4 General first order PDE’s

 4.1 Linear PDE, the transport equation
 4.2 Linear PDE
 4.3 Linear PDE, initial value problem
 4.4 Initial-boundary value problem
 4.5 Linear PDE, the transport equation with initial conditions
 4.6 First order wave PDE, with initial conditions (Haberman 12.2.2)
 4.7 First order wave PDE, with initial and boundary conditions (Haberman 12.2.4)
 4.8 First order wave PDE, with initial conditions, non homogeneous (Haberman 12.2.5 (a))
 4.9 First order wave PDE, with initial conditions, non homogeneous (Haberman 12.2.5 (d))
 4.10 General solution for a quasilinear first-order PDE
 4.11 quasilinear first-order PDE, scalar conservation law
 4.12 quasilinear first-order PDE, scalar conservation law with initial value
 4.13 nonlinear first-order PDE, the Clairaut equation
 4.14 nonlinear first-order PDE, the Clairaut equation with initial value
 4.15 Another example of nonlinear Clairaut equation
 4.16 Recover a function from its gradient vector
 4.17 General solution of a first order nonlinear PDE
 4.18 Nonlinear first order PDE
 4.19 first order PDE of three unknowns

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4.1 Linear PDE, the transport equation

problem number 1

Taken from Mathematica Symbolic PDE document

Solve for \(u(x,t)\) \[ \frac{ \partial u}{\partial t} + \frac{ \partial u}{\partial x} = 0 \]

Mathematica

\[ \left \{\left \{u(x,t)\to c_1(t-x)\right \}\right \} \]

Maple

\[ u \left ( x,t \right ) ={\it \_F1} \left ( -x+t \right ) \]

Hand solution

Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin{align} \frac{dt}{ds} & =1\tag{1}\\ \frac{dx}{ds} & =1\tag{2}\\ \frac{du}{ds} & =0\tag{3} \end{align}

With initial conditions\[ t_{s=0}=t_{1},x_{s=0}=t_{2},u_{s=0}=t_{3}\] Assuming \(u\left ( x,0\right ) =F\left ( x\right ) \) then this implies that \begin{equation} t_{3}=F\left ( t_{2}\right ) ,t_{1}=0\tag{4} \end{equation} Equation (1) gives\begin{align*} t & =s+t_{1}\\ & =s \end{align*}

Equation (2) gives\begin{align*} x & =s+t_{2}\\ t_{2} & =x-s\\ t_{2} & =x-t \end{align*}

Equation (3) gives\[ u=t_{3}\] Therefore (4) becomes\[ u\left ( x,t\right ) =F\left ( x-t\right ) \] The traditional way to solve this is as follows. Let \(u\left ( x,t\right ) =u\left ( x\left ( t\right ) ,t\right ) \). Then \begin{equation} \frac{du}{dt}=\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}\frac{dx}{dt}\tag{1} \end{equation} If \begin{equation} \frac{dx}{dt}=1\tag{2} \end{equation} Then comparing (1) with given PDE \(\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}=0\) shows that \(\frac{du}{dt}=0\) or \(u\) is constant on each characteristic line \(\frac{dx}{dt}=1\). Integrating (2) gives\begin{equation} x=t+x_{0}\tag{3} \end{equation} Let \[ u\left ( x,0\right ) =F\left ( x\right ) \] Since \(u\) is constant on each characteristic line, then \begin{align*} u\left ( x,t\right ) & =u\left ( x_{0},0\right ) \\ & =F\left ( x_{0}\right ) \end{align*}

But from (3) \(x_{0}=x-t\), hence the above solution becomes

\[ u\left ( x,t\right ) =F\left ( x-t\right ) \]

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4.2 Linear PDE

problem number 2

Taken from Mathematica help pages

Solve for \(u(x,y)\) \[ 3 u_x + 5 u_y = x \]

Mathematica

\[ \left \{\left \{u(x,y)\to \frac{1}{6} \left (6 c_1\left (\frac{1}{3} (3 y-5 x)\right )+x^2\right )\right \}\right \} \]

Maple

\[ u \left ( x,y \right ) =1/6\,{x}^{2}+{\it \_F1} \left ( -5/3\,x+y \right ) \]

Hand solution

Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin{align} \frac{dx}{ds} & =3\tag{1}\\ \frac{dy}{ds} & =5\tag{2}\\ \frac{du}{ds} & =x\tag{3} \end{align}

With initial conditions\[ x_{s=0}=t_{1},y_{s=0}=t_{2},u_{s=0}=t_{3}\] Assuming \(u\left ( 0,y\right ) =F\left ( y\right ) \) then this implies that \begin{equation} t_{3}=F\left ( t_{2}\right ) ,t_{1}=0\tag{4} \end{equation} Equation (1) gives\begin{align} x & =3s+t_{1}\nonumber \\ & =3s\nonumber \\ s & =\frac{x}{3}\tag{5} \end{align}

Equation (2) gives\begin{align} y & =5s+t_{2}\nonumber \\ t_{2} & =y-5s\nonumber \end{align}

From (5), the above becomes\begin{equation} t_{2}=y-\frac{5x}{3}\tag{5A} \end{equation} Equation (3) gives\begin{align*} \frac{du}{ds} & =x\\ & =3s \end{align*}

Hence by integrating and using \(u_{s=0}=t_{3}\) the above becomes\begin{align*} u & =\frac{3s^{2}}{2}+t_{3}\\ & =\frac{3s^{2}}{2}+F\left ( t_{2}\right ) \end{align*}

Substituting (5,5A) into the above gives\begin{align*} u & =\frac{3\left ( \frac{x}{3}\right ) ^{2}}{2}+F\left ( y-\frac{5x}{3}\right ) \\ & =\frac{1}{6}x^{2}+F\left ( y-\frac{5x}{3}\right ) \end{align*}

The traditional way to solve this is as follows. Let \(u\left ( x,y\right ) =u\left ( x,y\left ( x\right ) \right ) \). Then \begin{equation} \frac{du}{dx}=\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\frac{dy}{dx}\tag{1} \end{equation} Comparing to the PDE \(\frac{\partial u}{\partial x}+\frac{5}{3}\frac{\partial u}{\partial y}=\frac{1}{3}x\), then If \begin{equation} \frac{dy}{dx}=\frac{5}{3}\tag{2} \end{equation} Then comparing (1) with given PDE shows that \(\frac{du}{dx}=\frac{1}{3}x\) or \begin{align} u & =\frac{1}{6}x^{2}+C\nonumber \\ & =\frac{1}{6}x^{2}+u\left ( 0,y_{0}\right ) \tag{3} \end{align}

On each characteristic line. The characteristic lines are not constants in this case. Integrating (2) gives\begin{align} y & =\frac{5}{3}x+y_{0}\nonumber \\ y_{0} & =y-\frac{5}{3}x\tag{4} \end{align}

Let \[ u\left ( 0,y\right ) =F\left ( y\right ) \] Then \[ u\left ( 0,y_{0}\right ) =F\left ( y_{0}\right ) \] Substituting \(y_{0}\) found in (4) into the above gives\[ u\left ( 0,y_{0}\right ) =F\left ( y-\frac{5}{3}x\right ) \] Substituting the above into (3) gives the solution as\[ u\left ( x,y\right ) =\frac{1}{6}x^{2}+F\left ( y-\frac{5}{3}x\right ) \]

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4.3 Linear PDE, initial value problem

problem number 3

Taken from Mathematica help pages

Solve for \(u(x,y)\) \[ x \frac{ \partial u}{\partial x} + y \frac{ \partial u}{\partial y} = -4 x y u(x,y) \] with initial value \(u(x,0)=e^{-x^2}\)

Mathematica

\[ \left \{\left \{u(x,y)\to e^{-x^2-y^2}\right \}\right \} \]

Maple

\[ u \left ( x,y \right ) ={{\rm e}^{-{x}^{2}-{y}^{2}}} \]

Hand solution

Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin{align} \frac{dx}{ds} & =x\tag{1}\\ \frac{dy}{ds} & =y\tag{2}\\ \frac{du}{ds} & =-4xyu\tag{3} \end{align}

With initial conditions\[ x_{s=0}=t_{1},y_{s=0}=t_{2},u_{s=0}=t_{3}\] And \(u\left ( x,0\right ) =e^{-x^{2}}\) becomes \begin{equation} t_{3}=e^{-t_{1}^{2}},t_{2}=0\tag{4} \end{equation} Equation (1) gives\begin{align} \ln x & =s+t_{1}\nonumber \\ x & =t_{1}e^{s}\nonumber \\ t_{1} & =\frac{x}{e^{s}}\tag{5} \end{align}

Equation (2) gives\begin{align} \ln y & =s+t_{2}\nonumber \\ \ln y & =s\tag{5A} \end{align}

Substituting the above in (5) gives\begin{equation} t_{1}=\frac{x}{e^{\ln y}}=\frac{x}{y}\tag{5B} \end{equation} Equation (3) gives\begin{align*} \frac{du}{u} & =-4xyds\\ & =-4\left ( t_{1}e^{s}\right ) \left ( e^{s}\right ) ds \end{align*}

Hence by integrating and using \(u_{s=0}=t_{3}\) the above becomes\begin{align*} \ln u & =-\frac{4t_{1}e^{2s}}{2}+t_{3}\\ u & =t_{3}e^{-2t_{1}e^{2s}}\\ & =e^{-t_{1}^{2}}e^{-2t_{1}e^{2s}} \end{align*}

Substituting (5A,5B) into the above gives\begin{align*} u & =e^{-\left ( \frac{x}{y}\right ) ^{2}}e^{-2\left ( \frac{x}{y}\right ) e^{2\ln y}}\\ & =e^{-\left ( \frac{x}{y}\right ) ^{2}}e^{-2\left ( \frac{x}{y}\right ) y^{2}}\\ & =e^{-\left ( \frac{x}{y}\right ) ^{2}}e^{-2\left ( xy\right ) }\\ & =e^{-\frac{x^{2}}{y^{2}}-2xy} \end{align*}

To verify this solution, the LHS of the PDE is \begin{align*} xu_{x}+yu_{y} & =x\left ( -2\frac{x}{y^{2}}-2y\right ) \left ( e^{-\frac{x^{2}}{y^{2}}-2xy}\right ) +y\left ( -\frac{x^{2}}{y^{3}}-2x\right ) \left ( e^{-\frac{x^{2}}{y^{2}}-2xy}\right ) \\ & =x\left ( -2\frac{x}{y^{2}}-2y\right ) u+y\left ( 2\frac{x^{2}}{y^{3}}-2x\right ) u\\ & =u\left ( x\left ( -2\frac{x}{y^{2}}-2y\right ) +y\left ( 2\frac{x^{2}}{y^{3}}-2x\right ) \right ) \\ & =u\left ( -2\frac{x^{2}}{y^{2}}-2xy+2\frac{x^{2}}{y^{2}}-2yx\right ) \\ & =-4xyu \end{align*}

Which is the RHS of the PDE.

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4.4 Initial-boundary value problem

problem number 4

Taken from Mathematica help pages

Solve for \(u(x,t)\) \[ \frac{ \partial u}{\partial t} + \frac{ \partial u}{\partial x} = 0 \] with initial value \(u(x,0)=\sin x\) and boundary value \(u(0,t)=0\)

Mathematica

\[ \{\{u(x,t)\to (\theta (t-x)-1) \sin (t-x)\}\} \]

Maple

\[ u \left ( x,t \right ) =-\sin \left ( -x+t \right ){\it Heaviside} \left ( -t+x \right ) \]

Hand solution

Since initial and boundary conditions are given, the Laplace transform method will be used to solve this PDE. Let \(U\left ( x,s\right ) \) be the Laplace transform of \(u\left ( x,t\right ) \). Applying Laplace transform to the PDE gives\begin{align*} sU-u\left ( x,0\right ) +\frac{dU}{dx} & =0\\ \frac{dU}{dx}+sU & =\sin x \end{align*}

Integrating factor is \(\mu =e^{\int sdx}=e^{sx}\). Multiplying the above by \(\mu \) gives\[ \frac{d}{dx}\left ( Ue^{sx}\right ) =e^{sx}\sin x \] Integrating \begin{align*} Ue^{sx} & =\int e^{sx}\sin xdx+C\\ & =\frac{e^{sx}\left ( s\sin x-\cos x\right ) }{1+s^{2}}+C\\ U\left ( x,s\right ) & =\frac{s\sin x-\cos x}{1+s^{2}}+Ce^{-sx} \end{align*}

Applying boundary conditions \(U\left ( 0,s\right ) =0\) gives\begin{align*} 0 & =\frac{-1}{1+s^{2}}+C\\ C & =\frac{1}{1+s^{2}} \end{align*}

Hence \begin{align*} U\left ( x,s\right ) & =\frac{s\sin x-\cos x}{1+s^{2}}+\frac{e^{-sx}}{1+s^{2}}\\ & =\frac{s\sin x}{1+s^{2}}-\frac{\cos x}{1+s^{2}}+\frac{e^{-sx}}{1+s^{2}} \end{align*}

Applying inverse Laplace transform gives\begin{align*} u\left ( x,t\right ) & =\cos t\sin x-\cos x\sin t+\operatorname{Heaviside}\left ( t-x\right ) \sin \left ( t-x\right ) \\ & =-\sin \left ( t-x\right ) +\operatorname{Heaviside}\left ( t-x\right ) \sin \left ( t-x\right ) \\ & =\left ( \operatorname{Heaviside}\left ( t-x\right ) -1\right ) \sin \left ( t-x\right ) \end{align*}

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4.5 Linear PDE, the transport equation with initial conditions

problem number 5

Taken from Mathematica help pages

Solve for \(u(x,t)\) \[ \frac{ \partial u}{\partial t} + c \frac{ \partial u}{\partial x} = 0 \] With initial conditions \(u(x,0)=e^{-x^2}\)

Mathematica

\[ \left \{\left \{u(x,t)\to e^{-(x-c t)^2}\right \}\right \} \]

Maple

\[ u \left ( x,t \right ) ={{\rm e}^{- \left ( tc-x \right ) ^{2}}} \]

Hand solution

Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin{align} \frac{dt}{ds} & =1\tag{1}\\ \frac{dx}{ds} & =1\tag{2}\\ \frac{du}{ds} & =0\tag{3} \end{align}

With initial conditions\[ t_{s=0}=t_{1},x_{s=0}=t_{2},u_{s=0}=t_{3}\] And \(u\left ( x,0\right ) =e^{-x^{2}}\) becomes \begin{equation} t_{3}=e^{-t_{2}^{2}},t_{1}=0\tag{4} \end{equation} Equation (1) gives\begin{align} t & =s+t_{1}\nonumber \\ & =s\tag{5} \end{align}

Equation (2) gives\begin{equation} x=s+t_{2}\tag{6} \end{equation} From (5,6) solving for \(t_{2}\) gives\begin{align} t_{2} & =x-s\nonumber \\ & =x-t\tag{7} \end{align}

Equation (3) gives\begin{align*} u & =t_{3}\\ & =e^{-t_{2}^{2}} \end{align*}

Using (7) in the above gives the solution\[ u\left ( x,t\right ) =e^{-\left ( x-t\right ) ^{2}}\]

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4.6 First order wave PDE, with initial conditions (Haberman 12.2.2)

problem number 6

Added Nov 25, 2018.

Problem 12.2.2 from Richard Haberman applied partial differential equations book, 5th edition

Solve for \(u(x,t)\) \[ \frac{ \partial \omega }{\partial t} -3 \frac{ \partial \omega }{\partial x} = 0 \] With initial conditions \(\omega (x,0)=\cos x\).

See my HW 12, Math 322, UW Madison.

Mathematica

\[ \{\{w(x,t)\to \cos (3 t+x)\}\} \]

Maple

\[ w \left ( x,t \right ) =\cos \left ( x+3\,t \right ) \]

Hand solution

Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin{align} \frac{dt}{ds} & =1\tag{1}\\ \frac{dx}{ds} & =-3\tag{2}\\ \frac{du}{ds} & =0\tag{3} \end{align}

With initial conditions\[ t_{s=0}=t_{1},x_{s=0}=t_{2},u_{s=0}=t_{3}\] And \(u\left ( x,0\right ) =\cos x\) becomes \begin{equation} t_{3}=\cos \left ( t_{2}\right ) ,t_{1}=0\tag{4} \end{equation} Equation (1) gives\begin{align} t & =s+t_{1}\nonumber \\ & =s\tag{5} \end{align}

Equation (2) gives\begin{equation} x=-3s+t_{2}\tag{6} \end{equation} From (5,6) solving for \(t_{2}\) gives\begin{align} t_{2} & =x+3s\nonumber \\ & =x+3t\tag{7} \end{align}

Equation (3) gives\begin{align*} u & =t_{3}\\ & =\cos \left ( t_{2}\right ) \end{align*}

Using (7) in the above gives the solution\[ u\left ( x,t\right ) =\cos \left ( x+3t\right ) \]

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4.7 First order wave PDE, with initial and boundary conditions (Haberman 12.2.4)

problem number 7

Added Nov 25, 2018.

Problem 12.2.4 from Richard Haberman applied partial differential equations book, 5th edition

Solve for \(u(x,t)\) \[ \frac{ \partial \omega }{\partial t} +c \frac{ \partial \omega }{\partial x} = 0 \] With \(c>0\). For \(x>0,t>0\) if \(\omega (x,0)=f(x)\) and \(\omega (0,t)=h(t)\).

See my HW 12, Math 322, UW Madison.

Mathematica

\[ \text{DSolve}\left [\left \{c w^{(1,0)}(x,t)+w^{(0,1)}(x,t)=0,w(x,0)=f(x),w(0,t)=h(t)\right \},w(x,t),\{x,t\},\text{Assumptions}\to c>0\land x>0\land t>0\right ] \]

Maple

\[ w \left ( x,t \right ) ={\frac{1}{c} \left ({\it Heaviside} \left ({\frac{tc-x}{c}} \right ) h \left ({\frac{tc-x}{c}} \right ) c+{\it invlaplace} \left ({{\rm e}^{-{\frac{sx}{c}}}}\int \!f \left ( x \right ){{\rm e}^{{\frac{sx}{c}}}}\,{\rm d}x,s,t \right ) -{\it invlaplace} \left ({{\rm e}^{-{\frac{sx}{c}}}}\int ^{0}\!f \left ({\it \_a} \right ){{\rm e}^{{\frac{s{\it \_a}}{c}}}}{d{\it \_a}},s,t \right ) \right ) } \] Solution contains unresolved invlaplace calls

Hand solution

\begin{equation} \frac{\partial w}{\partial t}+c\frac{\partial w}{\partial x}=0\tag{1} \end{equation} Let \[ w\equiv w\left ( x\left ( t\right ) ,t\right ) \] Hence \begin{equation} \frac{dw}{dt}=\frac{\partial w}{\partial t}+\frac{\partial w}{\partial x}\frac{dx}{dt}\tag{2} \end{equation} Comparing given (1) and (2), we see that if we let \(\frac{dx}{dt}=c\) in (2), then we obtain (1). Hence we conclude that \(\frac{dw}{dt}=0\). Therefore, \(w\left ( x\left ( t\right ) ,t\right ) \) is constant. At \(t=0\), we are given that \begin{equation} w\left ( x\left ( t\right ) ,t\right ) =f\left ( x\left ( 0\right ) \right ) \qquad t=0\tag{3} \end{equation} We just now need to determine \(x\left ( 0\right ) \). This is found from \(\frac{dx}{dt}=c\), which has the solution \(x\left ( t\right ) =x\left ( 0\right ) +ct\,.\) Hence \(x\left ( 0\right ) =x\left ( t\right ) -ct\). Therefore (3) becomes\[ w\left ( x,t\right ) =f\left ( x-ct\right ) \] This is valid for \(x>ct.\) We now start all over again, and look at Let \[ w\equiv w\left ( x,t\left ( x\right ) \right ) \] Hence \begin{equation} \frac{dw}{dx}=\frac{\partial w}{\partial x}+\frac{\partial w}{\partial t}\frac{dt}{dx}\tag{4} \end{equation} Comparing (4) and (1), we see that if we let \(\frac{dt}{dx}=\frac{1}{c}\) in (4), then we obtain (1). Hence we conclude that \(\frac{dw}{dx}=0\). Therefore, \(w\left ( x,t\left ( x\right ) \right ) \) is constant. At \(x=0\), we are given that \begin{equation} w\left ( x,t\left ( x\right ) \right ) =h\left ( t\left ( 0\right ) \right ) \qquad x=0\tag{5} \end{equation} We just now need to determine \(t\left ( 0\right ) \). This is found from \(\frac{dt}{dx}=\frac{1}{c}\), which has the solution \(t\left ( x\right ) =t\left ( 0\right ) +\frac{1}{c}x\,.\) Hence \(t\left ( 0\right ) =t\left ( x\right ) -\frac{1}{c}x\). Therefore (5) becomes\[ w\left ( x,t\right ) =h\left ( t-\frac{1}{c}x\right ) \] Valid for \(t>\frac{x}{c}\) or \(x<ct\). Therefore, the solution is\[ w\left ( x,t\right ) =\left \{ \begin{array} [c]{ccc}f\left ( x-ct\right ) & & x>ct\\ h\left ( t-\frac{1}{c}x\right ) & & x<ct \end{array} \right . \]

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4.8 First order wave PDE, with initial conditions, non homogeneous (Haberman 12.2.5 (a))

problem number 8

Added Nov 25, 2018.

Problem 12.2.5 (a) from Richard Haberman applied partial differential equations book, 5th edition

Solve for \(u(x,t)\) \[ \frac{ \partial \omega }{\partial t} +c \frac{ \partial \omega }{\partial x} = e^{2 x} \]

With \(\omega (x,0)=f(x)\).

See my HW 12, Math 322, UW Madison.

Mathematica

\[ \left \{\left \{w(x,t)\to f(x-c t)+\frac{e^{2 x} \left (1-e^{-2 c t}\right )}{2 c}\right \}\right \} \]

Maple

\[ w \left ( x,t \right ) =1/2\,{\frac{2\,f \left ( -tc+x \right ) c-{{\rm e}^{-2\,tc+2\,x}}+{{\rm e}^{2\,x}}}{c}} \]

Hand solution

Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin{align} \frac{dt}{ds} & =1\tag{1}\\ \frac{dx}{ds} & =c\tag{2}\\ \frac{du}{ds} & =e^{2x}\tag{3} \end{align}

With initial conditions\[ t_{s=0}=t_{1},x_{s=0}=t_{2},u_{s=0}=t_{3}\] And \(u\left ( x,0\right ) =f\left ( x\right ) \) becomes \begin{equation} t_{3}=f\left ( t_{2}\right ) ,t_{1}=0\tag{4} \end{equation} Equation (1) gives\begin{align} t & =s+t_{1}\nonumber \\ & =s\tag{5} \end{align}

Equation (2) gives\begin{equation} x=cs+t_{2}\tag{6} \end{equation} From (5,6) solving for \(t_{2}\) gives\begin{align} t_{2} & =x-cs\nonumber \\ & =x-ct\tag{7} \end{align}

Equation (3) gives\begin{align*} du & =e^{2x}ds\\ & =e^{2\left ( cs+t_{2}\right ) }ds \end{align*}

Integrating \[ u=\frac{e^{2\left ( cs+t_{2}\right ) }}{2c}+t_{3}\] Using (7,4,5) in the above gives the solution\begin{align*} u\left ( x,t\right ) & =\frac{e^{2\left ( ct+\left ( x-ct\right ) \right ) }}{2c}+f\left ( x-ct\right ) \\ & =\frac{1}{2c}e^{2x}+f\left ( x-ct\right ) \end{align*}

My solution is not the same as CAS, but it was verified OK using Maple pdetest.

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4.9 First order wave PDE, with initial conditions, non homogeneous (Haberman 12.2.5 (d))

problem number 9

Added Nov 25, 2018.

Problem 12.2.5 (d) from Richard Haberman applied partial differential equations book, 5th edition

Solve for \(u(x,t)\) \[ \frac{ \partial \omega }{\partial t} +3 t \frac{ \partial \omega }{\partial x} = \omega (x,t) \]

with \(\omega (x,0)=f(x)\).

See my HW 12, Math 322, UW Madison.

Mathematica

\[ \left \{\left \{w(x,t)\to e^{-\sqrt{t^2}} f\left (x-\frac{3 t^2}{2}\right )\right \},\left \{w(x,t)\to e^{\sqrt{t^2}} f\left (x-\frac{3 t^2}{2}\right )\right \}\right \} \]

Maple

\[ w \left ( x,t \right ) =f \left ( -3/2\,{t}^{2}+x \right ){{\rm e}^{t}} \]

Hand solution

Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin{align} \frac{dt}{ds} & =1\tag{1}\\ \frac{dx}{ds} & =3t\tag{2}\\ \frac{dw}{ds} & =w\tag{3} \end{align}

With initial conditions\[ t_{s=0}=t_{1},x_{s=0}=t_{2},w_{s=0}=t_{3}\] And \(w\left ( x,0\right ) =f\left ( x\right ) \) becomes \begin{equation} t_{3}=f\left ( t_{2}\right ) ,t_{1}=0\tag{4} \end{equation} Equation (1) gives\begin{align} t & =s+t_{1}\nonumber \\ & =s\tag{5} \end{align}

Equation (2) gives, after replacing \(t\) by \(s\) from (5)\begin{align} \frac{dx}{ds} & =3s\nonumber \\ x & =\frac{3}{2}s^{2}+t_{2}\tag{6} \end{align}

Solving for \(t_{2}\) gives\begin{equation} t_{2}=x-\frac{3}{2}s^{2}\tag{7} \end{equation} Equation (3) gives\begin{align*} \ln w & =s+t_{3}\\ w & =t_{3}e^{s}\\ & =f\left ( t_{2}\right ) e^{s} \end{align*}

Using (7,5) in the above gives the solution\[ w\left ( x,t\right ) =f\left ( x-\frac{3}{2}t^{2}\right ) e^{t}\]

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4.10 General solution for a quasilinear first-order PDE

problem number 10

Taken from Mathematica help pages

Solve for \(u(x,y)\) \[ 2 \frac{ \partial u}{\partial x} + 5 \frac{ \partial u}{\partial y} = u^2(x,y) + 1 \]

Mathematica

\[ \left \{\left \{u(x,y)\to \tan \left (c_1\left (y-\frac{5 x}{2}\right )+\frac{x}{2}\right )\right \}\right \} \]

Maple

\[ u \left ( x,y \right ) =\tan \left ( x/2+1/2\,{\it \_F1} \left ( -5/2\,x+y \right ) \right ) \]

Hand solution

Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin{align} \frac{dx}{ds} & =2\tag{1}\\ \frac{dy}{ds} & =5\tag{2}\\ \frac{du}{ds} & =u^{2}+1\tag{3} \end{align}

With initial conditions\[ x_{s=0}=t_{1},y_{s=0}=t_{2},u_{s=0}=t_{3}\] Let \(u\left ( 0,y\right ) =F\left ( y\right ) \), where \(F\) is arbitray function. (we could also assume that \(u\left ( x,0\right ) =F\left ( x\right ) \), and both solutions should be valid), but CAS selected the first choice based on final result). Therefore\begin{equation} t_{3}=F\left ( t_{2}\right ) ,t_{1}=0\tag{4} \end{equation} Equation (1) gives\begin{align} x & =2s+t_{1}\nonumber \\ & =2s\tag{5} \end{align}

Equation (2) gives\begin{equation} y=5s+t_{2}\tag{6} \end{equation} Solving for \(t_{2}\) from (6,5) gives\begin{align} t_{2} & =y-5s\nonumber \\ & =y-\frac{5}{2}s\tag{7} \end{align}

Equation (3) gives\begin{align*} \frac{du}{1+u^{2}} & =ds\\ \arctan \left ( u\right ) & =s+t_{3}\\ u & =\tan \left ( s+t_{3}\right ) \end{align*}

Using (4,5,7) in the above gives the solution\[ u\left ( x,t\right ) =\tan \left ( \frac{x}{2}+F\left ( y-\frac{5}{4}x\right ) \right ) \]

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4.11 quasilinear first-order PDE, scalar conservation law

problem number 11

Taken from Mathematica Symbolic PDE document

Solve for \(u(x,y)\) \[ \frac{ \partial u}{\partial x} + u(x,y) \frac{ \partial u}{\partial y} = 0 \]

Mathematica

\[ \text{Solve}\left [u(x,y)=c_1\left (x-\frac{y}{u(x,y)}\right ),u(x,y)\right ] \] Implicit solution

Maple

\[ -y+xu \left ( x,y \right ) +{\it \_F1} \left ( u \left ( x,y \right ) \right ) =0 \]

Hand solution

Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin{align} \frac{dx}{ds} & =1\tag{1}\\ \frac{dy}{ds} & =u\tag{2}\\ \frac{du}{ds} & =0\tag{3} \end{align}

With initial conditions\[ x_{s=0}=t_{1},y_{s=0}=t_{2},u_{s=0}=t_{3}\] Let \(u\left ( x,0\right ) =F\left ( x\right ) \), where \(F\) is arbitray function. Therefore\begin{equation} t_{3}=F\left ( t_{1}\right ) ,t_{2}=0\tag{4} \end{equation} Equation (1) gives\begin{equation} x=s+t_{1}\tag{5} \end{equation} Equation (2) gives\begin{align} y & =su+t_{2}\nonumber \\ & =su\tag{7} \end{align}

Equation (3) gives \[ u=t_{3}\] Hence the solution is \begin{align*} u & =t_{3}\\ & =F\left ( t_{1}\right ) \\ & =F\left ( x-s\right ) \\ & =F\left ( x-\frac{y}{u}\right ) \end{align*}

I do not know now how Maple obtained the solution it  shows.

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4.12 quasilinear first-order PDE, scalar conservation law with initial value

problem number 12

Taken from Mathematica Symbolic PDE document

Solve for \(u(x,y)\) \[ u_x+ u(x,y) u_y = 0 \] With \(u(x,0)=\frac{1}{x+1}\)

Mathematica

\[ \left \{\left \{u(x,y)\to \frac{y+1}{x+1}\right \}\right \} \]

Maple

\[ u \left ( x,y \right ) ={\frac{y+1}{x+1}} \]

Hand solution

Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)\begin{align} \frac{dx}{ds} & =1\tag{1}\\ \frac{dy}{ds} & =u\tag{2}\\ \frac{du}{ds} & =0\tag{3} \end{align}

With initial conditions\[ x_{s=0}=t_{1},y_{s=0}=t_{2},u_{s=0}=t_{3}\] We are given that \(u\left ( x,0\right ) =\frac{1}{1+x}\). This initial condition translates to\begin{equation} t_{3}=\frac{1}{1+t_{1}},t_{2}=0\tag{4} \end{equation} Equation (1) gives\begin{equation} x=s+t_{1}\tag{5} \end{equation} Equation (2) gives\begin{align} y & =su+t_{2}\nonumber \\ & =su\tag{7} \end{align}

Equation (3) gives \[ u=t_{3}\] Hence the solution is \begin{align*} u & =t_{3}\\ & =\frac{1}{1+t_{1}}\\ & =\frac{1}{1+\left ( x-s\right ) }\\ & =\frac{1}{1+\left ( x-\frac{y}{u}\right ) } \end{align*}

Solving for \(u\) gives\begin{align*} u\left ( 1+\left ( x-\frac{y}{u}\right ) \right ) & =1\\ u+xu-y & =1\\ u\left ( 1+x\right ) & =1+y\\ u & =\frac{1+y}{1+x} \end{align*}

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4.13 nonlinear first-order PDE, the Clairaut equation

problem number 13

Taken from Mathematica Symbolic PDE document

Solve for \(u(x,y)\) \[ x \frac{ \partial u}{\partial x} + y \frac{ \partial u}{\partial y} + \frac{1}{2} \left ( \left ( \frac{ \partial u}{\partial x} \right )^2 + \left ( \frac{ \partial u}{\partial y} \right )^2 \right )=0 \]

Mathematica

\[ \left \{\left \{u(x,y)\to c_1 x+c_2 y+\frac{1}{2} \left (c_1^2+c_2^2\right )\right \}\right \} \]

Maple

\[ u \left ( x,y \right ) =-1/2\,{x}^{2}-1/2\,x\sqrt{{x}^{2}+2\,{\it \_c}_{{1}}}-{\it \_c}_{{1}}\ln \left ( x+\sqrt{{x}^{2}+2\,{\it \_c}_{{1}}} \right ) +{\it \_C1}-1/2\,{y}^{2}-1/2\,y\sqrt{{y}^{2}-2\,{\it \_c}_{{1}}}+{\it \_c}_{{1}}\ln \left ( y+\sqrt{{y}^{2}-2\,{\it \_c}_{{1}}} \right ) +{\it \_C2} \]

Hand solution

Assuming the solution is \(u\left ( x,y\right ) =X\left ( x\right ) +Y\left ( y\right ) \). Substituting this into the PDE gives\begin{align*} xX^{\prime }+yY^{\prime }+\frac{1}{2}\left ( \left ( X^{\prime }\right ) ^{2}+\left ( Y^{\prime }\right ) ^{2}\right ) & =0\\ \frac{1}{2}\left ( X^{\prime }\right ) ^{2}+xX^{\prime } & =-\frac{1}{2}\left ( Y^{\prime }\right ) ^{2}-yY^{\prime } \end{align*}

The above is possible when each side is equal to same constant, say \(C_{1}\). This gives two ODE’s\begin{align} \frac{1}{2}\left ( X^{\prime }\right ) ^{2}+xX^{\prime } & =C_{1}\tag{1}\\ \frac{1}{2}\left ( Y^{\prime }\right ) ^{2}+yY^{\prime } & =-C_{1}\tag{2} \end{align}

ODE (1) becomes\begin{align*} \left ( X^{\prime }\right ) ^{2}+2xX^{\prime }-2C_{1} & =0\\ X^{\prime } & =\frac{-b}{2a}\pm \frac{1}{2a}\sqrt{b^{2}-4ac}\\ & =\frac{-2x}{2}\pm \frac{1}{2}\sqrt{4x^{2}+8C_{1}}\\ & =-x\pm \sqrt{x^{2}+2C_{1}} \end{align*}

For the case \(X^{\prime }=-x+\sqrt{x^{2}+2C_{1}}\), the solution is \begin{align*} X\left ( x\right ) & =\int -x+\sqrt{x^{2}+2C_{1}}dx+C_{2}\\ & =-\frac{x^{2}}{2}+\frac{x\sqrt{x^{2}+2C_{1}}}{2}+C_{1}\ln \left ( x+\sqrt{x^{2}+2C_{1}}\right ) +C_{2} \end{align*}

For the case \(X^{\prime }=-x-\sqrt{x^{2}+2C_{1}}\), the solution is\begin{align*} X\left ( x\right ) & =\int -x-\sqrt{x^{2}+2C_{1}}dx+C_{2}\\ & =-\frac{x^{2}}{2}-\frac{x\sqrt{x^{2}+2C_{1}}}{2}-C_{1}\ln \left ( x+\sqrt{x^{2}+2C_{1}}\right ) +C_{2} \end{align*}

Combining the above two solutions to one gives\begin{equation} X\left ( x\right ) =-\frac{x^{2}}{2}\pm \frac{x\sqrt{x^{2}+2C_{1}}}{2}\pm C_{1}\ln \left ( x+\sqrt{x^{2}+2C_{1}}\right ) +C_{2}\tag{3} \end{equation} ODE (2) becomes\begin{align*} \left ( Y^{\prime }\right ) ^{2}+2yY^{\prime }+2C_{1} & =0\\ Y^{\prime } & =\frac{-b}{2a}\pm \frac{1}{2a}\sqrt{b^{2}-4ac}\\ & =\frac{-2y}{2}\pm \frac{1}{2}\sqrt{4y^{2}-8C_{1}}\\ & =-y\pm \sqrt{y^{2}-2C_{1}} \end{align*}

For the case \(Y^{\prime }=-y+\sqrt{x^{2}-2C_{1}}\), the solution is \begin{align*} Y\left ( y\right ) & =\int -y+\sqrt{y^{2}-2C_{1}}dy+C_{2}\\ & =\frac{-y^{2}}{2}+\frac{y\sqrt{y^{2}-2C_{1}}}{2}-C_{1}\ln \left ( y+\sqrt{y^{2}-2C_{1}}\right ) +C_{3} \end{align*}

For the case \(Y^{\prime }=-y-\sqrt{y^{2}-2C_{1}}\), the solution is\begin{align*} Y\left ( y\right ) & =\int -y-\sqrt{y^{2}+2C_{1}}dy+C_{2}\\ & =-\frac{y^{2}}{2}-\frac{y\sqrt{x^{2}-2C_{1}}}{2}+C_{1}\ln \left ( y+\sqrt{y^{2}-2C_{1}}\right ) +C_{3} \end{align*}

Combining the above two solutions to one gives\begin{equation} Y\left ( x\right ) =-\frac{y^{2}}{2}\pm \frac{y\sqrt{y^{2}-2C_{1}}}{2}\pm C_{1}\ln \left ( y+\sqrt{y^{2}-2C_{1}}\right ) +C_{3}\tag{4} \end{equation} From (3,4) the final solution is\begin{align*} u\left ( x,y\right ) & =X\left ( x\right ) +Y\left ( x\right ) \\ & =\left ( -\frac{x^{2}}{2}\pm \frac{x\sqrt{x^{2}+2C_{1}}}{2}\pm C_{1}\ln \left ( x+\sqrt{x^{2}+2C_{1}}\right ) +C_{2}\right ) +\left ( -\frac{y^{2}}{2}\pm \frac{y\sqrt{y^{2}-2C_{1}}}{2}\pm C_{1}\ln \left ( y+\sqrt{y^{2}-2C_{1}}\right ) +C_{3}\right ) \\ & =-\frac{x^{2}}{2}\pm \frac{x}{2}\sqrt{x^{2}+2C_{1}}\pm C_{1}\ln \left ( x+\sqrt{x^{2}+2C_{1}}\right ) -\frac{y^{2}}{2}\pm \frac{y}{2}\sqrt{y^{2}-2C_{1}}\pm C_{1}\ln \left ( y+\sqrt{y^{2}-2C_{1}}\right ) +C_{4} \end{align*}

Where \(C_{4}=C_{2}+C_{3}\).

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4.14 nonlinear first-order PDE, the Clairaut equation with initial value

problem number 14

Taken from Mathematica Symbolic PDE document

Solve for \(u(x,y)\) \[ x \frac{ \partial u}{\partial x} + y \frac{ \partial u}{\partial y} + \frac{1}{2} \left ( \left ( \frac{ \partial u}{\partial x} \right )^2 + \left ( \frac{ \partial u}{\partial y} \right )^2 \right ) = 0 \] With \(u(x,0)= \frac{1}{2} (1-x^2)\)

Mathematica

\[ \left \{\left \{u(x,y)\to \frac{1}{2} \left (-x^2-2 y+1\right )\right \}\right \} \]

Maple

\[ u \left ( x,y \right ) =-1/2\, \left ( x-y+1 \right ) \left ( x-y-1 \right ) \]

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4.15 Another example of nonlinear Clairaut equation

problem number 15

Taken from Mathematica DSolve help pages

Solve for \(u(x,y)\) \[ u(x,y)= x \frac{ \partial u}{\partial x} + y \frac{ \partial u}{\partial y} + \sin \left ( \frac{ \partial u}{\partial x} + \frac{\partial u}{\partial y} \right ) \]

Mathematica

\[ \left \{\left \{u(x,y)\to c_1 x+c_2 y+\sin \left (c_1+c_2\right )\right \}\right \} \]

Maple

\[ u \left ( x,y \right ) =x{\it \_c}_{{1}}+y{\it \_c}_{{2}}+\sin \left ({\it \_c}_{{1}}+{\it \_c}_{{2}} \right ) \]

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4.16 Recover a function from its gradient vector

problem number 16

Taken from Mathematica DSolve help pages

Solve for \(f(x,y)\) \begin{align*} \frac{ \partial f}{\partial x} &= x y \cos (x y)+ \sin (x y) \\ \frac{ \partial f}{\partial y} &= -e^{-y} +x^2 \cos (x y) \end{align*}

Mathematica

\[ \left \{\left \{f(x,y)\to c_1+x \sin (x y)+e^{-y}\right \}\right \} \]

Maple

\[ \left \{ f \left ( x,y \right ) =x\sin \left ( yx \right ) +{{\rm e}^{-y}}+{\it \_C1} \right \} \]

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4.17 General solution of a first order nonlinear PDE

problem number 17

Taken from Maple pdsolve help pages

Solve for \(f(x,y)\) \begin{align*} x \frac{ \partial f}{\partial y}- \frac{ \partial f}{\partial x} &= \frac{ f^2(x,y) g(x)}{h(y)} \end{align*}

Mathematica

\[ \text{DSolve}\left [x f^{(0,1)}(x,y)-f^{(1,0)}(x,y)=\frac{g(x) f(x,y)^2}{h(y)},f(x,y),\{x,y\}\right ] \]

Maple

\[ f \left ( x,y \right ) = \left ( \int ^{x}\!{\frac{g \left ({\it \_a} \right ) }{h \left ( -1/2\,{{\it \_a}}^{2}+1/2\,{x}^{2}+y \right ) }}{d{\it \_a}}+{\it \_F1} \left ( 1/2\,{x}^{2}+y \right ) \right ) ^{-1} \] Has unresolved integral in the answer

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4.18 Nonlinear first order PDE

problem number 18

Taken from Maple pdsolve help pages, probem 5

Solve for \(f(x,y,z)\) \begin{align*} f_x + (f_y)^2 &= f(x,y,z)+z \end{align*}

Mathematica

\[ \left \{\left \{f(x,y,z)\to \frac{1}{4} \left (c_1(z){}^2 \text{ProductLog}\left (-\frac{e^{\frac{y}{c_1(z)}+\frac{c_2(z)}{c_1(z)}+x-1}}{c_1(z)}\right ){}^2+2 c_1(z){}^2 \text{ProductLog}\left (-\frac{e^{\frac{y}{c_1(z)}+\frac{c_2(z)}{c_1(z)}+x-1}}{c_1(z)}\right )-4 z\right )\right \}\right \} \]

Maple

\[ f \left ( x,y,z \right ) =-{\frac{{{\rm e}^{-x}}z{{\it \_C5}}^{2}+{{\rm e}^{x}}{{\it \_C3}}^{2}+{\it \_C3}\,y{\it \_C5}+z{\it \_C4}\,{\it \_C5}+{\it \_C1}\,{\it \_C5}}{{{\it \_C5}}^{2}{{\rm e}^{-x}}}} \]

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4.19 first order PDE of three unknowns

problem number 19

From example 3.5.4, page 212 nonlinear pde’s by Lokenath Debnath, 3rd edition.

Solve for \(u(x,y,z)\) \begin{align*} (y-z) u_x + (z-x) u_y + (x-y) u_z = 0 \end{align*}

Mathematica

\[ \text{Timed out} \] Timed out

Maple

\[ u \left ( x,y,z \right ) ={{\rm e}^{1/2\,{\it \_C2}\,{x}^{2}}}{{\rm e}^{{\it \_C1}\,x}}{{\rm e}^{1/2\,{\it \_C2}\,{y}^{2}}}{{\rm e}^{{\it \_C1}\,y}}{\it \_C3}\,{\it \_C5}\,{\it \_C4}\,{{\rm e}^{1/2\,{\it \_C2}\,{z}^{2}}}{{\rm e}^{{\it \_C1}\,z}} \]