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In Henry C. Wente's paper https://msp.org/pjm/1980/88-2/pjm-v88-n2-p08-s.pdf, page $391$ to $392$, the following prescribed mean curvature equation is considered: $$\mathrm{div}(Tu)=nH(x,u),\qquad \text{where}~~Tu:=\frac{\nabla u}{\left(1+|\nabla u|^2\right)^{1/2}}\tag{1}$$ which can be written in the form using simple calculation: $$\frac{1}{W}\sum_{i=1}^{n}\partial_{ii}u-\frac{1}{W^3}\sum_{1\leq i,j\leq n}\partial_iu\partial_ju\partial_{ij}u=nH(x,u),\qquad\text{where}~~W^2=1+|\nabla u|^2\tag{2}$$ or equivalently $$\sum_{1\leq i,j\leq n}a_{ij}(x,u,\nabla u)\partial_{ij}u=nH(x,u),\qquad \text{where}~~a_{ij}=\frac{\delta_{ij}}{W}-\frac{1}{W^3}\partial_iu\partial_ju\tag{3}$$ Then he says: if $u,v$ are two solutions to $(1),$ then it's well known that $w:=u-v$ is a solution to a homogeneous linear elliptic P.D.E of the form $\widetilde{a_{ij}}(x)\partial_{ij}w+\widetilde{b_i}(x)\partial_iw+\widetilde{c}(x)w=0,$ where $\widetilde{a_{ij}}(x)$ are given by $$\widetilde{a_{ij}}(x)=\int_0^1a_{ij}(x,u+t(v-u),\nabla u+t(\nabla v-\nabla u))dt\tag{4}$$ with $a_{ij}$ given by $(3).$ But I don't know how does $(4)$ come about exactly, can someone help me?

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Yes, it follow from the fundamental theorem of calculus and it is 'well-known' amongst members of the PDE community.

Let $\Omega \subset \mathbb R^n$ be open, and $a^{ij},b \in C^1( \Omega \times \mathbb R\times \mathbb R^n ) $. Consider the quasilinear equation $$ a^{ij}(x,u,\nabla u ) \partial_{ij}u = b(x,u,\nabla u ) \qquad \text{in } \Omega. \tag{$\ast$}$$ and let $u,v\in C^2(\Omega)$ be solutions to ($\ast$). Here are throughout my answer I will always use the summation convention. Now the idea is to consider $$ f(t)= a^{ij}(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v ) \partial_{ij}\big (tu+(1-t)v \big ) - b(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v ) .$$ Since $u$ and $v$ satisfy ($\ast)$, $$f(1)=a^{ij}(x,u,\nabla u ) \partial_{ij}u - b(x,u,\nabla u )=0 \text{ and } f(0)=a^{ij}(x,v,\nabla v ) \partial_{ij}v - b(x,v,\nabla v )=0 ,$$ so $$ 0=f(1)-f(0)=\int_0^1 f'(t) \, dt$$ by the fundamental theorem of calculus. Since $$ \frac{d}{dt} \partial_{ij} (tu+(1-t)v)=\partial_{ij}(u-v)=\partial_{ij}w, $$ and, writing $a^{ij}=a^{ij}(x,z,p)$,\begin{align*} &\frac{d}{dt} a^{ij}(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v ) \\ &=\big ( \partial_z a^{ij}(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v ) \big ) (u-v) + \nabla_p a^{ij}(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v )\cdot \nabla (u-v) \\ &=\partial_z a^{ij}(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v ) w + \nabla_p a^{ij}(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v )\cdot \nabla w \end{align*} and similarly for $b$, we obtain \begin{align*} &f'(t)\\ &= a^{ij}(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v ) \partial_{ij}w \\ &\qquad + \big (t\partial_{ij}u+(1-t)\partial_{ij}v \big ) \bigg ( \partial_z a^{ij}(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v ) w + \nabla_p a^{ij}(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v )\cdot \nabla w\bigg ) \\ &- \bigg ( \partial_z b(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v ) w + \nabla_p b(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v )\cdot \nabla w\bigg ). \end{align*} Calling \begin{align*} &\tilde a^{ij}(x)\\ &=\int_0^1 a^{ij}(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v ) \, dt \\ &\tilde b^i(x)\\ &= \int_0^1 \big (\big (t\partial_{ij}u+(1-t)\partial_{ij}v \big )\nabla_{p_i} a^{ij}(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v )-\nabla_{p_i} b(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v )\big ) \,dt \\ &\tilde c(x) \\ &=\int_0^1 \big ( \big (t\partial_{ij}u+(1-t)\partial_{ij}v \big )\partial_z a^{ij}(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v ) - \partial_z b(x,tu+(1-t)v,t\nabla u +(1-t)\nabla v )\big ) \, dt \end{align*} we obtain $$ 0 =\int_0^1 f'(t) \, dt = \tilde a^{ij}(x)\partial_{ij}w(x) +{\tilde b}^i(x)\partial_iw(x)+\tilde c(x) w(x).$$ Notice that $\tilde a^{ij}$ is precisely what is given in the document you cite.

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  • $\begingroup$ I get it! Thank you for writing in such a great detail! $\endgroup$
    – Tiffany
    Commented Apr 9 at 14:18

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