Klasifikasi - Reduksi PDP Linear Orde Dua

Klasifikasi Persamaan Diferensial Parsial

Persamaan Diferensial Parsial Linear Orde Dua pada Bidang

Misalkan $\Gamma$ suatu kurva di $\displaystyle\mathbb{R}^2$, dan misalkan $f:\Gamma\rightarrow \mathbb{R}$ serta $g:\Gamma\rightarrow \mathbb{R}$ dua fungsi analitik. Kita meninjau Masalah Cauchy berbentuk

$$\begin{split} & L(x,y,\mathrm{D}) = a_{2,0}(x,y) \frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial x^{2}}(x,y) + 2a_{1,1}(x,y) \frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial x^{} \partial y^{}}(x,y) + a_{0,2}(x,y) \frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial y^{2}}(x,y) \\ &\qquad + a_{1,0}(x,y) \frac{\partial{u}}{\partial x}(x,y) + a_{0,1}(x,y) \frac{\partial{u}}{\partial y}(x,y) + a_{0,0}(x,y)u(x,y) + b(x,y) = 0 \end{split}$$(2.1.1) dengan syarat

$$u\big|_\Gamma = f \quad \text{dan} \quad \frac{\partial{u}}{\partial \nu}\bigg|_\Gamma = g \ ,$$(2.1.2) dengan $\displaystyle\nu\left(\boldsymbol{p}\right)$, untuk $\boldsymbol{p} \in \Gamma$, merupakan vektor normal satuan terhadap $\Gamma$ di $\boldsymbol{p}$. Persamaan (2.1.1) merupakan persamaan diferensial parsial linear orde dua

.

Mula-mula kita mereduksi Masalah Cauchy (2.1.1) dan (2.1.2) menjadi masalah dengan $\Gamma$ sebagai sumbu $x$. Misalkan $\boldsymbol{p}$ suatu titik pada $\Gamma$. Di suatu lingkungan $\boldsymbol{p}$, kita dapat mendefinisikan dua keluarga kurva ortogonal $\eta(x,y)=h$ dan $\xi(x,y)=k$ sedemikian sehingga $\Gamma$ merupakan kurva $\xi(x,y)=0$ (Gambar 2.1). Kita juga dapat mengasumsikan bahwa $\boldsymbol{p}$ merupakan titik perpotongan kurva $\xi(x,y)=0$ dan $\eta(x,y)=0$.

Syarat agar dua keluarga kurva bersifat transversal di suatu lingkungan $\boldsymbol{p}$ adalah

$$\frac{\partial (\eta, \xi)}{\partial (x,y)} = \det\begin{pmatrix} \displaystyle\frac{\partial{\eta}}{\partial x} & \displaystyle\frac{\partial{\eta}}{\partial y} \\[0.7em] \displaystyle\frac{\partial{\xi}}{\partial x} & \displaystyle\frac{\partial{\xi}}{\partial y} \end{pmatrix} \neq 0$$(2.1.3) di lingkungan $\boldsymbol{p}$. Memang, untuk $\boldsymbol{q}$ di dekat $\boldsymbol{p}$,

$$\begin{aligned} \frac{\partial (\eta, \xi)}{\partial (x,y)}\bigg|_{\boldsymbol{q}} = 0 & \Leftrightarrow \left(\frac{\partial{\eta}}{\partial x}(\boldsymbol{q}),\frac{\partial{\eta}}{\partial y}(\boldsymbol{q})\right) \cdot \left(-\frac{\partial{\xi}}{\partial y}(\boldsymbol{q}), \frac{\partial{\xi}}{\partial x}(\boldsymbol{q})\right) = 0 \\ & \Leftrightarrow \left(-\frac{\partial{\xi}}{\partial y}(\boldsymbol{q}), \frac{\partial{\xi}}{\partial x}(\boldsymbol{q})\right) \ \text{ortogonal terhadap} \ \left(\frac{\partial{\eta}}{\partial x}(\boldsymbol{q}),\frac{\partial{\eta}}{\partial y}(\boldsymbol{q})\right) \\ & \Leftrightarrow \left(\frac{\partial{\eta}}{\partial x}(\boldsymbol{q}),\frac{\partial{\eta}}{\partial y}(\boldsymbol{q})\right) \ \text{sejajar dengan} \ \left(\frac{\partial{\xi}}{\partial x}(\boldsymbol{q}),\frac{\partial{\xi}}{\partial y}(\boldsymbol{q})\right) \end{aligned}$$ karena $\displaystyle\left(-\frac{\partial{\xi}}{\partial y}(\boldsymbol{q}), \frac{\partial{\xi}}{\partial x}(\boldsymbol{q})\right)$ juga ortogonal terhadap $\displaystyle\left(\frac{\partial{\xi}}{\partial x}(\boldsymbol{q}),\frac{\partial{\xi}}{\partial y}(\boldsymbol{q})\right)$. Hal ini menunjukkan bahwa kurva $\xi(x,y)=k$ dan $\eta(x,y)=h$ memiliki garis normal yang sama di $\boldsymbol{q}$, sehingga kedua kurva tersebut bersinggungan di $\boldsymbol{q}$.

Berdasarkan (2.1.3), kita dapat menggunakan Teorema Fungsi Invers untuk menyatakan $x$ dan $y$ di suatu lingkungan $\boldsymbol{p}$ sebagai fungsi dari $\xi$ dan $\eta$ di suatu lingkungan titik asal. Kita akan menggunakan perubahan variabel ini untuk mereduksi (2.1.1).

Jika kita substitusikan

$$\begin{aligned} u(\eta,\xi) & = u(x,y) \ , \quad \ \frac{\partial{u}}{\partial x} = \frac{\partial{u}}{\partial \xi}\frac{\partial{\xi}}{\partial x} + \frac{\partial{u}}{\partial \eta}\frac{\partial{\eta}}{\partial x} \ , \quad \ \frac{\partial{u}}{\partial y} = \frac{\partial{u}}{\partial \xi}\frac{\partial{\xi}}{\partial y} + \frac{\partial{u}}{\partial \eta}\frac{\partial{\eta}}{\partial y} \ ,\\ \frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial x^{2}} &= \frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial \xi^{2}} \left(\frac{\partial{\xi}}{\partial x}\right)^2 + 2\frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial \eta^{} \partial \xi^{}}\frac{\partial{\eta}}{\partial x}\frac{\partial{\xi}}{\partial x} + \frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial \eta^{2}}\left(\frac{\partial{\eta}}{\partial x}\right)^2 + \frac{\partial{u}}{\partial \xi} \frac{\displaystyle \partial^{2}{\xi}}{\displaystyle \partial x^{2}} + \frac{\partial{u}}{\partial \eta}\frac{\displaystyle \partial^{2}{\eta}}{\displaystyle \partial x^{2}} \ ,\\ \frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial y^{2}} &= \frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial \xi^{2}} \left(\frac{\partial{\xi}}{\partial y}\right)^2 + 2\frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial \eta^{} \partial \xi^{}}\frac{\partial{\eta}}{\partial y}\frac{\partial{\xi}}{\partial y} + \frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial \eta^{2}}\left(\frac{\partial{\eta}}{\partial y}\right)^2 + \frac{\partial{u}}{\partial \xi} \frac{\displaystyle \partial^{2}{\xi}}{\displaystyle \partial y^{2}} + \frac{\partial{u}}{\partial \eta}\frac{\displaystyle \partial^{2}{\eta}}{\displaystyle \partial y^{2}} \\ &\text{dan}\\ \frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial x^{} \partial y^{}} &= \frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial \xi^{2}} \frac{\partial{\xi}}{\partial x}\frac{\partial{\xi}}{\partial y} + \frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial \eta^{} \partial \xi^{}}\frac{\partial{\eta}}{\partial y}\frac{\partial{\xi}}{\partial x} +\frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial \eta^{} \partial \xi^{}}\frac{\partial{\eta}}{\partial x}\frac{\partial{\xi}}{\partial y} +\frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial \eta^{2}}\frac{\partial{\eta}}{\partial x}\frac{\partial{\eta}}{\partial y} + \frac{\partial{u}}{\partial \xi} \frac{\displaystyle \partial^{2}{\xi}}{\displaystyle \partial y^{} \partial x^{}} + \frac{\partial{u}}{\partial \eta}\frac{\displaystyle \partial^{2}{\eta}}{\displaystyle \partial y^{} \partial x^{}} \end{aligned}$$ ke dalam (2.1.1), maka diperoleh

$$A_{2,0} \frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial \eta^{2}} + 2A_{1,1}\frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial \eta^{} \partial \xi^{}} + A_{0,2}\frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial \xi^{2}} + A_{1,0} \frac{\partial{u}}{\partial \eta} + A_{0,1} \frac{\partial{u}}{\partial \xi} + A_{0,0} u + B = 0 \ ,$$(2.1.4) dengan

$$\begin{aligned} A_{2,0}(\eta,\xi) &= a_{2,0}\, \left(\frac{\partial{\eta}}{\partial x}\right)^2 +2a_{1,1}\,\frac{\partial{\eta}}{\partial x}\frac{\partial{\eta}}{\partial y} + a_{0,2}\, \left(\frac{\partial{\eta}}{\partial y}\right)^2 \ , \\ A_{1,1}(\eta,\xi) &= a_{2,0}\, \frac{\partial{\eta}}{\partial x}\frac{\partial{\xi}}{\partial x} + a_{1,1}\, \left(\frac{\partial{\eta}}{\partial y}\frac{\partial{\xi}}{\partial x}+\frac{\partial{\eta}}{\partial x}\frac{\partial{\xi}}{\partial y}\right) + a_{0,2}\, \frac{\partial{\eta}}{\partial y}\frac{\partial{\xi}}{\partial y} \ , \\ A_{0,2}(\eta,\xi) &= a_{2,0}\,\left(\frac{\partial{\xi}}{\partial x}\right)^2 + 2a_{1,1}\, \frac{\partial{\xi}}{\partial x}\frac{\partial{\xi}}{\partial y} + a_{0,2}\,\left(\frac{\partial{\xi}}{\partial y}\right)^2 \ , \\ A_{1,0}(\eta,\xi) &= a_{2,0}\,\frac{\displaystyle \partial^{2}{\eta}}{\displaystyle \partial x^{2}} + 2a_{1,1}\,\frac{\displaystyle \partial^{2}{\eta}}{\displaystyle \partial x^{} \partial y^{}} + a_{0,2}\, \frac{\displaystyle \partial^{2}{\eta}}{\displaystyle \partial y^{2}} +a_{1,0} \, \frac{\partial{\eta}}{\partial x} + a_{0,1}\,\frac{\partial{\eta}}{\partial y} \ , \\ A_{0,1}(\eta,\xi) &= a_{2,0}\,\frac{\displaystyle \partial^{2}{\xi}}{\displaystyle \partial x^{2}} + 2a_{1,1}\,\frac{\displaystyle \partial^{2}{\xi}}{\displaystyle \partial x^{} \partial y^{}} + a_{0,2}\, \frac{\displaystyle \partial^{2}{\xi}}{\displaystyle \partial y^{2}} +a_{1,0} \, \frac{\partial{\xi}}{\partial x} + a_{0,1}\,\frac{\partial{\xi}}{\partial y} \ , \\ A_{0,0}(\eta,\xi) &= a_{0,0} \quad \text{dan} \quad B(\eta,\xi) = b(x,y) \end{aligned}$$ dengan ruas kanan setiap bentuk dievaluasi pada $(x,y) = \left(x(\eta,\xi),y(\eta,\xi)\right)$.

Kita memilih orientasi $\xi$ sehingga $\nu=\nabla\xi/\lVert\nabla\xi\rVert$ pada $\Gamma$. Dengan pilihan ini, syarat-syarat awal dalam (2.1.2) menjadi

$$u(\eta,0) = f\left(x(\eta,0),y(\eta,0)\right) \quad \text{dan} \quad \frac{\partial{u}}{\partial \xi}(\eta,0) = \frac{g\left(x(\eta,0),y(\eta,0)\right)} {\left\lVert\nabla\xi\left(x(\eta,0),y(\eta,0)\right)\right\rVert} \ .$$(2.1.5)

Sebagaimana akan kita lihat dalam Bab 3, untuk menentukan ekspansi deret solusi $u$ dari (2.1.4) dan (2.1.5) di suatu lingkungan titik asal pada bidang $\eta$,$\xi$ (jadi, di dekat $\boldsymbol{p}$ pada bidang $x$,$y$), kita perlu menghitung $\displaystyle\frac{\displaystyle \partial^{j}{u}}{\displaystyle \partial \xi^{j}}$ untuk $j\geq 2$ di titik asal. Untuk itu, kita mengasumsikan bahwa (2.1.4) dapat diselesaikan terhadap $\displaystyle\frac{\displaystyle \partial^{2}{u}}{\displaystyle \partial \xi^{2}}$ pada $(\eta,\xi)=(0,0)$. Hal ini hanya mungkin jika

$$A_{0,2}(0,0) = a_{2,0}(\boldsymbol{p})\, \left(\frac{\partial{\xi}}{\partial x}(\boldsymbol{p})\right)^2 +2a_{1,1}(\boldsymbol{p})\,\frac{\partial{\xi}}{\partial x}(\boldsymbol{p})\,\frac{\partial{\xi}}{\partial y}(\boldsymbol{p}) + a_{0,2}(\boldsymbol{p})\, \left(\frac{\partial{\xi}}{\partial y}(\boldsymbol{p})\right)^2 \neq 0 \ .$$(2.1.6) Syarat ini bersifat mendasar dan menjadi pokok bahasan bagian berikutnya.

Sebelum beralih ke bagian berikutnya, kita mencatat bahwa

$$A_{1,1}^2 - A_{2,0}\,A_{0,2} = \left(a_{1,1}^2 - a_{2,0}a_{0,2}\right) \left( \frac{\partial{\eta}}{\partial x}\frac{\partial{\xi}}{\partial y} - \frac{\partial{\eta}}{\partial y}\frac{\partial{\xi}}{\partial x} \right)^2$$(2.1.7) dengan ruas kiri dievaluasi pada $(\eta,\xi)$ dan ruas kanan dievaluasi pada $(x,y) = \left(x(\eta,\xi),y(\eta,\xi)\right)$. Identitas ini akan digunakan kemudian untuk mengklasifikasikan persamaan diferensial parsial linear orde dua. Pembaca diminta memverifikasi identitas ini.