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Let $m \geq 2$ and let $m'$ be its conjugate. Let $w_j$ for $j=1, ..., k$ be a basis of $H_1 \cap L^{m'}$. The task is to show that there is a $u(t) \in \text{span}(w_1, ..., w_k)=:A$ such that $$\langle\varphi(u(t)), w_j \rangle_{L^m, L^{m'}} = \langle \varphi(u_0), w_j \rangle_{L^m, L^{m'}} + \int_0^t \langle F(s, u(s)), w_j \rangle\mathrm{d}s$$ for all $j=1, ..., k$.

Here, $\varphi:L^{m'} \to L^m$ is $\varphi(u) = u^{m'-1}$ and $$\langle F(s,u), w_j \rangle = -\langle \nabla u, \nabla w_j \rangle.$$ In order to show existence, the author shows that $\langle F(s,u), w_j \rangle$ is locally Lipschitz continuous on the finite-dimensional subspace: $$|\langle F(s,u), w_j \rangle -\langle F(s,v), w_j \rangle| \leq C|u-v|_{H^1}$$ with $C$ depending on $u_0$.

And author states that "$\varphi$ has a locally Lipschitz continuous inverse" but the author only seems to prove $$|u-v|_A \leq C|\varphi(u) - \varphi(v)|_{L^m}.$$

Then it is stated

The integral equation has a solution due to Banach's fixed point theorem in analogy to the existence theorem of Picard-Lindelof.

From an answer to a thread I posted to MSE, it is suggested that we consider $$\langle y_{n+1}(t), w_j \rangle = \langle \varphi(u_0), w_j \rangle + \int_0^t \langle F(s, \varphi^{-1}(y_n)(s)), w_j \rangle\mathrm{d}s$$ But then $y_{n+1}$ need not be finite-dimensional and so I don't see how to use the Lipschitz condition on $\varphi^{-1}$. Even if we assume $y_{n}$ are finite-dimensional, why should $\varphi^{-1}(y_n) =: u_n$ be finite-dimensional as required? I don't know what the right way to do this is. Any help appreciated.

Here are links to the original text: http://imgur.com/1r2Ijil http://imgur.com/Frw3IsK http://imgur.com/drlwRgv

(I chose $p=2$ and the right hand side $f\equiv 0$).

EDIT: Source is this paper

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This is perhaps irrelevant, but isn't the inverse explicitly given by $\varphi^{-1}(z)=z^{\frac{1}{m'-1}}$, which is indedd locally Lipschitz since $m\geq 2\Rightarrow\frac{1}{m'-1}\geq 1$? Also, regarding your fixed point in $y$: you obviously have to project $y^0=\varphi(u_0)$ into your finite dimensional subspace $A$, and by construction your "laplacian" is also projected. So $y^{n+1}$ is really "finite-dimensional" in the end and by construction (the fixed point can be applied for example in $\mathcal{C}(0,T;A)$) –  leo monsaingeon May 27 at 13:19
    
Thanks for comment @leomonsaingeon. Well, $\varphi(u_0) \in L^m$, where $A$ is finite-dimensional subspace of $L^{m'}$. It's still sensible to project $\varphi(u_0)$? I know $L^m \subset L^{m'}$ but.. –  riem May 27 at 15:51
    
Yes, but then what are your brackets in your first equation $\langle\varphi(u(t)), w_j \rangle = \langle \varphi(u_0), w_j \rangle +\ldots$, are they $L^2$ scalar product, $H^1$ scalar product, or $L^m-L^{m'}$ duality pairing? Unless I'm mistaken you're trying to solve $\partial_t\varphi(u)=\Delta u$, so the typical parabolic regularity you expect is $\partial_t\varphi(u)\in L^2(0,T;H^{-1})$, $\varphi(u)\in \mathcal{C}(0,T;L^m)$ and $u\in L^2(0,T;H^1)$. I suggest you take a look at the classical paper [Alt-Luckhaus, quasilinear elliptic-parabolic differential equation, '83] –  leo monsaingeon May 27 at 16:20
    
@leomonsaingeon It is $L^m$ $L^{m'}$ duality pairing. The author explicitly states that we don't get time derivative at all. The PDE is only valid in the integral sense. –  riem May 27 at 17:05

2 Answers 2

up vote 3 down vote accepted

OK, I think I figured it out.

First let me define a few things: for any given basis $\{w_1\ldots w_k\}$ of $A\subset L^{m'}$ let $\{w'_1,\ldots w'_k\}$ be the canonical dual basis of $A'$ (the finite-dimensional dual), i-e $<w'_i,w_j>_{A',A}=\delta_{ij}$. In particular the $w'_i$'s can be seen as functions $w'_i(x)\in L^m$. Thus any linear form $v$ on $A$ can be written as $v(x)=\sum\limits_{i=1}^k\lambda_iw'_i(x)\in A'$, and for all $u(x)=\sum\limits_{i=1}^ky_iw_i(x)\in A$ we can compute the action $$ <v,u>_{A',A}=<v,u>_{L^m,L^{m'}}=\int_{\Omega}v(x)u(x)\,dx $$ by $L^m$ duality. For any $v\in L^m$ (and not necessarily in $A'$, this is important) we can moreover define the "projection" $P_A$ onto $A'$ by $$ \forall u=\sum_1^ky_i w_i\in A:\qquad <P_A v,u>_{A',A}=<v,u>_{L^m,L^{m'}}=\int v(x)u(x)\,dx. $$ Let also $\varphi_A=P_A\circ \varphi:A\to A'\subset L^m$. Now for given coefficients $\lambda_1,\ldots,\lambda_k$ the equation $$ \forall j=1\ldots k:\qquad <\varphi(u),w_j>_{L^m,L^{m'}}=\lambda_j\hspace{2cm}(1) $$ actually reads $$ \text{find }u\in A\text{ such that }\varphi_A(u)=v\text{ in }A', $$ where $v(x)=\sum_1^k\lambda_j w'_j(x)$ can be unambiguously considered either as a function in $L^m(\Omega)$ or as an element of $A'$.

Claim 1: given any $\lambda=\lambda_1,\ldots,\lambda_k$ equation (1) admits a unique solution $u(x)=\sum_{i=1}^k y_iw_i(x)\in A$. This comes from quite simple finite dimensional considerations by looking at the restriction $\Phi_A$ of $\Phi(u)=|u|_{L^{m'}}^{m'}$ to the subspace $A$, and noting that for all $u\in A$ the differential can be computed as $$ \forall\, h \in A:\qquad <D\Phi_A(u),h>_{A',A}=<\varphi(u),h>_{L^m,L^{m'}}=\int \varphi(u)(x)h(x)\,dx. $$ Please let me know if you need details (roughly, $\varphi_A$ is the gradient of a $\mathcal{C}^1(A)$ strictly convex and superlinear function $\Phi_A$, so in particular it is a bijection). This claim 1 defines in particular an inverse $\psi_A:=\varphi_A^{-1}:A'\to A$. This is just a fancy way to say that, when viewed as taking values in the dual $A'$ (by restricting the $L^m$ action to test functions in $A\subset L^{m'}$), the map $\varphi$ restricted to $A$ is invertible. This should come as no surprise since the map $\varphi(z)=z^{m'-1}$ is increasing, which is exactly what makes the PDE formally parabolic. Compared to the paper [Alt-Luckhaus] I already mentioned, this corresponds to the fact that $\varphi$ is the gradient of the convex function $\Phi$, hence a monotone vector-field (easier here since the setting is scalar, but this is really what it is).

Claim 2: the inverse $\psi_A$ form step 1 is locally Lipschitz. This is just a refinement of your paper, but with an extra subtlety. Let $v^1=\sum _{i=1}^k\lambda_i^1w'_i$ and $v^2=\sum_{i=1}^k \lambda_i^2w'_i$ be two given elements of $A'$, and denote the corresponding solutions to (1) by $u^1=\psi_A(v^1)$ and $u^2=\psi_A(v^2)$. Using the convexity inequality $$ |a-b|^2\leq C(a^{m'-1}-b^{m'-1})(a-b)(|a|+|b|)^{2-m'} $$ you get \begin{align*} |u^1-u^2|^2_{L^2} & \leq C\int_{\Omega}((u^1)^{m'-1}-(u^2)^{m'-1})(u^1-u^2)(|u^1|+|u^2|)^{2-m'}\,dx\\ & \leq C(essup |u^1|+essup |u^2|)^{2-m'}\int_{\Omega}((u^1)^{m'-1}-(u^2)^{m'-1})(u^1-u^2)\,dx. \end{align*} In the last line I used the fact that $m'>1$ (since $m<\infty$) and monotonicity of $z\mapsto z^{m'-1}$ so that the last integrand is really pointwise nonnegative a.e $x\in \Omega$. Now because $u^1,u^2$ actually lie in the finite dimensional subspace $A$, you can rewrite this as \begin{align*} |u^1-u^2|^2_{L^2} & \leq C(essup |u^1|+essup |u^2|)^{2-m'}\int_{\Omega}(\varphi(u^1)-\varphi(u^2))(u^1-u^2)\,dx\\ & =C(essup |u^1|+essup |u^2|)^{2-m'}\int_{\Omega}(\varphi_A(u^1)-\varphi_A(u^2))(u^1-u^2)\,dx. \end{align*} Indeed the $\varphi(u^1),\varphi(u^2)\in L^m$ terms above "see" test functions $u^1,u^2\in A$, so their action is the same as that of their projection $\varphi_A(u^1),\varphi_A(u^2)$. Applying Young's inequality you get $$ |u^1-u^2|^2_{L^2} \leq C(essup |u^1|+essup |u^2|)^{2-m'}|\varphi_A(u^1)-\varphi_A(u^2)|_{L^m}|u^1-u^2|_{L^{m'}}, $$ and by equivalence of the norms on $A$ you conclude that $$ \forall \,u^1,u^2\in A:\qquad |u^1-u^2|_A\leq C(essup |u^1|+essup |u^2|)^{2-m'}\varphi_A(u^1)-\varphi_A(u^2)|_{L^m}|, $$ which really means that the inverse $\psi_A=\varphi_A^{-1}$ is locally Lipschitz continuous if $u^1,u^2$ stay in a ball around some $u_0$. The trick was here essentially to control $|u^1-u^2|$ not by $|\varphi(u^1)-\varphi(u^)|$, which a priori takes values in the "large" $L^m$ space, but rather by their projection $|\varphi_A(u^1)-\varphi_A(u^2)|$ onto the dual $A'$, which is finite-dimensional ("small"). I guess this is what the authors had in mind in their paper, but it was well hidden and quite misleading.

Step 3: back to the fixed point in the integral formulation. $$ \forall j=1\ldots k:\qquad \langle\varphi(u(t)), w_j \rangle_{L^m, L^{m'}} = \langle \varphi(u_0), w_j \rangle_{L^m, L^{m'}} + \int_0^t \langle F(s, u(s)), w_j \rangle\mathrm{d}s $$ By the classical Lipschitz estimates in the OP the right hand-side defines, for given $u\in C(0,T;A)$ a Lipschitz continuous map $u\mapsto V[u]\in C(0,T;A')$. In other words, the above integral formulation reads: $$ \text{find }u\in C(0,T;A)\text{ such that}\qquad \phi_a[u]=V[u]\text{ in }C(0,T;A'). $$ Now because $\psi_A=\phi_A^{-1}$ is locally Lipschitz continuous you can easily rewrite the fixed point as $$ \text{find }u\in C(0,T;A)\text{ such that}\qquad u=\psi_A\circ V[u]\text{ in }C(0,T;A). $$ Finally tuning the maximal time $T$ (small) and initial condition (stay in a given large ball around the "initial datum" to take care of the $essup$'s) it should be straighforward to prove that $\psi_A\circ V$ is a contraction in $C(0,T;A)$, hence the desired projected solution.

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@riem: all of this immediately applies to the original context of the $p$-Laplacian, all you need is that the right-hand side is Lipschitz continuous (proved in your paper). –  leo monsaingeon Jun 2 at 13:55
    
Thanks so much for this, I appreciate the effort! –  riem Jun 3 at 17:40
    
You're welcome. –  leo monsaingeon Jun 3 at 17:55

Really an answer to the OP's last comment, but too long for the box.

Are you sure the author didn't mean time derivative in the "strong sense", i-e you don't get $\partial_t\varphi(u)\in L^p(0,T;L^q)$ but merely something like $\partial_t\varphi(u)\in L^2(0,T;H^{-1})$? Otherwise the final weak formulation looks like: for any (time independent and fixed) $w\in H^1\cap L^{m'}$ there holds $$ <\varphi(u(t)),w>_{L^m,L^{m'}}=<\varphi(u_0),w>_{L^m,L^{m'}}+\ldots, $$ but how do you understand this equality in time, pointwise a.e $t\in (0,T)$? what sense does the evaluation of $\varphi(u(t))\in L^{m}$ make at a specific time $t\in (0,T)$? in which sense does the initial condition hold? usually for these issues you work out the time continuity $\varphi(u(t))\in C([0,T);L^m)$, which is retrieved by non-linear Aubin-Lions type compactness arguments and really requires estimates on some time derivatives ($H^{-1}$ or negative Sobolev in space). Also, in order to take the limit $\operatorname{dim}(A)=k\to\infty$ in the Galerkin approximation and show that the projected solution $u_k$ converges to some solution $u$ (in a weak sense), you definitely need some compactness on $u_k$ both in space and time. Compactness in time usually involves estimates on some time derivative (here most likely $\partial_t\varphi(u)$). A related comment is also the following: even if you only ask $\varphi(u(t))$ to make sense in $L^{m}$ pointwise a.e. $t\in (0,T)$ in the end (what you called integral formulation I guess), you would at least have to show that $u_k(t)\rightharpoonup u(t)$ in $L^m$ for a.e. $t$. But this is quite strong and as far as I'm concerned almost impossible to get without compactness in time.

I would be curious to give a look at the paper, would you mind giving us the reference or sending me a PDF at: {leonard dot monsaingeon at "Gm@il.c0M"}?

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Sorry, you're right. I should have said you don't any time derivative on $u_k$ but you do get a bound on $(\varphi(u_k))'$ (by rearrangement of the equation). The paper is this: link.springer.com/article/10.1007%2Fs10492-012-0004-0 Let me know if you can't access it, I can send a copy. –  riem May 28 at 7:19
    
Check out equation (1.2). I think the duality pairing is still $L^m$ $L^{m'}$.. –  riem May 28 at 7:21
    
could you send it to me, mathscinet is down at my university? –  leo monsaingeon May 28 at 7:27
    
Sent, should be in your inbox. –  riem May 28 at 7:34
1  
got it, thanks. I'll look into that today or next week. –  leo monsaingeon May 28 at 7:55

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