On being able to write an arbitrary $ C^{3} $-solution of a particular third-order PDE in a special way

Question

Suppose that $ u: \mathbb{R}^{3} \to \mathbb{R} $ is a $ C^{3} $-function that satisfies the PDE $$ \forall (x,y,z) \in \mathbb{R}^{3}: \qquad (\partial_{2} \partial_{2} \partial_{1} u)(x,y,z) = 2 \sin(x). $$ Then $$ \forall (x,y,z) \in \mathbb{R}^{3}: \qquad (\partial_{2} \partial_{1} u)(x,y,z) = 2 \sin(x) y + f(x,z) $$ for some function $ f: \mathbb{R}^{2} \to \mathbb{R} $, which implies that $$ \forall (x,y,z) \in \mathbb{R}^{3}: \qquad (\partial_{1} u)(x,y,z) = \sin(x) y^{2} + f(x,z) y + g(x,z) $$ for some function $ g: \mathbb{R}^{2} \to \mathbb{R} $.

As $ u \in {C^{3}}(\mathbb{R}^{3}) $ and $$ \forall (x,z) \in \mathbb{R}^{2}: \qquad (\partial_{1} u)(x,0,z) = g(x,z), $$ we find that $ g \in {C^{2}}(\mathbb{R}^{2}) $. Then because $$ \forall (x,z) \in \mathbb{R}^{2}: \qquad (\partial_{1} u)(x,1,z) = \sin(x) + f(x,z) + g(x,z), $$ we find that $ f \in {C^{2}}(\mathbb{R}^{2}) $ as well.

Let $ F $ and $ G $ be, respectively, anti-derivatives of $ f $ and $ g $ with respect to their first arguments. Such anti-derivatives exist by the Fundamental Theorem of Calculus; for example, we could define \begin{align} \forall (x,z) \in \mathbb{R}^{2}: \qquad F(x,z) & \stackrel{\text{df}}{=} \int_{0}^{x} f(t,z) ~ \mathrm{d}{t}, \qquad (\clubsuit) \\ G(x,z) & \stackrel{\text{df}}{=} \int_{0}^{x} g(t,z) ~ \mathrm{d}{t}. \qquad (\spadesuit) \end{align} Then $$ \forall (x,y,z) \in \mathbb{R}^{3}: \qquad u(x,y,z) = - \cos(x) y^{2} + F(x,z) y + G(x,z) + h(y,z) \qquad (\star) $$ for some function $ h: \mathbb{R}^{2} \to \mathbb{R} $.

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Question. Can we find $ C^{3} $-functions $ F $, $ G $ and $ h $ so that $ u $ may be written as in $ (\star) $? We do not necessarily require $ F $ and $ G $ to be defined according to $ (\clubsuit) $ and $ (\spadesuit) $ respectively.

Why this is a non-trivial question can be explained as follows. Suppose that we defined $ F $ according to $ (\clubsuit) $ and then tried to take the partial derivative of $ F $ with respect to its second argument, twice. Using the fact that $ f \in {C^{2}}(\mathbb{R}^{2}) $ and differentiating under the integral sign, we would obtain $$ \forall (x,z) \in \mathbb{R}^{2}: \qquad (\partial_{2} \partial_{2} F)(x,z) = \int_{0}^{x} (\partial_{2} \partial_{2} f)(t,z) ~ \mathrm{d}{t}. $$ However, $ f $ is not regular enough to justify differentiation under the integral sign one more time to obtain a formula for $ \partial_{2} \partial_{2} \partial_{2} F $. Herein lies the difficulty.

Of course, if $ F $, $ G $ and $ h $ were arbitrary $ C^{3} $-functions, then the function $ u $ defined by $ (\star) $ would be a $ C^{3} $-solution of the given PDE. The question is thus asking if any $ C^{3} $-solution may be so expressed.

Thank you for your help!

Answer 1

I have managed to find a direct answer to my question above.

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Early on in my post, it was shown (rather easily) that there exist functions $ f,g: \mathbb{R}^{2} \to \mathbb{R} $ such that $$ \forall (x,y,z) \in \mathbb{R}^{3}: \qquad (\partial_{1} u)(x,y,z) = \sin(x) y^{2} + f(x,z) y + g(x,z). $$ It follows that \begin{align} \forall (x,z) \in \mathbb{R}^{2}: \qquad g(x,z) & = (\partial_{1} u)(x,0,z), \\ f(x,z) & = (\partial_{1} u)(x,1,z) - g(x,z) - \sin(x) \\ & = (\partial_{1} u)(x,1,z) - (\partial_{1} u)(x,0,z) - \sin(x). \end{align} Define anti-derivatives $ F $ and $ G $ of, respectively, $ f $ and $ g $ with respect to their first arguments by \begin{align} \forall (x,z) \in \mathbb{R}^{2}: \qquad G(x,z) & \stackrel{\text{df}}{=} u(x,0,z), \\ F(x,z) & \stackrel{\text{df}}{=} u(x,1,z) - u(x,0,z) + \cos(x). \end{align} Obviously, $ F,G \in {C^{3}}(\mathbb{R}^{2}) $. As mentioned above, there exists a function $ h: \mathbb{R}^{2} \to \mathbb{R} $ such that $$ \forall (x,y,z) \in \mathbb{R}^{3}: \qquad u(x,y,z) = - \cos(x) y^{2} + F(x,z) y + G(x,z) + h(y,z). $$ Obviously, $ h \in {C^{3}}(\mathbb{R}^{2}) $ as well, and the question therefore has an affirmative answer.

Answer 2

First of all, the right hand side is irrelevant; a general solution is $-y^2\cos x$ plus a general solution of the homogeneous PDE. So I'll consider the homogeneous case.

Let $h(y,z)=u(0,y,z)$, this is clearly $C^3$. Let $G(x,z) = u(x,0,z)-u(0,0,z)$, also $C^3$. Finally, let $$F(x,y,z) = \frac{u(x,y,z) - G(x,z) - h(y,z)}{y} \tag1$$ The numerator vanishes when $y=0$, so the quotient is defined there (as a partial derivative in $y$) and therefore is $C^2$ smooth. So far we have written $$ u(x,y,z) = yF(x,y,z) +G(x,z)+h(y,z) \tag2 $$ with $G,H$ in $C^3$ and $F\in C^2$. Note that $yF\in C^3$, which implies $F$ is $C^3$ except possibly where $y=0$.

Apply the PDE $\partial_2^2\partial_1u=0$ to (2) and deduce that $y\partial_1F$ is linear in $y$, hence $\partial_2\partial_1 F\equiv 0$. The latter implies $$ F(x,y,z) = F(x,0,z) + F(0, y, z) - F(0, 0, z) $$ because a $C^2$ function $u$ of two arguments with $\partial_2\partial_1 u\equiv 0$ is the sum of functions of one argument.

But $F(0, y, z)=0$, by plugging $x=0$ in (1). Thus, $F$ is independent of $y$. Since we already noted $F$ is $C^3$ except possibly at $y=0$, the final conclusion is that $F\in C^3$.