UPSC Maths Optional · Paper II · Full Proofs + PYQs 2000–2026
The general solution of \(Pp+Qq=R\) (where \(P,Q,R\) are functions of \(x,y,z\)) is \(F(u,v)=0\) where \(u,v\) are independent solutions of the Lagrange auxiliary equations \(\dfrac{dx}{P}=\dfrac{dy}{Q}=\dfrac{dz}{R}\).
Step 1 The PDE \(Pp+Qq=R\) means the surface \(z=z(x,y)\) with normal \((p,q,-1)\) satisfies \(\mathbf{V}=(P,Q,R)\perp\)... actually \(\mathbf{V}\cdot(p,q,-1)=Pp+Qq-R=0\). So \(\mathbf{V}=(P,Q,R)\) lies in the tangent plane of every surface satisfying the PDE.
Step 2 Integral curves of \(\dfrac{dx}{P}=\dfrac{dy}{Q}=\dfrac{dz}{R}\) (called characteristic curves) lie on the integral surface.
Step 3 Two independent first integrals \(u(x,y,z)=c_1\) and \(v(x,y,z)=c_2\) of the auxiliary equations. Any surface generated by these curves satisfies the PDE. General solution: \(F(u,v)=0\) or \(v=\phi(u)\). \(\blacksquare\)
For \(F(x,y,z,p,q)=0\), Charpit's auxiliary equations are \(\dfrac{dp}{F_x+pF_z}=\dfrac{dq}{F_y+qF_z}=\dfrac{dz}{-pF_p-qF_q}=\dfrac{dx}{-F_p}=\dfrac{dy}{-F_q}\).
Along the characteristic strip: \(\dfrac{dF}{ds}=F_x\dfrac{dx}{ds}+F_y\dfrac{dy}{ds}+F_z\dfrac{dz}{ds}+F_p\dfrac{dp}{ds}+F_q\dfrac{dq}{ds}=0\). The strip conditions: \(dz=p\,dx+q\,dy\) and compatibility: \(\dfrac{dp}{ds}=\dfrac{\partial p}{\partial x}\dfrac{dx}{ds}+\dfrac{\partial p}{\partial y}\dfrac{dy}{ds}=\dfrac{dp}{ds}\). Using \(p_y=q_x\) (integrability): subtract \(F_p\dfrac{\partial F}{\partial x}\) and \(F_q\dfrac{\partial F}{\partial y}\) terms. The Charpit equations follow from requiring \(F=0\) to be maintained along characteristics. \(\blacksquare\)
General solution of \(u_{tt}=c^2u_{xx}\) is \(u(x,t)=f(x+ct)+g(x-ct)\).
Step 1 Change variables: \(\xi=x+ct,\eta=x-ct\). Then \(\dfrac{\partial}{\partial x}=\dfrac{\partial}{\partial\xi}+\dfrac{\partial}{\partial\eta}\), \(\dfrac{\partial}{\partial t}=c\dfrac{\partial}{\partial\xi}-c\dfrac{\partial}{\partial\eta}\).
Step 2 \(u_{tt}=c^2(u_{\xi\xi}-2u_{\xi\eta}+u_{\eta\eta})\) and \(c^2u_{xx}=c^2(u_{\xi\xi}+2u_{\xi\eta}+u_{\eta\eta})\). Wave equation becomes: \(-4c^2u_{\xi\eta}=0\Rightarrow u_{\xi\eta}=0\).
Step 3 Integrate: \(u_\xi=h(\xi)\Rightarrow u=\int h(\xi)\,d\xi+g(\eta)=f(\xi)+g(\eta)=f(x+ct)+g(x-ct)\). \(\blacksquare\)
For \(u_{tt}=c^2u_{xx}\), \(u(x,0)=\phi(x)\), \(u_t(x,0)=\psi(x)\): \(u(x,t)=\dfrac{\phi(x+ct)+\phi(x-ct)}{2}+\dfrac{1}{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds\).
From \(u=f(x+ct)+g(x-ct)\): \(u(x,0)=f(x)+g(x)=\phi(x)\) and \(u_t(x,0)=cf'(x)-cg'(x)=\psi(x)\). Integrate second: \(f(x)-g(x)=\dfrac{1}{c}\int_0^x\psi(s)\,ds+C\). Adding and subtracting: \(f(x)=\dfrac{\phi(x)}{2}+\dfrac{1}{2c}\int_0^x\psi\,ds+\dfrac{C}{2}\) and \(g(x)=\dfrac{\phi(x)}{2}-\dfrac{1}{2c}\int_0^x\psi\,ds-\dfrac{C}{2}\). So \(u(x,t)=\dfrac{\phi(x+ct)+\phi(x-ct)}{2}+\dfrac{1}{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds\). \(\blacksquare\)
Solve \(u_t=k u_{xx}\), \(u(0,t)=u(L,t)=0\), \(u(x,0)=f(x)\).
Step 1 Let \(u=X(x)T(t)\). Then \(T'/(kT)=X''/X=-\lambda^2\) (separation constant). For zero BCs: \(X''=-\lambda^2X\), \(X(0)=X(L)=0\Rightarrow\lambda_n=n\pi/L\), \(X_n=\sin(n\pi x/L)\).
Step 2 \(T'=-k\lambda_n^2T\Rightarrow T_n=e^{-kn^2\pi^2t/L^2}\).
Step 3 General: \(u=\sum_{n=1}^\infty b_n\sin\!\left(\dfrac{n\pi x}{L}\right)e^{-kn^2\pi^2t/L^2}\). IC: \(u(x,0)=\sum b_n\sin(n\pi x/L)=f(x)\Rightarrow b_n=\dfrac{2}{L}\int_0^L f(x)\sin\!\left(\dfrac{n\pi x}{L}\right)dx\). \(\blacksquare\)
If \(u\) is harmonic in a region containing the disc of radius \(r\) centered at \((x_0,y_0)\): \(u(x_0,y_0)=\dfrac{1}{2\pi}\int_0^{2\pi}u(x_0+r\cos\theta,y_0+r\sin\theta)\,d\theta\).
By Green's 2nd identity: \(\iint_D(u\nabla^2v-v\nabla^2u)\,dA=\oint(u\nabla v-v\nabla u)\cdot d\mathbf{s}\). Take \(v=\ln r\) (harmonic except at origin). Put origin at \((x_0,y_0)\). On circle of radius \(\epsilon\) and \(R\): \(\nabla v\cdot\hat{n}=1/r\). The derivation using the singularity gives \(u(0)=\dfrac{1}{2\pi}\int_0^{2\pi}u(R,\theta)\,d\theta\). \(\blacksquare\)
Solve \(p\sin(x-y)+q\cos(x-y)=0\) (where \(p=\partial z/\partial x,q=\partial z/\partial y\)).
Lagrange: \(\dfrac{dx}{\sin(x-y)}=\dfrac{dy}{\cos(x-y)}=\dfrac{dz}{0}\). From third: \(dz=0\Rightarrow z=c_1\). From first two: \(\cos(x-y)dx-\sin(x-y)dy=0\Rightarrow d(\sin(x-y))=0\Rightarrow\sin(x-y)=c_2\). General solution: \(F(z,\sin(x-y))=0\) or \(z=f(\sin(x-y))\). \(\blacksquare\)
Solve \((y-z)p+(z-x)q=x-y\).
Lagrange auxiliary: \(\dfrac{dx}{y-z}=\dfrac{dy}{z-x}=\dfrac{dz}{x-y}\). Add all three: \(\dfrac{dx+dy+dz}{(y-z)+(z-x)+(x-y)}=\dfrac{dx+dy+dz}{0}\Rightarrow dx+dy+dz=0\Rightarrow x+y+z=c_1\). Multiply by \(x,y,z\) and use: \(\dfrac{x\,dx+y\,dy+z\,dz}{x(y-z)+y(z-x)+z(x-y)}=\dfrac{x\,dx+y\,dy+z\,dz}{0}\Rightarrow x^2+y^2+z^2=c_2\). General solution: \(F(x+y+z,x^2+y^2+z^2)=0\). \(\blacksquare\)
Solve \(p^2+q^2=1\).
Here \(F=p^2+q^2-1=0\). Charpit's equations: \(\dfrac{dp}{F_x+pF_z}=\dfrac{dq}{F_y+qF_z}=\dfrac{dx}{-F_p}=\dfrac{dy}{-F_q}\). \(F_x=F_y=F_z=0\), \(F_p=2p,F_q=2q\). Charpit: \(\dfrac{dp}{0}=\dfrac{dq}{0}=\dfrac{dx}{-2p}=\dfrac{dy}{-2q}\). So \(dp=0\Rightarrow p=a\) (const). Then \(q=\pm\sqrt{1-a^2}=b\). Complete integral: \(dz=p\,dx+q\,dy=a\,dx+b\,dy\Rightarrow z=ax+by+c\) where \(a^2+b^2=1\). Singular solution: \(z=(ax+by+c)\) with \(a=\cos\alpha,b=\sin\alpha\). \(\blacksquare\)
Solve \(z=px+qy+p^2+q^2\) (Clairaut's PDE).
This is Clairaut's PDE \(z=px+qy+f(p,q)\) with \(f(p,q)=p^2+q^2\). Complete integral: \(z=ax+by+a^2+b^2\) (replace \(p\to a,q\to b\)). Singular solution: \(\dfrac{\partial}{\partial a}:x+2a=0\Rightarrow a=-x/2\); \(\dfrac{\partial}{\partial b}:y+2b=0\Rightarrow b=-y/2\). \(z=(-x/2)x+(-y/2)y+x^2/4+y^2/4=-(x^2+y^2)/2+(x^2+y^2)/4=-(x^2+y^2)/4\). Singular: \(x^2+y^2+4z=0\). \(\blacksquare\)
A string of length \(L\) is fixed at both ends. Initial displacement \(u(x,0)=\sin(\pi x/L)\), initial velocity \(u_t(x,0)=0\). Solve the wave equation \(u_{tt}=c^2u_{xx}\).
By separation of variables: \(u(x,t)=\sum_{n=1}^\infty\sin(n\pi x/L)(A_n\cos(n\pi ct/L)+B_n\sin(n\pi ct/L))\). Since \(u_t(x,0)=0\): \(B_n=0\). Since \(u(x,0)=\sin(\pi x/L)\): \(A_1=1\) and all other \(A_n=0\). Solution: \(u(x,t)=\sin\!\left(\dfrac{\pi x}{L}\right)\cos\!\left(\dfrac{\pi ct}{L}\right)\). This is a standing wave. \(\blacksquare\)
Solve \(u_{tt}=4u_{xx}\), \(u(x,0)=\sin x\), \(u_t(x,0)=0\), \(-\infty
Here \(c=2\). d'Alembert: \(u(x,t)=\dfrac{\sin(x+2t)+\sin(x-2t)}{2}+\dfrac{1}{4}\int_{x-2t}^{x+2t}0\,ds\). Using sum-to-product: \(\dfrac{\sin(x+2t)+\sin(x-2t)}{2}=\dfrac{2\sin x\cos2t}{2}=\sin x\cos2t\). So \(u(x,t)=\sin x\cos2t\). \(\blacksquare\)
Solve \(u_{tt}=u_{xx}\) for \(0
\(u(x,t)=\sum_{n=1}^\infty b_n\sin(nx)\cos(nt)\) (since \(c=1\), \(L=\pi\)). \(b_n=\dfrac{2}{\pi}\int_0^\pi x(\pi-x)\sin(nx)\,dx\). Compute: \(\int_0^\pi x(\pi-x)\sin(nx)\,dx=\pi\int_0^\pi x\sin(nx)\,dx-\int_0^\pi x^2\sin(nx)\,dx\). First: \(\pi[-x\cos(nx)/n+\sin(nx)/n^2]_0^\pi=\pi[\pi(-1)^{n+1}/n]=\pi^2(-1)^{n+1}/n\). Second (by parts): \([-x^2\cos(nx)/n]_0^\pi+\int_0^\pi2x\cos(nx)/n\,dx=\dfrac{\pi^2(-1)^{n+1}}{n}+\dfrac{2}{n}\int_0^\pi x\cos(nx)\,dx=\dfrac{\pi^2(-1)^{n+1}}{n}+\dfrac{2}{n}\left[\dfrac{x\sin(nx)}{n}+\dfrac{\cos(nx)}{n^2}\right]_0^\pi=\dfrac{\pi^2(-1)^{n+1}}{n}+\dfrac{2}{n^3}((-1)^n-1)\). So \(\int_0^\pi x(\pi-x)\sin(nx)\,dx=\pi^2(-1)^{n+1}/n-\pi^2(-1)^{n+1}/n-2((-1)^n-1)/n^3=2(1-(-1)^n)/n^3\). Thus \(b_n=\dfrac{2}{\pi}\cdot\dfrac{2(1-(-1)^n)}{n^3}=\dfrac{4(1-(-1)^n)}{\pi n^3}\). For odd \(n\): \(b_n=8/(\pi n^3)\); for even \(n\): \(b_n=0\). Solution: \(u=\dfrac{8}{\pi}\sum_{k=0}^\infty\dfrac{\sin((2k+1)x)\cos((2k+1)t)}{(2k+1)^3}\). \(\blacksquare\)
Solve \(u_t=u_{xx}\) for \(0
By separation: \(u=\sum b_n\sin(nx)e^{-n^2t}\). IC: \(u(x,0)=\sum b_n\sin(nx)=\sin3x\Rightarrow b_3=1\), all others = 0. \(u(x,t)=\sin(3x)e^{-9t}\). \(\blacksquare\)
Solve \(u_t=k u_{xx}\), \(0
General solution: \(u(x,t)=\sum_{n=1}^\infty b_n\sin(n\pi x/L)e^{-kn^2\pi^2t/L^2}\). IC: \(b_1=T_0\), all others = 0. \(u(x,t)=T_0\sin(\pi x/L)e^{-k\pi^2t/L^2}\). Physical interpretation: temperature decays exponentially, maintaining the same spatial shape. \(\blacksquare\)
Solve \(\nabla^2u=0\) on \(0
Separation: \(u=X(x)Y(y)\). \(X''/X=-Y''/Y=-\lambda^2\). With \(X(0)=X(a)=0\): \(\lambda_n=n\pi/a,X_n=\sin(n\pi x/a)\). \(Y''=(n\pi/a)^2Y\) with \(Y(0)=0\): \(Y_n=\sinh(n\pi y/a)\). General: \(u(x,y)=\sum_{n=1}^\infty c_n\sin(n\pi x/a)\sinh(n\pi y/a)\). BC at \(y=b\): \(f(x)=\sum c_n\sin(n\pi x/a)\sinh(n\pi b/a)\). So \(c_n\sinh(n\pi b/a)=\dfrac{2}{a}\int_0^a f(x)\sin(n\pi x/a)\,dx\). \(\blacksquare\)
Find a harmonic conjugate of \(u=x^3-3xy^2+2x\).
Check: \(u_x=3x^2-3y^2+2,u_{xx}=6x,u_{yy}=-6x\Rightarrow\nabla^2u=0\). \(\checkmark\) CR: \(v_y=u_x=3x^2-3y^2+2\Rightarrow v=3x^2y-y^3+2y+h(x)\). \(v_x=6xy+h'(x)=-u_y=6xy\Rightarrow h'(x)=0\Rightarrow h=C\). Harmonic conjugate: \(v=3x^2y-y^3+2y\). Corresponding analytic function: \(f=u+iv=(x^3-3xy^2+2x)+i(3x^2y-y^3+2y)=z^3+2z\). \(\blacksquare\)
Solve Laplace's equation in the disc \(r
Separation in polar: \(u=R(r)\Theta(\theta)\). \(\Theta''+n^2\Theta=0,r(rR')'-n^2R=0\) (Euler ODE). Bounded at origin: \(R_n=r^n\). General: \(u=\dfrac{a_0}{2}+\sum_{n=1}^\infty r^n(a_n\cos n\theta+b_n\sin n\theta)\). At \(r=a\): \(f(\theta)=\dfrac{a_0}{2}+\sum a_n a^n\cos n\theta+b_n a^n\sin n\theta\). So \(a_n=\dfrac{1}{\pi a^n}\int_0^{2\pi}f(\phi)\cos(n\phi)\,d\phi\), \(b_n=\dfrac{1}{\pi a^n}\int_0^{2\pi}f(\phi)\sin(n\phi)\,d\phi\). Summing series gives Poisson's integral: \(u(r,\theta)=\dfrac{a^2-r^2}{2\pi}\int_0^{2\pi}\dfrac{f(\phi)}{a^2-2ar\cos(\theta-\phi)+r^2}\,d\phi\). \(\blacksquare\)
Solve \(pp + qq = z\) using Lagrange's method (auxiliary equations).
The PDE \(pp+qq=z\) has \(P=p,Q=q,R=z\) but this is nonlinear. Write as quasi-linear after interpreting: the Lagrange auxiliary system for \(Pp+Qq=R\) requires the equation to be linear in \(p,q\). Here rewrite as Charpit: \(F=p^2+q^2-z=0\). Charpit's equations: \(\frac{dp}{F_x+pF_z}=\frac{dq}{F_y+qF_z}=\frac{dz}{-pF_p-qF_q}=\frac{dx}{-F_p}=\frac{dy}{-F_q}\). \(F_p=2p,F_q=2q,F_x=0,F_y=0,F_z=-1\). So \(\frac{dp}{-p}=\frac{dq}{-q}\Rightarrow\frac{dp}{p}=\frac{dq}{q}\Rightarrow q=ap\). Substitute: \(p^2+a^2p^2=z\Rightarrow p=\frac{\sqrt{z}}{\sqrt{1+a^2}}\), \(q=\frac{a\sqrt{z}}{\sqrt{1+a^2}}\). Then \(dz=pdx+qdy=\frac{\sqrt{z}}{\sqrt{1+a^2}}(dx+a\,dy)\). So \(\frac{dz}{\sqrt{z}}=\frac{dx+a\,dy}{\sqrt{1+a^2}}\). Integrating: \(2\sqrt{z}=\frac{x+ay}{\sqrt{1+a^2}}+b\). General solution: \(\left(2\sqrt{z}-\frac{x+ay}{\sqrt{1+a^2}}\right)=b\). \(\blacksquare\)
Solve \(px + qy = z\) using Lagrange's auxiliary equations.
Lagrange's auxiliary equations: \(\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{z}\). From \(\frac{dx}{x}=\frac{dy}{y}\): \(u_1=\frac{x}{y}=c_1\). From \(\frac{dy}{y}=\frac{dz}{z}\): \(u_2=\frac{y}{z}=c_2\). General solution: \(\phi\!\left(\frac{x}{y},\frac{y}{z}\right)=0\), or equivalently \(z=y\,f\!\left(\frac{x}{y}\right)\) for arbitrary \(f\). \(\blacksquare\)
Find the complete integral of \(p^2+q^2=1\).
This is of type \(F(p,q)=0\). Try \(q=a\) (constant). Then \(p^2=1-a^2\Rightarrow p=\sqrt{1-a^2}\) (assuming \(|a|\le1\)). Complete integral: \(z=x\sqrt{1-a^2}+ay+b\). Singular solution: eliminate \(a\) from \(z=x\sqrt{1-a^2}+ay+b\), \(0=\frac{-ax}{\sqrt{1-a^2}}+y\Rightarrow y\sqrt{1-a^2}=ax\Rightarrow a=\frac{y}{\sqrt{x^2+y^2}}\), \(\sqrt{1-a^2}=\frac{x}{\sqrt{x^2+y^2}}\). Singular solution: \(z=\frac{x^2+y^2}{\sqrt{x^2+y^2}}+b=\sqrt{x^2+y^2}+b\), i.e., \((z-b)^2=x^2+y^2\). \(\blacksquare\)
Solve \((y-z)p+(z-x)q=x-y\).
Auxiliary equations: \(\frac{dx}{y-z}=\frac{dy}{z-x}=\frac{dz}{x-y}\). Adding all numerators and denominators: \(\frac{dx+dy+dz}{(y-z)+(z-x)+(x-y)}=\frac{dx+dy+dz}{0}\), so \(dx+dy+dz=0\Rightarrow x+y+z=c_1\). Multiply by \(x,y,z\) resp.: \(\frac{x\,dx}{x(y-z)}=\frac{y\,dy}{y(z-x)}=\frac{z\,dz}{z(x-y)}\); adding gives \(x\,dx+y\,dy+z\,dz=0\Rightarrow x^2+y^2+z^2=c_2\). General solution: \(\phi(x+y+z,\,x^2+y^2+z^2)=0\). \(\blacksquare\)
Solve \(z(xp-yq)=y^2-x^2\).
Rewrite: \(xzp-yzq=y^2-x^2\), so \(P=xz,Q=-yz,R=y^2-x^2\). Auxiliary system: \(\frac{dx}{xz}=\frac{dy}{-yz}=\frac{dz}{y^2-x^2}\). From \(\frac{dx}{xz}=\frac{dy}{-yz}\): \(\frac{dx}{x}=-\frac{dy}{y}\Rightarrow xy=c_1\). For second invariant, use \(x\,dx+y\,dy+z\,dz\) combination: from \(\frac{d(x^2)}{2x^2z}=\frac{d(y^2)}{-2y^2z}=\frac{dz}{y^2-x^2}\). Let \(u=x^2,v=y^2\): \(\frac{du}{2uz}=\frac{dv}{-2vz}=\frac{dz}{v-u}\). From first two: \(u\,du/\cdot=-v\), so \(du/u=-dv/v\Rightarrow uv=x^2y^2=c_1^2\) (same). Try \(\frac{du+dv}{2z(u-v)}=\frac{dz}{v-u}\Rightarrow (du+dv)(-1)=2z\,dz\Rightarrow -(x^2+y^2)+z^2=c_2'\Rightarrow x^2+y^2-z^2=c_2\). General solution: \(\phi(xy,\,x^2+y^2-z^2)=0\). \(\blacksquare\)
Solve \((x^2-yz)p+(y^2-zx)q=z^2-xy\).
Auxiliary system: \(\frac{dx}{x^2-yz}=\frac{dy}{y^2-zx}=\frac{dz}{z^2-xy}\). Subtract: \(\frac{dx-dy}{(x^2-yz)-(y^2-zx)}=\frac{dx-dy}{(x^2-y^2)+z(x-y)}=\frac{dx-dy}{(x-y)(x+y+z)}\). Also \(\frac{dy-dz}{(y-z)(x+y+z)}\). So \(\frac{d(x-y)}{x-y}=\frac{d(y-z)}{y-z}\Rightarrow\frac{x-y}{y-z}=c_1\). Similarly \(\frac{x-y}{x-z}=c_2'\). Equivalently: \(c_1=\frac{x-y}{y-z},\;c_2=\frac{y-z}{z-x}\). General solution: \(\phi\!\left(\frac{x-y}{y-z},\frac{y-z}{z-x}\right)=0\). \(\blacksquare\)
Solve \(pz-qz=z^2+(x+y)^2\).
Auxiliary system: \(\frac{dx}{z}=\frac{dy}{-z}=\frac{dz}{z^2+(x+y)^2}\). From \(\frac{dx}{z}=\frac{dy}{-z}\): \(dx=-dy\Rightarrow x+y=c_1\). Let \(u=x+y=c_1\). Then \(\frac{dx}{z}=\frac{dz}{z^2+c_1^2}\Rightarrow\frac{dz}{dx}=z+\frac{c_1^2}{z}\cdot\frac{z}{z}=z^2/z+c_1^2/z\). Actually \(\frac{dz}{z^2+c_1^2}=\frac{dx}{z}\Rightarrow z\,dz=\frac{z^2+c_1^2}{?}\). Let \(v=z^2+c_1^2\): \(\frac{dv}{2z}=(z^2+c_1^2)\frac{dx}{z}\Rightarrow\frac{dv}{2(z^2+c_1^2)}=dx\Rightarrow\frac{dv}{2v}=dx\Rightarrow \ln v=2x+C\Rightarrow z^2+(x+y)^2=e^{2x+C}\). So \(c_2=e^{-2x}(z^2+(x+y)^2)\). General solution: \(\phi(x+y,\,e^{-2x}(z^2+(x+y)^2))=0\). \(\blacksquare\)
Solve \(x(y^2-z^2)p+y(z^2-x^2)q=z(x^2-y^2)\).
Auxiliary system: \(\frac{dx}{x(y^2-z^2)}=\frac{dy}{y(z^2-x^2)}=\frac{dz}{z(x^2-y^2)}\). Multiplier \(\frac{1}{x}\): \(\frac{dx/x}{y^2-z^2}\). Adding all with multipliers \(1/x,1/y,1/z\): numerator \(\frac{dx}{x}+\frac{dy}{y}+\frac{dz}{z}\), denominator \((y^2-z^2)+(z^2-x^2)+(x^2-y^2)=0\). So \(\frac{d(\ln x)+d(\ln y)+d(\ln z)}{0}\Rightarrow d(\ln xyz)=0\Rightarrow xyz=c_1\). Adding with multipliers \(x,y,z\): numerator \(x\,dx+y\,dy+z\,dz\), denominator \(x^2(y^2-z^2)+y^2(z^2-x^2)+z^2(x^2-y^2)=0\). So \(x^2+y^2+z^2=c_2\). General solution: \(\phi(xyz,\,x^2+y^2+z^2)=0\). \(\blacksquare\)
Solve \(p+q=\sin x+\sin y\).
Auxiliary equations: \(\frac{dx}{1}=\frac{dy}{1}=\frac{dz}{\sin x+\sin y}\). From \(dx=dy\): \(x-y=c_1\). Along a characteristic, \(x=t+x_0,y=t+y_0\) with \(y_0=x_0-c_1\). \(\frac{dz}{dt}=\sin t+\sin(t-c_1)\). Integrate: \(z=-\cos t-\cos(t-c_1)+g(c_1)=-\cos x-\cos y+g(x-y)\). General solution: \(z=-\cos x-\cos y+f(x-y)\) for arbitrary \(f\). \(\blacksquare\)
Solve \(2(z+xp+yq)=yp\).
Rearrange: \(p(y-2x)-2yq=2z\Rightarrow\frac{P}{y-2x},Q=-2y,R=2z\). Auxiliary system: \(\frac{dx}{y-2x}=\frac{dy}{-2y}=\frac{dz}{2z}\). From \(\frac{dy}{-2y}=\frac{dz}{2z}\): \(\frac{dz}{z}=-\frac{dy}{y}\Rightarrow yz=c_1\). From \(\frac{dy}{-2y}=\frac{dx}{y-2x}\): \((y-2x)dy=-2y\,dx\Rightarrow y\,dy-2x\,dy+2y\,dx=0\Rightarrow y\,dy+2d(xy)=0\Rightarrow\frac{y^2}{2}+2xy=c_2'\Rightarrow y^2+4xy=c_2\). General solution: \(\phi(yz,\,y^2+4xy)=0\). \(\blacksquare\)
Find a complete integral of \(z=px+qy+p^2q^2\) (Clairaut form).
This is Clairaut's equation \(z=px+qy+f(p,q)\) with \(f(p,q)=p^2q^2\). Complete integral is obtained by replacing \(p=a,q=b\): \(z=ax+by+a^2b^2\). This is the complete integral with two arbitrary constants \(a,b\). Singular solution: differentiate w.r.t. \(a\): \(0=x+2ab^2\Rightarrow x=-2ab^2\). W.r.t. \(b\): \(0=y+2a^2b\Rightarrow y=-2a^2b\). Then \(xy=4a^3b^3\Rightarrow (xy)=4(ab)^3\). Also \(z=ax+by+a^2b^2=-2a^2b^2-2a^2b^2+a^2b^2=-3a^2b^2\). And \(xy=-4a^3b^3=-(ab)^3\cdot4\). So \(z^3=(-3a^2b^2)^3=-27a^6b^6\) and \((xy)^2=16a^6b^6\), giving \(27z^3+16(xy)^2\) ... singular solution satisfies \(27z^3+16x^2y^2=0\). \(\blacksquare\)
Find the complete integral of \(z=px+qy+\sqrt{1+p^2+q^2}\) (Clairaut).
Clairaut form with \(f(p,q)=\sqrt{1+p^2+q^2}\). Complete integral: replace \(p=a,q=b\): \(\boxed{z=ax+by+\sqrt{1+a^2+b^2}}\). Geometrically these are planes tangent to the sphere \(x^2+y^2+(z-c)^2=1\) for appropriate \(c\). Singular solution: \(\partial/\partial a: 0=x+\frac{a}{\sqrt{1+a^2+b^2}}\), \(\partial/\partial b: 0=y+\frac{b}{\sqrt{1+a^2+b^2}}\). So \(x=-\frac{a}{R},y=-\frac{b}{R}\) where \(R=\sqrt{1+a^2+b^2}\). Then \(x^2+y^2=(a^2+b^2)/R^2\), and \(1-x^2-y^2=1/R^2\Rightarrow R=1/\sqrt{1-x^2-y^2}\). Substituting back gives singular solution: \(z=\frac{-a^2-b^2}{R}+R=\frac{R^2-a^2-b^2}{R}=\frac{1}{R}=\sqrt{1-x^2-y^2}\), i.e., \(x^2+y^2+z^2=1\) (unit sphere). \(\blacksquare\)
Find a complete integral of \(p^2x+q^2y=z\) using Charpit's method.
\(F=p^2x+q^2y-z=0\). Charpit's equations: \(\frac{dp}{F_x+pF_z}=\frac{dq}{F_y+qF_z}\). \(F_x=p^2,F_y=q^2,F_z=-1\). So \(\frac{dp}{p^2-p}=\frac{dq}{q^2-q}\Rightarrow\frac{dp}{p(p-1)}=\frac{dq}{q(q-1)}\). Try compatible relation: assume \(q=ap\). Then \(p^2x+a^2p^2y=z\Rightarrow p^2(x+a^2y)=z\Rightarrow p=\sqrt{z/(x+a^2y)}\). Then \(dz=p\,dx+q\,dy=p(dx+a\,dy)\). Also \(\frac{dz}{z}=\frac{dx+a\,dy}{x+a^2y}\). This separable ODE gives \(\ln z=\ln(x+a^2y)+\ln b^2\cdot\text{(some const)}\). More directly: complete integral \(z=(ax+by)^2\) where Charpit gives \(a^2x+b^2y=z\) if we set \(p=a,q=b\) — but let us use standard form: answer is \(\sqrt{z}=\sqrt{x}+a\sqrt{y}+b\) after substitution \(u=\sqrt{x},v=\sqrt{y},w=\sqrt{z}\). Complete integral: \(\sqrt{z}=a\sqrt{x}+b\sqrt{y}\) (verify: \(p=\frac{a}{2\sqrt{x}\cdot\sqrt{z}}\cdot z=\frac{a\sqrt{z}}{2\sqrt{x}}\), \(p^2x=\frac{a^2z}{4}\), similarly \(q^2y=\frac{b^2z}{4}\); need \(\frac{(a^2+b^2)z}{4}=z\Rightarrow a^2+b^2=4\), constraint). Use Charpit properly: \(p=\sqrt{az/x},q=\sqrt{(1-a)z/y}\). Complete integral: \(2\sqrt{z}=2a_1\sqrt{x}+2a_2\sqrt{y}\) with \(a_1^2+a_2^2=1\). \(\blacksquare\)
Find the complete integral of \(p^2+q^2=z\).
Type \(F(z,p,q)=0\). Try \(q=ap\). Then \(p^2(1+a^2)=z\Rightarrow p=\frac{\sqrt{z}}{\sqrt{1+a^2}},q=\frac{a\sqrt{z}}{\sqrt{1+a^2}}\). \(dz=p\,dx+q\,dy=\frac{\sqrt{z}}{\sqrt{1+a^2}}(dx+a\,dy)\). \(\frac{dz}{\sqrt{z}}=\frac{dx+a\,dy}{\sqrt{1+a^2}}\). Integrate: \(2\sqrt{z}=\frac{x+ay}{\sqrt{1+a^2}}+b\). Complete integral: \(2\sqrt{z}=\frac{x+ay}{\sqrt{1+a^2}}+b\), or equivalently \(\left(2\sqrt{z}-b\right)^2=\frac{(x+ay)^2}{1+a^2}\). \(\blacksquare\)
Find the complete integral of \(z=px+qy-2pq\) (Clairaut).
Clairaut form \(z=px+qy+f(p,q)\) with \(f(p,q)=-2pq\). Complete integral: set \(p=a,q=b\): \(\boxed{z=ax+by-2ab}\). Singular solution: \(\partial_a: x=2b\Rightarrow b=x/2\); \(\partial_b: y=2a\Rightarrow a=y/2\). Substitute: \(z=\frac{y}{2}\cdot x+\frac{x}{2}\cdot y-2\cdot\frac{y}{2}\cdot\frac{x}{2}=xy+xy/2-xy/2=\frac{xy}{2}\). Wait: \(z=\frac{y}{2}x+\frac{x}{2}y-2\frac{x}{2}\frac{y}{2}=\frac{xy}{2}+\frac{xy}{2}-\frac{xy}{2}=\frac{xy}{2}\). Singular solution: \(z=\frac{xy}{2}\). \(\blacksquare\)
Solve \(p^2 - q^2 = 4\). Find the complete integral.
Type \(F(p,q)=0\). Set \(q=a\): \(p^2=4+a^2\Rightarrow p=\sqrt{4+a^2}\). Complete integral: \(z=x\sqrt{4+a^2}+ay+b\). General solution obtained by taking \(b=\phi(a)\) and eliminating \(a\). Singular solution: differentiate w.r.t. \(a\): \(0=\frac{ax}{\sqrt{4+a^2}}+y\Rightarrow y=-\frac{ax}{\sqrt{4+a^2}}\). So \(a^2x^2=(4+a^2)y^2\Rightarrow a^2(x^2-y^2)=4y^2\Rightarrow a^2=\frac{4y^2}{x^2-y^2}\). This gives the singular solution as \(x^2-y^2=\frac{4y^2}{a^2}\cdot a^2=4y^2\), i.e., \(x^2-5y^2=4\cdot\frac{x^2}{p^2}\cdot\ldots\) — checking: from \(y\sqrt{4+a^2}=-ax\) squaring gives \(y^2(4+a^2)=a^2x^2\Rightarrow 4y^2=a^2(x^2-y^2)\). Then \(z=\frac{(4+a^2)x-ax^2/y\cdot?}{}...\). Complete integral \(z=x\sqrt{4+a^2}+ay+b\) is the main answer. \(\blacksquare\)
Solve \(z^2(p^2+q^2)=x^2+y^2\). Find the complete integral.
Substitute \(Z=z^2,X=x^2,Y=y^2\): \(\frac{\partial Z}{\partial X}=\frac{dZ}{dz}\cdot\frac{dz}{dx}\cdot\frac{dx}{dX}=2z\cdot p\cdot\frac{1}{2x}=\frac{zp}{x}\), so \(P=zp/x,Q=zq/y\). Then \(z^2(p^2+q^2)=x^2P^2+y^2Q^2\)... try direct: set \(zp=P',zq=Q'\). Note \(z\,dz=z^2(p\,dx+q\,dy)/z=\). Try \(p=\frac{\partial z}{\partial x}\), substitute \(u=\frac{z^2}{2}\): \(u_x=zp,u_y=zq\). Then \(u_x^2+u_y^2=x^2+y^2\). Set \(u=X(x)+Y(y)\): \(X'^2+Y'^2=x^2+y^2\Rightarrow X'^2-x^2=y^2-Y'^2=a^2\). So \(X'=\sqrt{x^2+a^2},Y'=\sqrt{y^2-a^2}\). Thus \(X=\int\sqrt{x^2+a^2}\,dx=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln(x+\sqrt{x^2+a^2})\), similarly for \(Y\). Complete integral: \(\frac{z^2}{2}=\int_0^x\sqrt{t^2+a^2}\,dt+\int_0^y\sqrt{s^2-a^2}\,ds+b\). \(\blacksquare\)
Solve \(z=px+qy+p^2+q^2\). Find complete and singular integrals.
Clairaut form with \(f(p,q)=p^2+q^2\). Complete integral: set \(p=a,q=b\): \(z=ax+by+a^2+b^2\). Singular solution: \(\partial_a z=0\Rightarrow x+2a=0\Rightarrow a=-x/2\). \(\partial_b z=0\Rightarrow y+2b=0\Rightarrow b=-y/2\). Substitute: \(z=-\frac{x^2}{2}-\frac{y^2}{2}+\frac{x^2}{4}+\frac{y^2}{4}=-\frac{x^2}{4}-\frac{y^2}{4}\). Singular solution: \(\boxed{4z+x^2+y^2=0}\). \(\blacksquare\)
Find the singular solution of \(z=px+qy+p^2+q^2\) using the p-discriminant method.
Complete integral (from previous): \(F(x,y,z,a,b)=z-ax-by-a^2-b^2=0\). For singular solution via envelope, eliminate \(a,b\) from \(F=0,\partial F/\partial a=0,\partial F/\partial b=0\). \(\partial F/\partial a=-x-2a=0\Rightarrow a=-x/2\). \(\partial F/\partial b=-y-2b=0\Rightarrow b=-y/2\). Substitute: \(z=-(-x/2)x-(-y/2)y+x^2/4+y^2/4=x^2/2+y^2/2+x^2/4+y^2/4=\frac{3x^2}{4}+\frac{3y^2}{4}\). Wait: \(z=ax+by+a^2+b^2=(-x/2)x+(-y/2)y+x^2/4+y^2/4=-x^2/2-y^2/2+x^2/4+y^2/4=-x^2/4-y^2/4\). Singular solution: \(4z+x^2+y^2=0\). Verify: \(p=-x/2,q=-y/2,px+qy=-x^2/2-y^2/2,p^2+q^2=x^2/4+y^2/4\). LHS: \(-x^2/4-y^2/4\). RHS: \(z\). \(\checkmark\) \(\blacksquare\)
Find the singular solution (envelope) of \(z = ax + by + ab\) where \(a,b\) are parameters.
The one-parameter family (general solution with \(b=\phi(a)\)) gives envelope when \(\partial F/\partial a=0\). For the complete integral \(z=ax+by+ab\), the singular solution comes from eliminating \(a,b\): \(\partial_a: x+b=0\Rightarrow b=-x\). \(\partial_b: y+a=0\Rightarrow a=-y\). Substitute: \(z=(-y)x+(-x)y+(-y)(-x)=−xy-xy+xy=-xy\). Singular solution: \(\boxed{z+xy=0}\). Verify: \(p=\partial z/\partial x=-y=a\checkmark\), \(q=-x=b\checkmark\), \(px+qy+pq=-xy-xy+xy=-xy=z\checkmark\). \(\blacksquare\)
Classify the PDE \(u_{xx}+4u_{xy}+4u_{yy}=0\).
General second-order: \(Au_{xx}+Bu_{xy}+Cu_{yy}+\ldots=0\). Here \(A=1,B=4,C=4\). Discriminant \(\Delta=B^2-4AC=16-16=0\). Since \(\Delta=0\), the PDE is parabolic everywhere. Characteristics: \(\frac{dy}{dx}=\frac{B\pm\sqrt{\Delta}}{2A}=\frac{4}{2}=2\). One family: \(y-2x=c\) (repeated). \(\blacksquare\)
Classify the PDE \(u_{xx}-2u_{xy}+5u_{yy}=0\).
\(A=1,B=-2,C=5\). Discriminant \(\Delta=B^2-4AC=4-20=-16<0\). Since \(\Delta<0\), the PDE is elliptic everywhere. No real characteristics. Canonical form will be \(u_{\xi\xi}+u_{\eta\eta}+\ldots=0\) (Laplace-type). \(\blacksquare\)
Classify the PDE \(u_{xx}-y^2u_{yy}=0\) and identify where the type changes.
\(A=1,B=0,C=-y^2\). Discriminant \(\Delta=0-4(1)(-y^2)=4y^2\). For \(y\neq0\): \(\Delta>0\) → hyperbolic. For \(y=0\): \(\Delta=0\) → parabolic. The type changes along \(y=0\). Characteristics for \(y\neq0\): \(\frac{dy}{dx}=\pm\frac{\sqrt{\Delta}}{2A}=\pm y\Rightarrow\ln|y|=\pm x+c\Rightarrow ye^{\mp x}=c\). Two families: \(ye^{-x}=c_1,ye^x=c_2\). \(\blacksquare\)
Classify \(u_{xx}+xu_{xy}=0\) and find the regions of different types.
\(A=1,B=x,C=0\). Discriminant \(\Delta=x^2-0=x^2\). For \(x\neq0\): \(\Delta=x^2>0\) → hyperbolic. For \(x=0\): \(\Delta=0\) → parabolic. Type changes along \(x=0\). Characteristics: \(\frac{dy}{dx}=\frac{x\pm|x|}{2}\). For \(x>0\): \(dy/dx=x\) or \(dy/dx=0\), giving \(y-x^2/2=c_1\) and \(y=c_2\). For \(x<0\): \(dy/dx=0\) or \(dy/dx=x\), same families. \(\blacksquare\)
Reduce \(u_{xx}-4u_{xy}+4u_{yy}=0\) to canonical form.
\(A=1,B=-4,C=4\). \(\Delta=16-16=0\): parabolic. Single characteristic family: \(\frac{dy}{dx}=\frac{-B}{2A}=2\Rightarrow y-2x=c\). Set \(\xi=y-2x,\eta=x\) (or \(\eta=y\)). Then \(u_x=u_\xi\xi_x+u_\eta\eta_x=-2u_\xi+u_\eta\), \(u_y=u_\xi\). Compute: \(u_{xx}=4u_{\xi\xi}-4u_{\xi\eta}+u_{\eta\eta}\), \(u_{xy}=-2u_{\xi\xi}+u_{\xi\eta}\), \(u_{yy}=u_{\xi\xi}\). Substitute: \((4u_{\xi\xi}-4u_{\xi\eta}+u_{\eta\eta})-4(-2u_{\xi\xi}+u_{\xi\eta})+4u_{\xi\xi}=4u_{\xi\xi}-4u_{\xi\eta}+u_{\eta\eta}+8u_{\xi\xi}-4u_{\xi\eta}+4u_{\xi\xi}=16u_{\xi\xi}-8u_{\xi\eta}+u_{\eta\eta}\). Wait — for parabolic, the canonical form should eliminate \(u_{\xi\xi}\). With \(\xi=y-2x\): coefficient of \(u_{\xi\xi}\) in \(Au_{xx}+Bu_{xy}+Cu_{yy}\) after substitution \(= A\xi_x^2+B\xi_x\xi_y+C\xi_y^2=4+(-4)(1)+4\cdot0=0\checkmark\). Coefficient of \(u_{\eta\eta}\): \(A(-2\cdot0+1)^2+\ldots\). With \(\eta=x\): \(A\eta_x^2+B\eta_x\eta_y+C\eta_y^2=1\). Coefficient of \(u_{\xi\eta}\): \(2A\xi_x\eta_x+B(\xi_x\eta_y+\xi_y\eta_x)+2C\xi_y\eta_y=2(1)(-2)(1)+(-4)((-2)(0)+(1)(1))+0=-4-4=-8\ne0\). Choose \(\eta=y\): then \(A\eta_x^2+B\eta_x\eta_y+C\eta_y^2=0+0+4\). Cross: \(2(1)(-2)(0)+(-4)((-2)(1)+(1)(0))+2(4)(1)(1)=0+8+8=16\ne0\). Try \(\eta=2y+x\): \(2(1)(-2)(2)+(-4)((-2)(1)+(1)(2))+2(4)(1)(1)=-8+0+8=0\). So \(\eta=2y+x\) works. With \(\xi=y-2x,\eta=2y+x\): canonical form is \(u_{\eta\eta}=0\) (up to lower-order terms). Canonical form: \(\boxed{u_{\eta\eta}=0}\), i.e., parabolic. \(\blacksquare\)
Reduce \(u_{xx}+2u_{xy}+u_{yy}=x^2+y^2\) to canonical form and solve.
\(A=1,B=2,C=1\). \(\Delta=4-4=0\): parabolic. Characteristic: \(y-x=c_1\). Set \(\xi=y-x,\eta=x\). Then \(x=\eta,y=\xi+\eta\). \(u_{xx}+2u_{xy}+u_{yy}=(u_\eta-u_\xi)_x\cdot...\). Compute: \(u_x=-u_\xi+u_\eta,u_y=u_\xi\). \(u_{xx}=u_{\xi\xi}-2u_{\xi\eta}+u_{\eta\eta}\), \(u_{xy}=-u_{\xi\xi}+u_{\xi\eta}\), \(u_{yy}=u_{\xi\xi}\). Sum: \((u_{\xi\xi}-2u_{\xi\eta}+u_{\eta\eta})+2(-u_{\xi\xi}+u_{\xi\eta})+u_{\xi\xi}=u_{\eta\eta}\). RHS: \(x^2+y^2=\eta^2+(\xi+\eta)^2=2\eta^2+2\xi\eta+\xi^2\). Canonical form: \(u_{\eta\eta}=2\eta^2+2\xi\eta+\xi^2\). Integrate twice in \(\eta\): \(u_\eta=\frac{2\eta^3}{3}+\xi\eta^2+\xi^2\eta+f(\xi)\). \(u=\frac{\eta^4}{6}+\frac{\xi\eta^3}{3}+\frac{\xi^2\eta^2}{2}+f(\xi)\eta+g(\xi)\). Back-substitute \(\xi=y-x,\eta=x\). \(\blacksquare\)
Reduce \(u_{xx}-3u_{xy}+2u_{yy}=0\) to canonical form.
\(A=1,B=-3,C=2\). \(\Delta=9-8=1>0\): hyperbolic. Characteristics: \(\frac{dy}{dx}=\frac{3\pm1}{2}\), giving \(dy/dx=2\Rightarrow y-2x=c_1\) and \(dy/dx=1\Rightarrow y-x=c_2\). Set \(\xi=y-2x,\eta=y-x\). Then: \(u_x=-2u_\xi-u_\eta,u_y=u_\xi+u_\eta\). \(u_{xx}=4u_{\xi\xi}+4u_{\xi\eta}+u_{\eta\eta}\), \(u_{xy}=-2u_{\xi\xi}-3u_{\xi\eta}-u_{\eta\eta}\), \(u_{yy}=u_{\xi\xi}+2u_{\xi\eta}+u_{\eta\eta}\). Substitute: \((4-(-9)+2)u_{\xi\xi}+(4-(-9)+2\cdot2)u_{\xi\eta}+(1-(-3)+2)u_{\eta\eta}\). Carefully: \(u_{xx}-3u_{xy}+2u_{yy}=(4u_{\xi\xi}+4u_{\xi\eta}+u_{\eta\eta})-3(-2u_{\xi\xi}-3u_{\xi\eta}-u_{\eta\eta})+2(u_{\xi\xi}+2u_{\xi\eta}+u_{\eta\eta})=(4+6+2)u_{\xi\xi}+(4+9+4)u_{\xi\eta}+(1+3+2)u_{\eta\eta}=12u_{\xi\xi}+17u_{\xi\eta}+6u_{\eta\eta}\). That's not right — for hyperbolic canonical form should give \(u_{\xi\eta}=0\). The \(\xi,\eta\) are the characteristic coordinates so \(u_{\xi\xi}\) and \(u_{\eta\eta\)$ coefficients must vanish. Check: coeff of \(u_{\xi\xi}\): \(A\xi_x^2+B\xi_x\xi_y+C\xi_y^2=(1)(4)+(-3)(-2)(1)+(2)(1)=4+6+2=12\ne0\). So I need to re-do. Correct characteristic directions: \(\frac{dy}{dx}=\frac{B\pm\sqrt{\Delta}}{2A}=\frac{-3\pm1}{2}\). So \(dy/dx=-1\Rightarrow y+x=c_1\) and \(dy/dx=-2\Rightarrow y+2x=c_2\). Set \(\xi=y+x,\eta=y+2x\). Verify: coeff of \(u_{\xi\xi}\): \(A(1)^2+B(1)(1)+C(1)^2=1-3+2=0\checkmark\). Coeff of \(u_{\eta\eta}\): \(A(2)^2+B(2)(1)+C(1)^2=4-6+2=0\checkmark\). Coeff of \(u_{\xi\eta}\): \(2A(1)(2)+B(1\cdot1+2\cdot1)+2C(1)(1)=4+(-3)(3)+4=4-9+4=-1\). Canonical form: \(-u_{\xi\eta}=0\Rightarrow\boxed{u_{\xi\eta}=0}\). Solution: \(u=f(\xi)+g(\eta)=f(y+x)+g(y+2x)\). \(\blacksquare\)
Reduce \(3u_{xx}+10u_{xy}+3u_{yy}=0\) to canonical form and find the general solution.
\(A=3,B=10,C=3\). \(\Delta=100-36=64>0\): hyperbolic. Characteristics: \(\frac{dy}{dx}=\frac{-10\pm8}{6}\). So \(dy/dx=\frac{-2}{6}=-\frac{1}{3}\Rightarrow 3y+x=c_1\) and \(dy/dx=\frac{-18}{6}=-3\Rightarrow y+3x=c_2\). Set \(\xi=3y+x,\eta=y+3x\). Coeff of \(u_{\xi\xi}\): \(3(1)^2+10(1)(3)+3(3)^2=3+30+27=60\ne0\). Try \(\xi=x+3y,\eta=3x+y\): \(\xi_x=1,\xi_y=3\). Coeff: \(3(1)+10(3)+3(9)=3+30+27=60\ne0\). The correct coordinates come from the characteristic equations: \(A(dy)^2-B(dx)(dy)+C(dx)^2=0\Rightarrow 3\lambda^2-10\lambda+3=0\Rightarrow\lambda=3\) or \(\lambda=1/3\), where \(\lambda=dy/dx\). So \(y-3x=c_1\) and \(y-x/3=c_2\) i.e. \(3y-x=c_2\). Set \(\xi=y-3x,\eta=3y-x\). \(\xi_x=-3,\xi_y=1,\eta_x=-1,\eta_y=3\). Coeff of \(u_{\xi\xi}\): \(3(9)+10(-3)(1)+3(1)=27-30+3=0\checkmark\). Coeff of \(u_{\eta\eta}\): \(3(1)+10(-1)(3)+3(9)=3-30+27=0\checkmark\). Coeff of \(u_{\xi\eta}\): \(2(3)(-3)(-1)+10((-3)(3)+(-1)(1))+2(3)(1)(3)=-18+10(-10)+18=-18-100+18=-100\). Canonical: \(-100u_{\xi\eta}=0\Rightarrow u_{\xi\eta}=0\). General solution: \(\boxed{u=f(y-3x)+g(3y-x)}\). \(\blacksquare\)
Solve \(u_{tt}=c^2u_{xx}\) with \(u(x,0)=\sin x,\,u_t(x,0)=0\) on \(-\infty
d'Alembert's formula: \(u(x,t)=\frac{1}{2}[f(x+ct)+f(x-ct)]+\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)\,ds\) where \(f(x)=u(x,0)=\sin x\) and \(g(x)=u_t(x,0)=0\). So \(u(x,t)=\frac{1}{2}[\sin(x+ct)+\sin(x-ct)]=\frac{1}{2}\cdot2\sin x\cos(ct)=\sin x\cos(ct)\). Final answer: \(\boxed{u(x,t)=\sin x\cos(ct)}\). \(\blacksquare\)
Solve \(u_{tt}=c^2u_{xx}\) with initial conditions \(u(x,0)=x^2,\,u_t(x,0)=x\).
d'Alembert: \(u=\frac{(x+ct)^2+(x-ct)^2}{2}+\frac{1}{2c}\int_{x-ct}^{x+ct}s\,ds\). Compute: \(\frac{(x+ct)^2+(x-ct)^2}{2}=\frac{2x^2+2c^2t^2}{2}=x^2+c^2t^2\). \(\frac{1}{2c}\cdot\frac{s^2}{2}\Big|_{x-ct}^{x+ct}=\frac{(x+ct)^2-(x-ct)^2}{4c}=\frac{4xct}{4c}=xt\). Final answer: \(\boxed{u(x,t)=x^2+c^2t^2+xt}\). \(\blacksquare\)
Solve \(u_{tt}=c^2u_{xx}\) with \(u(0,t)=u(L,t)=0\) and \(u(x,0)=f(x),\,u_t(x,0)=g(x)\).
Assume \(u=X(x)T(t)\): \(XT''=c^2X''T\Rightarrow\frac{T''}{c^2T}=\frac{X''}{X}=-\lambda^2\). BCs give \(X(0)=X(L)=0\Rightarrow X_n=\sin\frac{n\pi x}{L},\lambda_n=\frac{n\pi}{L}\). Then \(T_n''+(c\lambda_n)^2T_n=0\Rightarrow T_n=A_n\cos\frac{n\pi ct}{L}+B_n\sin\frac{n\pi ct}{L}\). General solution: \(u=\sum_{n=1}^\infty\left(A_n\cos\frac{n\pi ct}{L}+B_n\sin\frac{n\pi ct}{L}\right)\sin\frac{n\pi x}{L}\). ICs: \(A_n=\frac{2}{L}\int_0^Lf(x)\sin\frac{n\pi x}{L}dx\), \(B_n=\frac{2}{n\pi c}\int_0^Lg(x)\sin\frac{n\pi x}{L}dx\). \(\blacksquare\)
Show that the total energy \(E=\frac{1}{2}\int_0^L[\rho u_t^2+Tu_x^2]\,dx\) is conserved for the vibrating string.
Differentiate: \(\frac{dE}{dt}=\int_0^L[\rho u_t u_{tt}+Tu_x u_{xt}]\,dx\). Since \(\rho u_{tt}=Tu_{xx}\): \(\frac{dE}{dt}=\int_0^L[Tu_{xx}u_t+Tu_xu_{xt}]\,dx=T\int_0^L\frac{\partial}{\partial x}(u_xu_t)\,dx=T[u_xu_t]_0^L\). With BCs \(u(0,t)=u(L,t)=0\Rightarrow u_t(0,t)=u_t(L,t)=0\). So \(\frac{dE}{dt}=T\cdot0=0\). Hence \(E=\text{constant}\). \(\blacksquare\)
Solve the heat equation \(u_t=ku_{xx}\) with \(u(0,t)=u(L,t)=0\) and \(u(x,0)=x(L-x)\).
Separation: \(u=X(x)T(t)\Rightarrow T'/kT=X''/X=-\lambda^2\). BCs: \(X_n=\sin\frac{n\pi x}{L},T_n=e^{-k(n\pi/L)^2t}\). Solution: \(u=\sum_{n=1}^\infty b_n e^{-k(n\pi/L)^2t}\sin\frac{n\pi x}{L}\). Coefficients from \(u(x,0)=x(L-x)\): \(b_n=\frac{2}{L}\int_0^Lx(L-x)\sin\frac{n\pi x}{L}dx\). Integrate by parts twice: \(b_n=\frac{2}{L}\cdot\frac{2L^3}{(n\pi)^3}(1-\cos n\pi)=\frac{4L^2}{(n\pi)^3}(1-(-1)^n)\). So \(b_n=\frac{8L^2}{(n\pi)^3}\) for odd \(n\), \(0\) for even \(n\). Final: \(u=\frac{8L^2}{\pi^3}\sum_{m=0}^\infty\frac{1}{(2m+1)^3}e^{-k(2m+1)^2\pi^2t/L^2}\sin\frac{(2m+1)\pi x}{L}\). \(\blacksquare\)
Solve \(u_t=ku_{xx}\) with \(u(0,t)=u(\pi,t)=0\) and \(u(x,0)=\sin x\).
Eigenfunctions: \(X_n=\sin nx,T_n=e^{-kn^2t}\). General solution: \(u=\sum_{n=1}^\infty b_n e^{-kn^2t}\sin nx\). IC: \(u(x,0)=\sin x=\sum b_n\sin nx\). By inspection (or Fourier): \(b_1=1,b_n=0\) for \(n\ge2\). Solution: \(\boxed{u(x,t)=e^{-kt}\sin x}\). \(\blacksquare\)
A rod of length \(L\) has temperatures \(T_1\) at \(x=0\) and \(T_2\) at \(x=L\). Find the steady-state temperature distribution.
Steady-state: \(u_t=0\Rightarrow ku_{xx}=0\Rightarrow u''=0\Rightarrow u=Ax+B\). BCs: \(u(0)=T_1\Rightarrow B=T_1\). \(u(L)=T_2\Rightarrow AL+T_1=T_2\Rightarrow A=\frac{T_2-T_1}{L}\). Steady-state: \(\boxed{u(x)=T_1+\frac{(T_2-T_1)x}{L}}\) (linear). For the transient problem write \(u=u_s(x)+v(x,t)\) where \(v\) satisfies homogeneous BCs and \(v_t=kv_{xx}\), \(v(x,0)=f(x)-u_s(x)\). \(\blacksquare\)
Solve \(u_t=ku_{xx}\), \(x>0,t>0\), with \(u(0,t)=0\) and \(u(x,0)=u_0\) (constant), using the error function.
Use similarity variable \(\eta=\frac{x}{2\sqrt{kt}}\). Assume \(u=f(\eta)\). Then \(u_t=f'(\eta)\cdot(-\frac{x}{4kt^{3/2}}\cdot2\sqrt{k})=f'\cdot(-\frac{\eta}{2t})\), \(u_{xx}=\frac{f''}{4kt}\). Equation: \(-\frac{\eta}{2t}f'=k\cdot\frac{f''}{4kt}\Rightarrow f''+2\eta f'=0\). This is \((e^{\eta^2}f')'=0\cdot e^{\eta^2}\), wait: \(f''+2\eta f'=0\Rightarrow\frac{d}{d\eta}[e^{\eta^2}f']=0\Rightarrow f'=Ce^{-\eta^2}\). Integrating: \(f=A+B\int_0^\eta e^{-s^2}ds=A+B\frac{\sqrt\pi}{2}\text{erf}(\eta)\). BCs: \(u(0,t)=0\Rightarrow f(0)=0\Rightarrow A=0\). As \(x\to\infty\): \(u\to u_0\Rightarrow f(\infty)=u_0\Rightarrow B\frac{\sqrt\pi}{2}=u_0\Rightarrow B=\frac{2u_0}{\sqrt\pi}\). Solution: \(\boxed{u(x,t)=u_0\,\text{erf}\!\left(\frac{x}{2\sqrt{kt}}\right)}\). \(\blacksquare\)
Solve \(u_{xx}+u_{yy}=0\) on \(0
Separate: \(u=X(x)Y(y)\). \(X''/X=-Y''/Y=-\lambda^2\). BCs \(X(0)=X(a)=0\Rightarrow X_n=\sin\frac{n\pi x}{a},\lambda_n=\frac{n\pi}{a}\). Then \(Y_n''-\lambda_n^2 Y_n=0\) with \(Y_n(0)=0\Rightarrow Y_n=\sinh\frac{n\pi y}{a}\). General: \(u=\sum_{n=1}^\infty c_n\sinh\frac{n\pi y}{a}\sin\frac{n\pi x}{a}\). At \(y=b\): \(f(x)=\sum c_n\sinh\frac{n\pi b}{a}\sin\frac{n\pi x}{a}\). So \(c_n=\frac{2}{a\sinh(n\pi b/a)}\int_0^a f(x)\sin\frac{n\pi x}{a}\,dx\). \(\blacksquare\)
Solve \(u_{xx}+u_{yy}=0\) on \([0,a]\times[0,b]\) with \(u(0,y)=u(a,y)=0\), \(u(x,0)=0\), \(u(x,b)=\sin(\pi x/a)\).
From the general rectangle solution: \(u=\sum c_n\sinh\frac{n\pi y}{a}\sin\frac{n\pi x}{a}\). At \(y=b\): \(\sin\frac{\pi x}{a}=\sum c_n\sinh\frac{n\pi b}{a}\sin\frac{n\pi x}{a}\). By orthogonality only \(n=1\) survives: \(c_1\sinh\frac{\pi b}{a}=1\Rightarrow c_1=\frac{1}{\sinh(\pi b/a)}\). Final: \(\boxed{u(x,y)=\frac{\sinh(\pi y/a)}{\sinh(\pi b/a)}\sin\frac{\pi x}{a}}\). \(\blacksquare\)
Solve \(u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}=0\) inside \(r
Separate: \(u=R(r)\Theta(\theta)\). \(\Theta''+\lambda\Theta=0\) with \(2\pi\)-periodicity \(\Rightarrow\lambda=n^2\), \(\Theta_n=A_n\cos n\theta+B_n\sin n\theta\). Euler equation for \(R\): \(r^2R''+rR'-n^2R=0\Rightarrow R_n=r^n\) (regular at \(r=0\)). Solution: \(u=\frac{a_0}{2}+\sum_{n=1}^\infty\left(\frac{r}{a}\right)^n(a_n\cos n\theta+b_n\sin n\theta)\). Fourier coefficients from BC. Summing the geometric series yields Poisson's formula: \(u(r,\theta)=\frac{a^2-r^2}{2\pi}\int_0^{2\pi}\frac{f(\phi)}{a^2-2ar\cos(\theta-\phi)+r^2}\,d\phi\). \(\blacksquare\)
Solve Laplace's equation inside the circle \(r
General interior solution: \(u=\frac{A_0}{2}+\sum_{n=1}^\infty r^n(A_n\cos n\theta+B_n\sin n\theta)\) (after normalising by \(a^n\)). At \(r=a\): \(\cos2\theta=\sum a^n(A_n\cos n\theta+B_n\sin n\theta)\). Match \(n=2\): \(a^2 A_2=1\Rightarrow A_2=1/a^2\); all other coefficients zero. So \(\boxed{u(r,\theta)=\frac{r^2}{a^2}\cos2\theta}\). \(\blacksquare\)
Solve Laplace's equation in the exterior region \(r>a\) with \(u(a,\theta)=\sin\theta\) and \(u\to0\) as \(r\to\infty\).
For exterior, bounded solutions are \(R_n=r^{-n}\). General: \(u=\sum_{n=1}^\infty r^{-n}(A_n\cos n\theta+B_n\sin n\theta)\). At \(r=a\): \(\sin\theta=\sum a^{-n}(A_n\cos n\theta+B_n\sin n\theta)\). Match \(n=1\): \(a^{-1}B_1=1\Rightarrow B_1=a\); all others zero. Solution: \(\boxed{u(r,\theta)=\frac{a}{r}\sin\theta}\). \(\blacksquare\)
State the Neumann problem for Laplace's equation. For \(\nabla^2 u=0\) inside \(r
Neumann problem: find \(u\) harmonic in \(\Omega\) with \(\partial u/\partial n=g\) on \(\partial\Omega\). Solvability condition (divergence theorem): \(\int_{\partial\Omega}g\,dS=0\). On disk, \(\partial u/\partial n=u_r|_{r=a}\). With expansion \(u=\frac{A_0}{2}+\sum r^n(A_n\cos n\theta+B_n\sin n\theta)\), we get \(u_r|_{r=a}=\sum na^{n-1}(A_n\cos n\theta+B_n\sin n\theta)=g(\theta)\). Uniqueness: up to an additive constant. \(A_n=\frac{1}{na^{n-1}}\cdot\frac{1}{\pi}\int_0^{2\pi}g(\phi)\cos n\phi\,d\phi\), similarly \(B_n\). The constant \(A_0\) is arbitrary (mean value). \(\blacksquare\)
Solve \(u_{xx}+u_{yy}=0\) for \(y>0\), \(-\infty
Take Fourier transform in \(x\): \(\hat{u}(\xi,y)=\int_{-\infty}^\infty u(x,y)e^{-i\xi x}dx\). Equation becomes \(\hat{u}_{yy}-\xi^2\hat{u}=0\Rightarrow\hat{u}=A(\xi)e^{-|\xi|y}\) (bounded as \(y\to\infty\)). BC: \(\hat{u}(\xi,0)=\hat{f}(\xi)\Rightarrow A=\hat{f}(\xi)\). So \(\hat{u}=\hat{f}(\xi)e^{-|\xi|y}\). Inverse transform (convolution): \(u(x,y)=\frac{y}{\pi}\int_{-\infty}^\infty\frac{f(t)}{(x-t)^2+y^2}\,dt\) (Poisson formula for half-plane). \(\blacksquare\)
Solve \(u_{xx}+u_{yy}=0\) on \(0
Separate: \(X''+\lambda^2 X=0\), \(X(0)=0\), \(X'(\pi)=0\). Eigenfunctions: \(X_n=\sin\frac{(2n-1)x}{2}\), \(\lambda_n=\frac{2n-1}{2}\), \(n=1,2,\ldots\). \(Y_n=e^{-\lambda_n y}\) (decays). Solution: \(u=\sum_{n=1}^\infty c_n e^{-\lambda_n y}\sin\lambda_n x\). IC: \(x=\sum c_n\sin\lambda_n x\) on \([0,\pi]\). Coefficients: \(c_n=\frac{2}{\pi}\int_0^\pi x\sin\frac{(2n-1)x}{2}dx=\frac{8(-1)^{n+1}}{\pi(2n-1)^2}\). Final: \(u=\frac{8}{\pi}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{(2n-1)^2}e^{-(2n-1)y/2}\sin\frac{(2n-1)x}{2}\). \(\blacksquare\)
State and prove the mean value property for harmonic functions. Hence show that a harmonic function cannot have an interior maximum unless it is constant.
Mean Value Property: If \(u\) is harmonic in \(\Omega\), then for any ball \(B(x_0,r)\subset\Omega\): \(u(x_0)=\frac{1}{|\partial B|}\int_{\partial B}u\,dS\). Proof (2D): Let \(h(r)=\frac{1}{2\pi}\int_0^{2\pi}u(x_0+r\cos\theta,y_0+r\sin\theta)\,d\theta\). Differentiating: \(h'(r)=\frac{1}{2\pi}\int_0^{2\pi}\frac{\partial u}{\partial r}\,d\theta=\frac{1}{2\pi r}\int_{\partial B_r}\nabla u\cdot\hat{n}\,ds=\frac{1}{2\pi r}\iint_{B_r}\nabla^2 u\,dA=0\). So \(h(r)\) is constant; at \(r=0\), \(h(0)=u(x_0)\). Maximum principle: If \(u\) attains interior maximum at \(x_0\), then \(u(x_0)\ge u(x)\) for all \(x\) near \(x_0\), but mean value = \(u(x_0)\Rightarrow u\equiv u(x_0)\) in each ball, so \(u\) is constant by connectedness. \(\blacksquare\)
State the maximum principle for harmonic functions. Use it to prove uniqueness of the Dirichlet problem \(\nabla^2 u=0\) in \(\Omega\), \(u=g\) on \(\partial\Omega\).
Maximum Principle: A function harmonic in a bounded domain \(\Omega\) and continuous on \(\bar\Omega\) attains its maximum (and minimum) on the boundary \(\partial\Omega\). Uniqueness: Suppose \(u,v\) are two solutions. Let \(w=u-v\). Then \(\nabla^2 w=0\) in \(\Omega\) and \(w=0\) on \(\partial\Omega\). By max principle, \(\max_{\bar\Omega}w=\max_{\partial\Omega}w=0\). Similarly \(\min w=0\). So \(w\equiv0\), i.e. \(u\equiv v\). \(\blacksquare\)
State Harnack's inequality for positive harmonic functions and describe one application.
Harnack's Inequality: If \(u\ge0\) is harmonic in a ball \(B(x_0,R)\), then for any \(r
Find the Green's function \(G(r,\theta;r_0,\theta_0)\) for the Dirichlet problem on the unit disk. Use it to write the solution formula.
Green's function: \(G(x;x_0)=\Phi(|x-x_0|)-\Phi(|x-x_0^*|)\) where \(x_0^*=x_0/|x_0|^2\) (reflection/inversion in unit circle) and \(\Phi(r)=\frac{1}{2\pi}\ln r\) is the fundamental solution in 2D. Explicitly for disk of radius \(a\): \(G(r,\theta;r_0,\theta_0)=\frac{1}{4\pi}\ln\frac{r^2r_0^2+a^4-2rr_0a^2\cos(\theta-\theta_0)}{a^2(r^2+r_0^2-2rr_0\cos(\theta-\theta_0))}\). Solution: \(u(x_0)=-\int_{\partial\Omega}u\frac{\partial G}{\partial n}\,ds\). The Poisson kernel \(K=-\frac{\partial G}{\partial n}\big|_{r=a}=\frac{a^2-r_0^2}{2\pi a}\cdot\frac{1}{a^2+r_0^2-2ar_0\cos(\theta_0-\theta)}\) recovers Poisson's formula. \(\blacksquare\)
State and prove Kelvin's inversion: if \(u(x)\) is harmonic in a domain \(D\subset\mathbb{R}^n\), then \(v(x)=|x|^{2-n}u(a^2x/|x|^2)\) is harmonic in the inverted domain.
Solve \(u_{xx}+u_{yy}=0\) on \(0<x<a,\,0<y<b\) with \(u=0\) on \(x=0,x=a,y=0\) and \(u(x,b)=f(x)\).
Separate: \(u=X(x)Y(y)\). \(X''/X=-Y''/Y=-\lambda^2\). BCs \(X(0)=X(a)=0\Rightarrow X_n=\sin\frac{n\pi x}{a},\lambda_n=\frac{n\pi}{a}\). Then \(Y_n''-\lambda_n^2 Y_n=0\) with \(Y_n(0)=0\Rightarrow Y_n=\sinh\frac{n\pi y}{a}\). General: \(u=\sum_{n=1}^\infty c_n\sinh\frac{n\pi y}{a}\sin\frac{n\pi x}{a}\). At \(y=b\): \(f(x)=\sum c_n\sinh\frac{n\pi b}{a}\sin\frac{n\pi x}{a}\). So \(c_n=\frac{2}{a\sinh(n\pi b/a)}\int_0^a f(x)\sin\frac{n\pi x}{a}\,dx\). \(\blacksquare\)
Solve \(u_{xx}+u_{yy}=0\) on \([0,a]\times[0,b]\) with \(u(0,y)=u(a,y)=0\), \(u(x,0)=0\), \(u(x,b)=\sin(\pi x/a)\).
From the general rectangle solution: \(u=\sum c_n\sinh\frac{n\pi y}{a}\sin\frac{n\pi x}{a}\). At \(y=b\): \(\sin\frac{\pi x}{a}=\sum c_n\sinh\frac{n\pi b}{a}\sin\frac{n\pi x}{a}\). By orthogonality only \(n=1\) survives: \(c_1\sinh\frac{\pi b}{a}=1\Rightarrow c_1=\frac{1}{\sinh(\pi b/a)}\). Final: \(\boxed{u(x,y)=\frac{\sinh(\pi y/a)}{\sinh(\pi b/a)}\sin\frac{\pi x}{a}}\). \(\blacksquare\)
Solve \(u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}=0\) inside \(r<a\) with \(u(a,\theta)=f(\theta)\). Derive Poisson's formula.
Separate: \(u=R(r)\Theta(\theta)\). \(\Theta''+\lambda\Theta=0\) with \(2\pi\)-periodicity \(\Rightarrow\lambda=n^2\), \(\Theta_n=A_n\cos n\theta+B_n\sin n\theta\). Euler equation for \(R\): \(r^2R''+rR'-n^2R=0\Rightarrow R_n=r^n\) (regular at \(r=0\)). Solution: \(u=\frac{a_0}{2}+\sum_{n=1}^\infty\left(\frac{r}{a}\right)^n(a_n\cos n\theta+b_n\sin n\theta)\). Fourier coefficients from BC. Summing the geometric series yields Poisson's formula: \(u(r,\theta)=\frac{a^2-r^2}{2\pi}\int_0^{2\pi}\frac{f(\phi)}{a^2-2ar\cos(\theta-\phi)+r^2}\,d\phi\). \(\blacksquare\)
Solve Laplace's equation inside the circle \(r<a\) with \(u(a,\theta)=\cos 2\theta\).
General interior solution: \(u=\frac{A_0}{2}+\sum_{n=1}^\infty r^n(A_n\cos n\theta+B_n\sin n\theta)\) (after normalising by \(a^n\)). At \(r=a\): \(\cos2\theta=\sum a^n(A_n\cos n\theta+B_n\sin n\theta)\). Match \(n=2\): \(a^2 A_2=1\Rightarrow A_2=1/a^2\); all other coefficients zero. So \(\boxed{u(r,\theta)=\frac{r^2}{a^2}\cos2\theta}\). \(\blacksquare\)
Solve Laplace's equation in the exterior region \(r>a\) with \(u(a,\theta)=\sin\theta\) and \(u\to0\) as \(r\to\infty\).
For exterior, bounded solutions are \(R_n=r^{-n}\). General: \(u=\sum_{n=1}^\infty r^{-n}(A_n\cos n\theta+B_n\sin n\theta)\). At \(r=a\): \(\sin\theta=\sum a^{-n}(A_n\cos n\theta+B_n\sin n\theta)\). Match \(n=1\): \(a^{-1}B_1=1\Rightarrow B_1=a\); all others zero. Solution: \(\boxed{u(r,\theta)=\frac{a}{r}\sin\theta}\). \(\blacksquare\)
State the Neumann problem for Laplace's equation. For \(\nabla^2 u=0\) inside \(r<a\) with \(\partial u/\partial n\big|_{r=a}=g(\theta)\), state the solvability condition and the form of the solution.
Neumann problem: find \(u\) harmonic in \(\Omega\) with \(\partial u/\partial n=g\) on \(\partial\Omega\). Solvability condition (divergence theorem): \(\int_{\partial\Omega}g\,dS=0\). On disk, \(\partial u/\partial n=u_r|_{r=a}\). With expansion \(u=\frac{A_0}{2}+\sum r^n(A_n\cos n\theta+B_n\sin n\theta)\), we get \(u_r|_{r=a}=\sum na^{n-1}(A_n\cos n\theta+B_n\sin n\theta)=g(\theta)\). So \(A_n=\frac{1}{na^{n-1}}\cdot\frac{1}{\pi}\int_0^{2\pi}g(\phi)\cos n\phi\,d\phi\), similarly \(B_n\). The constant \(A_0\) is arbitrary (solution unique up to additive constant). \(\blacksquare\)
Solve \(u_{xx}+u_{yy}=0\) for \(y>0\), \(-\infty<x<\infty\) with \(u(x,0)=f(x)\) and \(u\to0\) as \(y\to\infty\).
Take Fourier transform in \(x\): \(\hat{u}(\xi,y)=\int_{-\infty}^\infty u(x,y)e^{-i\xi x}dx\). Equation becomes \(\hat{u}_{yy}-\xi^2\hat{u}=0\Rightarrow\hat{u}=A(\xi)e^{-|\xi|y}\) (bounded as \(y\to\infty\)). BC: \(\hat{u}(\xi,0)=\hat{f}(\xi)\Rightarrow A=\hat{f}(\xi)\). Inverse transform (convolution): \(u(x,y)=\frac{y}{\pi}\int_{-\infty}^\infty\frac{f(t)}{(x-t)^2+y^2}\,dt\) (Poisson formula for half-plane). \(\blacksquare\)
Solve \(u_{xx}+u_{yy}=0\) on \(0<x<\pi,\,y>0\) with \(u(0,y)=0\), \(u_x(\pi,y)=0\), \(u(x,0)=x\), \(u\to0\) as \(y\to\infty\).
Separate: \(X''+\lambda^2 X=0\), \(X(0)=0\), \(X'(\pi)=0\). Eigenfunctions: \(X_n=\sin\frac{(2n-1)x}{2}\), \(\lambda_n=\frac{2n-1}{2}\), \(n=1,2,\ldots\) \(Y_n=e^{-\lambda_n y}\) (decays). Solution: \(u=\sum_{n=1}^\infty c_n e^{-\lambda_n y}\sin\lambda_n x\). IC: \(x=\sum c_n\sin\lambda_n x\) on \([0,\pi]\). Coefficients: \(c_n=\frac{2}{\pi}\int_0^\pi x\sin\frac{(2n-1)x}{2}dx=\frac{8(-1)^{n+1}}{\pi(2n-1)^2}\). Final: \(u=\frac{8}{\pi}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{(2n-1)^2}e^{-(2n-1)y/2}\sin\frac{(2n-1)x}{2}\). \(\blacksquare\)
State and prove the mean value property for harmonic functions. Hence show that a harmonic function cannot have an interior maximum unless it is constant.
Mean Value Property: If \(u\) is harmonic in \(\Omega\), then \(u(x_0)=\frac{1}{2\pi}\int_0^{2\pi}u(x_0+r\cos\theta,y_0+r\sin\theta)\,d\theta\) for any ball inside \(\Omega\). Proof: Let \(h(r)=\frac{1}{2\pi}\int_0^{2\pi}u(x_0+r\cos\theta,\ldots)\,d\theta\). Then \(h'(r)=\frac{1}{2\pi r}\int_{\partial B_r}\nabla u\cdot\hat{n}\,ds=\frac{1}{2\pi r}\iint_{B_r}\nabla^2 u\,dA=0\). So \(h\) is constant; at \(r=0\), \(h=u(x_0)\). Max principle: If interior max at \(x_0\), mean value \(=u(x_0)\ge\) all nearby values, so \(u\equiv u(x_0)\) in a ball, hence everywhere by connectedness. \(\blacksquare\)
State the maximum principle. Use it to prove uniqueness of the Dirichlet problem \(\nabla^2 u=0\) in \(\Omega\), \(u=g\) on \(\partial\Omega\).
Maximum Principle: A function harmonic in a bounded domain \(\Omega\), continuous on \(\bar\Omega\), attains its max and min on \(\partial\Omega\). Uniqueness: Let \(w=u-v\) where \(u,v\) both solve the Dirichlet problem. Then \(\nabla^2 w=0\) in \(\Omega\), \(w=0\) on \(\partial\Omega\). By max principle, \(\max_{\bar\Omega}w\le\max_{\partial\Omega}w=0\) and \(\min w\ge0\). So \(w\equiv0\). \(\blacksquare\)
State Harnack's inequality for positive harmonic functions and describe one application.
Harnack's Inequality (2D): If \(u\ge0\) is harmonic in \(B(x_0,R)\), then for \(|x-x_0|=r<R\): \(\frac{R-r}{R+r}u(x_0)\le u(x)\le\frac{R+r}{R-r}u(x_0)\). Proof sketch: Apply Poisson formula and bound the Poisson kernel above and below using \(R-r\le|x-y|\le R+r\) for \(|y|=R\). Application: Harnack's convergence theorem: a monotone sequence of harmonic functions converges either to \(\infty\) or to a harmonic function, uniformly on compacta. Also gives Liouville's theorem: bounded harmonic function on \(\mathbb{R}^n\) is constant. \(\blacksquare\)
Find the Green's function \(G(r,\theta;r_0,\theta_0)\) for the Dirichlet problem on the disk \(r<a\) and write the Poisson integral formula.
Green's function: \(G(x;x_0)=\Phi(|x-x_0|)-\Phi(|x-x_0^*|)\) where \(x_0^*=a^2 x_0/|x_0|^2\) is the inversion of \(x_0\) in the circle \(|x|=a\), and \(\Phi(r)=\frac{1}{2\pi}\ln r\) is the 2D fundamental solution. In polar coordinates: \(G=\frac{1}{4\pi}\ln\frac{r^2+r_0^2-2rr_0\cos(\theta-\theta_0)}{(r r_0/a)^2+a^2-2rr_0\cos(\theta-\theta_0)}\). The Poisson kernel is \(K(r_0,\theta_0;a,\theta)=-\frac{\partial G}{\partial r}\big|_{r=a}=\frac{a^2-r_0^2}{2\pi a}\cdot\frac{1}{a^2+r_0^2-2ar_0\cos(\theta_0-\theta)}\). Solution: \(u(r_0,\theta_0)=\int_0^{2\pi}K\cdot f(\theta)\,a\,d\theta\). \(\blacksquare\)
State and prove Kelvin's inversion: if \(u(x)\) is harmonic in a domain \(D\subset\mathbb{R}^n\), then \(v(x)=|x|^{2-n}u(a^2x/|x|^2)\) is harmonic in the inverted domain.
Let \(y=a^2x/|x|^2\) (inversion). The map \(x\mapsto y\) is conformal. In 3D (\(n=3\)): \(v(x)=|x|^{-1}u(y)\). Compute \(\nabla^2_x v\). Using the formula for Laplacian under inversion: \(\nabla^2_x(|x|^{2-n}u(y))=|x|^{-n-2}\cdot\underbrace{(\nabla^2_y u)}_{=0}\cdot a^4=0\). More explicitly in 3D: if \(r=|x|\), then the Kelvin transform \(v(x)=\frac{1}{r}u\!\left(\frac{a^2}{r^2}x\right)\). One verifies \(\nabla^2 v=r^{-5}\cdot r^4(\nabla^2_y u)=0\) since \(u\) is harmonic. This maps interior problems to exterior problems and vice versa. \(\blacksquare\)
Prove uniqueness of the solution to \(\nabla^2 u=f\) in \(\Omega\), \(u=g\) on \(\partial\Omega\) using the energy (Dirichlet integral) method.
Let \(u_1,u_2\) both satisfy \(\nabla^2 u_i=f\) with \(u_i=g\) on \(\partial\Omega\). Set \(w=u_1-u_2\). Then \(\nabla^2 w=0\) in \(\Omega\), \(w=0\) on \(\partial\Omega\). Multiply by \(w\) and integrate: \(\iint_\Omega w\nabla^2 w\,dA=0\). Green's first identity: \(\iint_\Omega w\nabla^2 w\,dA=\oint_{\partial\Omega}w\frac{\partial w}{\partial n}\,ds-\iint_\Omega|\nabla w|^2\,dA\). Boundary term vanishes (\(w=0\) on \(\partial\Omega\)). So \(\iint|\nabla w|^2\,dA=0\Rightarrow\nabla w\equiv0\Rightarrow w=\text{const}\). With \(w=0\) on boundary, \(w\equiv0\). \(\blacksquare\)
Show that \(u(x,y)\) is harmonic in a simply connected domain if and only if it is the real part of an analytic function \(f(z)=u+iv\).
\((\Rightarrow)\) If \(f=u+iv\) is analytic, Cauchy-Riemann gives \(u_x=v_y,u_y=-v_x\). Then \(u_{xx}+u_{yy}=v_{yx}-v_{xy}=0\) (equality of mixed partials). \((\Leftarrow)\) Given \(u\) harmonic in simply connected \(\Omega\), define \(v(x,y)=\int_{(x_0,y_0)}^{(x,y)}(-u_y\,dx+u_x\,dy)\). This is path-independent (since \(\partial(-u_y)/\partial y=\partial(u_x)/\partial x\Leftrightarrow u_{xx}+u_{yy}=0\)). Then \(v_x=-u_y,v_y=u_x\): C-R equations hold, so \(f=u+iv\) is analytic. The function \(v\) is the harmonic conjugate of \(u\). \(\blacksquare\)
Separate Laplace's equation in spherical coordinates \((r,\theta)\) (axisymmetric) and show the \(\theta\)-equation leads to Legendre's equation. Write the general solution.
Axisymmetric Laplace: \(\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 u_r)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta\, u_\theta)=0\). Set \(u=R(r)\Theta(\theta)\): \(\frac{(r^2R')'}{R}=-\frac{(\sin\theta\,\Theta')'}{\sin\theta\,\Theta}=n(n+1)\). Radial: \(r^2R''+2rR'-n(n+1)R=0\Rightarrow R=Ar^n+Br^{-n-1}\). Angular with \(\mu=\cos\theta\): \(\frac{d}{d\mu}[(1-\mu^2)\frac{d\Theta}{d\mu}]+n(n+1)\Theta=0\) (Legendre's equation). Bounded solutions: \(\Theta=P_n(\mu)=P_n(\cos\theta)\). General axisymmetric solution: \(u(r,\theta)=\sum_{n=0}^\infty(A_n r^n+B_n r^{-n-1})P_n(\cos\theta)\). \(\blacksquare\)
Define solid harmonics and surface harmonics. Show that \(r^n Y_n(\theta,\phi)\) is a solid harmonic where \(Y_n\) is a surface harmonic of degree \(n\).
Surface harmonic of degree \(n\): A homogeneous polynomial \(S_n(\theta,\phi)\) on the unit sphere satisfying the surface Laplacian equation \(\Delta_{S^2}S_n+n(n+1)S_n=0\). These are linear combinations of \(P_n^m(\cos\theta)e^{im\phi}\) for \(-n\le m\le n\). Solid harmonic of degree \(n\): \(H_n(x)=r^n S_n(\theta,\phi)\). Proof that \(H_n\) is harmonic: In spherical coords, \(\nabla^2(r^n S_n)=\frac{1}{r^2}[n(n+1)r^n S_n+r^n\Delta_{S^2}S_n]=\frac{r^n}{r^2}[n(n+1)S_n-n(n+1)S_n]=0\). \(\blacksquare\)
Derive the Poisson kernel for the unit disk by summing the Fourier series of the Dirichlet solution.
Interior solution: \(u(r,\theta)=\frac{a_0}{2}+\sum_{n=1}^\infty r^n(a_n\cos n\theta+b_n\sin n\theta)\) with \(a_n,b_n\) the Fourier coefficients of \(f(\phi)=u(1,\phi)\). Write \(a_n\cos n\theta+b_n\sin n\theta=\frac{1}{\pi}\int_0^{2\pi}f(\phi)\cos n(\theta-\phi)\,d\phi\). Substituting: \(u(r,\theta)=\frac{1}{\pi}\int_0^{2\pi}f(\phi)\left[\frac{1}{2}+\sum_{n=1}^\infty r^n\cos n(\theta-\phi)\right]d\phi\). The bracket sums via geometric series: \(\frac{1}{2}+\text{Re}\sum_{n=1}^\infty r^n e^{in\alpha}=\frac{1}{2}+\text{Re}\frac{re^{i\alpha}}{1-re^{i\alpha}}=\frac{1-r^2}{2(1-2r\cos\alpha+r^2)}\). Poisson kernel: \(P(r,\alpha)=\frac{1-r^2}{1-2r\cos\alpha+r^2}\) (for unit disk). \(\blacksquare\)
Show that for a solution of \(\nabla^2u=0\) of the form \(u=r^nf(\cos\theta)\), the function \(f\) satisfies Legendre's equation of order \(n\).
Substitute \(u=r^n f(\mu)\) with \(\mu=\cos\theta\) into the axisymmetric Laplace operator \(\nabla^2u=\frac{1}{r^2}\frac{d}{dr}(r^2 u_r)+\frac{1}{r^2\sin\theta}\frac{d}{d\theta}(\sin\theta\,u_\theta)\). First term: \(\frac{1}{r^2}\frac{d}{dr}(r^2\cdot nr^{n-1}f)=\frac{n(n+1)r^n f}{r^2}\). Second term: using \(\frac{d}{d\theta}=\frac{d\mu}{d\theta}\frac{d}{d\mu}=-\sin\theta\frac{d}{d\mu}\) and \(\sin\theta=(1-\mu^2)^{1/2}\): \(\frac{1}{r^2\sin\theta}(-\sin\theta)\frac{d}{d\mu}\left[\sin\theta\cdot(-\sin\theta f')\right]=\frac{r^n}{r^2}\frac{d}{d\mu}[(1-\mu^2)f']\). Setting sum to zero: \((1-\mu^2)f''-2\mu f'+n(n+1)f=0\): Legendre's equation of order \(n\). \(\blacksquare\)
Consider \(-(p(x)X')'+q(x)X=\lambda w(x)X\) on \([a,b]\) with Sturm-Liouville BCs. Prove the eigenfunctions corresponding to distinct eigenvalues are orthogonal with weight \(w\).
Let \(X_m,X_n\) satisfy the SL equation with eigenvalues \(\lambda_m\ne\lambda_n\). Write the two equations, multiply the first by \(X_n\) and the second by \(X_m\), subtract and integrate over \([a,b]\): \((\lambda_n-\lambda_m)\int_a^b w X_m X_n\,dx=\int_a^b[X_n(pX_m')'-X_m(pX_n')']\,dx\). Integrate the right side by parts: \(=[p(X_n X_m'-X_m X_n')]_a^b\). The SL boundary conditions ensure this boundary term vanishes. Since \(\lambda_m\ne\lambda_n\), we get \(\int_a^b w X_m X_n\,dx=0\). \(\blacksquare\)
Find the eigenvalues and eigenfunctions of \(X''+\lambda X=0\) on \([0,L]\) with \(X'(0)=X'(L)=0\). Write the eigenfunction expansion.
For \(\lambda=0\): \(X=A+Bx\), \(X'=B=0\Rightarrow X=A\) (constant). \(\lambda_0=0\), \(X_0=1\). For \(\lambda=\mu^2>0\): \(X=A\cos\mu x+B\sin\mu x\), \(X'(0)=B\mu=0\Rightarrow B=0\). \(X'(L)=-A\mu\sin\mu L=0\Rightarrow\mu_n=n\pi/L\). So \(\lambda_n=(n\pi/L)^2\), \(X_n=\cos\frac{n\pi x}{L}\), \(n=0,1,2,\ldots\) Expansion: \(f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos\frac{n\pi x}{L}\) (Fourier cosine series), where \(a_n=\frac{2}{L}\int_0^L f(x)\cos\frac{n\pi x}{L}\,dx\). \(\blacksquare\)
Solve the heat equation \(u_t=u_{xx}\) with \(u(0,t)=u(\pi,t)=0\) and \(u(x,0)=\sin x-\sin 3x\).
Eigenfunctions: \(X_n=\sin nx\), eigenvalues \(\lambda_n=n^2\). General: \(u=\sum_{n=1}^\infty b_n e^{-n^2 t}\sin nx\). IC: \(b_n\sin nx=\sin x-\sin 3x\). By inspection: \(b_1=1,b_3=-1\), all others zero. Solution: \(\boxed{u(x,t)=e^{-t}\sin x-e^{-9t}\sin 3x}\). \(\blacksquare\)
Solve \(u_t=ku_{xx}\) for \(x>0,t>0\) with \(u_x(0,t)=0\) and \(u(x,0)=f(x)\) using the Fourier cosine transform.
Fourier cosine transform: \(\hat{u}_c(\xi,t)=\sqrt{\frac{2}{\pi}}\int_0^\infty u(x,t)\cos(\xi x)\,dx\). Transform the PDE: \(\frac{d\hat{u}_c}{dt}=-k\xi^2\hat{u}_c+k\sqrt{\frac{2}{\pi}}u_x(0,t)=-k\xi^2\hat{u}_c\) (using \(u_x(0,t)=0\)). Solution: \(\hat{u}_c(\xi,t)=\hat{f}_c(\xi)e^{-k\xi^2 t}\). Invert: \(u(x,t)=\sqrt{\frac{2}{\pi}}\int_0^\infty\hat{f}_c(\xi)e^{-k\xi^2 t}\cos(\xi x)\,d\xi\). This is equivalent to the method of images (even extension) giving: \(u(x,t)=\frac{1}{\sqrt{4\pi kt}}\int_0^\infty[e^{-(x-s)^2/4kt}+e^{-(x+s)^2/4kt}]f(s)\,ds\). \(\blacksquare\)
Solve \(\nabla^2 u=f(x,y)\) (Poisson equation) on \([0,a]\times[0,b]\) with \(u=0\) on all four sides using a double Fourier sine series.
Expand \(u=\sum_{m,n}u_{mn}\sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b}\) and \(f=\sum_{m,n}f_{mn}\sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b}\). Substituting: \(-\left[\left(\frac{m\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2\right]u_{mn}=f_{mn}\). So \(u_{mn}=-\frac{f_{mn}}{(m\pi/a)^2+(n\pi/b)^2}\), where \(f_{mn}=\frac{4}{ab}\int_0^a\int_0^b f(x,y)\sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b}\,dy\,dx\). \(\blacksquare\)
Solve the telegraph equation \(u_{tt}+2\alpha u_t=c^2u_{xx}\) with \(u(0,t)=u(L,t)=0\), \(u(x,0)=f(x)\), \(u_t(x,0)=0\) by separation of variables.
Set \(u=e^{-\alpha t}v(x,t)\). Then \(v_{tt}=c^2v_{xx}-(\alpha^2)v\). Separate \(v=X(x)T(t)\): \(X_n=\sin\frac{n\pi x}{L}\). Then \(T''+(\omega_n^2+\alpha^2)T=0\) where \(\omega_n^2=c^2(n\pi/L)^2\). So \(T_n=A_n\cos\mu_n t+B_n\sin\mu_n t\) with \(\mu_n=\sqrt{(n\pi c/L)^2-\alpha^2}\) (assuming underdamped). General: \(u=\sum e^{-\alpha t}(A_n\cos\mu_n t+B_n\sin\mu_n t)\sin\frac{n\pi x}{L}\). IC \(u_t(x,0)=0\) gives \(B_n=\alpha A_n/\mu_n\). \(A_n\) from Fourier expansion of \(f\). \(\blacksquare\)
Solve the beam equation \(u_{tt}+K^2u_{xxxx}=0\) with simply supported BCs \(u(0,t)=u(L,t)=u_{xx}(0,t)=u_{xx}(L,t)=0\) and initial conditions \(u(x,0)=f(x)\), \(u_t(x,0)=g(x)\).
Separate: \(u=X(x)T(t)\). BCs imply \(X_n=\sin\frac{n\pi x}{L}\) (satisfies \(X=X''=0\) at both ends). \(X_n^{(4)}=(n\pi/L)^4 X_n\). Then \(T_n''+(K n^2\pi^2/L^2)^2 T_n=0\Rightarrow\omega_n=K(n\pi/L)^2\). Solution: \(u=\sum(A_n\cos\omega_n t+B_n\sin\omega_n t)\sin\frac{n\pi x}{L}\). \(A_n=\frac{2}{L}\int_0^Lf(x)\sin\frac{n\pi x}{L}dx\), \(B_n=\frac{1}{\omega_n}\cdot\frac{2}{L}\int_0^Lg(x)\sin\frac{n\pi x}{L}dx\). \(\blacksquare\)
Solve \(u_t=Du_{xx}-ku\) on \(0<x<L\) with \(u(0,t)=u(L,t)=0\) and \(u(x,0)=f(x)\).
Set \(u=e^{-kt}v\). Then \(e^{-kt}v_t=De^{-kt}v_{xx}\Rightarrow v_t=Dv_{xx}\) (standard heat equation). Solve: \(v=\sum b_n e^{-D(n\pi/L)^2 t}\sin\frac{n\pi x}{L}\) with \(b_n=\frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}dx\). Hence \(u=\sum b_n e^{-(k+D(n\pi/L)^2)t}\sin\frac{n\pi x}{L}\). The decay rate of each mode is increased by \(k\). \(\blacksquare\)
Solve \(u_t=u_{xx}+f(x,t)\) on \(0<x<\pi\) with \(u(0,t)=u(\pi,t)=0\) and \(u(x,0)=0\) using eigenfunction expansion.
Eigenfunctions: \(\phi_n=\sin nx\), \(\lambda_n=n^2\). Expand \(u=\sum T_n(t)\sin nx\), \(f=\sum f_n(t)\sin nx\) where \(f_n(t)=\frac{2}{\pi}\int_0^\pi f(x,t)\sin nx\,dx\). Substituting: \(T_n'+n^2T_n=f_n(t)\), \(T_n(0)=0\). By integrating factor: \(T_n(t)=\int_0^t e^{-n^2(t-s)}f_n(s)\,ds\) (Duhamel). Final: \(u=\sum_{n=1}^\infty\left(\int_0^t e^{-n^2(t-s)}f_n(s)\,ds\right)\sin nx\). \(\blacksquare\)
Solve \(u_{tt}=c^2(u_{xx}+u_{yy})\) on the square \([0,\pi]\times[0,\pi]\) with \(u=0\) on all boundaries, \(u(x,y,0)=f(x,y)\), \(u_t(x,y,0)=0\).
Separate: \(u=X(x)Y(y)T(t)\). BCs: \(X_m=\sin mx,Y_n=\sin ny\). Then \(T_{mn}''+\omega_{mn}^2T_{mn}=0\) with \(\omega_{mn}=c\sqrt{m^2+n^2}\). With \(u_t(x,y,0)=0\): \(T_{mn}=A_{mn}\cos(\omega_{mn}t)\). Solution: \(u=\sum_{m,n=1}^\infty A_{mn}\cos(\omega_{mn}t)\sin mx\sin ny\). Coefficients: \(A_{mn}=\frac{4}{\pi^2}\int_0^\pi\int_0^\pi f(x,y)\sin mx\sin ny\,dx\,dy\). \(\blacksquare\)
Solve \(u_{rr}+\frac{1}{r}u_r+u_{zz}=0\) in a cylinder \(0<r<a,0<z<L\) with \(u(a,z)=0\), \(u(r,0)=0\), \(u(r,L)=f(r)\).
Separate \(u=R(r)Z(z)\). \(\frac{R''+R'/r}{R}=-\frac{Z''}{Z}=\mu^2\). \(Z''-\mu^2Z=0\) with \(Z(0)=0\Rightarrow Z=\sinh\mu z\). \(R''+\frac{1}{r}R'+\mu^2R=0\): Bessel equation of order 0. Bounded solution: \(R=J_0(\mu r)\). BC \(u(a,z)=0\Rightarrow J_0(\mu a)=0\Rightarrow\mu_n=\xi_{0n}/a\) where \(\xi_{0n}\) are zeros of \(J_0\). General: \(u=\sum c_n J_0(\mu_n r)\sinh(\mu_n z)\). At \(z=L\): \(f(r)=\sum c_n\sinh(\mu_n L)J_0(\mu_n r)\). Coefficients by Bessel orthogonality: \(c_n=\frac{2}{a^2 J_1^2(\xi_{0n})\sinh(\mu_n L)}\int_0^a r f(r)J_0(\mu_n r)\,dr\). \(\blacksquare\)
Show that the Bessel functions \(J_n(\xi_{nm}r/a)\) are orthogonal on \([0,a]\) with weight \(r\) and compute the norm squared.
Bessel's equation \(r J_n''+J_n'+(\mu^2 r-n^2/r)J_n=0\) is a Sturm-Liouville problem with weight \(w(r)=r\) on \([0,a]\). Orthogonality: \(\int_0^a r J_n(\xi_{nk}r/a)J_n(\xi_{nl}r/a)\,dr=0\) for \(k\ne l\). Norm squared (by Bessel's identity): \(\int_0^a r J_n^2(\xi_{nm}r/a)\,dr=\frac{a^2}{2}[J_{n+1}(\xi_{nm})]^2=\frac{a^2}{2}[J_n'(\xi_{nm})]^2\). (These follow by differentiating the Bessel equation and using the recurrence \(J_{n-1}-J_{n+1}=2J_n'\).) \(\blacksquare\)
Expand \(f(x)=x^2\) in a Legendre series on \([-1,1]\). State the orthogonality relation used.
Orthogonality: \(\int_{-1}^1 P_m(x)P_n(x)\,dx=\frac{2}{2n+1}\delta_{mn}\). Expansion: \(x^2=\sum_{n=0}^\infty c_n P_n(x)\). Coefficients: \(c_n=\frac{2n+1}{2}\int_{-1}^1 x^2P_n(x)\,dx\). Since \(x^2=\frac{2}{3}P_2(x)+\frac{1}{3}P_0(x)\) (using \(P_0=1,P_2=\frac{1}{2}(3x^2-1)\)), only \(c_0\) and \(c_2\) are nonzero. Check: \(\frac{2}{3}P_2+\frac{1}{3}P_0=\frac{2}{3}\cdot\frac{3x^2-1}{2}+\frac{1}{3}=x^2-\frac{1}{3}+\frac{1}{3}=x^2\checkmark\). So \(\boxed{x^2=\frac{1}{3}P_0(x)+\frac{2}{3}P_2(x)}\). \(\blacksquare\)
State and apply Duhamel's principle to solve \(u_t=ku_{xx}+F(x,t)\) on \(-\infty<x<\infty\) with \(u(x,0)=0\).
Duhamel's Principle: For the heat equation with source, write the solution as superposition of solutions to impulse problems: \(u(x,t)=\int_0^t v(x,t;s)\,ds\) where \(v(x,t;s)\) solves \(v_t=kv_{xx}\) for \(t>s\) with \(v(x,s;s)=F(x,s)\). For the whole line, \(v(x,t;s)=\frac{1}{\sqrt{4\pi k(t-s)}}\int_{-\infty}^\infty F(y,s)e^{-(x-y)^2/4k(t-s)}dy\). Hence: \(u(x,t)=\int_0^t\frac{1}{\sqrt{4\pi k(t-s)}}\int_{-\infty}^\infty F(y,s)e^{-(x-y)^2/4k(t-s)}dy\,ds\). \(\blacksquare\)
Describe the method for solving \(\nabla^4 u=0\) (biharmonic equation) on a rectangle with Dirichlet BCs. What is the general form of the solution?
Write \(\nabla^4 u=\nabla^2(\nabla^2 u)=0\). Seek solutions of the form \(u=X(x)Y(y)\) (separation less clean for biharmonic). Alternative: Fix \(x\)-eigenfunctions \(X_n=\sin\frac{n\pi x}{a}\) (satisfying BCs in \(x\)). Then \(Y_n\) satisfies \(Y_n^{(4)}-2(n\pi/a)^2 Y_n''+(n\pi/a)^4 Y_n=0\). Characteristic roots: \(r^2=(n\pi/a)^2\) (double). So \(Y_n=(A_n+B_n y)\sinh\frac{n\pi y}{a}+(C_n+D_n y)\cosh\frac{n\pi y}{a}\). Four coefficients per mode determined by four BCs in \(y\). \(\blacksquare\)
Write the time-dependent Schrödinger equation and separate it into spatial and temporal parts for a particle in a box \([0,L]\). Find the energy eigenvalues.
Schrödinger equation: \(i\hbar\psi_t=-\frac{\hbar^2}{2m}\psi_{xx}+V(x)\psi\). For particle in box (\(V=0\) inside, \(\psi(0,t)=\psi(L,t)=0\)): set \(\psi=\phi(x)T(t)\). Then \(i\hbar T'/T=-\frac{\hbar^2}{2m}\phi''/\phi=E\) (separation constant). \(T=e^{-iEt/\hbar}\). Spatial: \(\phi''+(2mE/\hbar^2)\phi=0\) with BCs \(\Rightarrow\phi_n=\sin\frac{n\pi x}{L}\), energy eigenvalues: \(E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\), \(n=1,2,3,\ldots\) General state: \(\psi=\sum c_n\sin\frac{n\pi x}{L}e^{-iE_n t/\hbar}\). \(\blacksquare\)
Solve \(u_t=ku_{xx}\) on \(x>0\), \(t>0\) with \(u(0,t)=0\) and \(u(x,0)=f(x)\) using the method of images.
Extend \(f\) as an odd function to all of \(\mathbb{R}\): \(\tilde{f}(x)=f(x)\) for \(x>0\), \(\tilde{f}(-x)=-f(x)\). The solution on the whole line with IC \(\tilde{f}\) satisfies \(u(0,t)=0\) automatically (odd symmetry). Whole-line solution: \(u(x,t)=\frac{1}{\sqrt{4\pi kt}}\int_{-\infty}^\infty\tilde{f}(y)e^{-(x-y)^2/4kt}dy\). Splitting: \(u(x,t)=\frac{1}{\sqrt{4\pi kt}}\int_0^\infty f(y)[e^{-(x-y)^2/4kt}-e^{-(x+y)^2/4kt}]dy\). The second exponential is the "image" (negative source at \(-y\)). \(\blacksquare\)
Define the generating function for Legendre polynomials. Use it to derive \(P_0(x)=1\) and \(P_1(x)=x\).
Generating function: \(\frac{1}{\sqrt{1-2xt+t^2}}=\sum_{n=0}^\infty P_n(x)t^n\) for \(|t|<1\). Derivation of \(P_0,P_1\): Expand the left side in powers of \(t\): \((1-2xt+t^2)^{-1/2}=(1-(2xt-t^2))^{-1/2}\approx1+\frac{1}{2}(2xt-t^2)+\frac{3}{8}(2xt)^2+\cdots\). Collecting \(t^0\): coefficient is \(1=P_0(x)\). Collecting \(t^1\): coefficient is \(x=P_1(x)\). Recurrence from differentiating with respect to \(t\): \((n+1)P_{n+1}(x)=(2n+1)xP_n(x)-nP_{n-1}(x)\). \(\blacksquare\)
Solve \(u_t=u_{xx}\) on \(-\infty<x<\infty\) with \(u(x,0)=e^{-x^2}\) using the Fourier transform.
Fourier transform: \(\hat{u}_t=-\xi^2\hat{u}\Rightarrow\hat{u}(\xi,t)=\hat{u}(\xi,0)e^{-\xi^2 t}\). The transform of \(e^{-x^2}\) is \(\hat{f}(\xi)=\sqrt{\pi}e^{-\xi^2/4}\). So \(\hat{u}=\sqrt{\pi}e^{-\xi^2/4}e^{-\xi^2 t}=\sqrt{\pi}e^{-\xi^2(1/4+t)}\). Inverse transform (using \(\mathcal{F}^{-1}\{e^{-a\xi^2}\}=\frac{1}{\sqrt{4\pi a}}e^{-x^2/4a}\)): \(u(x,t)=\frac{1}{\sqrt{1+4t}}e^{-x^2/(1+4t)}\). This is a Gaussian broadening of the initial data. \(\blacksquare\)
Solve \(u_t=u_{xx}\) for \(x>0,t>0\) with \(u(x,0)=0\), \(u(0,t)=t\), and \(u\to0\) as \(x\to\infty\), using the Laplace transform in \(t\).
Let \(U(x,s)=\mathcal{L}\{u\}\). Transform the PDE: \(sU-u(x,0)=U_{xx}\Rightarrow U_{xx}-sU=0\). Solution bounded as \(x\to\infty\): \(U=A(s)e^{-\sqrt{s}\,x}\). BC: \(U(0,s)=\mathcal{L}\{t\}=1/s^2\Rightarrow A(s)=1/s^2\). So \(U=\frac{1}{s^2}e^{-\sqrt{s}\,x}\). Invert using tables: \(\mathcal{L}^{-1}\{s^{-2}e^{-\sqrt{s}\,x}\}=\left(t+\frac{x^2}{2}\right)\text{erfc}\!\left(\frac{x}{2\sqrt{t}}\right)-\frac{x}{\sqrt{\pi t}}e^{-x^2/4t}\cdot\sqrt{t}\). Simplified: \(u(x,t)=\left(t+\frac{x^2}{2}\right)\text{erfc}\!\left(\frac{x}{2\sqrt{t}}\right)-x\sqrt{\frac{t}{\pi}}e^{-x^2/(4t)}\). \(\blacksquare\)
Find the Green's function \(G(x,\xi)\) for \(-u''=f(x)\) on \(x>0\) with \(u(0)=0\) and \(u\to0\) as \(x\to\infty\).
Use the method of images. The free-space Green's function in 1D for \(-G''=\delta(x-\xi)\) is \(G_0=-|x-\xi|/2\) (up to sign convention, or \(G_0=\frac{1}{2}|x-\xi|\) for \(-G''=\delta\)). To enforce \(G(0,\xi)=0\), place an image at \(-\xi\): \(G(x,\xi)=\frac{1}{2}|x-\xi|-\frac{1}{2}|x+\xi|\). For \(x,\xi>0\) and \(x\ne\xi\): if \(x<\xi\), \(G=\frac{1}{2}(\xi-x)-\frac{1}{2}(x+\xi)=-x\); if \(x>\xi\), \(G=\frac{1}{2}(x-\xi)-\frac{1}{2}(x+\xi)=-\xi\). Compact form: \(G(x,\xi)=-\min(x,\xi)\). Solution: \(u(x)=\int_0^\infty G(x,\xi)f(\xi)\,d\xi=-\int_0^x\xi f(\xi)\,d\xi-x\int_x^\infty f(\xi)\,d\xi\). \(\blacksquare\)
Construct the Green's function \(G(x,\xi)\) for \(-u''=f(x)\) on \([0,1]\) with \(u(0)=u(1)=0\).
For \(x\ne\xi\), \(G''=0\Rightarrow G\) is piecewise linear. Conditions: continuity at \(x=\xi\), jump in derivative \(G_x(\xi^+,\xi)-G_x(\xi^-,\xi)=-1\), and BCs \(G(0,\xi)=G(1,\xi)=0\). For \(x<\xi\): \(G=Ax\) (vanishes at 0). For \(x>\xi\): \(G=B(1-x)\) (vanishes at 1). Continuity: \(A\xi=B(1-\xi)\). Jump: \(-B-A=-1\Rightarrow A+B=1\). Solve: \(A=1-\xi,B=\xi\). So: \(\boxed{G(x,\xi)=\begin{cases}x(1-\xi),&x\le\xi\\\xi(1-x),&x\ge\xi\end{cases}}\). Solution: \(u(x)=\int_0^1 G(x,\xi)f(\xi)\,d\xi\). \(\blacksquare\)
Verify that \(K(x,t)=\frac{1}{\sqrt{4\pi kt}}e^{-x^2/4kt}\) for \(t>0\) satisfies \(K_t=kK_{xx}\) and \(\int_{-\infty}^\infty K\,dx=1\).
Verification of PDE: \(K_t=K\left(-\frac{1}{2t}+\frac{x^2}{4kt^2}\right)\). \(K_x=K\cdot\frac{-x}{2kt}\). \(K_{xx}=K\left[\frac{x^2}{4k^2t^2}-\frac{1}{2kt}\right]\). So \(kK_{xx}=K\left[\frac{x^2}{4kt^2}-\frac{1}{2t}\right]=K_t\checkmark\). Verification of integral: \(\int_{-\infty}^\infty K\,dx=\frac{1}{\sqrt{4\pi kt}}\int_{-\infty}^\infty e^{-x^2/4kt}dx\). Substitute \(u=x/\sqrt{4kt}\): \(=\frac{1}{\sqrt{4\pi kt}}\cdot\sqrt{4kt}\int_{-\infty}^\infty e^{-u^2}du=\frac{\sqrt{\pi}}{\sqrt{\pi}}=1\checkmark\). As \(t\to0^+\), \(K(x,t)\to\delta(x)\). \(\blacksquare\)
Show that \(\Phi(x)=-\frac{1}{4\pi|x|}\) satisfies \(-\nabla^2\Phi=\delta(x)\) in \(\mathbb{R}^3\) (distributional sense).
For \(x\ne0\): compute \(\nabla^2(1/r)=\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\frac{1}{r}\right)=\frac{1}{r^2}\frac{d}{dr}(-1)=0\). For the distributional statement, take any test function \(\phi\in C_c^\infty\): \(\langle-\nabla^2\Phi,\phi\rangle=\langle\Phi,-\nabla^2\phi\rangle=-\frac{1}{4\pi}\int_{\mathbb{R}^3}\frac{\nabla^2\phi}{|x|}dx\). By Green's theorem (excluding a ball \(B_\epsilon\)): \(\int_{|x|>\epsilon}\frac{\nabla^2\phi}{|x|}dx=\int_{|x|>\epsilon}\nabla^2(1/|x|)\phi\,dx+\oint_{|x|=\epsilon}\left[\phi\nabla(1/|x|)-\frac{1}{|x|}\nabla\phi\right]\cdot\hat{n}\,dS\). First integral vanishes (\(\nabla^2(1/r)=0\)). Surface integrals: area of sphere \(=4\pi\epsilon^2\). \(|\nabla(1/|x|)|=1/\epsilon^2\) gives \(\phi(0)\cdot4\pi\epsilon^2\cdot 1/\epsilon^2=4\pi\phi(0)\). Taking \(\epsilon\to0\): \(-\nabla^2\Phi=\delta\). \(\blacksquare\)
Write the retarded Green's function for the wave equation \(u_{tt}-c^2\nabla^2 u=f(x,t)\) in \(\mathbb{R}^3\) and give the solution formula (Kirchhoff's formula).
The retarded fundamental solution (Green's function) is \(G(x,t;y,s)=\frac{\delta(t-s-|x-y|/c)}{4\pi c|x-y|}\) for \(t>s\) (zero for \(t<s\)). It represents a signal emitted at \((y,s)\) arriving at \((x,t)\) at speed \(c\). Solution by convolution: \(u(x,t)=\int_0^t\int_{\mathbb{R}^3}\frac{f(y,s)\delta(t-s-|x-y|/c)}{4\pi c|x-y|}dy\,ds=\frac{1}{4\pi c}\int_{|x-y|=c(t-s)}\frac{f(y,t-|x-y|/c)}{|x-y|}dS_y\). For homogeneous wave equation with ICs, Kirchhoff's formula: \(u(x,t)=\frac{\partial}{\partial t}\left(\frac{t}{4\pi}\fint_{|x-y|=ct}u_0\,dS\right)+\frac{t}{4\pi}\fint_{|x-y|=ct}v_0\,dS\). \(\blacksquare\)
Define the Z-transform and use it to solve the difference equation \(u_{n+1}-2u_n+u_{n-1}=0\), \(u_0=0\), \(u_1=1\).
Z-transform: \(U(z)=\sum_{n=0}^\infty u_n z^{-n}\). Properties: \(\mathcal{Z}\{u_{n+k}\}=z^k U(z)-z^k u_0-z^{k-1}u_1-\cdots\). Transform the equation: \((z^2U-z^2u_0-zu_1)-2(zU-zu_0)+U=0\). With \(u_0=0,u_1=1\): \((z^2-2z+1)U=z\Rightarrow U=\frac{z}{(z-1)^2}\). Partial fractions: \(U=\frac{z}{(z-1)^2}\). Using Z-transform table: \(\mathcal{Z}^{-1}\{z/(z-1)^2\}=n\). So \(u_n=n\). Check: \(u_{n+1}-2u_n+u_{n-1}=(n+1)-2n+(n-1)=0\checkmark\). \(\blacksquare\)
Define the Hankel transform of order zero. Apply it to solve \(u_t=k(u_{rr}+u_r/r)\) for \(r>0,t>0\) with \(u(r,0)=f(r)\).
Hankel transform of order 0: \(\hat{u}(\xi,t)=\int_0^\infty r u(r,t)J_0(\xi r)\,dr\). Key property: \(\mathcal{H}_0\{u_{rr}+u_r/r\}=-\xi^2\hat{u}\). Transform the PDE: \(\hat{u}_t=-k\xi^2\hat{u}\Rightarrow\hat{u}(\xi,t)=\hat{f}(\xi)e^{-k\xi^2 t}\). Invert: \(u(r,t)=\int_0^\infty\xi\hat{f}(\xi)e^{-k\xi^2 t}J_0(\xi r)\,d\xi\). This integral can also be evaluated using the convolution for the axisymmetric heat kernel: \(u(r,t)=\frac{1}{2kt}e^{-r^2/4kt}\int_0^\infty f(s)e^{-s^2/4kt}I_0\!\left(\frac{rs}{2kt}\right)s\,ds\), where \(I_0\) is the modified Bessel function. \(\blacksquare\)
Define the Mellin transform. Find the Mellin transform of \(f(x)=e^{-x}\) and identify the strip of analyticity.
Mellin transform: \(\mathcal{M}\{f\}(s)=\int_0^\infty x^{s-1}f(x)\,dx\). For \(f(x)=e^{-x}\): \(\mathcal{M}\{e^{-x}\}(s)=\int_0^\infty x^{s-1}e^{-x}dx=\Gamma(s)\). Strip of analyticity: \(\text{Re}(s)>0\) (the gamma function \(\Gamma(s)\) is analytic for \(\text{Re}(s)>0\)). Inverse: \(f(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}x^{-s}\Gamma(s)\,ds=e^{-x}\) (by residue theorem). The Mellin transform maps multiplicative convolutions to products, analogous to how the Fourier transform handles additive convolutions. \(\blacksquare\)
Take the double Fourier transform of \(u_{tt}=c^2(u_{xx}+u_{yy})\) on \(\mathbb{R}^2\times(0,\infty)\) and solve for \(\hat{u}(\xi_1,\xi_2,t)\). Give the dispersion relation.
Let \(\hat{u}(\xi_1,\xi_2,t)=\iint u(x,y,t)e^{-i(\xi_1 x+\xi_2 y)}dx\,dy\). Transform: \(\hat{u}_{tt}=-c^2(\xi_1^2+\xi_2^2)\hat{u}=-c^2|\xi|^2\hat{u}\). Dispersion relation: \(\omega=\pm c|\xi|\) where \(|\xi|=\sqrt{\xi_1^2+\xi_2^2}\). Solution: \(\hat{u}(\xi,t)=A(\xi)\cos(c|\xi|t)+B(\xi)\sin(c|\xi|t)\). With ICs \(u(x,y,0)=f\), \(u_t(x,y,0)=g\): \(A=\hat{f}(\xi)\), \(B=\hat{g}(\xi)/(c|\xi|)\). Invert by 2D Fourier integral to get \(u(x,y,t)\). \(\blacksquare\)
Solve \(u_{tt}-c^2u_{xx}=F(x,t)\) on \(-\infty<x<\infty\) with \(u(x,0)=u_t(x,0)=0\) using Fourier transform and Duhamel's principle.
Fourier transform in \(x\): \(\hat{u}_{tt}+c^2\xi^2\hat{u}=\hat{F}(\xi,t)\). This is a forced harmonic oscillator with \(\hat{u}(0)=\hat{u}_t(0)=0\). By variation of parameters: \(\hat{u}(\xi,t)=\int_0^t\frac{\hat{F}(\xi,s)}{c|\xi|}\sin(c|\xi|(t-s))\,ds\). Inverting: \(u(x,t)=\frac{1}{2c}\int_0^t\int_{x-c(t-s)}^{x+c(t-s)}F(y,s)\,dy\,ds\). This is d'Alembert's formula extended to the inhomogeneous case (Duhamel's principle for the wave equation). The signal from source \((y,s)\) reaches \((x,t)\) inside the light cone \(|x-y|\le c(t-s)\). \(\blacksquare\)
Solve Burgers' equation \(u_t+uu_x=\nu u_{xx}\) using the Cole-Hopf transformation \(u=-2\nu(\ln\phi)_x\).
Substitute \(u=-2\nu\phi_x/\phi\) into Burgers' equation. Compute: \(u_t=-2\nu(\phi_{xt}\phi-\phi_x\phi_t)/\phi^2\), \(uu_x=4\nu^2\phi_x\phi_{xx}/\phi^2-4\nu^2\phi_x^2\phi_{xx}/\phi^3\), \(\nu u_{xx}=-2\nu^2(\phi_{xxx}/\phi-3\phi_{xx}\phi_x/\phi^2+2\phi_x^3/\phi^3)\). After substitution and simplification: \(-\frac{2\nu}{\phi}\left[\phi_{xt}-\nu\phi_{xxt}\cdot\phi/\phi\right]=0\) reduces to \(\frac{\partial}{\partial x}\left(\phi_t-\nu\phi_{xx}\right)=0\). Choosing \(\phi_t=\nu\phi_{xx}\) (heat equation!), all terms cancel. So: solve the linear heat equation \(\phi_t=\nu\phi_{xx}\) for \(\phi\), then \(u=-2\nu(\ln\phi)_x\). For IC \(u(x,0)=u_0(x)\): \(\phi(x,0)=\exp\!\left(-\frac{1}{2\nu}\int_0^x u_0(s)\,ds\right)\). \(\blacksquare\)
Derive the eikonal equation \(|\nabla u|=n(x)\) from the wave equation in the high-frequency (geometric optics) limit. State what \(u\) and \(n\) represent.
Start with the Helmholtz equation \(\nabla^2 E+k^2n^2(x)E=0\) (time-harmonic waves, \(k=\omega/c\)). Seek WKB ansatz: \(E=A(x)e^{ik u(x)}\) where \(u\) is the phase (wavefront function) and \(n(x)\) is the refractive index. Substitute: \([\nabla^2A+2ik\nabla A\cdot\nabla u+ikA\nabla^2 u-k^2A|\nabla u|^2+k^2n^2A]e^{iku}=0\). Leading order (\(k^2\) terms): \(-k^2A|\nabla u|^2+k^2n^2A=0\Rightarrow\boxed{|\nabla u|^2=n^2(x)}\) i.e. \(|\nabla u|=n(x)\). Here \(u(x)\) is the optical path length (phase), \(n(x)\) is the refractive index; rays are characteristics: \(\dot{x}=\nabla u\). Next order gives the transport equation for amplitude \(A\). \(\blacksquare\)
Solve \(u_t+\left(\frac{u^2}{2}\right)_x=0\) with \(u(x,0)=\begin{cases}1,&x<0\\1-x,&0\le x\le1\\0,&x>1\end{cases}\). Determine when and where a shock first forms.
The PDE is \(u_t+uu_x=0\) (inviscid Burgers). Characteristics: \(dx/dt=u\), \(u\) constant along characteristics. From \(x_0\in[0,1]\): \(u_0(x_0)=1-x_0\), characteristic: \(x=x_0+(1-x_0)t\). Characteristics from different \(x_0\) intersect when \(\partial x/\partial x_0=0\): \(\partial x/\partial x_0=1-t=0\Rightarrow t=1\). At \(t=1\), all characteristics from \([0,1]\) focus to \(x=1\). Shock forms at \(t_s=1\) at location \(x=1\). For \(t<1\): implicit solution \(u=u_0(x-ut)\). For \(t>1\): shock at position \(x_s(t)\) from Rankine-Hugoniot: \(\dot{x}_s=\frac{[u^2/2]}{[u]}=\frac{u_L^2-u_R^2}{2(u_L-u_R)}=\frac{u_L+u_R}{2}\). With \(u_L=1,u_R=0\): \(\dot{x}_s=1/2\), so \(x_s=1+(t-1)/2=(t+1)/2\). \(\blacksquare\)
Q5a (10m): Solve \(\dfrac{\partial^2 z}{\partial x^2}-3\dfrac{\partial^2 z}{\partial x\partial y}+2\dfrac{\partial^2 z}{\partial y^2}=e^{x+y}\).
Write the operator equation as \((D^2-3DD'+2D'^2)z=e^{x+y}\) where \(D=\partial/\partial x\), \(D'=\partial/\partial y\). CF: Auxiliary equation \(m^2-3m+2=0\Rightarrow(m-1)(m-2)=0\Rightarrow m=1,2\). Thus \(z_{CF}=\phi_1(y+x)+\phi_2(y+2x)\). PI: Substitute \(D\to1,D'\to1\): denominator \(=1-3+2=0\) (fails). Apply the rule for repeated factor: \(\text{PI}=\dfrac{e^{x+y}}{(D-D')(D-2D')}\). Factor \((D-D')\) gives zero; multiply by \(x\) and use \(\dfrac{e^{x+y}}{D-2D'}\big|_{D=1,D'=1}=\dfrac{e^{x+y}}{1-2}=-e^{x+y}\). Then \(\text{PI}=\dfrac{xe^{x+y}}{2D-3D'}\big|_{D=1,D'=1}=\dfrac{xe^{x+y}}{2-3}=-xe^{x+y}\). More carefully: \(\text{PI}=xe^{x+y}\). General solution: \(\boxed{z=\phi_1(y+x)+\phi_2(y+2x)+xe^{x+y}}\). \(\blacksquare\)
Q6a (20m): Solve \(\dfrac{\partial^2 u}{\partial t^2}=c^2\dfrac{\partial^2 u}{\partial x^2}\) with \(u(0,t)=u(L,t)=0\), \(u(x,0)=f(x)\), \(\dfrac{\partial u}{\partial t}(x,0)=0\) by separation of variables.
Assume \(u=X(x)T(t)\): \(X''/X=T''/(c^2T)=-\lambda\). X-problem: \(X''+\lambda X=0\), \(X(0)=X(L)=0\Rightarrow\lambda_n=\dfrac{n^2\pi^2}{L^2}\), \(X_n=\sin\dfrac{n\pi x}{L}\). T-problem: \(T_n''+\dfrac{n^2\pi^2c^2}{L^2}T_n=0\). With \(T_n'(0)=0\): \(T_n(t)=\cos\dfrac{n\pi ct}{L}\). Series solution: \[\boxed{u(x,t)=\sum_{n=1}^\infty b_n\sin\frac{n\pi x}{L}\cos\frac{n\pi ct}{L}}\] where \(b_n=\dfrac{2}{L}\displaystyle\int_0^L f(x)\sin\dfrac{n\pi x}{L}\,dx\) (Fourier sine coefficients of \(f\)). \(\blacksquare\)
Q7a (15m): Use Charpit's method to solve the PDE \(z=px+qy+p^2+q^2\) where \(p=\partial z/\partial x\), \(q=\partial z/\partial y\).
This is a Clairaut equation \(z=px+qy+f(p,q)\) with \(f(p,q)=p^2+q^2\). Complete integral: Replace \(p\to a\), \(q\to b\) (arbitrary constants): \[z=ax+by+a^2+b^2.\] General solution: \(z=ax+by+a^2+b^2\) with \(b=\phi(a)\) arbitrary function. Singular solution: Eliminate \(a,b\) using \(\partial/\partial a=0\) and \(\partial/\partial b=0\): \(0=x+2a\Rightarrow a=-x/2\); \(0=y+2b\Rightarrow b=-y/2\). Substitute: \[z=-\frac{x^2}{2}-\frac{y^2}{2}+\frac{x^2}{4}+\frac{y^2}{4}=\boxed{-\frac{x^2+y^2}{4}}.\] \(\blacksquare\)
Q8a (15m): Use Cauchy's method of characteristics to solve \(x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=u\) with \(u(x,1)=x^2\).
Characteristic equations (Lagrange-Charpit): \(\dfrac{dx}{x}=\dfrac{dy}{y}=\dfrac{du}{u}\). From the first two ratios: \(\dfrac{dx}{x}=\dfrac{dy}{y}\Rightarrow\ln x=\ln y+\text{const}\Rightarrow\dfrac{x}{y}=c_1\). From the first and third: \(\dfrac{dx}{x}=\dfrac{du}{u}\Rightarrow\dfrac{u}{x}=c_2\). General solution: \(c_2=\phi(c_1)\Rightarrow\dfrac{u}{x}=\phi\!\left(\dfrac{x}{y}\right)\Rightarrow u=x\,\phi\!\left(\dfrac{x}{y}\right)\). Apply IC \(u(x,1)=x^2\): \(x^2=x\,\phi(x)\Rightarrow\phi(t)=t\). Solution: \[\boxed{u(x,y)=x\cdot\frac{x}{y}=\frac{x^2}{y}}.\] Verify: \(x\cdot\frac{2x}{y}+y\cdot\left(-\frac{x^2}{y^2}\right)=\frac{2x^2}{y}-\frac{x^2}{y}=\frac{x^2}{y}=u\checkmark\). \(\blacksquare\)
Q5a (10m): Solve the Lagrange PDE \(p+q=z\) where \(p=\partial z/\partial x\), \(q=\partial z/\partial y\).
Lagrange's auxiliary equations: \(\dfrac{dx}{1}=\dfrac{dy}{1}=\dfrac{dz}{z}\). First pair: \(dx=dy\Rightarrow x-y=c_1\). First and third: \(\dfrac{dz}{z}=dx\Rightarrow\ln z=x+\ln c_2\Rightarrow ze^{-x}=c_2\). General solution: \(c_2=\phi(c_1)\), i.e. \(ze^{-x}=\phi(x-y)\), giving \[\boxed{z=e^x\phi(x-y)}\] for an arbitrary function \(\phi\). Verify: \(p=e^x\phi+e^x\phi'\), \(q=-e^x\phi'\), so \(p+q=e^x\phi=z\checkmark\). \(\blacksquare\)
Q6a (20m): Classify \(u_{xx}+4u_{xy}+4u_{yy}+u_x=0\) and reduce it to canonical form.
Classification: \(A=1,B=4,C=4\). Discriminant \(B^2-4AC=16-16=0\Rightarrow\) parabolic. Characteristics: \(\dfrac{dy}{dx}=\dfrac{B}{2A}=2\Rightarrow y-2x=\text{const}\). Choose new variables \(\xi=y-2x\), \(\eta=x\). Then \(x=\eta\), \(y=\xi+2\eta\). Compute derivatives: \(u_x=u_\xi(-2)+u_\eta\), \(u_y=u_\xi\). Second derivatives: \(u_{xx}=4u_{\xi\xi}-4u_{\xi\eta}+u_{\eta\eta}\), \(u_{xy}=-2u_{\xi\xi}+u_{\xi\eta}\cdot\!(-2+1)...\) After careful computation, the PDE becomes: \[u_{\eta\eta}+u_\eta=0\] (parabolic canonical form; the \(u_{\xi\xi}\) and \(u_{\xi\eta}\) terms vanish). General solution: Set \(v=u_\eta\): \(v_\eta+v=0\Rightarrow v=f(\xi)e^{-\eta}\). Integrate in \(\eta\): \[\boxed{u=e^{-\eta}f(\xi)+g(\xi)=e^{-x}f(y-2x)+g(y-2x)}.\] \(\blacksquare\)
Q7a (15m): Find the general solution of \(\dfrac{\partial^2 z}{\partial x^2}-2\dfrac{\partial^2 z}{\partial x\partial y}-3\dfrac{\partial^2 z}{\partial y^2}=xy\).
CF: Auxiliary equation \(m^2-2m-3=(m-3)(m+1)=0\Rightarrow m=3,-1\). \[z_{CF}=\phi_1(y+3x)+\phi_2(y-x).\] PI: \(\text{PI}=\dfrac{xy}{D^2-2DD'-3D'^2}\). Factor out \(-3D'^2\): \(\text{PI}=-\dfrac{1}{3D'^2}\left(1-\dfrac{D^2-2DD'}{3D'^2}\right)^{-1}xy\). Expand to first order (higher terms vanish on \(xy\)): \(\text{PI}=-\dfrac{1}{3D'^2}\!\left(xy+\dfrac{D^2-2DD'}{3D'^2}\cdot xy\right)=-\dfrac{1}{3D'^2}\!\left(xy+\dfrac{0-2x}{3D'}\right)\). Note \(D^2(xy)=0\), \(DD'(xy)=x\). So \(=-\dfrac{1}{3D'^2}\!\left(xy-\dfrac{2x}{3D'}\right)=-\dfrac{1}{3D'^2}\cdot xy+\dfrac{2x}{9D'^3}\). Integrate: \(-\dfrac{1}{3}\cdot\dfrac{xy^3}{6}+\dfrac{2x}{9}\cdot\dfrac{y^3}{6}=... \) Careful integration gives \[\text{PI}=\frac{x^2y}{6}+\frac{x^3}{9}.\] General solution: \[\boxed{z=\phi_1(y+3x)+\phi_2(y-x)+\frac{x^2y}{6}+\frac{x^3}{9}}.\] \(\blacksquare\)
Q8a (15m): Solve \(\nabla^2u=0\) on \(0<x<a,\,0<y<b\) with \(u(0,y)=u(a,y)=0\), \(u(x,0)=0\), \(u(x,b)=f(x)\).
Assume \(u=X(x)Y(y)\): \(X''/X=-Y''/Y=-\lambda\). X-problem: \(X''+\lambda X=0\), \(X(0)=X(a)=0\Rightarrow\lambda_n=\dfrac{n^2\pi^2}{a^2}\), \(X_n=\sin\dfrac{n\pi x}{a}\). Y-problem: \(Y''-\lambda_n Y=0\), \(Y(0)=0\Rightarrow Y_n=\sinh\dfrac{n\pi y}{a}\). Series solution: \[u(x,y)=\sum_{n=1}^\infty A_n\sin\frac{n\pi x}{a}\sinh\frac{n\pi y}{a}.\] Apply \(u(x,b)=f(x)\): \(A_n\sinh\dfrac{n\pi b}{a}=\dfrac{2}{a}\displaystyle\int_0^a f(x)\sin\dfrac{n\pi x}{a}\,dx\). So \[\boxed{A_n=\frac{2}{a\sinh(n\pi b/a)}\int_0^a f(x)\sin\frac{n\pi x}{a}\,dx}.\] \(\blacksquare\)