Solving heat equation — separation of variables

WHAT are we solving?

HOW: the derivation from scratch

Step 1 — The separation guess. Assume $u(x,t)=X(x)\,T(t)$ (space-part times time-part). Why this step? If it works, derivatives split nicely; we lose nothing by trying — if no solution of this form exists, we'll find out.

Then $u_t = X T'$ and $u_{xx}=X'' T$. Substitute: $X T' = \alpha^2 X'' T.$

Step 2 — Separate the variables. Divide by $\alpha^2 X T$: $\frac{T'}{\alpha^2 T} = \frac{X''}{X}.$ Why this step? Now the left side depends only on $t$, the right side only on $x$. The only way a function of $t$ equals a function of $x$ for all $x,t$ is if both equal the same constant. Call it $-\lambda$. Why the minus sign? Foresight: we want decaying-in-time solutions, which needs $-\lambda$ with $\lambda>0$. We'll verify the other sign cases fail.

$\frac{X''}{X} = -\lambda,\qquad \frac{T'}{\alpha^2 T}=-\lambda.$

Step 3 — Solve the space ODE (an eigenvalue problem). $X'' + \lambda X = 0,\qquad X(0)=0,\ X(L)=0.$ Why these BCs? Since $u(0,t)=X(0)T(t)=0$ for all $t$, and $T\not\equiv0$ (else trivial), we need $X(0)=0$. Same at $L$.

Test the sign of $\lambda$ (Forecast-then-Verify):

• $\lambda<0$, say $\lambda=-\mu^2$: $X=A e^{\mu x}+Be^{-\mu x}$. BCs force $A=B=0$. Only trivial. ✗
• $\lambda=0$: $X=A+Bx$. $X(0)=0\Rightarrow A=0$; $X(L)=0\Rightarrow B=0$. Trivial. ✗
• $\lambda>0$, say $\lambda=k^2$: $X=A\cos kx + B\sin kx$. $X(0)=0\Rightarrow A=0$, so $X=B\sin kx$. $X(L)=0\Rightarrow \sin kL=0\Rightarrow kL=n\pi$.

So the eigenvalues and eigenfunctions are $\lambda_n=\left(\frac{n\pi}{L}\right)^2,\qquad X_n(x)=\sin\frac{n\pi x}{L},\quad n=1,2,3,\dots$

Step 4 — Solve the time ODE. $T' = -\lambda\alpha^2 T \ \Rightarrow\ T_n(t)=e^{-\lambda_n \alpha^2 t}=e^{-\alpha^2 (n\pi/L)^2 t}.$ Why this step? This is first-order linear; its solution is a pure exponential decay — exactly the "fading knob."

Step 5 — Build product solutions and superpose. Each $u_n=\sin\frac{n\pi x}{L}\,e^{-\alpha^2(n\pi/L)^2 t}$ solves the PDE + BCs. The equation is linear and homogeneous, so sums of solutions are solutions: $\boxed{\,u(x,t)=\sum_{n=1}^{\infty} B_n \sin\frac{n\pi x}{L}\;e^{-\alpha^2 (n\pi/L)^2 t}\,}$

Step 6 — Fit the initial condition with Fourier coefficients. At $t=0$ the exponentials are $1$: $f(x)=\sum_{n=1}^\infty B_n \sin\frac{n\pi x}{L}.$ This is a Fourier sine series. Multiply by $\sin\frac{m\pi x}{L}$, integrate $0$ to $L$, use orthogonality $\int_0^L \sin\frac{n\pi x}{L}\sin\frac{m\pi x}{L}\,dx=\frac{L}{2}\delta_{nm}$: $\boxed{\,B_n=\frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx\,}$

Worked examples

Common mistakes

Flashcards

Connections

• Fourier Series — coefficients $B_n$ are exactly Fourier sine coefficients.
• Sturm-Liouville Theory — $X''+\lambda X=0$ is a Sturm–Liouville eigenvalue problem; orthogonality is guaranteed.
• Wave Equation — same separation method, but time ODE is oscillatory ($T''=-\lambda c^2 T$) not decaying.
• Laplace Equation — steady-state heat ($u_t=0$) gives $u_{xx}=0$.
• Boundary Conditions — Dirichlet vs Neumann — Neumann (insulated) ends give cosine series instead.
• Superposition Principle — why summing product solutions is legal (linear homogeneous PDE).

Concept Map

Hinglish (regional understanding)