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Suppose I have the following PDEs: (for $u(t,x)$)

(1) $u_{t}-u_{x}=-x$

(2) $u_{t}+2u_{x}=1$

(3) $u_{t}+u_{x}+u=e^{x+3t}$

(4) $2u_{t}+u_{x}=sin(x-t)$

For all equations, it is easy to find the homogenous part of the general solution because they are all transport/transport with decay equations. However, I am having trouble finding particular solutions for (1) and (2).

For equations (3) and (4), the method of undetermined coefficients used in ODEs seems to work to find particular solutions. However, for equations (1) and (2), it doesn't. Am I supposed to find the particular solutions by intuition in those cases?

Taking $u_{p}(t,x)=x-t$ for equation (2) works, but according to wolframalpha, it is not the particular solution ($u_{p}=t$ is). Are there many kinds of particular solutions? When solving an equation, should I find all of them or does one suffice?

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    In (1) try making a substitution $u = v + f(x)$ for some suitable function $f(x)$ so that $-f'(x) = -x$. Same trick works in (2): try $v = u + g(x)$ or $v = u + h(t)$ and pick $g$ or $h$ such that you can cancel the right hand side. – 2017-02-03
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    @Winther, thank for your answer. For (1), I get $u(t,x)=A(x+t)+\frac{x^2}{2}+c$. Is this correct? (wolframlpha provides me with the following, different solution: $u(t,x)=A(x+t)-\frac{t^2}{2}-tx$) – 2017-02-03
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    No Wolfram-Alpha gives you $B(x+t) - \frac{t^2}{2} - tx$ (it does not have to be the same function as your $A$). Is there a way of picking $B(z)$ relative to $A(z)$ such that we have $A(z) = B(z) + g(z)$? Since $A$ represents a general function this would mean that it's the same solution. This is the same issue as your question of $x-t$ relative to $t$. There are always infinitely many possible particular solutions to these kind of equations and it does not matter which one you pick. – 2017-02-03
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    @Winther, I see. So the important thing is to include at least one function, different from the general function A, in our general solution? I thought general solution meant to include ALL possible solutions. – 2017-02-03
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    Yes and remember that a general function is a general fuction so it could really be anything until you impose initial conditions for which it usually uniquely determines the solution. As an exericise to convince yourself that this makes sense you could try to impose an initial condition like $u(x,0) = x^2$ using both your solution and WAs solution to determine $A$ and $B$ and see that this leads to the same solution in both cases. – 2017-02-03
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    @Winther, I see, the problem I am solving does include an initial condition but I didn't think it mattered. My bad. Last question, what if I just write the solution as C(x+t), and nothing else. Without initial conditions, would that be considered correct? – 2017-02-03
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    Can you write $A(x+t) + \frac{x^2}{2}$ on the form $C(x+t)$ valid for all $x,t$? Try to first take $t=0$ and $x=y$ and then $x=2y$ and $t=-y$ and see that you get a contradiction. The role of the particular solution is to "take care of" any additional coordinate-dependence that cannot be encaptulated by the $A(x+t)$ function. – 2017-02-03
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    $A(y)+\frac{y^2}{2}=C(y)$ and $A(y)+2y^2=C(y)$. This would imply $\frac{y^2}{2}=2y^2$, only possible if $y=0$, is this the contradiction? – 2017-02-03
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    Try to use this to answer the question I asked above: can you write $A(x+t)+\frac{x^2}{2}$ on the form $C(x+t)$ ${\bf \text{valid for all}}$ $x,t$? – 2017-02-03
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    No, because there is a condition on $x$ and $t$ if $y=0$. – 2017-02-03
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    Yes exactly. Thus you need to keep the $\frac{x^2}{2}$ term or equivalently $-\frac{t^2}{2} - tx$ as WA did it. – 2017-02-03
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    I see. Thank you so much for the clear explanation @Winther. – 2017-02-03

1 Answers 1

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I've tried the following variant of the method of characteristics and works well with some type of quasi-linear pde's.

For $(1)\;u_{t}-u_{x}=-x$ we write this sistem of equalities.

$$\frac{\mathrm dt}{1}=\frac{-\mathrm dx}{1}=\frac{-\mathrm du}{x}$$

With the first and second ratio:

$$\frac{\mathrm dt}{1}=\frac{-\mathrm dx}{1}\implies t=-x+c_1\;;t+x=c_1$$

With the second and the third:

$$\frac{-\mathrm dx}{1}=\frac{-\mathrm du}{x};\;x\mathrm dx=\mathrm du;\;\frac{x^2}{2}+c_2=u$$

Now, the intersection of the two surfaces obtained are the characteristics, so we need a relation to spread these curves to form a new surface, the solution of the pde, so is $c_2=A(c_1)$ with $A(u)$ any differentiable single variable function. The general solution is:

$$u(t,x)=A(x+t)+\frac{x^2}{2}$$

Chosing $A(x+t)=B(x+t)-(1/2)(x+t)^2$, $u(t,x)=B(x+t)+x^2/2-x^2/2-xt-t^/2=B(x+t)-xt-t^2/2$

For $(2)\;u_{t}+u_{x}+u=\exp(x+3t)$ we get

$$\frac{\mathrm dt}{1}=\frac{\mathrm dx}{1}=\frac{\mathrm du}{e^{x+3t}-u}$$

First and second:

$$\frac{\mathrm dt}{1}=\frac{\mathrm dx}{1}\implies t=x+c_1\;;t-x=c_1$$

With this, $\exp(x+3t)=\exp(4x+3c_1)$ So,

$$\frac{\mathrm dx}{1}=\frac{\mathrm du}{e^{4x+3c_1}-u};\;(e^{4x+3c_1}-u)\mathrm dx-\mathrm du=0$$

We can solve it finding an integrating factor: $M(x)=\exp(x)$

$$e^x(e^{4x+3c_1}-u)\mathrm dx-e^x\mathrm du=(e^{5x-3c_1}-e^xu)\mathrm dx-e^x\mathrm du=0\;\text{is now exact}\;\implies$$

$$\implies \mathrm d\left(\frac{1}{5}e^{5x+3c_1}-ue^x\right)=0$$

$$u=c_2e^{-x}+\frac{1}{5}e^{4x+3c_1}=c_2e^{-x}+\frac{1}{5}e^{x+3t}$$

And, as in the previous case, with $c_2=f(c_1)$, the general solution is:

$$u(t,x)=f(t-x)e^{-x}+\frac{1}{5}e^{x+3t}$$

For (4), $2u_{t}+u_{x}=sin(x-t)$, the system is:

$$\frac{\mathrm dt}{2}=\frac{\mathrm dx}{1}=\frac{\mathrm du}{\sin(x-t)}$$

First and second:

$$\frac{\mathrm dt}{2}=\frac{\mathrm dx}{1}\implies t=2x+c_1\;;t-2x=c_1$$

Second and third:

$$\frac{\mathrm dx}{1}=\frac{\mathrm du}{\sin(x-t)}\;\text{with}\;\sin(x-t)=\sin(-x-c_1)$$

$$\frac{\mathrm dx}{1}=\frac{\mathrm du}{\sin(-x-c_1)}\;;\sin(-x-c_1)\mathrm dx=\mathrm du$$

$$\cos(-x-c_1)+c_2=u\;;u=\cos(x-t)+c_2$$

And with $c_2=g(c_1)$ as before:

$$u(t,x)=cos(x-t)+g(t-2x)$$