Solutions to the PDE :$2V\frac{\partial I}{\partial V}+2W\frac{\partial I}{\partial W}=I$

Question

While working on engineering problem, I came across this PDE: Let $c_1,c_2$ be two real numbers. Find a continuous function $I:\mathbb{R}_{\geq 0}\times\mathbb{R}_{\geq 0}\rightarrow \mathbb{R}$ such that

1) For every $V,W$, we have: $$I(V,0)=c_1\sqrt{V},I(0,W)=c_2\sqrt{W}$$

2) $I|\mathbb{R}_{>0}\times\mathbb{R}_{>0}$ is a differentiable function that satisfies:

$$2V\frac{\partial I}{\partial V}+2W\frac{\partial I}{\partial W}=I$$

After some educated guesses, I know that for every $\lambda\in \mathbb{R}$, the follwoing is a solution:

$$I(V,W)=\lambda(c_1\sqrt{V}+c_2\sqrt{W})+(1-\lambda)(\sqrt{c_1^2V+c_2^2W})$$

Question: Are there any other solutions ?

It would also be great if someone can get all solutions.

Thank you

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I will award a bounty worth 500 points for finding the set of all solutions.

Answer 1

Introduce polar coordinates in the $(v,w)$-plane: $$v=r\cos\phi,\quad w=r\sin\phi\qquad(0\leq r<\infty, \ 0\leq\phi\leq{\pi\over2})\ .$$ Then $$2r\ I_r=2r\>(I_vv_r+I_w w_r)=2r(I_v\cos\phi+I_w\sin\phi)=2v\>I_v+2w\>I_w=I\ .$$ In short: In the new variables $r$ and $\phi$ the original PDE appears as $$2r\>I_r=I\ .$$ The solutions to this are the functions $$I(r,\phi)=c(\phi)\sqrt{r}\ ,\tag{1}$$ where the function $$\phi\mapsto c(\phi)\qquad(0\leq\phi\leq{\pi\over2})$$ is arbitrary. In order to satisfy the given boundary conditions you have to impose the conditions $c(0)=c_1$, $\>c\bigl({\pi\over2}\bigr)=c_2$. In terms of $v$ and $w$ the functions $(1)$ appear as $$I(v,w)=c\bigl({\rm arg}(v,w)\bigr)\>(v^2+w^2)^{1/4}\ .$$

It follows that there are many more solutions than you thought.

Answer 2

Given

$$ 2 V \frac{\partial I}{\partial V} + 2 W \frac{\partial I}{\partial W} = I.\tag{1} $$

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The PDE (1) has the property that if $I$ is a solution and $a$ is a function of $V/W$, then $a(V/W) I$ is also a solution, as

$$ \begin{eqnarray} \left[2 V \frac{\partial}{\partial V} + 2 W \frac{\partial}{\partial W} \right] a(V/W) I &=& a(V/W) \left[2 V \frac{\partial}{\partial V} + 2 W \frac{\partial}{\partial W} \right] I\\ && + I \left[2 V \frac{\partial}{\partial V} + 2 W \frac{\partial}{\partial W} \right] a(V/W)\\ && a(V/W) I + I \left[ \frac{2V}{W} - \frac{2VW}{W^2} \right] a'(V/W)\\ &=& a(V/W) I.\tag{2} \end{eqnarray} $$

So factors can be a function of $V/W$.

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It is clear that PDE (1) is linear, but it has an even stronger property. Let $I_1$ and $I_2$ be solutions of the PDE (1), then

$$ \Big( a_1 I_1^{\zeta} + a_2 I_2^{\zeta} \Big)^{1/\zeta} $$

is also a solution. We find

$$ \begin{eqnarray} \left[2 V \frac{\partial}{\partial V} + 2 W \frac{\partial}{\partial W} \right] \Big( a_1 I_1^{\zeta} + a_2 I_2^{\zeta} \Big)^{1/\zeta} &=& \Big( a_1 I_1^{2\zeta} + a_2 I_2^{2\zeta} \Big)^{1/\zeta-1} \times\\ && \hspace{1em} \left\{ a_1 \left[2 V \frac{\partial I_1}{\partial V} + 2 W \frac{\partial I_1}{\partial W} \right] I_1^{\zeta-1} \right.\\ && \hspace{2em} \left. + a_2 \left[2 V \frac{\partial I_2}{\partial V} + 2 W \frac{\partial I_2}{\partial W} \right] I_2^{\zeta-1} \right\}\\ &=& \Big( a_1 I_1^{\zeta} + a_2 I_2^{\zeta} \Big)^{1/\zeta-1} \times \Big( a_1 I_1^{\zeta} + a_2 I_2^{\zeta} \Big)\\ &=& \Big( a_1 I_1^{\zeta} + a_2 I_2^{\zeta} \Big).\tag{3} \end{eqnarray} $$

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To solve the PDE (1), we consider the co-ordinates

$$ \begin{eqnarray} p &=& \ln\Big(\sqrt[4]{V}\Big) + \ln\Big(\sqrt[4]{W}\Big),\\ q &=& \ln\Big(\sqrt[4]{V}\Big) - \ln\Big(\sqrt[4]{W}\Big), \end{eqnarray} $$

whence

$$ \begin{eqnarray} \frac{\partial}{\partial V} &=& \frac{\partial p}{\partial V} \frac{\partial}{\partial p} + \frac{\partial q}{\partial V} \frac{\partial}{\partial q}\\ &=& \frac{1}{4V} \frac{\partial}{\partial p} + \frac{1}{4V} \frac{\partial}{\partial q},\\ \frac{\partial}{\partial W} &=& \frac{\partial p}{\partial W} \frac{\partial}{\partial p} + \frac{\partial q}{\partial W} \frac{\partial}{\partial q}\\ &=& \frac{1}{4W} \frac{\partial}{\partial p} - \frac{1}{4W} \frac{\partial}{\partial q}, \end{eqnarray} $$

and therefore

$$ 2 V \frac{\partial I}{\partial V} + 2 W \frac{\partial I}{\partial W} =\frac{\partial I}{\partial p}, $$

so we obtain the equation

$$ \frac{\partial I}{\partial p} = I, $$

and the solution is given by

$$ I = a(q) \exp\Big( p + b q \Big), $$

where $a(q)$ is a function of $q$, i.e. constant with respect to $p$.

Filling in the co-ordinates $p$ and $q$, we get

$$ \begin{eqnarray} I &=& a\left[ \ln\Big(\sqrt[4]{V}\Big) - \ln\Big(\sqrt[4]{W}\Big) \right] \times\\ && \hspace{1em} \exp\left[ \ln\Big(\sqrt[4]{V}\Big) + \ln\Big(\sqrt[4]{W}\Big) + 4 \xi \ln\Big(\sqrt[4]{V}\Big) - 4 \xi \ln\Big(\sqrt[4]{W}\Big) \right]\\ &=& a\left[ \ln\Big(\sqrt[4]{V/W}\Big) \right] \sqrt[4]{V^{1+\xi} W^{1-\xi}}.\tag{4} \end{eqnarray} $$

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However, due to (3) there are many solutions to build from (4). Let us write

$$ I_k(V,W) = \sqrt[4]{V^{1+\xi_k} W^{1-\xi_k}}, $$

then the general solution is given by

$$ J_k(V,W) = \left( \sum a_{k_p}(V/W) I_{k_p}(U,V)^\zeta + \sum b_{k_p}(V/W)^\zeta J_{k_q}(U,V)^\zeta \right)^{1/\zeta}, $$

where we have used (2). This solution is a recursion.

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The boundary condition

$$ \begin{eqnarray} I(V,0) &=& c_1 \sqrt{V},\\ I(0,W) &=& c_2 \sqrt{W}, \end{eqnarray} $$

is a restriction to the factors $a_k$ and $b_k$.

So we do find solutions like

$$ \begin{eqnarray} I(V,W) &=& c_1 \sqrt{V} + c_2 \sqrt{W},\\ I(V,W) &=& \lambda c_1 \sqrt{V} + \lambda c_2 \sqrt{W}\\ && + (1-\lambda) \sqrt{ c_1^2 V + c_2^2 W }, \end{eqnarray} $$

but also

$$ \begin{eqnarray} I(V,W) &=& c_1 \sqrt{V} + c_2 \sqrt{W} + c_3 \sqrt[4]{VW},\\ I(V,W) &=& \lambda c_1 \sqrt{V} + \lambda c_2 \sqrt{W} + c_3 c\sqrt[8]{V^3W^5}\\ && + (1-\lambda) \sqrt{ c_1^2 V + c_2^2 W + c_3 \sqrt[4]{VW^3} } + c_4 \sqrt[8]{ V W ^3 + V^3 W }. \end{eqnarray} $$

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The general solution is given by

$$ J_k(V,W) = \left( \sum a_{k_p}(V/W) I_{k_p}(U,V)^\zeta + \sum b_{k_p}(V/W)^\zeta J_{k_q}(U,V)^\zeta \right)^{1/\zeta}, $$

where

$$ I_k(V,W) = \sqrt[4]{V^{1+\xi_k} W^{1-\xi_k}}, $$

and the factors follow from the boundary condition.