit{(4)}$, and use the (family of) solution(s) to find solutions to $\textit{(2)}$ and $\textit{(3)}$. Then use those to find $h$ and $g$ such that $\textit{(1)}$ is satisfied as well as your initial conditions. To go about solving $\textit{(4)}$ I'd suggest using the method of undetermined coefficients.

Now $\textit{(4)}$ is a second-order inhomogeneous linear PDE. The discriminant is, \begin{align*} discr=c_2^2-4c_1c_3&=[b_2(b_1-a_1)+b_1(b_2-a_2)]^2-4b_1b_2(b_1-a_1)(b_2-a_2)\\ &=b_2^2(b_1-a_1)^2-2b_1b_2(b_1-a_1)(b_2-a_2)+b_1^2(b_2-a_2)^2\\ &=[b_2(b_1-a_1)-b_1(b_2-a_2)]^2\\ &=(a_1b_2-b_1a_2)^2 \geq 0 \end{align*} so this is either a parabolic ($discr=0$) or hyperbolic ($discr>0$) PDE (http://en.wikipedia.org/wiki/Partial_differential_equations#Equations_of_second_order). The equation can now be transformed into the wave equation when $discr>0$ or $h_{11}=c$ when $discr=0$, in which case $h$ is quadratic.

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  • $\begingroup$ Thanks. I'll complete the final steps and add some parts to the end of your solution... and I'll mark it as the official solution tomorrow after giving some time if someone else wants to add anything. $\endgroup$ – user103828 Aug 5 '14 at 10:25
  • $\begingroup$ Let's say we complete your solution the way you described. Is it immediate that these are the only solutions? $\endgroup$ – user103828 Aug 5 '14 at 10:37
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    $\begingroup$ I'm a bit rusty and have not studied a lot of PDE's so I can't think of any uniqueness theorems off the top of my head. However, if we take the last equation above, (4), and we make the substitution h(x,y) = H(kx+ly,mx+ny) for some constants k,l,m, and n, then I believe we can choose those constants to cancel out the cross term, c_2, and perhaps even normalize the other two constants c_1 and c_3 to one. Ending up with H_11 + H_22 = 2. Where H_11 and H_22 are the non-mixed second order partials of H. This is Poisson's Equation. Which definitely has at least one uniqueness theorem. $\endgroup$ – DAS Aug 6 '14 at 0:01
  • $\begingroup$ Yes, the final equation can be transformed into the heat or wave equation (based on the discriminant and then transformation of variables) and the solutions follow... but I think the initial condition $h(0,0)=0$ is still not enough to show uniqueness. $\endgroup$ – user103828 Aug 6 '14 at 8:18
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    $\begingroup$ I think this is a safe guess. The initial conditions are somewhat scarce. If you can provide two distinct solutions then you can prove that the solution is not unique. For example, if you take the sum of a particular solution and a homogeneous solution then that is also a solution. So see if you can find several distinct homogeneous solutions to (4) which will allow you to obtain a solution to (4) and have h(0,0)=g(0,0)=f(0,0)=0. If you look for a homogeneous solution to (4), say some linear combination of exponential functions, then I think that might be a good start. $\endgroup$ – DAS Aug 6 '14 at 19:16

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