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I want to solve the following partial differential equation:

$$\frac{\partial h}{\partial t} + (y+t)\frac{\partial h}{\partial y} + h = 0$$

So far, the only methods I am familiar with solving PDEs are separation of variables, Laplace transforms, Fourier transforms, change of variables, method of characteristics, and a little bit of Fourier series, though I would like a closed form for the solution. I tried using the method of change of variables to solve the PDE and got the following general answer:

$$h(t,y)=ce^{-t}g(t-\ln\left | y+t \right |)$$

However, I don't think this is correct since when I plugged this in, it turned out to not be the solution. Wolfram Alpha gives the answer as:

$$h(t,y) = ce^{-2t}(t+y+1)$$

From the PDEs I've solved before, it seems obvious that this isn't the general solution. I am pretty sure that Wolfram Alpha chose a specific function $g(c)$, where $c$ is a variable of $t$ and $y$, in order to get a certain solution. Either that or Wolfram chose an initial value to apply. How would I get the general solution to this PDE? Any method would be acceptable. From there, how do you get the specific solution that Wolfram Alpha got? Preferably, I would like to avoid solving the following ODE:

$$\frac{\mathrm{d} y}{\mathrm{d} t}=y+t$$

in the process, since the PDE comes from solving $\frac{\mathrm{d} y}{\mathrm{d} t}=y+t$ as an inexact differential equation. If more work is needed to be shown, I will do so without hesitation.

Eliot
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  • But $y'=y+t$ is a linear ode. I don't understand why you set up a pde to solve this simple ode. – Jacky Chong Jan 26 '20 at 05:58
  • @Jacky Chong Because I challenged myself to do it through an inexact differential equation. Very unnecessary, and it is a simple ode to solve, but I like to challenge myself with stuff like this. The context isn't really necessary, I just want to solve this PDE. – Eliot Jan 26 '20 at 06:24
  • You could solve this equation by inexact differential, but you need to make assumptions on the integrating factor to reduce the problem to an ode. – Jacky Chong Jan 26 '20 at 06:32
  • @JackyChong if your wondering, I know, I am a weird person. – Eliot Jan 26 '20 at 06:33
  • @JackyChong what assumption would I have to make? As far as I see, the pde is not separable and change of variables didn't work for me. – Eliot Jan 26 '20 at 06:35
  • This is very standard. You either assume the integrating factor depends only on $y$ or only on $t$. – Jacky Chong Jan 26 '20 at 06:36
  • @JackyChong That's why I'm solving the PDE, because none of those integrating factors work. – Eliot Jan 26 '20 at 06:38
  • check again. This is standard. – Jacky Chong Jan 26 '20 at 06:39
  • @JackyChong Checked again. It doesn't work. Even if I did something wrong, can we forget the context and try to solve the PDE? I'm just curious about what I'm missing in that certain aspect, I don't care about anything else. Treat it as if only the PDE was given. – Eliot Jan 26 '20 at 06:54
  • It's easy if I am allowed to solve the ode. But since I am not allowed to solve the ode then I have no clue. – Jacky Chong Jan 26 '20 at 07:07
  • @JackyChong yeah, I think you're right. I tried with the method of characteristics but it basically gave the original ode. How would you solve it if you were allowed to solve the ode? By that I mean what's your prefered method to solve this pde? – Eliot Jan 26 '20 at 07:14

1 Answers1

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Suppose $h$ is a solution your original PDE, then consider \begin{align} u(t, y) = e^{t}h(t, y). \end{align} Observe \begin{align} u_t = e^{t}h+e^{t}h_t \end{align} and \begin{align} u_y = e^{t}h_y \end{align} which means \begin{align} u_t+(t+y)u_y = e^t(h_t+(t+y)h_y+h) = 0. \end{align} Hence it suffices to solve \begin{align} u_t+(t+y)u_y = 0. \end{align} which is a good old transport equation.

Using the method of characteristics, observe \begin{align} \frac{d}{dt}u(t, y(t)) = \partial_t u + y'(t)\partial u_y = u_t+(t+y)u_y =0 \end{align} which means \begin{align} y' = t+y \ \ \implies \ \ y(t) = (y_0+1)e^{t}-t-1. \end{align} Hence \begin{align} u(t, (y_0+1)e^{t}-t-1) = \text{ const} = u(0, y_0) = g(y_0) \end{align} which means \begin{align} u(t, y) = g(e^{-t}(y+t+1)-1) \ \ \implies \ \ h(t, y) = e^{-t}g(e^{-t}(y+t+1)-1) \end{align}

Jacky Chong
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  • I didn't realize that $e^{t}h(t,y)$ would be a solution of the PDE. Even though it uses the ode, I realized how to approach a PDE like this. I really tried a lot of things and it didn't work out. Thanks for the post! – Eliot Jan 26 '20 at 07:25
  • $\begin{align} y' = t+y \ \ \implies \ \ y(t) = (y_0+1)e^{t}-t-1. \end{align}$ Jacky it's just $$\begin{align} y' = t+y \ \ \implies \ \ y(t) = ce^{t}-t-1. \end{align}$$ – user577215664 Jan 26 '20 at 20:36
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    @Isham $y_{0} + 1$ is still just a constant. $(y_{0} + 1)e^{t} - t - 1$ is only the solution of the ODE with the initial contion $y(0)=y_{0}$. $y_{0} + 1$ can be treated as if it were a constant $c$. – Eliot Jan 28 '20 at 02:19
  • ok thanks Eliot – user577215664 Jan 28 '20 at 02:24