CHAPTER 3
Partial Differential Equations
Temperature in a plate, heat in a bar, a vibrating guitar string. Each is a function of two variables obeying a PDE. Separation of variables turns one PDE into two ODEs, and Fourier series stitch the answer together.
3.0 Vocabulary: order, linear, quasilinear, homogeneous
Definitions 3.3–3.6
OrderThe highest derivative that appears.
Linear\(u\) and its derivatives appear only to the first power, are never multiplied together, and their coefficients depend only on \(x,y\).
Quasilinear (order m)The highest-order derivatives appear linearly. Their coefficients may contain \(u\) and lower derivatives.
HomogeneousA linear PDE that can be written \(L[u]=0\). If it is \(L[u]=g\) with \(g\ne0\), it is nonhomogeneous.
Classify these (Example 3.7). Tap a card to check.
\(u_{xx}+u_{yy}=0\)tap
linear · homogeneous · 2nd order
\(xu\,u_{xx}+xyu^2u_{xy}+yu\,u_{yy}+u_x^2+u_y^2+u^2=0\)tap
nonlinear · quasilinear · 2nd order
\(u_{xx}+u_{yy}=(u_x^2+u_y^2)u\)tap
nonlinear · quasilinear · 2nd order
\((x^2+y^2)u_x+u_{xy}-3u=0\)tap
linear · homogeneous · 2nd order
3.1 Classification: elliptic, parabolic, hyperbolic
Only the principal part \(Au_{xx}+Bu_{xy}+Cu_{yy}\) matters
Discriminant testFor \(Au_{xx}+Bu_{xy}+Cu_{yy}+Du_x+Eu_y+Fu+G=0\):
\(B^2-4AC<0\): elliptic (Laplace) · \(B^2-4AC=0\): parabolic (heat) · \(B^2-4AC>0\): hyperbolic (wave)
\(B^2-4AC<0\): elliptic (Laplace) · \(B^2-4AC=0\): parabolic (heat) · \(B^2-4AC>0\): hyperbolic (wave)
Why those names?It is the same test as for the conic \(Ax^2+Bxy+Cy^2=1\). Change A, B, C below and watch the curve change into an ellipse, a parabola-like pair of lines, or a hyperbola.
Discriminant classifier
Variable coefficients: the type can change from region to region
Type map over the plane
ellipticparabolic (boundary)hyperbolic
3.2 Solving by direct integration
Integrate as in an ODE, but the "constant" is a function of the other variable
The key ruleWhen you integrate with respect to \(x\), the constant of integration is an arbitrary function of \(y\) (or \(t\)). Do not write \(+C\).
\(u_{xx}=12x^2(t+1)\), with \(u=\cos2t\) and \(u_x=\sin t\) at \(x=0\).
Integrate in \(x\): \(u_x=4x^3(t+1)+f(t)\). At \(x=0\): \(f(t)=\sin t\).
Again: \(u=x^4(t+1)+x\sin t+g(t)\). At \(x=0\): \(g(t)=\cos 2t\).
\(\boxed{u=x^4(t+1)+x\sin t+\cos2t}\)
\(u_{xy}=\sin(x+y)\), with \(u_x=1\) at \(y=0\) and \(u=(y-1)^2\) at \(x=0\).
Integrate in \(y\): \(u_x=-\cos(x+y)+g(x)\). At \(y=0\): \(g(x)=1+\cos x\).
Integrate in \(x\): \(u=-\sin(x+y)+x+\sin x+h(y)\). At \(x=0\): \(h(y)=(y-1)^2+\sin y\).
\(\boxed{u=\sin x+\sin y-\sin(x+y)+x+(y-1)^2}\). The handwritten notes drop the \(+x\). Check: \(u_x=\cos x-\cos(x+y)+1\), which is 1 at \(y=0\) ✔.
Tutorial Q3\(u_{xx}=24x^2(t-2)\), \(u=e^{2t}\) and \(u_x=4t\) at \(x=0\) gives \(u=2x^4(t-2)+4tx+e^{2t}\).
3.3 Separation of variables and the three cases
2021 exam Q3(a), 12 marks: "consider all three cases"
Heat equation \(u_t=c^2u_{xx}\). Try \(u(x,t)=X(x)T(t)\). Then \(XT'=c^2X''T\).
Divide by \(c^2XT\): \(\dfrac{T'}{c^2T}=\dfrac{X''}{X}\). The left side depends only on \(t\) and the right only on \(x\), so both equal a constant \(k\).
\(X''-kX=0\) and \(T'-c^2kT=0\).
\(X''=0\Rightarrow X=Ax+B\), and \(T'=0\Rightarrow T=C\). So \(u=(Ax+B)C\), a steady straight-line temperature.
\(X=Ae^{px}+Be^{-px}\) and \(T=Ce^{c^2p^2t}\). The temperature grows without bound in time, which is not physical for a cooling bar.
\(X=A\cos px+B\sin px\) and \(T=Ce^{-c^2p^2t}\). It decays in time, which is the physically useful case:
\[u=(A\cos px+B\sin px)\,Ce^{-c^2p^2t}\]
With \(u(0,t)=u(L,t)=0\): \(A=0\) and \(\sin pL=0\), so \(p=\frac{n\pi}{L}\). Superpose: \(u=\sum B_n\sin\frac{n\pi x}{L}e^{-c^2n^2\pi^2t/L^2}\).
Why only k < 0 survives the boundary conditions X(0) = X(L) = 0
3.4 Steady-state heat in a plate: Laplace's equation
\(\theta_{xx}+\theta_{yy}=0\) means no time dependence. The edges decide everything.
AssumptionsUniform conductivity. Thin plate with insulated faces. Steady state, so nothing depends on time.
Notes' result, top edge at 100 °C\[\theta=\sum_{n\,odd}\frac{400}{n\pi\sinh\frac{n\pi b}{a}}\sin\frac{n\pi x}a\sinh\frac{n\pi y}{a}\]
Temperature map: hover to read θ(x, y)coldhot
2025 exam Q3(a) in 5 steps (13 marks)
\(U=X(x)Y(y)\Rightarrow\frac{X''}{X}=-\frac{Y''}{Y}=-k^2\), so \(X''+k^2X=0\) and \(Y''-k^2Y=0\).
\(U(0,y)=U(1,y)=0\Rightarrow X=\sin n\pi x\) with \(k=n\pi\).
\(U(x,1)=0\Rightarrow\) choose \(Y=\sinh\big(n\pi(1-y)\big)\), which is already zero at \(y=1\).
\(U=\sum A_n\sin n\pi x\,\sinh n\pi(1-y)\). At \(y=0\): \(\sum A_n\sinh(n\pi)\sin n\pi x=\sinh\pi\sin\pi x\).
Compare coefficients: \(A_1=1\) and every other \(A_n=0\). \(\boxed{U(x,y)=\sin(\pi x)\sinh\big(\pi(1-y)\big)}\)
3.5 Laplace's equation in polar coordinates
\(r^2T_{rr}+rT_r+T_{\theta\theta}=0\): semicircular plate, Exercise 3.14
\(T=R(r)\Theta(\theta)\Rightarrow\frac{r^2R''+rR'}{R}=-\frac{\Theta''}{\Theta}=h\).
\(\Theta(0)=\Theta(\pi)=0\Rightarrow h=n^2\), \(\Theta_n=\sin n\theta\).
Radial equation \(r^2R''+rR'-n^2R=0\) is Euler–Cauchy: \(R=A_nr^n+B_nr^{-n}\). It must stay finite at \(r=0\), so \(B_n=0\).
\(T(a,\theta)=T_0\): the sine series of a constant gives \(A_na^n=\frac{2T_0}{n\pi}(1-(-1)^n)\).
\[T=\frac{4T_0}{\pi}\sum_{m=0}^\infty\frac1{2m+1}\left(\frac ra\right)^{2m+1}\sin(2m+1)\theta\]
Semicircular plate: hot rim, cold diameter
3.6 Heat flow in one dimension
\(\theta_{xx}=\frac1k\theta_t\): each sine mode dies at its own rate \(e^{-kn^2\pi^2t/l^2}\)
Example 3.15, ends held at 0
\[\theta(x,t)=\sum_{r=1}^\infty B_r\sin\frac{r\pi x}l\,e^{-r^2\pi^2kt/l^2},\qquad B_r=\frac2l\int_0^lf(x)\sin\frac{r\pi x}{l}dx\]
High harmonics decay fastest, so sharp corners smooth out almost immediately.
Cooling bar: press playt =
2021 exam Q3(b) (13 marks)
\(u_t=u_{xx}\) on \(0<x<\pi\), \(u(0,t)=u(\pi,t)=0\), \(u(x,0)=100\).
From 3.3, the solution that satisfies the BCs is \(u=\sum B_n\sin nx\,e^{-n^2t}\).
\(B_n=\frac2\pi\int_0^\pi100\sin nx\,dx=\frac{200}{n\pi}\big(1-(-1)^n\big)\). That is \(\frac{400}{n\pi}\) for odd \(n\) and 0 for even \(n\).
\[\boxed{u(x,t)=\frac{400}{\pi}\sum_{m=0}^\infty\frac{\sin(2m+1)x}{2m+1}e^{-(2m+1)^2t}}\]
Ex 3.16 · non-zero endsFirst find the new steady state \(u_s=50-\frac{40x}{L}\). Then \(v=u-u_s\) has zero ends and starts at \(v(x,0)=(20+\frac{20x}L)-u_s=-30+\frac{60x}L\). Solve for \(v\) as before: \(B_n=-\frac{60(1+(-1)^n)}{n\pi}\), so only even \(n\) survive.
3.7 The vibrating string: the wave equation
\(y_{tt}=c^2y_{xx}\): every mode is a standing wave \(\sin\frac{n\pi x}L\cos\frac{n\pi ct}L\)
\(y=X(x)T(t)\Rightarrow\frac{X''}X=\frac{T''}{c^2T}=-p^2\). This gives \(X=A\cos px+B\sin px\) and \(T=C\cos cpt+D\sin cpt\).
Fixed ends \(y(0,t)=y(L,t)=0\Rightarrow A=0\) and \(p=\frac{n\pi}L\).
\[y=\sum_{n=1}^\infty\sin\frac{n\pi x}L\left(C_n\cos\frac{n\pi ct}L+D_n\sin\frac{n\pi ct}L\right)\]
Initial shape \(y(x,0)=f(x)\Rightarrow C_n=\frac2L\int_0^Lf\sin\frac{n\pi x}Ldx\).
Initial velocity \(y_t(x,0)=g(x)\Rightarrow D_n\frac{n\pi c}{L}=\frac2L\int_0^Lg\sin\frac{n\pi x}Ldx\).
Initial velocity \(y_t(x,0)=g(x)\Rightarrow D_n\frac{n\pi c}{L}=\frac2L\int_0^Lg\sin\frac{n\pi x}Ldx\).
\(f=200\sin\frac{4\pi x}l\) and \(g=0\). The shape is already mode 4, so \(C_4=200\) and every other coefficient is 0: \(y=200\sin\frac{4\pi x}l\cos\frac{4\pi ct}l\) ✔.
\(f=0\), \(g=100\sin\frac{\pi x}l\). Only \(D_1\): \(D_1\frac{\pi c}l=100\), so \(y=\frac{100l}{\pi c}\sin\frac{\pi x}l\sin\frac{\pi ct}l\).
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