Finite element method for radially symmetric solution of a multidimensional semilinear heat equation

Toru Nakanishi; Norikazu Saito

doi:10.1007/s13160-019-00393-z

Japan Journal of Industrial and Applied Mathematics

pp 1–27 | Cite as

Finite element method for radially symmetric solution of a multidimensional semilinear heat equation

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Toru Nakanishi
Norikazu Saito

Original Paper

First Online: 26 October 2019

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Abstract

This study was conducted to present error analysis of a finite element method for computing the radially symmetric solutions of semilinear heat equations. Particularly, this study establishes optimal order error estimates in $L^\infty $ and weighted $L^2$ norms, respectively, for the symmetric and nonsymmetric formulation. Some numerical examples are presented to validate the obtained theoretical results.

Keywords

Finite element method Numerical analysis Radially symmetric solution Semilinear parabolic equation

Mathematics Subject Classification

65M60 35K58

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Notes

Acknowledgements

This work was supported by JST CREST Grant no. JPMJCR15D1, Japan, and JSPS KAKENHI Grant no. 15H03635, Japan. In addition, the first author was supported by the Program for Leading Graduate Schools, MEXT, Japan.

Proofs of (41a) and (41b)

Proofs of (41a) and (41b) are stated in this appendix using the same notation as that used in Sect. 4.

Proof of (41a)

By application of (39), (37a), and (37b), we derived the expression

$$\begin{aligned} \frac{1}{\tau _n}\left( |||\theta ^{n+1}|||^2-|||\theta ^{n+1}|||\cdot |||\theta ^{n}|||\right)\le & {} M(|||\theta ^n|||+Ch^2\Vert u_{xx}\Vert _{L^\infty (Q_T)})\cdot |||\theta ^{n+1}|||\\&+\, M\tau _n \Vert u_t\Vert _{L^\infty (Q_T)}\cdot |||\theta ^{n+1}|||\\&+\, \tau _{n} \Vert u_{tt}\Vert _{L^{\infty }(Q_T)}\cdot |||\theta ^{n+1}|||\\&+\, Ch^2\Vert u_{xxt}\Vert _{L^\infty (Q_T)}\cdot |||\theta ^{n+1}|||. \end{aligned}$$

Consequently, we have

$$\begin{aligned} |||\theta ^{n+1}||| \le (1+\tau _n M) |||\theta ^n|||+C\tau _n (h^2+\tau _n). \end{aligned}$$

Therefore, similarly to the derivation of (31), we obtain from (22) the expression of

$$\begin{aligned} |||\theta ^n|||\le C\left( h^2+\tau \right) \end{aligned}$$

to complete the proof. $\square $

Proof of (41b)

First, we prove the case of $n=0$. Substituting (38) for $n=0$ and $\chi =\theta ^{1}$, we obtain

$$\begin{aligned} \left\langle \frac{\theta ^{1}-\theta ^0}{\tau _{0}},\theta ^{1}\right\rangle +B(\theta ^{1},\theta ^{1})\le & {} \left\langle f(u_{h}^{0})-f(u^0),\theta ^1\right\rangle \\&-\, \left\langle f(u(t_{1}))-f(u^0),\theta ^1\right\rangle \\&-\,\left\langle \partial _{\tau _0} u(t_{1})-u_{t}(t_{1}),\theta ^1\right\rangle \\&-\,\left\langle \frac{\rho ^{1}-\rho ^0}{\tau _{0}},\theta ^{1}\right\rangle . \end{aligned}$$

Because $\theta ^0=0$, we apply (37b) to get

$$\begin{aligned} \frac{1}{\tau _0}|||\theta ^{1}|||^2&\le M |||\rho ^0|||\cdot |||\theta ^1||| +M\tau _0\Vert u_t\Vert _{L^\infty (Q_T)}|||\theta ^1|||\\&{ }\quad +\tau _0\Vert u_{tt}\Vert _{L^\infty (Q_T)}|||\theta ^1||| +|||\partial _{\tau _0}\rho ^{1}|||\cdot |||\theta ^1|||\\&\le C(\tau _0+h^2)|||\theta ^1|||. \end{aligned}$$

Repeatedly using $\theta ^0=0$, we obtain

$$\begin{aligned} |||\partial _{\tau _0}\theta ^1|||\le C(\tau _0+h^2). \end{aligned}$$

(44)

Next we assume $n\ge 0$ and $t_{n+2}\le T$. Consequently, from (38), we derive

$$\begin{aligned}&\left\langle \partial _{\tau _{n+1}}\theta ^{n+2}-\partial _{\tau _n}\theta ^{n+1},\chi \right\rangle +B(\theta ^{n+2}-\theta ^{n+1},\chi ) \nonumber \\&\quad =\langle \underbrace{f(u_{h}^{n+1})-f(u(t_{n+1}))-f(u_{h}^{n})+f(u(t_{n}))}_{=J_1},\chi \rangle \nonumber \\&\qquad -\langle \underbrace{f(u(t_{n+2}))-f(u(t_{n+1}))-f(u(t_{n+1}))+f(u(t_{n}))}_{=J_2},\chi \rangle \nonumber \\&\qquad -\langle \underbrace{\partial _{\tau _{n+1}}u(t_{n+2})-u_{t}(t_{n+2}) -\partial _{\tau _{n}}u(t_{n+1}) + u_{t}(t_{n+1})}_{=J_3} ,\chi \rangle \nonumber \\&\qquad -\langle \underbrace{\partial _{\tau _{n+1}}\rho ^{n+2}-\partial _{\tau _{n}}\rho ^{n+1}}_{=J_4},\chi \rangle \end{aligned}$$

(45)

for any $\chi \in S_h$. Substituting this expression for $\chi =\partial _{\tau _{n+1}}\theta ^{n+2}$, we obtain

$$\begin{aligned}&|||\partial _{\tau _{n+1}}\theta ^{n+2}|||^2- |||\partial _{\tau _n}\theta ^{n+1}|||\cdot |||\partial _{\tau _{n+1}}\theta ^{n+2}||| \\&\quad \le |||\partial _{\tau _{n+1}}\theta ^{n+2}|||\sum _{j=1}^4|||J_j|||. \end{aligned}$$

Here, we accept the following estimates:

$$\begin{aligned} |||J_1|||&\le C\tau _n(1+\tau _n) |||\partial _{\tau _n}\theta ^{n+1}||| +C\tau _n(h^2+\tau _n+\tau _nh^2), \end{aligned}$$

(46a)

$$\begin{aligned} |||J_2|||,|||J_3|||&\le C\tau _{n+1}(\tau _{n+1}+\tau _{n})+ C|\tau _{n+1}-\tau _{n}|, \end{aligned}$$

(46b)

$$\begin{aligned} |||J_4|||&\le C (\tau _{n+1}+\tau _n)h^2. \end{aligned}$$

(46c)

In view of the quasi-uniformity of time partition (23), we have

$$\begin{aligned} \tau _{n+1}=\tau _n\frac{\tau _{n+1}}{\tau _n}\le \gamma \tau _n. \end{aligned}$$

Summing up, we deduce

$$\begin{aligned} b_{n+1}-b_n\le C \tau _n b_n +C\tau _n\left( h^2+\tau +\frac{\delta }{\tau _{\min }}\right) , \end{aligned}$$

(47)

where $b_n=|||\partial _{\tau _n}\theta ^{n+1}|||$. Therefore,

$$\begin{aligned} b_{n}\le e^{CT}b_0+C(e^{CT}-1)\left( h^2+\tau +\frac{\delta }{\tau _{\min }}\right) , \end{aligned}$$

which, together with (44), implies the desired inequality (41b).

We now prove (46a)–(46c).

Estimation for$J_1$. We apply Taylor’s theorem to obtain

$$\begin{aligned} J_1&=f'(s_{1})(u_{h}^{n+1}-u_{h}^{n})-f'(s_{2})(u(t_{n+1})-u(t_{n}))\\&= f'(s_{1})[(\theta ^{n+1}+\rho ^{n+1})-(\theta ^{n}+\rho ^{n})] \\&\quad +\frac{f'(s_{1})-f'(s_{2})}{s_{1}-s_{2}}(s_{1}-s_{2})(u(t_{n+1})-u(t_{n})), \end{aligned}$$

where $s_{1}=u_{h}^{n+1}-\mu _{1}(u_{h}^{n+1}-u_{h}^{n})$ and $s_{2}=u(t_{n+1})-\mu _{2}[u(t_{n+1})-u(t_{n})]$ for some $\mu _{1},\mu _2\in [0,1]$. In view of (37a), (37b), and (41a), we find the following estimates

$$\begin{aligned} |||J_1|||&\le \tau _nM|||\partial _{\tau _n}\theta ^{n+1}||| +\tau _nM|||\partial _{\tau _n}\rho ^{n+1}|||\\&\quad +\, \left| \left| \left| \frac{f'(s_{1})-f'(s_{2})}{s_{1}-s_{2}}(s_{1}-s_{2})\right| \right| \right| \cdot \tau _n\Vert u_t\Vert _{L^\infty (Q_T)},\\&\le \tau _nM|||\partial _{\tau _n}\theta ^{n+1}||| +C\tau _n Mh^2\Vert u_{txx}\Vert _{L^\infty (Q_T)}\\&\quad +\, \left| \left| \left| \frac{f'(s_{1})-f'(s_{2})}{s_{1}-s_{2}}(s_{1}-s_{2})\right| \right| \right| \cdot \tau _n\Vert u_t\Vert _{L^\infty (Q_T)}, \end{aligned}$$

and

$$\begin{aligned}&\left| \left| \left| \displaystyle {\frac{f'(s_{1})-f'(s_{2})}{s_{1}-s_{2}}(s_{1}-s_{2})}\right| \right| \right|&\\&\quad \le M_{2}|||\theta ^{n+1}+\rho ^{n+1}-\mu _{1}(\theta ^{n+1}+\rho ^{n+1}-\theta ^{n}-\rho ^{n})\\&\qquad +\,(\mu _{2}-\mu _{1})(u(t_{n+1})-u(t_{n}))|||\\&\quad \le M_{2}\{|||\theta ^{n+1}|||+|||\rho ^{n+1}|||+ \tau _n|||\partial _{\tau _n}\theta ^{n+1}|||\\&\qquad +\,\tau _n|||\partial _{\tau _n}\rho ^{n+1}|||+\tau _n\Vert u_t\Vert _{L^\infty (Q_T)}\}&\\&\quad \le M_{2}\{C(h^2+\tau )+Ch^2\Vert u_{xx}\Vert _{L^\infty (Q_T)}+ \tau _n|||\partial _{\tau _n}\theta ^{n+1}||| \\&\qquad +\,C\tau _n h^2\Vert u_{txx}\Vert _{L^\infty (Q_T)}+\tau _n\Vert u_t\Vert _{L^\infty (Q_T)}\}. \end{aligned}$$

Estimation for$J_2$. We begin with

$$\begin{aligned} J_2&=f'(s_{3})(u(t_{n+2})-u(t_{n+1}))-f'(s_{4})(u(t_{n+1})-u(t_{n}))\\&=\frac{f'(s_{3})-f'(s_{4})}{s_{3}-s_{4}}(s_{3}-s_{4})\tau _{n+1}u_{t}(\eta _{1})+f'(s_{4})(\tau _{n+1}u_{t}(\eta _{1})-\tau _{n}u_{t}(\eta _{2}))\\&=\frac{f'(s_{3})-f'(s_{4})}{s_{3}-s_{4}}(s_{3}-s_{4})\tau _{n+1}u_{t}(\eta _{1}) \\&\quad +\,f'(s_{4})\tau _{n+1}(u_{t}(\eta _{1})-u_{t}(\eta _{2})) +f'(s_{4})(\tau _{n+1}-\tau _n)u_{t}(\eta _{2}), \end{aligned}$$

where $s_{3}=u(t_{n+1})+\mu _{3}(u(t_{n+2})-u(t_{n+1}))$ and $s_{4}=u(t_{n+1})+\mu _{4}(u(t_{n})-u(t_{n+1}))$ for some $\mu _3,\mu _4\in [0,1]$, $\eta _1\in [t_{n+1},t_{n+2}]$, and $\eta _2\in [t_{n},t_{n+1}]$. Next, we obtain the following estimate:

$$\begin{aligned} |||J_2|||&\le \tau _{n+1} \left| \left| \left| \frac{f'(s_{3})-f'(s_{4})}{s_{3}-s_{4}}(s_{3}-s_{4})\right| \right| \right| \cdot \Vert u_{t}\Vert _{L^\infty (Q_T)} \\&\quad +\, M_1\tau _{n+1}(\tau _{n+1}+\tau _{n})\Vert u_{tt}\Vert _{L^\infty (Q_T)} +M_1|\tau _{n+1}-\tau _n|\cdot \Vert u_t\Vert _{L^\infty (Q_T)}; \end{aligned}$$

$$\begin{aligned} \left| \left| \left| \frac{f'(s_{3})-f'(s_{4})}{s_{3}-s_{4}}(s_{3}-s_{4})\right| \right| \right|&\le CM_{2}(\tau _{n+1}+\tau _n)\Vert u_t\Vert _{L^\infty (Q_T)}. \end{aligned}$$

Estimation for$J_3$. We express $J_3$ as

$$\begin{aligned} J_3&= \frac{\tau _{n+1}u_t(t_{n+2})-\frac{1}{2}\tau _{n+1}^2u_{tt}(s_5)}{\tau _{n+1}} -u_{t}(t_{n+2}) \\&\quad -\,\left( \frac{\tau _{n}u_t(t_{n+1})-\frac{1}{2}\tau _{n}^2u_{tt}(s_6)}{\tau _{n}} -u_{t}(t_{n+1})\right) \\&= -\frac{1}{2}\tau _{n+1}u_{tt}(s_5) +\frac{1}{2}\tau _{n}u_{tt}(s_6)\\&= \frac{1}{2}\tau _{n+1}(u_{tt}(s_6)-u_{tt}(s_5))- \frac{1}{2}(\tau _{n+1}-\tau _{n})u_{tt}(s_6)\\&=\frac{1}{2} \tau _{n+1}u_{ttt}(s_{7})(s_{5}-s_{6})-\frac{1}{2} (\tau _{n+1}-\tau _{n})u_{tt}(s_{6}) \end{aligned}$$

for some $s_5\in [t_{n+1},t_{n+2}]$, $s_6\in [t_{n},t_{n+1}]$ and $s_7\in [s_6,s_5]\subset [t_{n},t_{n+2}]$. Therefore,

$$\begin{aligned} |||J_3|||\le \frac{1}{2}\tau _{n+1}(\tau _{n+1}+\tau _{n})\Vert u_{ttt}\Vert _{L^\infty (Q_T)}+ \frac{1}{2}|\tau _{n+1}-\tau _{n}|\cdot \Vert u_{tt}\Vert _{L^\infty (Q_T)}. \end{aligned}$$

Estimation for$J_4$. For some $s_8\in [t_{n+1},t_{n+2}]$, $s_9\in [t_{n},t_{n+1}]$, and $s_{10}\in [s_9,s_8]$, we obtain the expression

$$\begin{aligned} \frac{\rho ^{n+2}-\rho ^{n+1}}{\tau _{n+1}}-\frac{\rho ^{n+1}-\rho ^{n}}{\tau _{n}} =\rho _{t}(s_{8})-\rho _{t}(s_{9}) =(s_8-s_9)\rho _{tt}(s_{10}) \end{aligned}$$

Therefore, using (35),

$$\begin{aligned} |||J_4|||\le C (\tau _{n+1}+\tau _n)h^2\Vert u_{ttxx}\Vert _{L^\infty (Q_T)}. \end{aligned}$$

$\square $

Proof of Proposition 5.1

Proof

Substituting $\chi =\partial _{\tau _n}u_{h}^{n+1}$ for (7), we have

$$\begin{aligned} \Vert \partial _{\tau _n}u_{h}^{n+1}\Vert _h^{2}=-\left( (u^{n+1}_{h})_x,\frac{(u^{n+1}_{h})_x-(u^{n}_{h})_x}{\tau _{n}}\right) +\left( u_{h}^{n}|u_{h}^{n}|^{\alpha },\frac{u_{h}^{n+1}-u_{h}^{n}}{\tau _{n}}\right) . \end{aligned}$$

Therefore, for the conditions

$$\begin{aligned}&\left( (u^{n+1}_{h})_x,\frac{(u^{n+1}_{h})_x-(u^{n}_{h})_x}{\tau _{n}}\right) \ge \frac{1}{2}\left( (u^{n}_{h})_x+(u^{n+1}_{h})_x,\frac{(u^{n+1}_{h})_x-(u^{n}_{h})_x}{\tau _{n}}\right) , \end{aligned}$$

(48a)

$$\begin{aligned}&\left( u_{h}^{n}|u_{h}^{n}|^{\alpha },\frac{u_{h}^{n+1}-u_{h}^{n}}{\tau _{n}}\right) \le \frac{1}{\tau _{n}(\alpha +2)}\left[ \int _Ix^{N-1} (|u_{h}^{n+1}|^{\alpha +2}-|u_{h}^{n}|^{\alpha +2})~dx\right] , \end{aligned}$$

(48b)

we obtain

$$\begin{aligned} \left\| \partial _{\tau _n}u_{h}^{n+1}\right\| _h^{2} \le -\frac{1}{\tau _{n}}(J_{h}(n+1)-J_{h}(n)), \end{aligned}$$

(49)

which implies that $J_{h}(n+1)\le J_{h}(n)$.

We can validate (48a) and (48b). Also, (48a) is derived readily. To prove (48b), we set $g(s)=\frac{1}{\alpha +2}|s|^{\alpha +2}$, and apply the mean value theorem to deduce

$$\begin{aligned} g(u_h^{n+1})- g(u_h^{n})=w|w|^{\alpha }(u_h^{n+1}-u_h^{n}), \end{aligned}$$

where $w=w(x)=u_{h}^{n}+\sigma (u_{h}^{n+1}-u_{h}^{n})$ and $\sigma =\sigma (x)\in (0,1)$. Consequently,

$$\begin{aligned} J\equiv & {} \frac{1}{\tau _{n}(\alpha +2)}\left[ \int _Ix^{N-1} (|u_{h}^{n+1}|^{\alpha +2}-|u_{h}^{n}|^{\alpha +2})~dx\right] \\&-\, \int _Ix^{N-1}u_{h}^{n}|u_{h}^{n}|^{\alpha }\frac{u_{h}^{n+1}-u_{h}^{n}}{\tau _{n}}~dx\\= & {} \frac{1}{\tau _n}\int _Ix^{N-1}\left[ w|w|^\alpha -u_{h}^{n}|u_{h}^{n}|^{\alpha }\right] (u_h^{n+1}-u_h^{n})~dx. \end{aligned}$$

Then we repeat the mean value theorem to resolve

$$\begin{aligned} w|w|^\alpha -u_{h}^{n}|u_{h}^{n}|^{\alpha }=(\alpha +1)|{\tilde{w}}|^\alpha (w-u^n_h)=(\alpha +1)|{\tilde{w}}|^\alpha {\tilde{\sigma }}(u^{n+1}_h-u_h^n), \end{aligned}$$

where ${\tilde{w}}=u^n_h+{\tilde{\sigma }}(w-u_h^n)$ and ${\tilde{\sigma }}={\tilde{\sigma }}(x)\in (0,1)$. Therefore,

$$\begin{aligned} J=\frac{1}{\tau _n}\int _I x^{N-1}(\alpha +1)|{\tilde{w}}|^\alpha {\tilde{\sigma }} (u_h^{n+1}-u_h^{n})^2~dx\ge 0, \end{aligned}$$

which gives (48b). $\square $

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Toru Nakanishi
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Norikazu Saito
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1.Graduate School of Mathematical SciencesThe University of TokyoTokyoJapan

Finite element method for radially symmetric solution of a multidimensional semilinear heat equation

Abstract

Keywords

Mathematics Subject Classification

Notes

Acknowledgements

References

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