Find all quadratic polynomial solutions of the 3D Laplace equation \(\frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}+\frac{\partial ^{2}u}{\partial z^{2}}=0\)
Solution
A quadratic polynomial in variables \(x,y,z\) is \begin{equation} u=a_{1}+a_{2}x+a_{3}y+a_{4}z+a_{5}x^{2}+a_{6}y^{2}+a_{7}z^{2}+a_{8}xy+a_{9}xz+a_{10}yz \tag{1} \end{equation} Hence \(u_{x}=a_{2}+2a_{5}x+a_{8}y+a_{9}z\) which implies that \(u_{xx}=2a_{5}\). Similarly \(u_{y}=a_{3}+2a_{6}y+a_{8}x+a_{10}z\), therefore \(u_{yy}=2a_{6}\). And finally \(u_{z}=a_{4}+2a_{7}z+a_{9}x+a_{10}y\) and \(u_{zz}=2a_{7}\). Substituting these results in the Laplace equation gives above result in\begin{align*} 2a_{5}+2a_{6}+2a_{7} & =0\\ a_{5}+a_{6}+a_{7} & =0 \end{align*}
Therefore \(a_{5}=-\left ( a_{6}+a_{7}\right ) \). Using this relation back in (1) gives\begin{align*} u & =a_{1}+a_{2}x+a_{3}y+a_{4}z-\left ( a_{6}+a_{7}\right ) x^{2}+a_{6}y^{2}+a_{7}z^{2}+a_{8}xy+a_{9}xz+a_{10}yz\\ & =a_{1}+a_{2}x+a_{3}y+a_{4}z+a_{6}\left ( -x^{2}+y^{2}\right ) +a_{7}\left ( -x^{2}+z^{2}\right ) +a_{8}xy+a_{9}xz+a_{10}yz \end{align*}
Which can be written as\[ u\left ( x,y,z\right ) =A_{1}+A_{2}x+A_{3}y+A_{4}z+A_{5}\left ( y^{2}-x^{2}\right ) +A_{6}\left ( z^{2}-x^{2}\right ) +A_{7}xy+A_{8}xz+A_{9}yz \]
Find all real solutions to 2D Laplace equation \(u_{xx}+u_{yy}=0\) of the form \(u=\log \left ( p\left ( x,y\right ) \right ) \) where \(p\left ( x,y\right ) \) is a quadratic polynomial.
Solution
A quadratic polynomial \(p\left ( x,y\right ) \) in variables \(x,y\) is\[ p\left ( x,y\right ) =a_{1}+a_{2}x+a_{3}y+a_{4}x^{2}+a_{5}y^{2}+a_{6}xy \] Therefore\[ u\left ( x,y\right ) =\log \left ( a_{1}+a_{2}x+a_{3}y+a_{4}x^{2}+a_{5}y^{2}+a_{6}xy\right ) \] Hence \[ u_{x}=\frac{a_{2}+2a_{4}x+a_{6}y}{p\left ( x,y\right ) }\] and \begin{equation} u_{xx}=\frac{2a_{4}}{p\left ( x,y\right ) }-\frac{\left ( a_{2}+2a_{4}x+a_{6}y\right ) ^{2}}{p\left ( x,y\right ) ^{2}} \tag{1} \end{equation} Similarly\[ u_{y}=\frac{a_{3}+2a_{5}y+a_{6}x}{p\left ( x,y\right ) }\] And\begin{equation} u_{yy}=\frac{2a_{5}}{p\left ( x,y\right ) }-\frac{\left ( a_{3}+2a_{5}y+a_{6}x\right ) ^{2}}{p\left ( x,y\right ) ^{2}} \tag{2} \end{equation} Substituting (1,2) into \(u_{xx}+u_{yy}=0\) gives\begin{align*} \left ( \frac{2a_{4}}{p\left ( x,y\right ) }-\frac{\left ( a_{2}+2a_{4}x+a_{6}y\right ) ^{2}}{p\left ( x,y\right ) ^{2}}\right ) +\left ( \frac{2a_{5}}{p\left ( x,y\right ) }-\frac{\left ( a_{3}+2a_{5}y+a_{6}x\right ) ^{2}}{p\left ( x,y\right ) ^{2}}\right ) & =0\\ 2a_{4}-\frac{\left ( a_{2}+2a_{4}x+a_{6}y\right ) ^{2}}{p\left ( x,y\right ) }+2a_{5}-\frac{\left ( a_{3}+2a_{5}y+a_{6}x\right ) ^{2}}{p\left ( x,y\right ) } & =0\\ 2a_{4}+2a_{5}-\frac{\left ( a_{2}+2a_{4}x+a_{6}y\right ) ^{2}+\left ( a_{3}+2a_{5}y+a_{6}x\right ) ^{2}}{p\left ( x,y\right ) } & =0 \end{align*}
Or\[ \left ( 2a_{4}+2a_{5}\right ) p\left ( x,y\right ) =\left ( a_{2}+2a_{4}x+a_{6}y\right ) ^{2}+\left ( a_{3}+2a_{5}y+a_{6}x\right ) ^{2}\] But \(p\left ( x,y\right ) =a_{1}+a_{2}x+a_{3}y+a_{4}x^{2}+a_{5}y^{2}+a_{6}xy\). Hence the above becomes\[ \left ( 2a_{4}+2a_{5}\right ) \left ( a_{1}+a_{2}x+a_{3}y+a_{4}x^{2}+a_{5}y^{2}+a_{6}xy\right ) =\left ( a_{2}+2a_{4}x+a_{6}y\right ) ^{2}+\left ( a_{3}+2a_{5}y+a_{6}x\right ) ^{2}\] Expanding and comparing coefficients gives
\begin{multline*} 2x^{2}a_{4}^{2}+2x^{2}a_{4}a_{5}+2a_{6}a_{4}xy+2a_{6}a_{5}xy+2a_{2}a_{4}x+2a_{2}xa_{5}+2y^{2}a_{4}a_{5}+2y^{2}a_{5}^{2}+2a_{3}a_{4}y+2a_{3}a_{5}y+2a_{1}a_{4}+2a_{1}a_{5}=\\ 4x^{2}a_{4}^{2}+x^{2}a_{6}^{2}+4a_{4}a_{6}xy+4a_{5}a_{6}xy+4xa_{2}a_{4}+2a_{3}a_{6}x+4y^{2}a_{5}^{2}+y^{2}a_{6}^{2}+2a_{2}a_{6}y+4a_{3}a_{5}y+a_{2}^{2}+a_{3}^{2} \end{multline*} Simplifying\begin{multline*} 2a_{4}a_{5}x^{2}+2a_{2}a_{5}x+2a_{4}a_{5}y^{2}+2a_{3}a_{4}y+2a_{1}a_{4}+2a_{1}a_{5}=\\ 2x^{2}a_{4}^{2}+a_{6}^{2}x^{2}+2a_{4}a_{6}xy+2a_{5}a_{6}xy+2a_{2}a_{4}x+2a_{3}a_{6}x+2a_{5}^{2}y^{2}+a_{6}^{2}y^{2}+2a_{2}a_{6}y+2a_{3}a_{5}y+a_{2}^{2}+a_{3}^{2} \end{multline*} Comparing coefficients of terms that contain no \(x,y\) and coefficients of \(x,y,xy,x^{2},y^{2}\) gives the following equations in order\begin{align*} 2a_{1}a_{4}+2a_{1}a_{5} & =a_{2}^{2}+a_{3}^{2}\\ 2a_{2}a_{5} & =2a_{2}a_{4}+2a_{3}a_{6}\\ 2a_{3}a_{4} & =2a_{2}a_{6}+2a_{3}a_{5}\\ 0 & =4a_{4}a_{6}\\ 2a_{4}a_{5} & =2a_{4}^{2}+a_{6}^{2}\\ 2a_{4}a_{5} & =2a_{5}^{2}+a_{6}^{2} \end{align*}
Equation \(0=4a_{4}a_{6}\) above implies that \(a_{4}=0\) or \(a_{6}=0\) or both are zero. But if both are zero, there is no solution. On the other hand, if \(a_{4}=0,\) then this also leads to no solution as all equations reduce to \(0=0\). Therefore only choice left is \(a_{6}=0\). Now the above equations become\begin{align*} 2a_{1}a_{4}+2a_{1}a_{5} & =a_{2}^{2}+a_{3}^{2}\\ 2a_{2}a_{5} & =2a_{2}a_{4}\\ 2a_{3}a_{4} & =2a_{3}a_{5}\\ 0 & =0\\ 2a_{4}a_{5} & =2a_{4}^{2}\\ 2a_{4}a_{5} & =2a_{5}^{2} \end{align*}
Or\begin{align*} 2a_{1}a_{4}+2a_{1}a_{5} & =a_{2}^{2}+a_{3}^{2}\\ a_{5} & =a_{4}\\ a_{4} & =a_{5}\\ 0 & =0\\ a_{5} & =a_{4}\\ a_{4} & =a_{5} \end{align*}
Hence\begin{align} a_{4} & =a_{5}\tag{3}\\ a_{6} & =0\tag{4}\\ 2a_{1}a_{4}+2a_{1}a_{5} & =a_{2}^{2}+a_{3}^{2}\nonumber \end{align}
Since \(a_{4}=a_{5}\) then\begin{align} 2a_{1}a_{5}+2a_{1}a_{5} & =a_{2}^{2}+a_{3}^{2}\nonumber \\ a_{5} & =\frac{a_{2}^{2}+a_{3}^{2}}{2a_{1}} \tag{5} \end{align}
Using (3,4,5) in \(p\left ( x,y\right ) =a_{1}+a_{2}x+a_{3}y+a_{4}x^{2}+a_{5}y^{2}+a_{6}xy\) gives\begin{align*} p\left ( x,y\right ) & =a_{1}+a_{2}x+a_{3}y+a_{5}x^{2}+a_{5}y^{2}\\ & =a_{1}+a_{2}x+a_{3}y+a_{5}\left ( x^{2}+y^{2}\right ) \\ & =a_{1}+a_{2}x+a_{3}y+\frac{a_{2}^{2}+a_{3}^{2}}{2a_{1}}\left ( x^{2}+y^{2}\right ) \end{align*}
Only three arbitrary constants are needed. Let \(a_{1}=a,a_{2}=b,a_{3}=c\) the above becomes\[ p\left ( x,y\right ) =a+bx+cy+\frac{b^{2}+c^{2}}{2a}\left ( x^{2}+y^{2}\right ) \] And the solution becomes\[ u\left ( x,y\right ) =\log \left ( a+bx+cy+\frac{b^{2}+c^{2}}{2a}\left ( x^{2}+y^{2}\right ) \right ) \]
Find all solutions \(u=f\left ( r\right ) \) of the 3D Laplace equation \(u_{xx}+u_{yy}+u_{zz}=0\) that depends only on radial coordinates \(r=\sqrt{x^{2}+y^{2}+z^{2}}\)
Solution
The Laplacian in 3D in spherical coordinates is\[ \nabla ^{2}u\left ( r,\theta ,\phi \right ) =u_{rr}+\frac{2}{r}u_{r}+\frac{1}{r^{2}}\left ( \frac{\cos \theta }{\sin \theta }u_{\theta }+u_{\theta \theta }\right ) +\frac{1}{r^{2}\sin ^{2}\theta }u_{\phi \phi }\] The above shows that the terms that depend only on \(r\) makes the laplacian \[ \nabla ^{2}u\left ( r\right ) =u_{rr}+\frac{2}{r}u_{r}\] Hence the PDE \(\nabla ^{2}u\left ( r\right ) =0\) becomes an ODE now since there is only one dependent variable giving\[ u^{\prime \prime }\left ( r\right ) +\frac{2}{r}u^{\prime }\left ( r\right ) =0 \] Let \(v=u^{\prime }\left ( r\right ) \) and the above becomes\[ v^{\prime }\left ( r\right ) +\frac{2}{r}v\left ( r\right ) =0 \] This is linear first order ODE. The integrating factor is \(I=e^{\int \frac{2}{r}dr}=e^{2\ln r}=r^{2}\). Therefore the above becomes \(\frac{d}{dr}\left ( vr^{2}\right ) =0\) or \(vr^{2}=C_{1}\) or \(v\left ( r\right ) =\frac{C_{1}}{r^{2}}\). Therefore\begin{align*} u^{\prime } & =\frac{C_{1}}{r^{2}}\\ du & =\frac{C_{1}}{r^{2}}dr \end{align*}
Integrating gives the solution\[ u=-\frac{C_{1}}{r}+C_{2}\] The above is the required solution. Hence \[ \fbox{$f\left ( r\right ) =-\frac{C_1}{r}+C_2$}\] Where \(C_{1},C_{2}\) are arbitrary constants.
The displacement \(u\left ( t,x\right ) \) of a forced violin string is modeled by the PDE \(u_{tt}=4u_{xx}+F\left ( t,x\right ) \). When the string is subjected to the external force \(F\left ( t,x\right ) =\cos x\), the solution is \(u\left ( t,x\right ) =\cos \left ( x-2t\right ) +\frac{1}{4}\cos x\), while when \(F\left ( t,x\right ) =\sin x\), the solution is \(u\left ( t,x\right ) =\sin \left ( x-2t\right ) +\frac{1}{4}\sin x\). Find a solution when the forcing function is (a) \(\cos x-5\sin x\), (b) \(\sin \left ( x-3\right ) \)
Solution
Since the PDE is linear, superposition can be used. When the input is \(F\left ( t,x\right ) =\cos x-5\sin x\) then the solution is\begin{align*} u\left ( t,x\right ) & =\left ( \cos \left ( x-2t\right ) +\frac{1}{4}\cos x\right ) -5\left ( \sin \left ( x-2t\right ) +\frac{1}{4}\sin x\right ) \\ & =\cos \left ( x-2t\right ) +\frac{1}{4}\cos x-5\sin \left ( x-2t\right ) -\frac{5}{4}\sin x \end{align*}
Since the PDE is linear, superposition can be used. When the input is \(F\left ( t,x\right ) =\sin \left ( x-3\right ) \) then the solution same as when the input is \(\sin x\) but shifted by \(3\). Hence\[ u\left ( t,x\right ) =\sin \left ( \left ( x-3\right ) -2t\right ) +\frac{1}{4}\sin \left ( x-3\right ) \]
Solve the following inhomogeneous linear ODE \(5u^{\prime \prime }-4u^{\prime }+4u=e^{x}\cos x\)
Solution
First the homogeneous solution \(u_{h}\) is found, then a particular solution \(u_{p}\) is found. The general solution will be the sum of both \(u=u_{h}+u_{p}\). Since this is a constant coefficient ODE, the characteristic equation is \(5\lambda ^{2}-4\lambda +4=0\). The roots are \(\lambda _{1}=\frac{2}{5}+\frac{4}{5}i,\lambda _{1}=\frac{2}{5}-\frac{4}{5}i\), which implies the solution is \[ u_{h}\left ( x\right ) =e^{\frac{2}{5}x}\left ( c_{1}\cos \left ( \frac{4}{5}x\right ) +c_{2}\sin \left ( \frac{4}{5}x\right ) \right ) \] Using the method of undetermined coefficients, and since the forcing function is \(e^{x}\cos x\), then let \begin{equation} u_{p}=Ae^{x}\left ( B\cos x+C\sin x\right ) \tag{1} \end{equation} Hence\begin{align} u_{p}^{\prime } & =Ae^{x}\left ( B\cos x+C\sin x\right ) +Ae^{x}\left ( -B\sin x+C\cos x\right ) \tag{2}\\ u_{p}^{\prime \prime } & =Ae^{x}\left ( B\cos x+C\sin x\right ) +Ae^{x}\left ( -B\sin x+C\cos x\right ) +Ae^{x}\left ( -B\sin x+C\cos x\right ) +Ae^{x}\left ( -B\cos x-C\sin x\right ) \nonumber \\ & =Ae^{x}\left ( B\cos x+C\sin x-B\sin x+C\cos x-B\sin x+C\cos x-B\cos x-C\sin x\right ) \nonumber \\ & =Ae^{x}\left ( -B\sin x+C\cos x-B\sin x+C\cos x\right ) \nonumber \\ & =Ae^{x}\left ( -2B\sin x+2C\cos x\right ) \tag{3} \end{align}
Substituting (1,2,3) back into the original ODE gives\begin{align*} 5Ae^{x}\left ( -2B\sin x+2C\cos x\right ) -4\left ( Ae^{x}\left ( B\cos x+C\sin x\right ) +Ae^{x}\left ( -B\sin x+C\cos x\right ) \right ) +4Ae^{x}\left ( B\cos x+C\sin x\right ) & =e^{x}\cos x\\ Ae^{x}\left ( -10B\sin x+10C\cos x\right ) -Ae^{x}\left ( 4B\cos x+4C\sin x\right ) -Ae^{x}\left ( -4B\sin x+4C\cos x\right ) +Ae^{x}\left ( 4B\cos x+4C\sin x\right ) & =e^{x}\cos x\\ Ae^{x}\left ( -10B\sin x+10C\cos x-4B\cos x-4C\sin x+4B\sin x-4C\cos x+4B\cos x+4C\sin x\right ) & =e^{x}\cos x \end{align*}
Hence\[ Ae^{x}\left ( 6C\cos x-6B\sin x\right ) =e^{x}\cos x \] Comparing coefficients shows that \begin{align*} A & =1\\ B & =0\\ C & =\frac{1}{6} \end{align*}
Hence from (1) \[ u_{p}=e^{x}\frac{\sin x}{6}\] Therefore the general solution is \begin{align*} u\left ( x\right ) & =u_{h}\left ( x\right ) +u_{p}\left ( x\right ) \\ & =e^{\frac{2}{5}x}\left ( c_{1}\cos \left ( \frac{4}{5}x\right ) +c_{2}\sin \left ( \frac{4}{5}x\right ) \right ) +e^{x}\frac{\sin x}{6} \end{align*}
Solve the PDE \(\frac{\partial ^{2}u}{\partial x\partial y}=0\) for \(u\left ( x,y\right ) \)
Solution
Integrating once w.r.t \(x\) gives\[ \frac{\partial u}{\partial y}=F\left ( y\right ) \] Where \(F\left ( y\right ) \) acts as the constant of integration, but since this is a PDE, it becomes an arbitrary function of \(y\) only. Integrating the above again w.r.t. \(y\) gives\[ u=\int F\left ( y\right ) dy+G\left ( x\right ) \] Where \(G\left ( x\right ) \) is an arbitrary function of \(x\) only. If we let \(\int F\left ( y\right ) dy=H\left ( y\right ) \) where \(H\left ( y\right ) \) is the antiderivative for the indefinite integral which depends on \(y\) only. Then the above can be written as\[ \fbox{$u\left ( x,y\right ) =H\left ( y\right ) +G\left ( x\right ) $}\] To verify, from the above \(\frac{\partial u}{\partial y}=H^{\prime }\left ( y\right ) \) and hence \begin{align*} \frac{\partial ^{2}u}{\partial x\partial y} & =\frac{d}{dx}\left ( H^{\prime }\left ( y\right ) \right ) \\ & =0 \end{align*}
Solve the following initial value problems and graph the solutions at \(t=1,2,3\)
Solution
Let \(\xi \) be the characteristic variable defined such that \(\xi =x-ct\). Where characteristic lines are given by \(x=x_{0}+ct\). But \(c=-3\) in this problem. Hence characteristic lines are \[ x=x_{0}-3t \] Where \(x_{0}\) means the same as \(x\left ( 0\right ) \), i.e. \(x\left ( t\right ) \) at time \(t=0\). Since \(c=-3\) then \[ \xi =x+3t \] Let \[ u\left ( t,x\right ) \equiv v\left ( t,\xi \right ) \] \(u_{t}-3u_{x}=0\) is now transformed to \(v\left ( t,\xi \right ) \) as follows\begin{align} \frac{\partial u}{\partial t} & =\frac{\partial v}{\partial t}\frac{\partial t}{\partial t}+\frac{\partial v}{\partial \xi }\frac{\partial \xi }{\partial t}\nonumber \\ & =\frac{\partial v}{\partial t}+3\frac{\partial v}{\partial \xi } \tag{1} \end{align}
And\begin{align} \frac{\partial u}{\partial x} & =\frac{\partial v}{\partial t}\frac{\partial t}{\partial x}+\frac{\partial v}{\partial \xi }\frac{\partial \xi }{\partial x}\nonumber \\ & =0+\frac{\partial v}{\partial \xi }\nonumber \\ & =\frac{\partial v}{\partial \xi } \tag{2} \end{align}
Substituting (1,2) in \(u_{t}-3u_{x}=0\) gives the transformed PDE as\begin{align*} \frac{\partial v}{\partial t}+3\frac{\partial v}{\partial \xi }-3\frac{\partial v}{\partial \xi } & =0\\ \frac{\partial v}{\partial t} & =0 \end{align*}
Integrating w.r.t \(\xi \) gives the solution in \(v\left ( t,\xi \right ) \) space as\[ v\left ( t,\xi \right ) =F\left ( \xi \right ) \] Where \(F\left ( \xi \right ) \) is an arbitrary continuous function of \(\xi \). Transforming back to \(u\left ( t,x\right ) \) gives\begin{equation} u\left ( t,x\right ) =F\left ( x+3t\right ) \tag{3} \end{equation} At \(t=0\) the above becomes\[ e^{-x_{0}^{2}}=F\left ( x_{0}\right ) \] This means that (3) becomes (since \(x=x_{0}+ct\) or \(x=x_{0}-3t\) or \(x_{0}=x+3t\))\[ u\left ( t,x\right ) =e^{-\left ( x+3t\right ) ^{2}}\]
\begin{align*} u_{t}+2u_{x} & =0\\ u\left ( -1,x\right ) & =\frac{x}{1+x^{2}} \end{align*}
Let \(\xi \) be the characteristic variable defined such that \(\xi =x-ct\). Where characteristic lines are given by \(x=x_{0}+ct\). But \(c=2\) in this problem. Hence characteristic lines are \[ x=x_{0}+2t \] And\[ \xi =x-2t \] Let \(u\left ( t,x\right ) \equiv v\left ( t,\xi \right ) \). Then \(u_{t}+2u_{x}=0\) is transformed to \(v\left ( t,\xi \right ) \) as was done in part (a) (will not be repeated) which results in\[ \frac{\partial v}{\partial t}=0 \] Integrating w.r.t \(\xi \) gives the solution\[ v\left ( t,\xi \right ) =F\left ( \xi \right ) \] Where \(F\left ( \xi \right ) \) is an arbitrary continuous function of \(\xi \). Transforming back to \(u\left ( t,x\right ) \) results in\begin{equation} u\left ( t,x\right ) =F\left ( x-2t\right ) \tag{3} \end{equation} At \(t=-1\) the above becomes\[ \frac{x_{0}}{1+x_{0}^{2}}=F\left ( x_{0}+2\right ) \] Let \(x_{0}+2=z\). Then \(x_{0}=z-2\). And the above becomes\[ \frac{z-2}{1+\left ( z-2\right ) ^{2}}=F\left ( z\right ) \] This means that (3) becomes\begin{align*} u\left ( t,x\right ) & =\frac{\left ( x-2t\right ) -2}{1+\left ( \left ( x-2t\right ) -2\right ) ^{2}}\\ & =\frac{x-2t-2}{1+\left ( x-2t-2\right ) ^{2}} \end{align*}
\begin{align} u_{t}+u_{x}+\frac{1}{2}u & =0\tag{1}\\ u\left ( 0,x\right ) & =\arctan \left ( x\right ) \nonumber \end{align}
Let \(\xi \) be the characteristic variable defined such that \(\xi =x-ct\). Where characteristic lines are given by \(x=x_{0}+ct\). But \(c=1\) in this problem. Hence characteristic lines are given by solution to \begin{align*} \frac{dx}{dt} & =1\\ x\left ( t\right ) & =x_{0}+t \end{align*}
And\begin{align*} \xi & =x-ct\\ & =x-t \end{align*}
Then \(u_{t}+u_{x}\) are transformed to \(v\left ( t,\xi \right ) \) as was done in part (a) (will not be repeated) which results in\[ u_{t}+u_{x}=\frac{\partial v}{\partial t}\] Substituting the above into (1) gives (where now \(v\) is used in place of \(u\)).\[ \frac{\partial v}{\partial t}+\frac{1}{2}v=0 \] This is now first order ODE since it only depends on \(t\). Therefore \(v^{\prime }+\frac{1}{2}v=0\). This is linear in \(v\). Hence the solution is \(\frac{d}{dt}\left ( ve^{\int \frac{1}{2}dt}\right ) =0\) or \(ve^{\frac{1}{2}t}=F\left ( \xi \right ) \) where \(F\) is arbitrary function of \(\xi \). Hence\[ v\left ( t,\xi \right ) =e^{\frac{-1}{2}t}F\left ( \xi \right ) \] Converting back to \(u\left ( t,x\right ) \) gives\begin{equation} u\left ( t,x\right ) =e^{\frac{-t}{2}}F\left ( x-t\right ) \tag{2} \end{equation} At \(t=0\) the above becomes\[ \arctan \left ( x_{0}\right ) =F\left ( x_{0}\right ) \] From the above then (2) can be written as\[ u\left ( t,x\right ) =e^{\frac{-t}{2}}\arctan \left ( x-t\right ) \]
\begin{align*} u_{t}-4u_{x}+u & =0\\ u\left ( 0,x\right ) & =\frac{1}{1+x^{2}} \end{align*}
Let \(\xi \) be the characteristic variable defined such that \(\xi =x-ct\). Where characteristic lines are given by \(x=x_{0}+ct\). But \(c=-4\) in this problem. Hence characteristic lines are \[ x=x_{0}-4t \] And\[ \xi =x+4t \] Then \(u_{t}-4u_{x}\) are transformed to \(v\left ( t,\xi \right ) \) as was done in part (a) (will not be repeated) which results in\[ u_{t}-4u_{x}=\frac{\partial v}{\partial t}\] Substituting the above into (1) gives (where now \(v\) is used in place of \(u\)).\[ \frac{\partial v}{\partial t}+v=0 \] This is now first order ODE since it only depends on \(t\). Therefore \(v^{\prime }+v=0\). This is linear in \(v\). Hence the solution is \(\frac{d}{dt}\left ( ve^{\int dt}\right ) =0\) or \(ve^{t}=F\left ( \xi \right ) \) where \(F\) is arbitrary function of \(\xi \). Hence\[ v\left ( t,\xi \right ) =e^{-t}F\left ( \xi \right ) \] Converting to \(u\left ( t,x\right ) \) gives\begin{equation} u\left ( t,x\right ) =e^{-t}F\left ( x+4t\right ) \tag{2} \end{equation} At \(u\left ( 0,x\right ) =\frac{1}{1+x^{2}}\) the above becomes\[ \frac{1}{1+x_{0}^{2}}=F\left ( x_{0}\right ) \] From the above then (2) can be written as\[ u\left ( t,x\right ) =\frac{e^{-t}}{1+\left ( x+4t\right ) ^{2}}\]
Graph some of the characteristic lines for the following equation and write down the formula for the general solution
(b) \(u_{t}+5u_{x}=0\,\), (d) \(u_{t}-4u_{x}+u=0\)
Solution
\[ u_{t}+5u_{x}=0 \]
Let \(\xi \) be the characteristic variable defined such that \(\xi =x-ct\). Where characteristic lines are given by \(x=x_{0}+ct\). But \(c=5\) in this problem. Hence characteristic lines are \begin{equation} \fbox{$x\left ( t\right ) =x_0+5t$} \tag{1} \end{equation} And\[ \xi =x-5t \] Then \(u_{t}-5u_{x}=0\) is transformed to \(v\left ( t,\xi \right ) \) as was done in earlier (will not be repeated) which results in\[ u_{t}-5u_{x}=\frac{\partial v}{\partial t}\] Therefore \(\frac{\partial v}{\partial t}=0\) which has the general solution \(v\left ( t,\xi \right ) =F\left ( \xi \right ) \) where \(F\) is arbitrary function of \(\xi \). Transforming back to \(u\left ( t,x\right ) \) gives\[ \fbox{$u\left ( t,x\right ) =F\left ( x-5t\right ) $}\] On the characteristic lines given by (1) the solution \(u\left ( t,x\right ) \) is constant. The slope of the characteristic lines is \(5\) and intercept is \(x_{0}\). The following is a plot of few lines using different values of \(x_{0}\).
\[ u_{t}-4u_{x}+u=0 \] Let \(\xi \) be the characteristic variable defined such that \(\xi =x-ct\). Where characteristic lines are given by \(x=x_{0}+ct\). But \(c=-4\) in this problem. Hence characteristic lines are \begin{equation} \fbox{$x\left ( t\right ) =x_0-4t$} \tag{1} \end{equation} And\[ \xi =x+4t \] Then \(u_{t}-4u_{x}\) is transformed to \(v\left ( t,\xi \right ) \) as was done in earlier (will not be repeated) which results in\[ u_{t}-4u_{x}=\frac{\partial v}{\partial t}\] Therefore the original PDE becomes \(\frac{\partial v}{\partial t}+v=0\), where \(u\) is replaced by \(v\). This is linear first order ODE which has the solution \(v\left ( t,\xi \right ) =e^{-t}F\left ( \xi \right ) \) where \(F\) is arbitrary function of \(\xi \). Transforming back to \(u\left ( t,x\right ) \) gives the general solution as\[ u\left ( t,x\right ) =e^{-t}F\left ( x+4t\right ) \] The following is a plot of few characteristic lines \(x=x_{0}-4t\) using different values of \(x_{0}\).
Solve \(u_{t}+2u_{x}=\sin x,u\left ( 0,x\right ) =\sin x\)
Solution
Let \(\xi \) be the characteristic variable defined such that \(\xi =x-ct\). Where characteristic lines are given by \(x=x_{0}+ct\). But \(c=2\) in this problem. Hence characteristic lines are \begin{equation} \fbox{$x=x_0+2t$} \tag{1} \end{equation} And\[ \xi =x-2t \] Then \(u_{t}+2u_{x}\) is transformed to \(v\left ( t,\xi \right ) \) as was done in earlier (will not be repeated) which results in\[ u_{t}+2u_{x}=\frac{\partial v}{\partial t}\] Substituting this into the original PDE gives\[ \frac{\partial v\left ( t,\xi \right ) }{\partial t}=\sin \left ( \xi +2t\right ) \] Integrating w.r.t \(t\) gives\begin{align*} v\left ( t,\xi \right ) & =\int \sin \left ( \xi +2t\right ) dt+F\left ( \xi \right ) \\ & =-\frac{\cos \left ( \xi +2t\right ) }{2}+F\left ( \xi \right ) \end{align*}
Transforming back to \(u\left ( t,x\right ) \) gives\begin{align} u\left ( t,x\right ) & =-\frac{\cos \left ( x-2t+2t\right ) }{2}+F\left ( x-2t\right ) \nonumber \\ & =\frac{-1}{2}\cos \left ( x\right ) +F\left ( x-2t\right ) \tag{1} \end{align}
When \(t=0\), \(u\left ( 0,x\right ) =\sin x\), therefore the above becomes\begin{align*} \sin x_{0} & =F\left ( x_{0}\right ) -\frac{1}{2}\cos x_{0}\\ F\left ( x_{0}\right ) & =\sin x_{0}+\frac{1}{2}\cos x_{0} \end{align*}
Therefore the solution (1) becomes\begin{align*} u\left ( t,x\right ) & =\left ( \sin \left ( x-2t\right ) +\frac{1}{2}\cos \left ( x-2t\right ) \right ) -\frac{1}{2}\cos x\\ & =\sin \left ( x-2t\right ) +\frac{1}{2}\cos \left ( x-2t\right ) -\frac{1}{2}\cos x \end{align*}
(a) Prove that if the initial data is bounded, \(\left \vert f\left ( x\right ) \right \vert \leq M\) for all \(x\in \mathbb{R} \), then the solution to the damped transport equation (2.14) \(u_{t}+cu_{x}+au=0\) with \(a>0\) satisfies \(u\left ( t,x\right ) \rightarrow 0\) as \(t\rightarrow \infty \). (b) Find a solution to (2.14) that is defined for all \(\left ( t,x\right ) \) but does not satisfy \(u\left ( t,x\right ) \rightarrow 0\) as \(t\rightarrow \infty \).
Solution
\(u_{t}+cu_{x}+au=0\) is solved to show what is required. Let \(\xi \) be the characteristic variable defined such that \(\xi =x-ct\). Where characteristic lines are given by \(x=x_{0}+ct\). Hence characteristic lines are \begin{equation} x=x_{0}+ct \tag{1} \end{equation} And\[ \xi =x-ct \] Then \(u_{t}+cu_{x}\) is transformed to \(v\left ( t,\xi \right ) \) as was done in earlier (will not be repeated) which results in\[ u_{t}+cu_{x}=\frac{\partial v}{\partial t}\] Substituting this into the original PDE gives\[ \frac{\partial v}{\partial t}+av=0 \] Where \(u\) is replaced by \(v\). This can be viewed as first order linear ODE since it depends on \(t\) only. Its solution is \(v\left ( t,\xi \right ) =e^{-at}F\left ( \xi \right ) \,\) where \(F\) is arbitrary function of \(\xi \). Transforming back to \(u\left ( t,x\right ) \) gives\begin{equation} u\left ( t,x\right ) =e^{-at}F\left ( x-ct\right ) \tag{1} \end{equation} At \(t=0\) initial data is \(f\left ( x\right ) \). Hence the above becomes at \(t=0\)\[ f\left ( x\right ) =F\left ( x\right ) \] Hence (1) now becomes\begin{equation} u\left ( t,x\right ) =e^{-at}f\left ( x-ct\right ) \tag{2} \end{equation} But since \(\left \vert f\left ( x\right ) \right \vert \) is bounded, and since \(a>0\) then \(e^{-at}\rightarrow 0\) as \(t\rightarrow \infty \). Which implies the solution itself \(u\left ( t,x\right ) \) goes to zero as well. This is the reason why initial data needed to be bounded for this to happen.
Keeping \(a>0\). If initial data have the form \(f\left ( x\right ) e^{-bx}\) where \(\left \vert b\right \vert >a\), then at \(t=0\) the solution found in (1) becomes\[ f\left ( x_{0}\right ) e^{-bx_{0}}=F\left ( x_{0}\right ) \] Then the solution (2) now becomes, after replacing \(x_{0}\) by \(x-ct\) \begin{align*} u\left ( t,x\right ) & =e^{-at}e^{-b\left ( x-ct\right ) }f\left ( x-ct\right ) \\ & =e^{-at+bct}e^{-bx}f\left ( x-ct\right ) \\ & =e^{\left ( bc-a\right ) t}e^{-bx}f\left ( x-ct\right ) \end{align*}
The problem is asking to show that this does not go to zero for all \(x\in \mathbb{R} \) as \(t\rightarrow \infty \). Since \(\left \vert b\right \vert >a\) then \(bc-a\) is positive quantity (\(c\) is assumed positive)1 .
Therefore \(e^{\left ( bc-a\right ) t}\) will blow up as \(t\rightarrow \infty \). And therefore the whole solution will not go to zero. For any \(x\), no matter how large \(x\) is, a large enough \(t\) can be found to make the product \(e^{\left ( bc-a\right ) t}e^{-bx}\) blow up.