Chapter J Problem Sheet 10 (Lecture 20)
J.1
Solve
\[ \partial _{xx}u+3\partial _{xy}u+2\partial _{yy}u=0,\qquad u(x,0)=\cos (x),\qquad \partial _y u(x,0)=x, \]
using the method of characteristics. Note that
\[ \partial _{xx}+3\partial _{xy}+2\partial _{yy}=(\partial _{x}+2\partial _y)(\partial _{x}+\partial _y). \]
-
Solution. We define
\[ w:=\partial _{x}u+\partial _yu, \]
so that \(w\) satisfies
\[ \partial _{x}w+2\partial _yw=0. \]
We further have
\[ w(x,0)=\partial _{x}u(x,0)+\partial _yu(x,0)=-\sin (x)+x. \]
We therefore first consider
\[ \partial _{x}w+2\partial _yw=0,\qquad w(x,0)=x-\sin (x). \]
The corresponding characteristic equations are
\[ \dot {x}=1,\qquad \dot {y}=2,\qquad \dot {z}=0, \]
with initial conditions
\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=r-\sin (r). \]
The solution is
\[ x(r,t)=t+r,\qquad y(r,t)=2t,\qquad z(r,t)=r-\sin (r). \]
Solving for \(r\) and \(t\) in terms of \(x\) and \(y\) gives \(t=\frac {y}{2}\) and \(r=x-\frac {y}{2}\). Substituting in \(z\) gives
\[ w(x,y)=x-\frac {y}{2}-\sin \left (x-\frac {y}{2}\right ). \]
We now solve for \(u\) using that \(\partial _{x}u+\partial _yu=w\). This gives
\[ \partial _{x}u+\partial _yu=x-\frac {y}{2}-\sin \left (x-\frac {y}{2}\right ),\qquad u(x,0)=\cos (x). \]
The corresponding characteristic equations are
\[ \dot {x}=1,\qquad \dot {y}=1,\qquad \dot {z}=x-\frac {y}{2}-\sin \left (x-\frac {y}{2}\right ), \]
with initial conditions
\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=\cos (r). \]
The solution for \(x\) and \(y\) is
\[ x(r,t)=t+r,\qquad y(r,t)=t. \]
Substituting this in the equation for \(z\) gives
\[ \dot {z}=\frac {t}{2}+r-\sin \left (\frac {t}{2}+r\right ),\qquad z(r,0)=\cos (r), \]
the solution of which is
\[ z(r,t)=\frac {t^2}{4}+rt+2\cos \left (\frac {t}{2}+r\right )-\cos (r). \]
Solving for \(r\) and \(t\) in terms of \(x\) and \(y\) gives \(t=y\) and \(r=x-y\). Substituting in \(z\) gives
\[ u(x,y)=\frac {y^2}{4}+(x-y)y+2\cos \left (x-\frac {y}{2}\right )-\cos (x-y), \]
which can be simplified to
\[ u(x,y)=xy-\frac {3y^2}{4}+2\cos \left (x-\frac {y}{2}\right )-\cos (x-y). \]
□
J.2
Solve
\[ \partial _{xx}u+3\partial _{xy}u-4\partial _{yy}u=0,\qquad u(0,y)=0,\qquad \partial _x u(0,y)=y, \]
using the method of characteristics. Note that
\[ \partial _{xx}+3\partial _{xy}-4\partial _{yy}=(\partial _{x}+4\partial _y)(\partial _{x}-\partial _y). \]
-
Solution. We define
\[ w:=\partial _{x}u-\partial _yu, \]
so that \(w\) satisfies
\[ \partial _{x}w+4\partial _yw=0. \]
We further have
\[ w(0,y)=\partial _{x}u(0,y)-\partial _yu(0,y)=y. \]
We therefore first consider
\[ \partial _{x}w+4\partial _yw=0,\qquad w(0,y)=y. \]
The corresponding characteristic equations are
\[ \dot {x}=1,\qquad \dot {y}=4,\qquad \dot {z}=0, \]
with initial conditions
\[ x(r,0)=0,\qquad y(r,0)=r,\qquad z(r,0)=r. \]
The solution is
\[ x(r,t)=t,\qquad y(r,t)=4t+r,\qquad z(r,t)=r. \]
Solving for \(r\) and \(t\) in terms of \(x\) and \(y\) gives \(t=x\) and \(r=y-4x\). Substituting in \(z\) gives
\[ w(x,y)=y-4x. \]
We now solve for \(u\) using that \(\partial _{x}u-\partial _yu=w\). This gives
\[ \partial _{x}u-\partial _yu=y-4x,\qquad u(0,y)=0. \]
The corresponding characteristic equations are
\[ \dot {x}=1,\qquad \dot {y}=-1,\qquad \dot {z}=y-4x, \]
with initial conditions
\[ x(r,0)=0,\qquad y(r,0)=r,\qquad z(r,0)=0. \]
The solution for \(x\) and \(y\) is
\[ x(r,t)=t,\qquad y(r,t)=-t+r. \]
Substituting this in the equation for \(z\) gives
\[ \dot {z}=-5t+r,\qquad z(r,0)=0, \]
the solution of which is
\[ z(r,t)=-\frac {5}{2}t^2+rt. \]
Solving for \(r\) and \(t\) in terms of \(x\) and \(y\) gives \(t=x\) and \(r=x+y\). Substituting in \(z\) gives
\[ u(x,y)=\frac {-5}{2}x^2+(x+y)x, \]
which can be simplified to
\[ u(x,y)=\frac {-3}{2}x^2+xy. \]
□
J.3
For given \(\alpha \in \mR \backslash \{0\}\) and \(\phi \) and \(\psi \) solve
\[ \partial _{xx}u+2\alpha \partial _{xy}u+\alpha ^2\partial _{yy}u=0,\qquad u(x,0)=\phi (x),\qquad \partial _y u(x,0)=\psi (x), \]
using the method of characteristics. Note that
\[ \partial _{xx}+2\alpha \partial _{xy}+\alpha ^2\partial _{yy}=(\partial _x+\alpha \partial _y)^2. \]
-
Solution. We define \(w:=\partial _xu+\alpha \partial _yu\), so that \(w\) satisfies
\[ \partial _xw+\alpha \partial _yw=0. \]
We further have
\[ w(x,0)=\partial _xu(x,0)+\alpha \partial _yu(x,0)=\phi '(x)+\alpha \psi (x). \]
We therefore first consider
\[ \partial _xw+\alpha \partial _yw=0,\qquad w(x,0)=\phi '(x)+\alpha \psi (x). \]
The corresponding characteristic equations are
\[ \dot {x}=1,\qquad \dot {y}=\alpha ,\qquad \dot {z}=0, \]
with initial conditions
\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=\phi '(r)+\alpha \psi (r). \]
The solution is
\[ x(r,t)=t+r,\qquad y(r,t)=\alpha t,\qquad z(r,t)=\phi '(r)+\alpha \psi (r). \]
Solving for \(r\) and \(t\) in terms of \(x\) and \(y\) gives \(t=\frac {y}{\alpha }\) and \(r=x-\frac {y}{\alpha }\). Substituting in \(z\) gives
\[ w(x,y)=\phi '\left (x-\frac {y}{\alpha }\right )+\alpha \psi \left (x-\frac {y}{\alpha }\right ). \]
We now solve for \(u\). The equation for \(u\) is
\[ \partial _xu+\alpha \partial _yu=\phi '\left (x-\frac {y}{\alpha }\right )+\alpha \psi \left (x-\frac {y}{\alpha }\right ),\qquad u(x,0)=\phi (x). \]
The corresponding characteristic equations are
\[ \dot {x}=1,\qquad \dot {y}=\alpha ,\qquad \dot {z}=\phi '\left (x-\frac {y}{\alpha }\right )+\alpha \psi \left (x-\frac {y}{\alpha }\right ), \]
with initial conditions
\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=\phi (r). \]
Solving for \(x\) and \(y\) gives
\[ x(r,t)=t+r,\qquad y(r,t)=\alpha t. \]
Substituting in the equation for \(z\) gives
\[ \dot {z}=\phi '(r)+\alpha \psi (r),\qquad z(r,0)=\phi (r). \]
Solving this gives
\[ z(r,t)=\left [\phi '(r)+\alpha \psi (r)\right ]t+\phi (r). \]
Solving for \(r\) and \(t\) in terms of \(x\) and \(y\) gives \(t=\frac {y}{\alpha }\) and \(r=x-\frac {y}{\alpha }\). Substituting in \(z\) gives
\[ u(x,y)=\frac {y}{\alpha }\phi '\left (x-\frac {y}{\alpha }\right ) +\frac {y}{\alpha }\alpha \psi \left (x-\frac {y}{\alpha }\right ) +\phi \left (x-\frac {y}{\alpha }\right ). \]
This simplifies to
\[ u(x,y)=\frac {y}{\alpha }\phi '\left (x-\frac {y}{\alpha }\right ) +y\psi \left (x-\frac {y}{\alpha }\right ) +\phi \left (x-\frac {y}{\alpha }\right ). \]
□
J.4
For given \(\alpha \in \mR \) and \(\phi \) and \(\psi \) solve
\[ \alpha ^2\partial _{xx}u+2\alpha \partial _{xy}u+\partial _{yy}u=0,\qquad u(x,0)=\phi (x),\qquad \partial _y u(x,0)=\psi (x), \]
using the method of characteristics. Note that
\[ \alpha ^2\partial _{xx}+2\alpha \partial _{xy}+\partial _{yy}=(\alpha \partial _x+\partial _y)^2. \]
-
Solution. We define \(w:=\alpha \partial _xu+\partial _yu\), so that \(w\) satisfies
\[ \alpha \partial _xw+\partial _yw=0. \]
We further have
\[ w(x,0)=\alpha \partial _xu(x,0)+\partial _yu(x,0)=\alpha \phi '(x)+\psi (x). \]
We therefore first consider
\[ \alpha \partial _xw+\partial _yw=0,\qquad w(x,0)=\alpha \phi '(x)+\psi (x). \]
The corresponding characteristic equations are
\[ \dot {x}=\alpha ,\qquad \dot {y}=1,\qquad \dot {z}=0, \]
with initial conditions
\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=\alpha \phi '(r)+\psi (r). \]
The solution is
\[ x(r,t)=\alpha t+r,\qquad y(r,t)=t,\qquad z(r,t)=\alpha \phi '(r)+\psi (r). \]
Solving for \(r\) and \(t\) in terms of \(x\) and \(y\) gives \(t=y\) and \(r=x-\alpha y\). Substituting in \(z\) gives
\[ w(x,y)=\alpha \phi '\left (x-\alpha y\right )+\psi \left (x-\alpha y\right ). \]
We now solve for \(u\). The equation for \(u\) is
\[ \alpha \partial _xu+\partial _yu=\alpha \phi '\left (x-\alpha y\right )+\psi \left (x-\alpha y\right ),\qquad u(x,0)=\phi (x). \]
The corresponding characteristic equations are
\[ \dot {x}=\alpha ,\qquad \dot {y}=1,\qquad \dot {z}=\alpha \phi '\left (x-\alpha y\right )+\psi \left (x-\alpha y\right ), \]
with initial conditions
\[ x(r,0)=r,\qquad y(r,0)=0,\qquad z(r,0)=\phi (r). \]
Solving for \(x\) and \(y\) gives
\[ x(r,t)=\alpha t+r,\qquad y(r,t)=t. \]
Substituting in the equation for \(z\) gives
\[ \dot {z}=\alpha \phi '(r)+\psi (r),\qquad z(r,0)=\phi (r). \]
Solving this gives
\[ z(r,t)=\left [\alpha \phi '(r)+\psi (r)\right ]t+\phi (r). \]
Solving for \(r\) and \(t\) in terms of \(x\) and \(y\) gives \(t=y\) and \(r=x-\alpha y\). Substituting in \(z\) gives
\[ u(x,y)=\alpha y\phi '\left (x-\alpha y\right ) +y\psi \left (x-\alpha y\right ) +\phi \left (x-\alpha y\right ). \]
□
J.5
Consider the wave equation \(\partial _{yy}u-c^2\partial _{xx}u=0\). Use d’Alembert’s formula to determine \(u\) in the following cases
-
(i) \(u(x,0)=\sin (x)\), \(\partial _y u(x,0)=1\);
-
(ii) \(u(x,0)=\phi (x)\), \(\partial _yu(x,0)=0\).
-
Solution. (i) In the notation of the notes we have \(\phi (x)=\sin (x)\) and \(\psi (x)=1\). Therefore \(\Psi (x)=x\) and d’Alembert’s formula is
\[ u(x,y)=\frac {-1}{2c}(x-cy)+\frac {1}{2}\sin (x-cy)+\frac {1}{2c}(x+cy)+\frac {1}{2}\sin (x+cy) =y+\sin (x)\cos (cy). \]
(ii) We have \(\psi =0\) and therefore \(\Psi =0\) so that d’Alembert’s formula gives
\[ u(x,y)=\frac {\phi (x-cy)+\phi (x+cy)}{2}. \]
□
Further reading
The theory for Sturm–Liouville problems in these notes is condensed from [Weidmann87]. The underlying Hilbert space theory (including versions of the spectral theorem) can be found in [Weidmann80]. Various versions of the spectral theorem can also be found in [Simon].
The more computational aspects of separation of variables can be found in for example [Arfken] and [Haberman] and many other books with similar titles.
The theory for the Fourier transform in these notes is in part condensed from [Staffans].
An approachable introduction to the theory of distributions is given in [Sasane, Chapter 6].
Bibliography
-
[Weidmann87] Weidmann, Joachim (1987). Spectral theory of ordinary differential operators.
-
[Weidmann80] Weidmann, Joachim (1980). Linear operators in Hilbert spaces.
-
[Simon] Simon, Barry (2015). A Comprehensive Course in Analysis, Part 4: Operator theory.
-
[Arfken] Arfken, George B., Weber, Hans J. and Harris, Frank E. (2013 seventh edition). Mathematical methods for physicists.
-
[Haberman] Haberman, Richard (2013 fifth edition). Applied partial differential equations with Fourier series and boundary value problems.
-
[Staffans] Staffans, Olof (2009). Fourier Analysis. http://web.abo.fi/fak/mnf/mate/kurser/fourieranalys/fourier.pdf
-
[Sasane] Sasane, Amol (2017). A Friendly Approach to Functional Analysis.