Applying a substitution to a PDE

In summary, the problem involves solving the equation $$\frac{\partial v}{\partial t} = \frac{\partial^{2} v}{\partial x^2} + \frac{2v}{t+1}$$ with boundary conditions and using the substitution ##v=(t+1)^{2}u##. The solution involves using the product rule and treating ##u## as a function of ##x## and ##t##. A complete solution involves guessing ##u(x,t) = A(x)B(t)## and substituting in the boundary conditions.
  • #1
etotheipi
Homework Statement
Please see below
Relevant Equations
N/A
Problem: Consider the equation $$\frac{\partial v}{\partial t} = \frac{\partial^{2} v}{\partial x^2} + \frac{2v}{t+1}$$ where ##v(x,t)## is defined on ##0 \leq x \leq \pi## and is subject to the boundary conditions ##v(0,t) = 0##, ##v(\pi, t) = f(t)##, ##v(x,0) = h(x)## for some functions ##f(t)## and ##h(x)##. Using the substitution ##v=(t+1)^{2}u##, show that ##u## satisfies $$\frac{\partial u}{\partial t} = \frac{\partial^{2} u}{\partial x^2}$$ Attempt: I'm not sure if I'm doing the differentiation correctly. I did $$\frac{\partial v}{\partial t} = 2u(t+1)$$ $$\frac{\partial^{2} v}{\partial x^{2}} = \frac{\partial}{\partial x} \frac{\partial v}{\partial u} \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (t+1)^{2} \frac{\partial u}{\partial x} = (t+1)^{2} \frac{\partial^2 u}{\partial x ^2}$$ Plugging this in doesn't appear to give the result. My suspicion is that I was supposed to use the product rule for the first derivative, however I don't think that is right since I thought we were supposed to hold everything else constant during the differentiation? If I try this for the sake of it, I get $$\frac{\partial v}{\partial t} = 2u(t+1) + (t+1)^{2} \frac{\partial u}{\partial t}$$ $$\frac{\partial^{2} v}{\partial x^{2}} = \frac{\partial}{\partial x} \frac{\partial v}{\partial u} \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} [(t+1)^{2} \frac{\partial u}{\partial x} + 2u(t+1)][\frac{\partial u}{\partial x}]$$ This seems even more wrong. So I wondered whether anyone could give me a pointer? Thanks!
 
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  • #2
etotheipi said:
Homework Statement:: Please see below
Relevant Equations:: N/A

$$\frac{\partial v}{\partial t} = 2u(t+1)$$ $$\frac{\partial^{2} v}{\partial x^{2}} = \frac{\partial}{\partial x} \frac{\partial v}{\partial u} \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (t+1)^{2} \frac{\partial u}{\partial x} = (t+1)^{2} \frac{\partial^2 u}{\partial x ^2}$$
The second partial differentiation you could have simply written down, as ##t## and ##x## are independent variables.

For the first one you need to treat ##u## as a function of ##x## and ##t##.
 
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  • #3
etotheipi said:
Using the substitution ##v=(t+1)^{2}u##

If you get stuck you should always write out functions with their arguments. In this case we have:
$$v(x, t) = (t+1)^2u(x, t)$$
 
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  • #4
PeroK said:
The second partial differentiation you could have simply written down, as ##t## and ##x## are independent variables.

For the first one you need to treat ##u## as a function of ##x## and ##t##.

Ah okay that works. Then I get

$$\frac{\partial v}{\partial t} = 2u(t+1) + (t+1)^{2}\frac{\partial u}{\partial t}$$ $$\frac{\partial^{2} v}{\partial x^{2}} = (t+1)^{2} \frac{\partial^{2} u}{\partial x^{2}}$$

and that checks out. The take away is that you only hold constant the variables which the one you're differentiating is a function of. The others you still have to treat with product rules etc.

PeroK said:
If you get stuck you should always write out functions with their arguments. In this case we have:
$$v(x, t) = (t+1)^2u(x, t)$$

Yeah, I might start doing this. It makes it a little clearer. Thanks!
 
  • #5
I thought I'd write up the solution I got for completeness, using the technique of guessing ##u(x,t) = A(x)B(t)## and then setting both sides equal to ##-\lambda##. One side turns out to be a "radioactive decay" one and the other is a "SHM" one: $$\frac{dB}{dt} = -\lambda B \implies B = Ce^{- \lambda t}$$ $$\frac{d^{2}A}{dx^{2}} = -\lambda A \implies A = D\cos{(\sqrt{\lambda} x + \phi)}$$ So I get ##v(x,t) = E(t+1)^{2}e^{- \lambda t}\cos{(\sqrt{\lambda} x + \phi)}##, and I guess I also need to substitute in the boundary conditions, though this should be fine since I have three unknowns and three conditions.

There are a few warning signs, since the question says that I "may find it helpful to use the substitution ##u(x,t) = w(x,t) + \gamma x## for a suitably chosen constant ##\gamma##". I've no idea what this is supposed to be implying...
 

Related to Applying a substitution to a PDE

1. What is a substitution in the context of solving a PDE?

A substitution in the context of solving a PDE (partial differential equation) refers to the process of replacing the dependent variable and/or its derivatives with new variables in order to simplify the equation and make it easier to solve. This is often done to transform the PDE into a more familiar form or to eliminate certain terms.

2. When should a substitution be used to solve a PDE?

A substitution should be used to solve a PDE when the equation is too complex to be solved using traditional methods such as separation of variables or the method of characteristics. It can also be used when the PDE is in a non-standard form and needs to be transformed into a more familiar form.

3. What are the common substitutions used in solving PDEs?

Some common substitutions used in solving PDEs include the change of variables, the chain rule, and the method of characteristics. These substitutions involve replacing the dependent variable and/or its derivatives with new variables such as u, v, or ξ, η.

4. What are the benefits of applying a substitution to a PDE?

Applying a substitution to a PDE can make the equation easier to solve by reducing its complexity or transforming it into a more familiar form. It can also help to identify any symmetries or patterns in the equation, which can provide insight into the behavior of the solution.

5. Are there any limitations to using substitutions in solving PDEs?

While substitutions can be a powerful tool in solving PDEs, they may not always lead to a solution. In some cases, the substitution may not simplify the equation enough to make it solvable, or it may introduce new complexities. It is important to carefully choose the appropriate substitution for a given PDE and to check the validity of the solution obtained.

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