UPSC Mathematics Optional · Paper II · Classification to Fourier Methods
\(p=\dfrac{\partial z}{\partial x}\), \(q=\dfrac{\partial z}{\partial y}\), \(r=\dfrac{\partial^2 z}{\partial x^2}\), \(s=\dfrac{\partial^2 z}{\partial x\partial y}\), \(t=\dfrac{\partial^2 z}{\partial y^2}\).
Formation: eliminate arbitrary constants/functions by differentiation.
\(Pp+Qq=R\) where \(P,Q,R\) are functions of \(x,y,z\).
Lagrange's auxiliary (subsidiary) equations: \(\dfrac{dx}{P}=\dfrac{dy}{Q}=\dfrac{dz}{R}\)
Solve for two independent integrals \(u(x,y,z)=a\) and \(v(x,y,z)=b\). General solution: \(\phi(u,v)=0\) or \(u=f(v)\).
Tricks for auxiliary equations: use multipliers \((l,m,n)\) s.t. \(lP+mQ+nR=0\) → \(l\,dx+m\,dy+n\,dz=0\).
For \(F(x,y,z,p,q)=0\). Charpit's auxiliary equations:
\[\frac{dp}{F_x+pF_z}=\frac{dq}{F_y+qF_z}=\frac{dz}{-pF_p-qF_q}=\frac{dx}{-F_p}=\frac{dy}{-F_q}\]
Find one integral involving \(p\) or \(q\) → substitute back → integrate for complete integral.
Discriminant \(\Delta=B^2-4AC\):
Canonical forms obtained by transforming to characteristic coordinates.
CF of \(f(D,D')z=0\) where \(D=\partial/\partial x\), \(D'=\partial/\partial y\).
Auxiliary: \(f(m,1)=0\) gives \(m_1,m_2,\ldots\) (put \(D\to m\), \(D'\to 1\)).
PI: \(\dfrac{1}{f(D,D')}e^{ax+by}=\dfrac{e^{ax+by}}{f(a,b)}\) (if \(f(a,b)\ne 0\)).
\(\dfrac{1}{f(D,D')}\sin(ax+by)=\dfrac{\sin(ax+by)}{f(-a^2,-ab,-b^2)}\) (replace \(D^2\to-a^2\), \(D D'\to-ab\), \(D'^2\to-b^2\)).
D'Alembert's solution: \(u(x,t)=f(x+ct)+g(x-ct)\).
With \(u(x,0)=\phi(x)\), \(u_t(x,0)=\psi(x)\):
\[u(x,t)=\frac{\phi(x+ct)+\phi(x-ct)}{2}+\frac{1}{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds\]
Separation of variables (finite string, \(0\le x\le L\)):
\(u(x,t)=\sum_{n=1}^\infty\sin\dfrac{n\pi x}{L}\left(A_n\cos\dfrac{n\pi ct}{L}+B_n\sin\dfrac{n\pi ct}{L}\right)\)
where \(A_n,B_n\) from Fourier series of \(\phi,\psi\).
Separation of variables: \(u(x,t)=X(x)T(t)\) → \(X''/X=T'/(kT)=-\lambda^2\).
With \(u(0,t)=u(L,t)=0\), \(u(x,0)=f(x)\):
\[u(x,t)=\sum_{n=1}^\infty B_n\sin\frac{n\pi x}{L}\cdot e^{-k(n\pi/L)^2 t}\]
\(B_n=\dfrac{2}{L}\int_0^L f(x)\sin\dfrac{n\pi x}{L}\,dx\).
Steady state: As \(t\to\infty\), transient terms die out; only constant remains if non-zero BCs.
Elliptic. Solutions = harmonic functions.
Rectangular domain: \(u(x,y)=\sum(A_n\sinh\dfrac{n\pi y}{L}+B_n\cosh\dfrac{n\pi y}{L})\sin\dfrac{n\pi x}{L}+\cdots\)
Polar coordinates: \(u=\sum(A_n r^n+B_n r^{-n})(C_n\cos n\theta+D_n\sin n\theta)\).
\[f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty\left(a_n\cos\frac{n\pi x}{L}+b_n\sin\frac{n\pi x}{L}\right)\]
\[a_0=\frac{1}{L}\int_{-L}^L f(x)\,dx,\quad a_n=\frac{1}{L}\int_{-L}^L f(x)\cos\frac{n\pi x}{L}\,dx,\quad b_n=\frac{1}{L}\int_{-L}^L f(x)\sin\frac{n\pi x}{L}\,dx\]
Dirichlet conditions: \(f\) piecewise smooth → Fourier series converges to \(f(x)\) at continuity points, to \(\frac{f(x^+)+f(x^-)}{2}\) at jumps.
Half-range expansions: Even → cosine series; Odd → sine series.
Parseval's identity: \(\dfrac{1}{L}\int_{-L}^L[f(x)]^2\,dx=\dfrac{a_0^2}{2}+\sum_{n=1}^\infty(a_n^2+b_n^2)\)
\[\hat f(\xi)=\int_{-\infty}^\infty f(x)e^{-2\pi i\xi x}\,dx, \qquad f(x)=\int_{-\infty}^\infty\hat f(\xi)e^{2\pi i\xi x}\,d\xi\]
(Convention varies: UPSC typically uses \(e^{i\omega x}\) form with \(\frac{1}{2\pi}\) factor.)
Key properties: Linearity, shifting, scaling, differentiation (\(\mathcal{F}[f']=i\omega\mathcal{F}[f]\)), convolution (\(\mathcal{F}[f*g]=\mathcal{F}[f]\cdot\mathcal{F}[g]\)).
Parseval: \(\int|f|^2\,dx=\int|\hat f|^2\,d\xi\).
Fourier sine/cosine transforms: used when \(f\) defined on \([0,\infty)\).