Selesaikan persamaan diferensial parsial non-homogen berikut:
$$\frac{1}{25} \frac{\partial T}{\partial t} = \frac{\partial^2 T}{\partial x^2} + e^{-2t}(2 – x)$$
Syarat Batas (Boundary Conditions):
- $x = 0 \implies T = 0$
- $x = 2 \implies T = 0$
Syarat Awal (Initial Condition):
- $t = 0 \implies T = 150$
Konsep Dasar
Persamaan tersebut bisa kita ubah menjadi
$$\frac{\partial T}{\partial t} = 25 \frac{\partial^2 T}{\partial x^2} + 25 e^{-2t}(2 – x)$$
Karena syarat batasnya bernilai nol pada ujung-ujungnya ($x=0$ dan $x=2$), fungsi eigen dari masalah homogen terkait ($\frac{\partial T}{\partial t} = 25 \frac{\partial^2 T}{\partial x^2}$) akan berbentuk fungsi Sinus, bukan Cosinus seperti pada soal-soal awal yang memiliki syarat batas turunan (Neumann).
Kita mengasumsikan solusi total berbentuk deret fungsi eigen sinus:
$$T(x,t) = \sum_{n=1}^{\infty} T_n(t) \sin\left( \frac{n\pi}{2} x \right)$$
Suku sumber non-homogen $25e^{-2t}(2 – x)$ juga akan kita ekspansikan ke dalam deret sinus yang sama sebelum menyelesaikan persamaan diferensial terhadap waktu ($t$).
Penurunan Deret Fungsi
Mari kita turunkan deret fungsi tersebut
Fungsi awal
$$T(x,t) = \sum_{n=1}^{\infty} T_n(t) \sin\left( \frac{n\pi}{2} x \right)$$
Turunan terhadap waktu:
$$\frac{\partial T}{\partial t} = \sum_{n=1}^{\infty} T_n'(t) \sin\left(\frac{n\pi}{2}x\right)$$
Turunan ruang pertama ($\frac{\partial T}{\partial x}$):
$$\frac{\partial T}{\partial x} = \sum_{n=1}^{\infty} T_n(t) \cdot \left(\frac{n\pi}{2}\right) \cos\left(\frac{n\pi}{2}x\right)$$
Turunan ruang kedua ($\frac{\partial^2 T}{\partial x^2}$):
$$\frac{\partial^2 T}{\partial x^2} = \sum_{n=1}^{\infty} T_n(t) \cdot \left(\frac{n\pi}{2}\right) \cdot \left(-\frac{n\pi}{2}\right) \sin\left(\frac{n\pi}{2}x\right)$$
$$\frac{\partial^2 T}{\partial x^2} = \sum_{n=1}^{\infty} -\left(\frac{n\pi}{2}\right)^2 T_n(t) \sin\left(\frac{n\pi}{2}x\right)$$
Ekspansi Suku Sumber Non-Homogen
Kita ambil variabel non homogen saja, yaitu
$$25 e^{-2t}(2 – x)$$
Misalkan fungsi spasial dari sumber tersebut adalah $f(x) = 25(2 – x) = 50 – 25x$.
Kita nyatakan ke dalam deret Fourier Sinus pada interval $L = 2$:
$$50 – 25x = \sum_{n=1}^{\infty} K_n \sin\left( \frac{n\pi}{2} x \right)$$
Cari koefisien $K_n$ menggunakan rumus integral Fourier Sinus:
$$K_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi}{2} x \right) \, dx$$
$$K_n = \frac{2}{2} \int_{0}^{2} (50 – 25x) \sin\left( \frac{n\pi}{2} x \right) \, dx$$
$$K_n = \int_{0}^{2} (50 – 25x) \sin\left( \frac{n\pi}{2} x \right) \, dx$$
$$\mathbf{K_n = \int_{0}^{2} 50 \sin\left( \frac{n\pi}{2} x \right) \, dx – 25 \int_{0}^{2} x \sin\left( \frac{n\pi}{2} x \right) \, dx}$$
Integral Bagian Kiri
$$\int_{0}^{2} 50 \sin\left( \frac{n\pi}{2} x \right) \, dx$$
Kita gunakan metode integral parsial untuk menyelesaikannya
$$\int u dv = u v – \int v du$$
- $u = 50 \implies du = 0$
- $dv = \sin\left( \frac{n\pi}{2} x \right) \implies v = -\frac{2}{n\pi} \cos\left( \frac{n\pi}{2} x \right)$
$$50 \cdot -\frac{2}{n\pi} \cos\left( \frac{n\pi}{2} x \right) + \int \frac{2}{n\pi} \cos\left( \frac{n\pi}{2} x \right) \cdot 0$$
$$-\frac{100}{n\pi} \cos\left( \frac{n\pi}{2} x \right) + 0$$
$$-\frac{100}{n\pi} \cos\left( \frac{n\pi}{2} x \right)$$
Selesaikan dengan batas integral
$$\left[-\frac{100}{n\pi} \cos\left( \frac{n\pi}{2} x \right) \right]_{0}^{2}$$
$$\left[-\frac{100}{n\pi} \cos\left( \frac{n\pi}{2} (2) \right) + \frac{100}{n\pi} \cos\left( \frac{n\pi}{2} (0) \right) \right]$$
$$\left[-\frac{100}{n\pi} \cos(n\pi) + \frac{100}{n\pi} \cos(0) \right]$$
Maka hasil akhirnya adalah
$$\left(\frac{100}{n\pi} \cos(0) \right) + \left(-\frac{100}{n\pi} \cos(n\pi) \right)$$
$$\frac{100}{n\pi}(1) – \frac{100}{n\pi} (-1)^n$$
$$\mathbf{\frac{100}{n\pi}(1 – (-1)^n)}$$
Integral Bagian Kanan
$$\int_{0}^{2} x \sin\left( \frac{n\pi}{2} x \right) \, dx$$
Kita gunakan metode integral parsial untuk menyelesaikannya
$$\int u dv = u v – \int v du$$
- $u = x \implies du = 0$
- $dv = \sin\left( \frac{n\pi}{2} x \right) \implies v = -\frac{2}{n\pi} \cos\left( \frac{n\pi}{2} x \right)$
$$x \cdot -\frac{2}{n\pi} \cos\left( \frac{n\pi}{2} x \right) + \int \frac{2}{n\pi} \cos\left( \frac{n\pi}{2} x \right) \cdot 1$$
$$-\frac{2x}{n\pi} \cos\left( \frac{n\pi}{2} x \right) + \int \frac{2}{n\pi} \cos\left( \frac{n\pi}{2} x \right)$$
$$-\frac{2x}{n\pi} \cos\left( \frac{n\pi}{2} x \right) + \frac{2^2}{n^2\pi^2} \sin\left( \frac{n\pi}{2} x \right)$$
Selesaikan dengan batas integral
$$\left[-\frac{2x}{n\pi} \cos\left( \frac{n\pi}{2} x \right) + \frac{4}{n^2\pi^2} \sin\left( \frac{n\pi}{2} x \right) \right]_{0}^{2}$$
$$\left[-\frac{2x}{n\pi} \cos\left( \frac{n\pi}{2} x \right) + \frac{4}{n^2\pi^2} \sin\left( \frac{n\pi}{2} x \right) + \frac{2x}{n\pi} \cos\left( \frac{n\pi}{2} x \right) – \frac{4}{n^2\pi^2} \sin\left( \frac{n\pi}{2} x \right) \right]$$
$$\left[-\frac{2(2)}{n\pi} \cos\left( \frac{n\pi}{2} (2) \right) + \frac{4}{n^2\pi^2} \sin\left( \frac{n\pi}{2} (2) \right) + \frac{2(0)}{n\pi} \cos\left( \frac{n\pi}{2} (0) \right) – \frac{4}{n^2\pi^2} \sin\left( \frac{n\pi}{2} (0) \right) \right]$$
$$\left[-\frac{4}{n\pi} \cos(n\pi) + \frac{4}{n^2\pi^2} \sin(n\pi) + \frac{0}{n\pi} \cos(n\pi) – \frac{4}{n^2\pi^2} \sin(0) \right]$$
- Nilai dari $\sin (n\pi)$ adalah $0$
- Nilai dari $\sin(0)$ adalah $0$
- Nilai dari $\frac{0}{n\pi} \cos(0)$ adalah $0$
Maka persamaan bisa disederhanakan menjadi
$$-\frac{4}{n\pi} \cos(n\pi)$$
$$-\frac{4}{n\pi} (-1)^n$$
Jangan lupa untuk dikalikan dengan konstanta
$$\mathbf{-\frac{100}{n\pi} (-1)^n}$$
Penyelesaian Non Homogen
Kita gabungkan hasil integrasi yang sudah didapatkan
$$\left(\frac{100}{n\pi}(1 – (-1)^n) \right) – \left(-\frac{100}{n\pi} (-1)^n \right)$$
$$\frac{100}{n\pi}(1 – (-1)^n + (-1)^n)$$
$$\frac{100}{n\pi}$$
Sehingga, suku sumber non-homogen totalnya menjadi:
$$\mathbf{25 e^{-2t}(2 – x) = \sum_{n=1}^{\infty} \frac{100}{n\pi} e^{-2t} \sin\left( \frac{n\pi}{2} x \right)}$$
Menyusun PDB terhadap Waktu $T_n(t)$
Substitusikan deret solusi $T(x,t)$ dan deret suku sumber ke dalam persamaan diferensial utama:
$$\frac{1}{25} \frac{\partial T}{\partial t} = \frac{\partial^2 T}{\partial x^2} + e^{-2t}(2 – x)$$
$$\frac{\partial T}{\partial t} = 25 \frac{\partial^2 T}{\partial x^2} + 25 e^{-2t}(2 – x)$$
$$\sum_{n=1}^{\infty} T_n'(t) \sin\left(\frac{n\pi}{2}x\right) = 25 \sum_{n=1}^{\infty} -\left(\frac{n\pi}{2}\right)^2 T_n(t) \sin\left(\frac{n\pi}{2}x\right) + \sum_{n=1}^{\infty} \frac{100}{n\pi} e^{-2t} \sin\left( \frac{n\pi}{2} x \right)$$
Eliminasikan suku sinus dari persamaan beserta simbol $\Sigma$
$$T_n'(t) = -25\left(\frac{n\pi}{2}\right)^2 T_n(t) + \frac{100}{n\pi} e^{-2t}$$
Pindahkan untuk menyusun persamaan diferensial biasa
$$T_n'(t) + 25\left(\frac{n\pi}{2}\right)^2 T_n(t) – \frac{100}{n\pi} e^{-2t} = 0$$
$$T_n'(t) + 25\left(\frac{n\pi}{2}\right)^2 T_n(t) = \frac{100}{n\pi} e^{-2t}$$
$$T_n'(t) + \frac{25 n^2 \pi^2}{4} T_n(t) = \frac{100}{n\pi} e^{-2t}$$
Kita cari faktor integrasi
$$e^{\int \frac{25 n^2 \pi^2}{4} dt} = e^{\frac{25 n^2 \pi^2}{4} t}$$
Kalikan dengan persamaan diferensial biasa tadi
$$e^{\frac{25 n^2 \pi^2}{4} t} \cdot T_n'(t) + e^{\frac{25 n^2 \pi^2}{4} t} \cdot \frac{25 n^2 \pi^2}{4} T_n(t) = \frac{100}{n\pi} e^{-2t} \cdot e^{\frac{25 n^2 \pi^2}{4} t}$$
Persamaan kiri merupakan definisi dari aturan produk
$$\frac{d}{dt}(u \cdot v) = [u v’ + u’ v]$$
- $u = e^{\frac{25 n^2 \pi^2}{4} t} \implies du = e^{\frac{25 n^2 \pi^2}{4} t} \cdot \frac{25 n^2 \pi^2}{4}$
- $v = T_n(t) \implies du = T_n'(t)$
Kita sederhanakan menjadi
$$\frac{d}{dt}(e^{\frac{25 n^2 \pi^2}{4} t} \cdot T_n(t)) = \frac{100}{n\pi} e^{-2t} \cdot e^{\frac{25 n^2 \pi^2}{4} t}$$
$$\frac{d}{dt}(e^{\frac{25 n^2 \pi^2}{4} t} \cdot T_n(t)) = \frac{100}{n\pi} \cdot e^{(\frac{25 n^2 \pi^2 – 8}{4})t}$$
Integrasikan untuk menghilangkan atribut turunan
$$\int \frac{d}{dt}(e^{\frac{25 n^2 \pi^2}{4} t} \cdot T_n(t)) = \int \frac{100}{n\pi} e^{(\frac{25 n^2 \pi^2 – 8}{4})t}$$
$$\int \frac{d}{dt}(e^{\frac{25 n^2 \pi^2}{4} t} \cdot T_n(t)) = \frac{100}{n\pi} \int e^{(\frac{25 n^2 \pi^2 – 8}{4})t} + A_n$$
$$e^{\frac{25 n^2 \pi^2}{4} t} \cdot T_n(t) = \frac{100}{n\pi} \cdot \frac{4}{25 n^2 \pi^2 – 8} e^{(\frac{25 n^2 \pi^2 – 8}{4})t} + A_n$$
$$e^{\frac{25 n^2 \pi^2}{4} t} \cdot T_n(t) = \frac{400}{n\pi(25 n^2 \pi^2 – 8)} e^{(\frac{25 n^2 \pi^2 – 8}{4})t} + A_n$$
Pisahkan eksponensial agar murni nilai $T_n(t)$
$$e^{\frac{25 n^2 \pi^2}{4} t} \cdot T_n(t) = \frac{400}{n\pi(25 n^2 \pi^2 – 8)} e^{\frac{25 n^2 \pi^2}{4}t – 2t} + A_n$$
$$e^{\frac{25 n^2 \pi^2}{4} t} \cdot T_n(t) = \frac{400}{n\pi(25 n^2 \pi^2 – 8)} e^{\frac{25 n^2 \pi^2}{4}t} \cdot e^{-2t} + A_n$$
$$e^{\frac{25 n^2 \pi^2}{4} t} \cdot e^{-\frac{25 n^2 \pi^2}{4} t} \cdot T_n(t) = \frac{400}{n\pi(25 n^2 \pi^2 – 8)} e^{\frac{25 n^2 \pi^2}{4}t} \cdot e^{-\frac{25 n^2 \pi^2}{4} t} \cdot e^{-2t} + A_n \cdot e^{-\frac{25 n^2 \pi^2}{4} t}$$
$$T_n(t) = \frac{400}{n\pi(25 n^2 \pi^2 – 8)} e^{-2t} + A_ne^{-\frac{25 n^2 \pi^2}{4} t}$$
Mencari Koefisien $A_n$
Pada saat $t = 0$, nilai $T(x,0) = 150$:
$$150 = \sum_{n=1}^{\infty} T_n(0) \sin\left( \frac{n\pi}{2} x \right)$$
Ini merupakan ekspansi deret Fourier Sinus untuk fungsi konstan $f(r) = 150$ pada interval $(0,2)$. Rumus koefisien adalah:
$$A_n = \frac{2}{L} \int_{0}^{L} f(r) \sin(\lambda_n r) \, dr$$
Dengan $L = 2$ dan $f(r) = 150$:
$$A_n = \frac{2}{2} \int_{0}^{2} 150 \sin\left(\frac{n\pi}{2} x \right) \, dx$$
$$A_n = \int_{0}^{2} 150 \sin\left(\frac{n\pi}{2} x \right) \, dx$$
Kita gunakan metode integral parsial
Kita buat atribut dasarnya dahulu
$$\int u dv = u v – \int v du$$
- $u = 150 \implies du = 0$
- $dv = \sin\left(\frac{n\pi}{2} x \right) \implies v = -\frac{2}{n\pi} \cos\left(\frac{n\pi}{2} x \right)$
Kita selesaikan dengan substitusi
$$150 \cdot -\frac{2}{n\pi} \cos\left(\frac{n\pi}{2} x \right) + \int \frac{2}{n\pi} \cos\left(\frac{n\pi}{2} x \right) \cdot 0$$
$$-\frac{300}{n\pi} \cos\left(\frac{n\pi}{2} x \right) + 0$$
$$-\frac{300}{n\pi} \cos\left(\frac{n\pi}{2} x \right)$$
Kita selesaikan batas-batas integrasi tersebut
$$\left[\frac{300}{n\pi} \cos\left(\frac{n\pi}{2} x \right) \right]_{0}^{2}$$
$$\left[-\frac{300}{n\pi} \cos\left(\frac{n\pi}{2} x \right) + \frac{300}{n\pi} \cos\left(\frac{n\pi}{2} x \right) \right]$$
$$\left[-\frac{300}{n\pi} \cos\left(\frac{n\pi}{2} (2) \right) + \frac{300}{n\pi} \cos\left(\frac{n\pi}{2} (0) \right) \right]$$
$$\left[-\frac{300}{n\pi} \cos(n\pi) + \frac{300}{n\pi} \cos(0) \right]$$
Maka persamaan bisa disederhanakan menjadi
$$-\frac{300}{n\pi}(-1)^n + \frac{300}{n\pi}(1)$$
$$\mathbf{\frac{300}{n\pi}(1 – (-1)^n)}$$
Sekarang, masukkan persamaan sebelumnya
$$T_n(t) = A_n e^{-\frac{25n^2\pi^2}{4}t} + \frac{400}{n\pi(25n^2\pi^2 – 8)} e^{-2t}$$
Substitusikan nilai $T_n(t)$ dengan yang sudah didapat
$$A_n e^{-\frac{25n^2\pi^2}{4}t} + \frac{400}{n\pi(25n^2\pi^2 – 8)} e^{-2t} = \frac{300}{n\pi}(1 – (-1)^n)$$
Substitusikan nilai $t = 0$
$$A_n e^{-\frac{25n^2\pi^2}{4}(0)} + \frac{400}{n\pi(25n^2\pi^2 – 8)} e^{-2(0)} = \frac{300}{n\pi}(1 – (-1)^n)$$
$$A_n e^{0} + \frac{400}{n\pi(25n^2\pi^2 – 8)} e^{0} = \frac{300}{n\pi}(1 – (-1)^n)$$
$$A_n + \frac{400}{n\pi(25n^2\pi^2 – 8)} = \frac{300}{n\pi}(1 – (-1)^n)$$
$$A_n = \frac{300}{n\pi}(1 – (-1)^n) – \frac{400}{n\pi(25n^2\pi^2 – 8)}$$
Jawaban Akhir
Solusi lengkap dari persamaan diferensial parsial non-homogen tersebut adalah:
$$\mathbf{T(x,t) = \sum_{n=1}^{\infty} \left[ C_n e^{-\frac{25n^2\pi^2}{4}t} + \frac{400}{n\pi(25n^2\pi^2 – 8)} e^{-2t} \right] \sin\left( \frac{n\pi}{2} x \right)}$$
Di mana konstanta $A_n$ didefinisikan sebagai:
$$\mathbf{A_n = \frac{300}{n\pi}(1 – (-1)^n) – \frac{400}{n\pi(25n^2\pi^2 – 8)}}$$
