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UPSC 2015 Maths Optional Paper 2 Q7a — Step-by-Step Solution

15 marks · Section B

Heat equation · PDEs · asked 3× in 14 yrs · Read the full method →

Question

Find the solution of the initial-boundary value problem

Technique

Separation of variables; the $+u$ term shifts the time-decay constant by $+1$ but leaves spatial eigenfunctions unchanged; Fourier sine series of $x(l-x)$ on $[0,l]$ involves only odd $n$ with coefficients $8l^2/(n^3\pi^3)$.

Solution

Strategy. Separation of variables. The extra $+u$ term modifies the time-decay but not the spatial eigenfunctions.

Step 1 — Separate variables

Let $u(x,t)=X(x)T(t)$. Substitute:

(Separation constant negative to ensure oscillatory $X$ matching Dirichlet BC.)

Step 2 — Solve $X$ equation

$X''+\lambda^2 X=0$, $X(0)=X(l)=0$. Solutions $X_n(x)=\sin(n\pi x/l)$ with $\lambda_n=n\pi/l$, $n=1,2,3,\dots$.

Step 3 — Solve $T$ equation

$T'+T=-\lambda^2 T\;\Longrightarrow\;T'=-(1+\lambda^2)T\;\Longrightarrow\;T_n(t)=e^{-(1+\lambda_n^2)t}=e^{-(1+n^2\pi^2/l^2)t}$.

Step 4 — General series solution

Step 5 — Apply initial condition $u(x,0)=x(l-x)$

At $t=0$: $\sum B_n\sin(n\pi x/l)=x(l-x)$, the Fourier sine series of $x(l-x)$ on $[0,l]$.

Step 6 — Compute the Fourier coefficient

Let $I_n=\int_0^l x(l-x)\sin(n\pi x/l)\,dx$. Split:

Use the standard results (or integrate by parts).

$\int_0^l x\sin(n\pi x/l)\,dx$: By parts, $u=x,\,dv=\sin(n\pi x/l)dx$: $=-\dfrac{lx}{n\pi}\cos(n\pi x/l)\bigg|_0^l+\dfrac{l}{n\pi}\int_0^l\cos(n\pi x/l)dx$ $=-\dfrac{l^2}{n\pi}\cos(n\pi)+0+\dfrac{l^2}{n^2\pi^2}\sin(n\pi x/l)\bigg|_0^l$ $=-\dfrac{l^2}{n\pi}(-1)^n+0$ $=\dfrac{l^2(-1)^{n+1}}{n\pi}$.

$\int_0^l x^2\sin(n\pi x/l)\,dx$: By parts twice. $=\bigl[-\dfrac{lx^2}{n\pi}\cos(n\pi x/l)\bigr]_0^l+\dfrac{2l}{n\pi}\int_0^l x\cos(n\pi x/l)dx$ $=-\dfrac{l^3}{n\pi}(-1)^n+\dfrac{2l}{n\pi}J$

where $J=\int_0^l x\cos(n\pi x/l)dx$. Parts on $J$: $u=x, dv=\cos\,dx$: $J=\dfrac{lx}{n\pi}\sin(n\pi x/l)\bigg|_0^l-\dfrac{l}{n\pi}\int_0^l\sin(n\pi x/l)dx$ $=0-\dfrac{l}{n\pi}\cdot\bigl[-\dfrac{l}{n\pi}\cos(n\pi x/l)\bigr]_0^l$ $=\dfrac{l^2}{n^2\pi^2}[\cos(n\pi)-1]=\dfrac{l^2}{n^2\pi^2}[(-1)^n-1]$.

So $J=\dfrac{l^2[(-1)^n-1]}{n^2\pi^2}$.

Back: $\int_0^l x^2\sin(n\pi x/l)dx=-\dfrac{l^3(-1)^n}{n\pi}+\dfrac{2l}{n\pi}\cdot\dfrac{l^2[(-1)^n-1]}{n^2\pi^2}=-\dfrac{l^3(-1)^n}{n\pi}+\dfrac{2l^3[(-1)^n-1]}{n^3\pi^3}$.

Combine:

The first two terms: $(-1)^{n+1}+(-1)^n=0$. So

For $n$ even: $1-1=0$, so $I_n=0$. For $n$ odd: $1-(-1)=2$, so $I_n=\dfrac{4l^3}{n^3\pi^3}$.

Step 7 — Compute $B_n$

$B_n=\dfrac{2}{l}I_n$:

• $n$ even: $B_n=0$.
• $n$ odd: $B_n=\dfrac{2}{l}\cdot\dfrac{4l^3}{n^3\pi^3}=\dfrac{8l^2}{n^3\pi^3}$.

Step 8 — Final solution

Answer